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8086 PINS and SIGNALS

The 8086 pins and signals are shown below. Unless otherwise

indicated, all 8086 pins are TTL compatible. The 8086 can

operate in two modes. These are minimum mode (uniprocessor

system – single 8086) and maximum mode (multiprocessor

system system – more than one 8086). Pin Diagram for the 8086

is given below:

REFER DIAGRAM 3:1.1

Pin(s) Symbol Input/ Output

Description

1 GND - Ground 2-16 AD14-

ADO I/O-3 Output address during

the first part of the bus cycle and inputs or outputs data during the remaining part of the bus cycle.

17 NMI I Nonmaskable interrupt request level triggered

18 INTR I Maskable interrupt request level triggered

19 CLK I Generates clock signals that synchronize the operation of processor.

20 GND Ground 21 RESET I Terminates activity,

clears PSW, IP, DS,SS,ES, and the instruction queue, and

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sets CS to FFFF; IP to 0000H; SS to 0000H; DS to 0000H; PSW to 0000H. Processing begins at FFFFO when signal is dropped. Signal must be 1 for at least 4 clock cycles.

22 READY I Acknowledgment from memory or I/O interface that cpu can complete the current bus cucle.

23 TEST I Used in conjuction with the WAIT instruction in multiprocessing environments. A WAIT instruction will cause the cpu to idle, except for processing interruptsm, until a 0 is applied to this pin see chp-11

24-31 Definition depends on mode

32 RD 0-3 Indicates a memory or I/O read is to be performed

33 MN/MX I Cpu is in minimum mode when strapped to +5 v and in maximum mode when grounded

34 BHE/s7 0-3 If o during first of bus cycle this pin indicates that at least one byte of

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the current transfer is to be made on pins AD15-AD8 if 1 the transfer is made on AD7-AD0. Status s7 is output during the latter part of bus assigned a meaning

35-38 A19/s6- A16/s3

0-3 During the first part of the bus cycle the upper 4 bits of the address are output and during the remainder of the bus cycle status is output. S3 and S4 indicates the segment register being used as follows; S4 s3 Register 0 0 ES 0 1 SS

1 0 CS or more 1 1 DS s5 gives the current setting of IF. S6 is always 0.

39 AD15 I/0-3 Same as AD14-AD0

40 Vcc Supply voltage - +5 v 10%

1. MN/MX is an input pin used to select one of these modes.

When MN/MX is HIGH, the 8086 operates in the minimum

mode. When MN/MX is LOW, 8086 is configured to support

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multiprocessor systems. In this case, the Intel 8288 bus

controller is added to the 8086 to provide bus controls.

2. Pins 2 through 16 and 39 (AD15 - AD0) are a 16-bit

multiplexed address/data bus. During the first clock cycle

AD0-AD15 are the low order 16 bits of address. The 8086

have a total of 20 address lines. The upper four lines (35 –

38)are multiplexed with the status signals for the 8086. These

are the A16/S3, A17/S4, A18/S5, and A19/S6. During the first

clock period of a bus cycle, the entire 20-bit address is

available on these lines. During all other clock cycles for

memory and I/O operations, AD15-AD0 contains the 16-bit

data, and S3, S4, S5, and S6 become status lines. S3 and S4

lines are decoded as follows:

A17/S4 A16/S3 Function

0 0 Extra segment

0 1 Stack segment

1 0 Code or no segment

1 1 Data Segment

Status bits S3 and S4 indicate the segment register that is

being used to generate the address the address and bit S5

reflects the contents of the IF flag. S6 is always held at 0 and

indicates that an 8086 is controlling the system bus.

3. Pins 1 and 20 are grounded. Pin 17 is NMI. This is the

Non Maskable Interrupt input activated by a leading edge. Pin

18 is INTR. INTR is the maskable interrupt input. Pin 19 is

CLK is for supplying the clock signal that synchronizes the

activity within the CPU.

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4. Pin 21 (RESET) is the system reset input signal. When the

8086 detects the positive going edge of a pulse on RESET, it

stops all activities until the signal goes LOW. When the reset

is low, the 8086 initializes as follows:

8086 Component Content

Falgs Clear

IP 0000H

CS FFFFH

DS 0000H

SS 0000H

ES 0000H

Queue Empty

5. Pin 22 (READY) is for inputting an acknowledge from a

memory or I/O interface that input data will be put on the data

bus or output data will be accepted from the data bus within

the next clock cycle. In either case, the CPU and its bus

control logic can complete the current bus cycle after the next

clock cycle.

6. Pin 23 TEST is an input pin and is only used by the WAIT

instruction. The 8086 enter a wait state after execution of the

WAIT instruction until a LOW is seen on the TEST pin.

7. Pins 24 to 31 are mode dependent and are considered

later.

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8. Pin 32 (RD) is LOW whenever the 8086 is reading data

from memory or an I/O location.

9. Pin 34 (BHE/S7) is used as BHE (BUS High Enable) during

the first clock cycle of an instruction execution. BHE can be

used in conjunction with AD0 to select memory banks. During

all other clock cycles BHE/S7 is used as S7.

Operation BHE AD0 Data pins used

Write/Read a word at an even

Address 0 0 AD15 – AD0

Write/Read a byte at an even

Address 1 0 AD7 – AD0

Write/Read a byte at an odd

Address 0 1 AD15 – AD8

Write/Read a word at an odd

Address 0 1 AD15 – AD8

1 0 AD7 - AD0

10. Pin 40 (VCC) receives the supply voltage, which must be

+5 V.

TIME, SPEED and DISTANCE - Worksheet

1. A person crosses a 600 m long street in 5 minutes. What is his

speed in km per hour?

(a).3.6 (b).7.2 (c).8.4 (d).10

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2. An aeroplane covers a certain distance at a speed of 240 kmph

in 5 hours. To cover the same distance in 1 hours, it must travel at

a speed of:

(a).300 kmph (b).360 kmph (c).600 kmph (d).720

kmph

3. A train can travel 50% faster than a car. Both start from point A

at the same time and reach point B 75 kms away from A at the

same time. On the way, however, the train lost about 12.5 minutes

while stopping at the stations. The speed of the car is:

(a).100 kmph (b).110 kmph (c).120 kmph (d).130

kmph

4. A swimmer swims from a point A against a current for 5min and

then in favour of the current for next 5 min and comes to the point

B. if AB=100mts find the speed of the current?

(a)0.5 km/hr (b)0.6km/hr (c)0.7km/hr (d)0.8

km/hr

5. If the current flows at 2 mph and it takes me 3 hrs. to row 9

miles upstream, how long will it take to return?

(a) 1 2/7 hrs (b) 2 5/7 hrs (c)2 2/3 hrs (d)1 5/7

hrs

6. A train 110 metres long travels at 60 km/hr. In what time will it

pass a man walking at 6 km/hr?

(i). against it

(ii). in the same direction

(a) 6 secs and 7 secs (b) 6 secs and 7 1/3 secs

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(c) 6 1/3 secs and 7 secs (d) 6 secs and 7.1 secs

7. The ratio between the speeds of two trains is 7: 8. If the second

train runs 400 kms in 4 hours, then the speed of the first train is:

(a).70 km/hr (b).75 km/hr (c).84 km/hr (d).87.5

km/hr

8. A farmer travelled a distance of 61 km in 9 hours. He travelled

partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The

distance travelled on foot is:

(a).14 km (b).15 km (c).16 km (d).17 km

9. Excluding stoppages, the speed of a bus is 54 kmph and

including stoppages, it is 45 kmph. For how many minutes does

the bus stop per hour?

(a).9mins (b).10mins (c).12mins (d).20mins

10. Robert is travelling on his cycle and has calculated to reach

point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12

noon if he travels at 15 kmph. At what speed must he travel to

reach A at 1 P.M.?

(a).8 kmph (b).11 kmph (c).12 kmph (d).14 kmph

11. It takes eight hours for a 600 km journey, if 120 km is done by

train and the rest by car. It takes 20 minutes more, if 200 km is

done by train and the rest by car. The ratio of the speed of the

train to that of the cars is:

(a).2 : 3 (b).3 : 2 (c).3 : 4 (d).4 : 3

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12. There are two towns A and B. Anil goes from A to B at 40

kmph and comes back to the starting point at 60 kmph. What is his

average speed during the whole journey?

(a). 84 kmph (b). 4.8 kmph (c). 48 kmph (d). 8.4

kmph

13. A man is standing on a railway bridge which is 180 meter long;

he finds that a train crosses a bridge in 20 secs, but himself in 8

secs. Find the length of the train and its speed?

(a). 120m and 54 kmph (b). 120 kmph and 54m (c). 120m and

48 kmph

(d). 1.20m and 5.4 kmph

14. Two trains 137 meters and 163 meters in length are running

towards each other on parallel lines, one at the rate of 42 kmph

and another at 48 kmph. In what time will they be clear of each

other from the moment they meet?

(a). 12 mins (b). 1.2 mins (c). 12 secs (d). 24 hrs

15. A train 100 meters long takes 6 secs to cross a man walking at

5 kmph in a direction opposite to that of the train. Find the speed

of the train?

(a). 5 kmph (b). 5.5 kmph (c). 60 kmph (d). 55

kmph

16. A train running at 54 kmph takes 20 secs to pass a platform,

next it takes 12 secs to pass a man walking at 6 kmph in the same

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direction in which the trains is going. Find the length of the train

and the length of the platform?

(a). 160m and 100m (b). 160m and 140m (c). 130m and 140m

(d). 160m and 130m

17. If a boy walks from house to school at the rate of 4 kmph, he

reaches the school 10 mins earlier than the scheduled time.

However if he walks at the rate of 3 kmph, he reaches 10 mins

late. The distance of the school from his house is

(a). 6 km (b). 4.5 km (c). 4 km (d). 3 km

18. A man drives 150 km from A to B in 3 hrs 20 mins and return

to A in 4 hrs 10 mins. Then the average speed from A to B

exceeds the average speed for the entire trip by

(a). 5 kmph (b). 4.5 kmph (c). 4 kmph (d). 2.5

kmph

19. The ratio between the speeds of A and B is 4:3 and therefore

A takes 10 mins more than the time taken by B to reach a

destination. If A had walked at double the speed he would have

covered the distance in?

(a). 30 mins (b). 25 mins (c). 20 mins (d). 15

mins

20. A car travelling with of its actual speed covers 42 km in 1 hr

40 min 48 sec. Find the actual speed of the car.

(a). 17 6/7 km/hr (b). 25 km/hr (c). 30 km/hr (d). 35

km/hr

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SET THEORY and CUBES - WORKSHEET

SET THEORY

1). Which of the following is/are true?

1. A set of all natural numbers is a well defined set.

2. A null set contains an element ‘0’

3. S1 = ( a, b, c ) and S2 = ( a, b, c), S2 is a subset of S1.

4. S1 = (a, b, c, d, e) and S2 = (a, b, c, m). S2 is an improper

subset of S1.

A. 1 only B. 1and 3 only C. 2 and 4 only D. 1, 2 and 4

only

2). Which of the following is not an empty set?

1) A = ( x/x є I, x2 is not positive )

2) B = ( x/x є N, 2x + 1 is even )

3) C = ( x/x є N, x is odd and x2

is even )

4) D = ( x/x є R, x2 + 1 = 0 )

3). If A = ( x: x = 2n – 2, n ≤ 3, n є N ); B = (4n-1: n ≤ 5, n є N ),

find A ∩ B.

A. (0, 2, 4) B. Φ C. (3, 7) D. (2, 3)

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4). DIRECTIONS for questions 1 and 2: choose the correct

alternative.

1. Set A, B and C are such that A = ( x2 – 5x + 6 = 0); B = ( y

2

– 8y + 15 = 0 )

and C = ( z2 – 7x + 10 = 0 ). Find ( A ∩ B∩ C ) ∩ ( A U B U C ).

A. ( ) B. (2, 5) C. ( 2,3,5) D. (2,5)

2. If A = (6x2 + x – 15 = 0 ), B = ( 2x

2 – 5X +3 = 0), and C =

(2x2 + x -3 = 0 ); find

A ∩ B∩ C.

A. ( 1 ) B. (3/2) C. ( 3/2, -5/3, 1) D. None of

these

5). In a class of 100, 64% of the students have taken politics and

56% of the students have taken history, how many students have

taken both subjects if all the students take atleast one of these

subjects?

A. 18 B. 20 C. 0 D. 25

6). After recent examination every candidate took French or Latin.

75.8% took French and 49.4% took latin. If the total number of

candidates was 2500, how many took both French and latin?

A. 460 B. 560 C. 1999 D. 640

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7). Describe the shaded region?

8). In a class of 120 students numbered 1 to 120, all even

numbered students opt for Physics, whose numbers are divisible

by 5 opt for Chemistry and those whose numbers are divisible by

7 opt for Math. How many opt for none of the three subjects?

A. 19 B. 41 C. 21 D. 57 E. 26

9). Of the 200 candidates who were interviewed for a position at a

call center, 100 had a two-wheeler, 70 had a credit card and 140

had a mobile phone. 40 of them had both, a two-wheeler and a

credit card, 30 had both, a credit card and a mobile phone and 60

had both, a two wheeler and mobile phone and 10 had all three.

How many candidates had none of the three?

A. 0 B. 20 C. 10 D. 18 E. 25

10). In a class of 40 students, 12 enrolled for both English and

German. 22 enrolled for German. If the students of the class

enrolled for at least one of the two subjects, then how many

students enrolled for only English and not German?

A. 30 B. 10 C. 18 D. 28 E. 32

11). In a class 40% of the students enrolled for Math and 70%

enrolled for Economics. If 15% of the students enrolled for both

Math and Economics, what % of the students of the class did not

enroll for either of the two subjects?

A B

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A. 5% B. 15% C. 0% D. 25% E. None of these

12).There are three different cable channels namely Ahead, Luck

and Bang. In a survey it was found that 850 viewers respond to

Bang, 200 to Luck, and 300 to Ahead. 200 viewers respond to

exactly two channels and 50 to none. What percentage of the

viewers responded to all three?

A. 10 B. 12 C. 14 D. None of these

13). Assuming 20% respond to Ahead and Bang, and 16%

respond to Bang and Luck, what is the percentage of viewers who

watch only Luck?

A. 20 B. 10 C. 16 D. None of these

14). In a survey about the popularity of magazines A, B and C

among 200 pupils, 78 pupils like A, 72 like B, 90 like C, 28 like A

and B only, 22 like B and C only, 10 like A and C only. Find the

maximum number of pupils who do not like any of the magazines?

A. 20 B. 24 C. 16 D. None of these

CUBES

Directions to Solve

The following questions are based on the information given below:

1. A cuboid shaped wooden block has 6 cm length, 4 cm breadth

and 1 cm height.

2. Two faces measuring 4 cm x 1 cm are coloured in black.

3. Two faces measuring 6 cm x 1 cm are coloured in red.

4. Two faces measuring 6 cm x 4 cm are coloured in green.

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5. The block is divided into 6 equal cubes of side 1 cm (from 6 cm

side), 4 equal cubes of side 1 cm(from 4 cm side).

1. How many cubes having red, green and black colours on at

least one side of the cube will be formed?

A.16 B.12 C.10 D.4

2. How many small cubes will be formed?

A.6 B.12 C.16 D.24

3. How many cubes will have 4 coloured sides and two non-

coloured sides ?

A.8 B.4 C.16 D.10

4. How many cubes will have green colour on two sides and rest of

the four sides having no colour?

A.12 B.10 C.8 D.4

5. How many cubes will remain if the cubes having black and

green coloured are removed?

A.4 B.8 C.12 D.16

Directions to Solve

The following questions are based on the information given below:

1. All the faces of cubes are painted with red colour.

2. The cubes is cut into 64 equal small cubes.

1. How many small cubes have only one face coloured?

A.4 B.8 C.16 D.24

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2. How many small cubes have no faces coloured ?

A.24 B.8 C.16 D.0

3. How many small cubes are there whose three faces are

coloured ?

A.4 B.8 C.16 D.24

4. How many small cubes are there whose two adjacent faces are

coloured red ?

A.0 B.8 C.16 D.24

Directions to Solve

All the six faces of a cube of a cube are coloured with six different

colours - black, brown, green, red, white and blue.

1. Red face is opposite to the black face.

2. Green face is between red and black faces.

3. Blue face is adjacent to white face.

4. Brown face is adjacent to blue face.

5. Red face is in the bottom.

1. The upper face is

A. White B. Black C. Brown D. None of these

2. The face opposite to brown is _________

A. Blue B. White C. Green D. Red

3. Which of the following is adjacent to green?

A. Black, white, brown, red B. Blue, black, red, white

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C. Red, black, blue, white D. None of these

4. Which face is opposite to green?

A. Red B. White C. Blue D. Brown

Ratio Proportion and Mixtures Solutions

Work Sheet

Ratio Proportion

1. If a:b=3:7,find the value of (5a+b):(4a+5b)

(a)15:44 (b)22:35 (c)15:49 (d)22:47

2. Which of the following ratios is greatest?

(a) 7:15 (b)22:35 (c)15:49 (d)21:29

3. If 76 is divided in to four parts proportional to 7,5,3,4 the smallest part

is

(a) 12 (b) 15 (c)16 (d)19

4. Suppose x varies inversely as square of y and when y=3 and x=4. Find

x, when y=6

(a) 1 (b)2 (c)3 (d)4

5. Divide Rs. 1162 among A, B, C in the ratio 35:28:20

(a) 120 (b)280 (c)170 (d)210

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6. A bag contains one rupee, 50 paisa and 25 paisa coins in the ratio

1:2:4.If the total amount is Rs. 75,then find the number of 50 paisa coins

in the bag

(a)25 (b)50 (c)75 (d)100

7. If a is 75% of b,b is 150% of c and d is 25% of c,then find a:d

(a)9:1 (b)9:2 (c)8:3 (d)8:1

8. The present ages of two persons are in the ratio 7:8.Twenty years ago

the ratio of their ages was 9:11.Find the present age of the older son

(a)64 years (b)72 years (c)56 years (d)40 years

9. A certain sum is divided among A,B and C in a manner that for every

rupee that A gets, B gets 75 paisa and for every rupee that B gets,C gets

50 paisae.If B's share in the total sum is Rs 840.Find the share of A

(a)Rs 2380 (b)Rs 2240 (c)Rs 1750 (d)Rs 1120

10.A certain amount of money is divided among A,B and C such that A

gets half of what B and C gets together. B gets one third of what A and C

together get.If A got Rs 500 more than B, then how much money was

divided?

(a)Rs4500 (b)Rs. 6000 (c)Rs. 8000 (d)none of these

11. Annual income of A and B are in the ratio 4:3 and their expenses, as

3:2. If each saves Rs 600 at the end of the year, find the annual income of

each

(a) 1200, 900 (b) 1500, 1125

(c) 2400, 1800 (d) 3600, 2700

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12. The value of a silver coin varies directly as the square of its diameter

when the thickness remains the same and directly as its thickness while

the diameter is constant.

Two silver coins have their diameter as 4:3. Find the ratio of their

thickness if the value of the first is 4 times the value of the second.

(a) 4:3 (b) 3:2 (c) 2:1 (d) 9:4

Mixture Solutions

1.15 litres of mixture contains 20% alcohol and the rest water. If 3 litres

of water be mixed with it, the percentage of alcohol in the new mixture

would be

(a)15 b)16 2/3 (c)17 (d)none of these

2.85 kg of a mixture contains milk and water in the ratio 27:7.How much

more water is to be added to get a new mixture containing milk and water

in the ratio 3:1?

(a)5kg (b)6.5kg (c)7.25kg (d)8kg

3. Two liquids are mixed in the ratio 3:2 and the solution is sold at Rs.11

per liter at 10%

Profit. If the first liquid costs Rs.2 more than the second, what is the cost

price of the first liquid?

(a) Rs.8.80 (b) Rs.10.80 (c) Rs.9.80 (d) none

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4. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third

variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 kg, the price

of the third variety per kg will be:

(a) Rs. 169.50 (b) Rs. 170 (c)Rs. 175.50 (d) Rs. 180

5. In what ratio must a person mix three kinds of tea costing Rs.60/kg,

Rs.75/kg and

Rs.100 /kg, so that the resultant mixture when sold at Rs.96/kg yields a

profit of 20%?

(a)1 : 2 : 4 (b)3 : 7 : 6 (c)1 : 4 : 2 (d)None of these

6. Two liquids A and B are in the ratio 5:1 in container 1 and in container

2, they are in the ratio 1:3. In what ratio should the contents of the two

containers be mixed so as to obtain a mixture of A and B in the ratio 1:1?

(a) 2 : 3 (b) 4 : 3 (c) 3 : 2 (d) 3 : 4

7. 85% kg of a mixture contains milk and water in the ratio 27:7. How

much more water is to be added to get a new mixture containing milk and

water in the ratio 3:1?

(a) 5 kg (b) 6.5 kg (c) 7.25 kg (d) 8 kg

8. Find the proportion in which 2 kinds of tea costing Rs.3 and Rs.3.30

per pound are mixed up to produce a mixture of cost Rs. 3.20 per pound?

(a) 2:1 (b) 1:2 (c) 3:2 (d) 2:3

9. A grocer sells one kind of tea at Rs.2.70 per kg and loss is 10% and

another at Rs.4.50 per kg and gains 12 ½ %. How should the two

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quantities of tea be mixed so that the mixture may be sold at Rs.3.95 per

kg at the profit of 25%?

(a) 42:8 (b) 16:2 (c) 21:4 (d) 4:21

10. In a mixture 60 litres, the ratio of milk and water 2 : 1. If the this ratio

is to be 1 : 2, then the quanity of water to be further added is:

(a) 20 litres (b) 30 litres (c) 40 litres (d) 60 litres

PROFIT AND LOSS

WORK SHEET

1. By selling a book for Rs.115.20, a man losses 10%. At what price

should he sell it to gain?

a). Rs.117.50 b). Rs.134.40 c). Rs.120 d). Rs.5

2. Ashok purchased a radio set and sold it to Shyam at a profit of 25%

and Shyam sold it to Mohan at a loss of 10% and Mohan paid Rs.675 for

it. For how much did Ashok purchase it?

a). 599 b). 674 c). 600 d). 300

3. A man sold two houses for Rs.6,75,958 each. On one he gains 16%

while on the other he loses 16%. How much does he gain or lose in the

whole transaction?

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a). 2.56% b). 25.6% c). 15.6% d). 1.56%

4. A grain dealer cheats to the extent of 10% while buying as well as

selling by using false weights. His total gain is

a). 200% b). 8.34% c). 21% d). 12%

5. A dishonest dealer professes to sell his goods at cost price but uses a

weight of 960 gms, for a kg weight. Find his gain percent.

a). 6 ¼% b). 8.5% c). 4 2/3% d). 4 1/6%

6. Find a single discount equivalent to series discount of 20%, 10% and

5%?

a). 31.6% b). 3.16% c). 1.8% d). 0.31%

7. If C.P of 21 oranges is equal to S.P of 18 oranges, then profit percent

is____?

a). 15 2/3% b). 16 2/3% c). 21/18% d). 3%

8. A man sold two horses for Rs.3000 each gaining 25% on the one and

losing 25% on the other. His loss percent is___?

a). 6.15% b). 5.25% c). 6.25% d). 25%

9. A man buys lemon at the rate of 9 for 80 paisa and sells them at 11 for

120 paisa. His gain per lemon is

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a). 200/99 b). 100/99 c). 99/200 d). 99/100

10. A man sells an article at 5% above its C.P. If he had bought it at 5%

less than what he paid for it and sold it for Rs.2 less, he would have

gained 10%. Find the C.P of the article?

a). 401 b). 399 c). 400 d). 991

11. If a person makes a profit of 10% on one fourth of the quantity sold

and a loss of 20% on the rest, then what is his average percent profit or

loss?

a). 12.5% loss b). 12.5 % profit c). 11.9% profit d). 12% loss

12. An article was sold at a profit of 10%. Had it been sold at a loss of

20%, the S.P would have been Rs.90 less. Find the C.P of the article.

a). 299 b). 399 c). 328.5 d). 300

13. A person sold his watch for Rs.75 and got a % profit equal to C.P.

Find the C.P of the watch?

a). 50 b). 75 c). 25 d). 100

14. If goods be purchased for Rs.450 and one-third be sold at a loss of

10%, at what gain % should the remainder be sold as to gain 20% on the

whole transaction?

a). 28% b). 35% c). 14.8% d). 100%

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15. A man loses Rs.20 by selling a toy at Rs.3 per piece and gains Rs. 30

by selling the same piece at Rs. 3.25 per piece. The number of pieces sold

by the man is?

a). 800 b). 5 c). 200 d). 199

16. A sold an article to B at a gain of 15%. B sold it to C at a gain of

20%. If C paid Rs.897, then A would have paid?

a). 6,500 b). 1,500 c). 65 d). 650

17. A man bought a horse and sold it at a gain of 10%. If he had bought

the horse at 20% less and sold it for Rs.100 more, he would have a profit

of 40%. The C.P of the horse is?

a). 5000 b). 50,000 c). 500 d). 499

18. A merchant blends two varieties of tea from two different tea gardens,

one costing Rs.45 per kg and other Rs.60 per kg in the ratio of 7:3

respectively. He sells the blended variety at Rs.54.45 per kg. His profit %

is?

a). no profit b). 1% c). 10% d). 12%

19. A shopkeeper professes to sell his articles on C.P. But he uses false

weight of 900 gm for 1 kg. His gain percentage is?

a). 11% b). 1/9% c). 9% d). 11 1/9%

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20. A publisher sells book to retail dealer at Rs.5 a copy but allows 25

copies to count as 24. If the retailer sells each of the 25 copies at Rs.6, the

profit percent made by him is?

a). 0.25% b). 25% c). 2.5% d). 250%

21. After getting two successive discounts, a shirt with a list price of

Rs.150 is available at Rs.105. If the second discount is 12.5%, find the

first discount?

a). 18.7% b). 45% c). 20% d). 2%

22. A merchant sold two dresses for Rs 2400 each. He made a 25% profit

on one and lost 20% on the other. What was his net gain or loss on the

sale of two dresses?

a). 120 b). 100 c). 130 d). 150

PROFIT AND LOSS

WORK SHEET

1. By selling a book for Rs.115.20, a man losses 10%. At what price

should he sell it to gain?

a). Rs.117.50 b). Rs.134.40 c). Rs.120 d). Rs.5

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2. Ashok purchased a radio set and sold it to Shyam at a profit of 25%

and Shyam sold it to Mohan at a loss of 10% and Mohan paid Rs.675 for

it. For how much did Ashok purchase it?

a). 599 b). 674 c). 600 d). 300

3. A man sold two houses for Rs.6,75,958 each. On one he gains 16%

while on the other he loses 16%. How much does he gain or lose in the

whole transaction?

a). 2.56% b). 25.6% c). 15.6% d). 1.56%

4. A grain dealer cheats to the extent of 10% while buying as well as

selling by using false weights. His total gain is

a). 200% b). 8.34% c). 21% d). 12%

5. A dishonest dealer professes to sell his goods at cost price but uses a

weight of 960 gms, for a kg weight. Find his gain percent.

a). 6 ¼% b). 8.5% c). 4 2/3% d). 4 1/6%

6. Find a single discount equivalent to series discount of 20%, 10% and

5%?

a). 31.6% b). 3.16% c). 1.8% d). 0.31%

7. If C.P of 21 oranges is equal to S.P of 18 oranges, then profit percent

is____?

a). 15 2/3% b). 16 2/3% c). 21/18% d). 3%

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8. A man sold two horses for Rs.3000 each gaining 25% on the one and

losing 25% on the other. His loss percent is___?

a). 6.15% b). 5.25% c). 6.25% d). 25%

9. A man buys lemon at the rate of 9 for 80 paisa and sells them at 11 for

120 paisa. His gain per lemon is

a). 200/99 b). 100/99 c). 99/200 d). 99/100

10. A man sells an article at 5% above its C.P. If he had bought it at 5%

less than what he paid for it and sold it for Rs.2 less, he would have

gained 10%. Find the C.P of the article?

a). 401 b). 399 c). 400 d). 991

11. If a person makes a profit of 10% on one fourth of the quantity sold

and a loss of 20% on the rest, then what is his average percent profit or

loss?

a). 12.5% loss b). 12.5 % profit c). 11.9% profit d). 12% loss

12. An article was sold at a profit of 10%. Had it been sold at a loss of

20%, the S.P would have been Rs.90 less. Find the C.P of the article.

a). 299 b). 399 c). 328.5 d). 300

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13. A person sold his watch for Rs.75 and got a % profit equal to C.P.

Find the C.P of the watch?

a). 50 b). 75 c). 25 d). 100

14. If goods be purchased for Rs.450 and one-third be sold at a loss of

10%, at what gain % should the remainder be sold as to gain 20% on the

whole transaction?

a). 28% b). 35% c). 14.8% d). 100%

15. A man loses Rs.20 by selling a toy at Rs.3 per piece and gains Rs. 30

by selling the same piece at Rs. 3.25 per piece. The number of pieces sold

by the man is?

a). 800 b). 5 c). 200 d). 199

16. A sold an article to B at a gain of 15%. B sold it to C at a gain of

20%. If C paid Rs.897, then A would have paid?

a). 6,500 b). 1,500 c). 65 d). 650

17. A man bought a horse and sold it at a gain of 10%. If he had bought

the horse at 20% less and sold it for Rs.100 more, he would have a profit

of 40%. The C.P of the horse is?

a). 5000 b). 50,000 c). 500 d). 499

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18. A merchant blends two varieties of tea from two different tea gardens,

one costing Rs.45 per kg and other Rs.60 per kg in the ratio of 7:3

respectively. He sells the blended variety at Rs.54.45 per kg. His profit %

is?

a). no profit b). 1% c). 10% d). 12%

19. A shopkeeper professes to sell his articles on C.P. But he uses false

weight of 900 gm for 1 kg. His gain percentage is?

a). 11% b). 1/9% c). 9% d). 11 1/9%

20. A publisher sells book to retail dealer at Rs.5 a copy but allows 25

copies to count as 24. If the retailer sells each of the 25 copies at Rs.6, the

profit percent made by him is?

a). 0.25% b). 25% c). 2.5% d). 250%

21. After getting two successive discounts, a shirt with a list price of

Rs.150 is available at Rs.105. If the second discount is 12.5%, find the

first discount?

a). 18.7% b). 45% c). 20% d). 2%

22. A merchant sold two dresses for Rs 2400 each. He made a 25% profit

on one and lost 20% on the other. What was his net gain or loss on the

sale of two dresses?

a). 120 b). 100 c). 130 d). 150

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PERCENTAGE

WORKSHEET I

1. The length and breadth of a rectangle are changed by +20% and by -10%

respectively. What is the percentage change in the area of the rectangle?

(a) 4% decrease (b) 4 % increase (c) 8 % decrease (d) 8%

increase

2. If A’s income is 33% more than that of B, then how much % is B’s

income is less than that of A?

(a) 17.8% (b) 24.8% (c) 23.8% (d) 21%

3. If the price of tea is increased by 20%, find by how much percent must a

house holder reduce her consumption of tea, so as not to increase the

expenditure?

(a) 4% (b) 13 3/2% (c) 16 2/3 % (d) 16%

4. The population of a town is 1, 76,400. If it increases at the rate of 5% per

annum, what will be its population 2 years hence? What was it 2 years ago?

(a) 1,94,481 and 1,60,000 (b) 2,00,000 and 1,28,500

(c)1,55,800 and 2,78,000 (d) 1,10,100 and 1,76,400

5. A man spent 20% of his monthly income on house rent. Out of the

remaining, he spent 70% on food. If he had balance of Rs.250 at month

end, the monthly income of the man is approximately:

(a) 1042 rupees (b) 2000 rupees (c) 1800 rupees (d) 900 rupees

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6. A woman has a certain number of apples, of which 13% are bad. She

gives 75% of the remainder to charity, and then has 261 left. How many did

she have initially?

(a) 1300 (b) 1200 (c) 1250 (d) 610

7. In an election between 2 candidates, Bhiku gets 65% of the total valid

votes. If the total number of votes is 12000, what is the number of votes

that the other candidate Sourabh gets if 25% of the total votes polled were

declared invalid?

(a) 3000 (b) 4150 (c) 5150 (d) 3150

8. 80 litres of milk and water contains 5% water in it. How much more

water must be added to the mixture to make it 25% in the resulting mixture?

(a) 10 litres (b) 21.33 litres (c) 20 litres (d) 25 litres

This is the required quantity of water.

9. A’s salary is 20% lower than B’s salary which is 25% lower than C’s

salary. By what percent is C’s salary more than A’s salary?

(a) 60% (b)65% (c)66.67% (d)70%

10. A’s salary is first decreased by 30% and then increased by 40%. The

result

is the same as B’s salary being first increased by 30% and then decreased

by 40%. Find the ratio of B’s salary to A’s salary.

(a) 49:39 (b) 47:37 (c) 19:17 (d) 7:5

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11. A water melon weighs 5000gms. 99% of its weight is water. It is kept in

a drying room and after some time it turns out that only 98% of its weight is

water. What is its weight now?

(a) 4500 (b) 2500 (c) 4950 (d) none

12. In an examination 65% of the students passed in Science, 68% passed in

Mathematics and 25% failed in both. Find the pass percentage?

(a) 55% (b) 58% (c) 60% (d) 75%

13. In a class of 52 students, 25% are rich and the others are poor. There

are 20 female members in the class, of whom 55% are poor. How many

rich male members are there in the class?

(a) 4 (b) 8 (c) 5 (d) 6

14. At King’s college, 60% of the students are boys and the rest are girls.

Further, 15% of the boys and 7.5% of the girls are getting a fee waiver. If

the number of those getting a fee waiver is 90, find the total number of

students getting 50% fee-concession, if it is given that 10% of those not

getting a fee-waiver are eligible for half-fee concession?

(a) 64 (b) 66 (c) 68 (d) 63

15. A person fails in an exam for want of 12 marks. Had he got 27 marks

more he would have passed by scoring a clear 50% in the exam. Find the

maximum marks given that the pass percentage is 40?

(a) 100 (b) 175 (c) 300 (d) 150

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16. A pile of coins is to be divided amongst 6 people A,B,C,D,E,F in that

order such that each gets twice the other . If A gets the least and F the

highest, what is the % increase in number of coins received by (a)F over C

and (b)E over B?

(a) 700%, 700% (b) 600%, 500% (c) 400%, 300% (d) 500%, 400%

17. Anindo’s project report consists of 25 pages each of 60 lines with 75

characters on each line. In case, the number of lines is reduced to 50, but

the number of characters is increased to 100 per line, what is the percentage

change in the number of pages?

(a) 10% decrease (b) 12% decrease (c) 20% increase (d) 5% decrease

ASSIGNMENT SHEET –I

1. At an election where there are only two candidates. The candidate who

gets 32% of the votes is elected by a majority of 144 votes. Find the total

number of votes recorded assuming that no vote is void.

(a) 500 (b) 300 (c) 450 (d) 600

2. A pie-chart giving the marks distribution in a certain exam shows that

20% got 80%, 25%got 70%, 25%got 60% and the rest got 50%. Find the

average percentage of marks scored by the students.

(a) 65 (b) 60 (c) 61 (d) 63.5

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3. A man buys a truck for Rs.2,50,000. The annual repair cost comes to

2.0% of the price of purchase. Besides, he has to pay an annual tax of

Rs.2000. At what monthly rent must he rent out the truck to get a return of

24% on his net annual investment?

(a) 4000 rupees (b) 3140 rupees (c) 5140 rupees (d) 4140 rupees

4. The value of a machine depreciates at 20 % per annum. If the present

value of the machine is Rs.10, 000, find its value after three years.

(a) 5000 rupees (b) 5120 rupees (c) 5400 rupees (d) 6000 rupees

5. In an exam, 20% failed in English and 15% failed in Maths. If 12%

failed in both the exams, find the pass percentage.

(a) 70% (b) 72% (c) 77% (d) 75%

6. The side of a square is increased by 10%. Find the percentage change in

its area.

(a) 21% (b) 30% (c) 10% (d) 100%

7. Vicky’s salary is 50% more than Rahul’s. Vicky got a raise of 40% on

his salary while Rahul got a 25% raise on his salary. By what percent is

Vicky’s new salary more than Rahul’s?

(a) 56% (b) 68% (c) 72% (d) 84%

8. An ore contains 25% of an alloy that has 80% iron. Other than this, in

the remaining 75% of the ore, there is no iron. How many kg of the ore are

needed to obtain 60 kgs of pure iron?

(a) 250 kg. (b) 275 kg. (c) 350 kg. (d) 300 kg.

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9. Sandeep spends 20% of his salary on house rent, and 20% of the rest he

spends on his children’s education and 14% of his total salary he spends on

clothes. After his expenditure, he is left with Rs.7000. What is his total

salary?

(a) 15000 rupees (b) 14000 rupees (c) 9000 rupees (d) 12000

rupees

10. Fresh fruit contains 68% water and dry fruit contains 20% water. How

much

dry fruit can be obtained from 100 kgs of fresh fruits?

(a) 25 kg (b) 40 kg (c) 30 kg (d) 20 kg

11. If A’s height is 24% less than that of B, then how much percent is B’s

height more than that of A?

(a) 91.0% (b) 31.57% (c) 29.1% (d) 13.8%

12. If the price of sugar decreases by 20%, find by how much percent must

a house holder increases her consumption of sugar so as not to decrease the

expenditure?

(a) 27% (b) 26.5% (c) 25% (d) 25.5%

13. The value of the machine depreciates at the rate of 10% per annum, if

its present value is Rs. 16,200 what would be its worth after 2 years? What

was the value of the machine 2 years ago?

(a) 1,94,481 and 1,60,000 (b) 2,00,000 and 1,28,500

(c) 1,55,800 and 2,78,000 (d) 1,31,220 and 2,00,000

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LOGICAL REASONING - Worksheet

CODING - DECODING

1. If a meaningful word can be formed by rearranging the letters

USCALA, the first letter so formed is the answer. If no such letter can be

formed, the answer is X.

(a) C (b) A (c) S (d) L

2. If Cushion is called Pillow, Pillow is called mat, mat is called bed

sheet, and bed sheet is called cover, which will be spread on the floor?

(a) Pillow (b) Mat (c) Bed sheet (d) Cover

3. If ROTARY is coded as 36 and ROTARIAN is coded as 64, how will

you code ROTARACT?

(a) 64 (b) 26 (c) 72 (d) 36

4. In a certain code language

‘tom na rod’ means ‘give me sweet’

‘jo ta rod’ means ‘you and me’

‘pot ta noc’ means ‘you are good’

‘jo mit noc’ means ‘good and bad’

Which of the following represents ‘bad’ in the language?

(a)mit (b)noc (c)jo (d)rod

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5. In a certain code language LANDMINE is written as PYRBQGRC.

How will HOMEMADE be written in that code language?

(a)LMPCPZID (b)LMPCOYHC (c)LMQCQYHC (d)LMOCOZID

6. If in the word ‘DISTURBANCE’ the first letter is interchanged with

the last letter, the second letter is interchanged with the tenth letter and so

on, which letter would come after the letter “T” in the newly formed

word?

(a) I (b) N (c) S (d) U

7. In a certain code language-

‘743’ means ‘Mangoes are good’.

‘657’ means ‘Eat good food’

‘934’ means ‘Mangoes are ripe’.

Which digit means ‘ripe’ in that language ?

(a) 5 (b) 4 (c) 9 (d) 7

BLOOD RELATIONSHIP

8. James goes to the local church and sees a man sitting to his left. He

finds that the man is his relative. The man is the husband of the sister of

his mother. How is the man related to James?

(a) Brother (b) Uncle (c) Nephew (d) Father

9. Pointing to a photograph a man said “He is my brother’s mother’s

brother’s only sibling’s son.” Whose photograph was it?

(a) His own (b) Brother (c) Cousin (d) Cannot say

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10. C is A’s father’s nephew. D is A’s cousin but not the brother of C.

How is D related to C?

(a) Cousin (b) Sister (c) Father (d) Cannot say

11. When Anil saw Mani, he recalled ‘He is the son of the father of the

mother of my daughter.’ How is Mani related to Anil?

(a) Brother-in-law (b) Brother (c) Cousin (d) Uncle

12. Pointing out to a lady, a girl said,” She is the daughter-in-law of the

grandmother of my father’s only son.” How is the lady related to the girl?

(a) Sister-in-law (b)Aunt (c) Mother (d) mother-in-law

SYLLOGISM

Directions Q 11 to Q 13

Each question consists of two statements followed by four options

consisting of two options put together in a specific order. Choose the

option which indicates a valid argument, that is, the third statement is a

conclusion drawn from the preceding two statements.

13. All shares are debentures. No debentures are deposits.

(a) No shares are deposits

(b) Some shares are deposits

(c) All shares are deposits

(d) None of the above

14. Some students are smart. All students are hardworking

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(a) some hardworking students are smart

(b) some smart students are hardworking

(c) Both (a) and (b)

(d) None of the above

15.Some actors are singers. All the singers are dancers.

1. Some actors are dancers.

2. No singer is actor.

A. Only (1) conclusion follows

B. Only (2) conclusion follows

C. Either (1) or (2) follows

D. Neither (1) nor (2) follows

E. Both (1) and (2) follow

16. All scientists are fools. All fools are literates.

(a) All scientists are literates

(b) All literates are scientists

(c) No scientists are literates

(d) Both (a) and (b) are correct

17. Statements:

some roses are red

some red are black.

Conclusions:

(I) No black is a rose.

(II) No rose is a black

(a) I (b) II (c) Both (d) None of these

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ODD ONE OUT:

19 (a) Paradeep (b) Kalpakkam (c) Kota (d) Tarapur

20 (a) Silk (b) Cotton (c) Nylon (d) wool

21 (a) Gold (b) Silver (c) Bronze (d) Iron

22 (a) Motorcar (b) Tractor (c) Bus (d) Tram

23 (a) triangle (b) Pentagon (c) square (d) tangent

24. Find the odd one out

(a) Cubic metre (b) cubic centimetre (c) Gallons (d) Square metre

VENN-DIAGRAM:

25. Read the following Venn-diagram carefully and answer the question

given below:

Which letters represent girls who are athletes but not singers?

(a) C (b) A (c) D (d) E

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26. Study the following figure and answer the questions given below.

i). How many educated people are employed ?

A. 9 B. 18

C. 20 D. 15

ii). How many backward people are educated ?

A. 9 B. 28

C. 14 D. 6

iii). How many backward uneducated people are employed ?

A. 14 B. 5

C. 7 D. 11

iv). How many backward people are not educated ?

A. 3 B. 14 C. 22 D. 25

NUMBER SERIES

27. Find the next of the series. 83, 82, 81, 77, 69, 60, ?

(a) 54 (b) 33 (c) 35 (d) 52

28. Find the missing term of the series. 0, 2, 6, ?, 20, 30, 42.

(a) 8 (b) 10 © 12 (d) 14

29. Look at this series: 36, 34, 30, 28, 24, ... What number should come

next?

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A.20 B.22 C.23 D.26

30. Look at this series: 22, 21, 23, 22, 24, 23, ... What number should

come next?

A.22 B.24 C.25 D.26

31. Look at this series: 53, 53, 40, 40, 27, 27, ... What number should

come next?

A.12 B.14 C.27 D.53

33. Raji’s house is to the right of Vel’s house at a distance of 20 metres in

the same row facing north. Shakthi’s house is in the north-east direction

of Vel’s house at a distance of 25 metres. Determine Raji’s house is in

which direction with respect to Shakthi’s house.

(a) South (b) South-east (c) South-west (d) None of these

34. In a row of boys, Rajan is tenth from the right and Suraj is tenth from

the left. When Rajan and Suraj interchange their positions, Suraj will be

twenty-seven from the left. Which of the following will be Rajan’s

position from the right?

(a) Twenty fifth (b) Twenty sixth (c) Twenty seven (d)

Twenty eight

1. CUBOID

Let length = l, breadth = b and height = h units. Then

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i. Volume = (l x b x h) cubic units.

ii. Surface area = 2(lb + bh + lh) sq. units.

iii. Diagonal = l2 + b

2 + h

2 units.

2. CUBE

Let each edge of a cube be of length a. Then,

i. Volume = a3 cubic units.

ii. Surface area = 6a2 sq. units.

iii. Diagonal = 3a units.

3. CYLINDER

Let radius of base = r and Height (or length) = h. Then,

i. Volume = ( r2h) cubic units.

ii. Curved surface area = (2 rh) sq. units.

iii. Total surface area = 2 r(h + r) sq. units.

4. CONE

Let radius of base = r and Height = h. Then,

i. Slant height, l = h2 + r

2 units.

ii. Volume = r2h cubic units.

iii. Curved surface area = ( rl) sq. units.

iv. Total surface area = ( rl + r2) sq. units.

5. SPHERE

Let the radius of the sphere be r. Then,

i. Volume = r3

cubic units.

ii. Surface area = (4 r2) sq. units.

6. HEMISPHERE

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Let the radius of a hemisphere be r. Then,

i. Volume = r3

cubic units.

ii. Curved surface area = (2 r2) sq. units.

iii. Total surface area = (3 r2) sq. units.

Note: 1 litre = 1000 cm3.

FUNDAMENTAL CONCEPTS

1. Results on Triangles:

i. Sum of the angles of a triangle is 180°.

ii. The sum of any two sides of a triangle is greater

than the third side.

iii. Pythagoras Theorem:

In a right-angled triangle, (Hypotenuse)2 =

(Base)2 + (Height)2.

iv. The line joining the mid-point of a side of a triangle

to the positive vertex is called the median.

v. The point where the three medians of a triangle

meet, is called centroid.The centroid divided each

of the medians in the ratio 2 : 1.

vi. In an isosceles triangle, the altitude from the vertex

bisects the base.

vii. The median of a triangle divides it into two

triangles of the same area.

viii. The area of the triangle formed by joining the mid-

points of the sides of a given triangle is one-fourth

of the area of the given triangle.

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2. Results on Quadrilaterals:

i. The diagonals of a parallelogram bisect each other.

ii. Each diagonal of a parallelogram divides it into

triangles of the same area.

iii. The diagonals of a rectangle are equal and bisect

each other.

iv. The diagonals of a square are equal and bisect each

other at right angles.

v. The diagonals of a rhombus are unequal and bisect

each other at right angles.

vi. A parallelogram and a rectangle on the same base

and between the same parallels are equal in area.

vii. Of all the parallelogram of given sides, the

parallelogram which is a rectangle has the greatest

area.

IMPORTANT FORMULAE

I. 1. Area of a rectangle = (Length x Breadth).

Length = Area

and Breadth = Area

. Breadth Length

II. 2. Perimeter of a rectangle = 2(Length + Breadth)

III. Area of a square = (side)2 = (diagonal)2.

IV. Area of 4 walls of a room = 2 (Length + Breadth) x

Height.

V. 1. Area of a triangle = x Base x Height. (Or) x any

side x length of perpendicular dropped on that side

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2. Area of a triangle = s(s-a)(s-b)(s-c)

where a, b, c are the sides of the triangle and s =

(a + b + c).

3. Area of an equilateral triangle = 3

x (side)2. 4

4. Radius of incircle of an equilateral triangle of

side a =

a .

23

5. Radius of circumcircle of an equilateral triangle of side a =

a .

3

6. Radius of incircle of a triangle of area and semi-

perimeter s =

s

In a right angled triangle, the radius of the incircle is given by the

following relation

.

VI. 1. Area of parallelogram = (Base x Height).

2. Area of a rhombus = x (Product of diagonals).

3. Area of a trapezium = x (sum of parallel sides) x

distance between them.

VII. 1. Area of a circle = R2, where R is the radius.

2. Circumference of a circle = 2 R.

3. Length of an arc =

2 R

, where is the central angle.

360

4. Area of a sector = 1 (arc x R) =

R2 .

2 360

VIII. 1. Circumference of a semi-circle = R.

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2. Area of semi-circle = R2

. 2

MENSURATION – Worksheet

1). A sizes of a triangle are 25m, 39m and 66m

respectively. Find the perpendicular from the opposite

angles and the greatest side?

A. 13m B. 15m C. 9.5m D. 6m

2). The perimeter of an equilateral triangle is 15m. Find

the length of the perpendicular and the area of the

triangle?

3). The parallel sides of a trapezoid are 8m and 6m and

its altitude is 3m. Find the area of the trapezoid?

A. 21m2 B. 31m2 C. 12m2 D. 13m2

4). Find the area and the circumference of a circle whose

radius is 3.5m?

5). A piece of wire is bent in the shape of an equilateral

triangle of each side 6.6m, it is re-bent to form a circular

ring. Find the diameter of the ring?

A. 3.6m B. 6.6m C. 6.3m D. 6m

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6). A right triangle with sides 3 cm, 4 cm and 5 cm is

rotated the side of 3 cm to form a cone. The volume of

the cone so formed is:

A. 12∏cm3 B. 15∏cm3 C. 16∏cm3 D. 20∏cm3

7). In a shower, 5 cm of rain falls. The volume of water

that falls on 1.5 hectares of ground is:

A. 75 cu. m B. 750 cu. m C. 7500 cu. m D.75000 cu.

m

8). The slant height of a right circular cone is 10 m and its

height is 8 m. Find the area of its curved surface.

A. 30∏m2 B. 40∏m2 C. 60∏m2 D. 80∏m2

9). The curved surface area of a cylindrical pillar is 264

m2 and its volume is 924 m3. Find the ratio of its

diameter to its height.

A. 3 : 7 B. 7 : 3 C. 6 : 7 D. 7 : 6

10). What is the total surface area of a right circular cone

of height 14 cm and base radius 7 cm?

A. 344.35 cm2 B. 462 cm2 C. 498.35 cm2 D.

None of these

11). A boat having a length 3 m and breadth 2 m is

floating on a lake. The boat sinks by 1 cm when a man

gets on it. The mass of the man is:

A. 12 kg B. 60 kg C. 72 kg D. 96 kg

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12). 50 men took a dip in a water tank 40 m long and 20

m broad on a religious day. If the average displacement

of water by a man is 4 m3, then the rise in the water level

in the tank will be:

A. 20 cm B. 25 cm C. 35 cm D. 50 cm

13). How many bricks, each measuring 25 cm x 11.25 cm

x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5

cm?

A. 5600 B. 6000 C. 6400 D. 7200

14). A hall is 15 m long and 12 m broad. If the sum of the

areas of the floor and the ceiling is equal to the sum of

the areas of four walls, the volume of the hall is:

A. 720m3 B. 900m3 C. 1200m3 D. 1800m3

15). The ratio between the length and the breadth of a

rectangular park is 3 : 2. If a man cycling along the

boundary of the park at the speed of 12 km/hr completes

one round in 8 minutes, then the area of the park (in sq.

m) is:

A. 15360m2 B. 153600m2 C. 30720m2 D. 307200m2

16). The percentage increase in the area of a rectangle, if

each of its sides is increased by 20% is:

A. 40% B. 42% C. 44% D. 46%

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17). A towel, when bleached, was found to have lost 20%

of its length and 10% of its breadth. The percentage of

decrease in area is:

A. 10% B. 10.08% C. 20% D. 28%

18). The diagonal of a rectangle is 41 cm and its area is

20 sq. cm. The perimeter of the rectangle must be:

A. 9 cm B. 18 cm C. 20 cm D. 41 cm

19). The difference between the length and bradth of a

rectangle is 23 m. If its perimeter is 206 m, then its area

is:

A. 1520 m2 B. 2420 m2 C. 2480 m2 D. 2520 m2

20). The length of a rectangle is halved, while its breadth

is tripled. What is the percentage change in area?

A. 25% increase B. 50% increase C. 50% decrease

D. 75% decrease

GEOMETRY– Worksheet

1. Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5),

C(9, 9) and D(5, 4). What is the shape of the

quadrilateral?

A. Square B.Rectangle but not a square C.Rhombus

D.Parallelogram but not a rhombus

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2. What is the radius of the incircle of the triangle whose

sides measure 5, 12 and 13 units?

A. 2 units B. 12 units C. 6.5 units D. 6 units

3. How many diagonals does a 63-sided convex polygon

have?

A. 3780 B. 1890 C. 3843 D. 3906

4. What is the area of the following square, if the length

of BD is ?

A. 1 B.2 C.3 D.4

5. In the figure below, what is the value of y ?

A. 40 B.60 C.100 D.120

6. The square ABCD touches the circle at 4 points. The

length of the side of the square is 2 cm. Find the area of

the shaded region.

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A. π – 4 B. 2π – 4 C.3 π – 4 D. 4π – 4

7. Find the length of the hypotenuse of a right triangle if

the lengths of the other two sides are 6 inches and 8

inches.

8. Find the length of one side of a right triangle if the

length of the hypotenuse is 15 inches and the length of

the other side is 12 inches.

9. Find the length of the hypotenuse of a right triangle if

the lengths of the other two sides are both 3 inches.

10. How many triangles of perimeter 14 can be formed if

the sides measure integral values only?

A. 1 B.2 C.3 D.4

11. In the given figure, AD is the bisector of BAC , AB = 6 cm, AC = 5

cm and

BD = 3 cm. Find DC

A. 11.3 cm B. 2.5 cm C. 3.5 cm D. None of these

12. A toy has a hemisphere base and a conical top as

shown in the figure. The perpendicular height of the cone

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is 10 cm and radius of the hemisphere is 4 cm. Find the

volume of the toy.

A. 302 cm3 B. 408 cm3 C. 364 cm3 D. None of these

13. A rectangular tank is 50 m long and 29 m deep. If

2000 m3 of water is drawn off the tank, the level of the

water in the tank goes down by 4m, how many cubic

metre of the water can the tank hold?

A. 18,000 m3 B. 12,500 m3 C. 13,000 m3

D. 14,500 m3

14. The base of a prism is a triangle of which the sides

are 17, 28 and 25 m respectively. Volume of the prism is

2,100 m3. Find its height.

A. 14 m B. 10 m C. 12 m D. 15 m

15. Two sides of a triangle are 22 and 18. Which of the

following cannot be the area of the triangle?

A. 196 B. 197 C. 198 D. 199

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DATA SUFFICIENCY – WORKSHEET

Each of the questions given below consists of a statement and / or a

question and two statements numbered I and II given below it. You have

to decide whether the data provided in the statement(s) is / are sufficient

to answer the given question. Read the both statements and

Give answer (A) if the data in Statement I alone are sufficient to answer

the question, while the data in Statement II alone are not sufficient to

answer the question.

Give answer (B) if the data in Statement II alone are sufficient to answer

the question, while the data in Statement I alone are not sufficient to

answer the question.

Give answer (C) if the data either in Statement I or in Statement II alone

are sufficient to answer the question.

Give answer (D) if the data even in both Statements I and II together are

not sufficient to answer the question.

Give answer(E) if the data in both Statements I and II together are

necessary to answer the question.

1. What is the volume of 32 metre high cylindrical tank?

I. The area of its base is 154 m2.

II. The diameter of the base is 14 m.

A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

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D. Both I and II are not sufficient to answer

E. Both I and II are necessary to answer

2. What is the capacity of a cylindrical tank?

I. Radius of the base is half of its height which is 28 metres.

II. Area of the base is 616 sq. metres and its height is 28 metres.

( Same options from first problem)

3. What is the capacity of the cylindrical tank?

I. The area of the base is 61,600 sq. cm.

II. The height of the tank is 1.5 times the radius

III. The circumference of base is 880 cm.

A. Only I and II

B. Only II and III

C. Only I and III

D. Any two of the three

E. Only II and either I or III

4. Two cars pass each other in opposite direction. How long would they

take to be 500 km apart?

I. The sum of their speeds is 135 km/hr.

II. The difference of their speed is 25 km/hr.

( Same options from first problem)

5.How much time did X take to reach the destination?

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I. The ratio between the speed of X and Y is 3 : 4.

II. Y takes 36 minutes to reach the same destination.

( Same options from first problem)

6.The towns A, B and C are on a straight line. Town C is between A and

B. The distance from A to B is 100 km. How far is A from C?

I. The distance from A to B is 25% more than the distance from C to B.

II. The distance from A to C is 1/4 of the distance C to B.

( Same options from first problem)

Directions for 7 and 8

The problems consist of a question and two statements, labelled (I) and

(II), in which certain data are given. You have to decide whether the data

given in the statements are sufficient for answering the question.

Mark (a) if statement (I) alone is sufficient, but statement (II) alone is not

sufficient to answer the question;

Mark (b) if statement (II) alone is sufficient, but statement (I) alone is not

sufficient to answer the question;

Mark (c) if both statements (I) and (II) together are sufficient to answer

the question but neither statement alone is sufficient;

Mark (d) if each statement alone is sufficient to answer the question;

Mark if both statements (I) and (II) together are not sufficient to answer

the question and additional data specific to the problem are needed.

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7.The distance from Madhavi’s office to her house is 45 miles. On

Monday Madhavi went to the office for a while but returned home early.

what was the total time spent in the travelling?

I. She travelled at an uniform rate of 40 mph both in the onward as

well as the return journey.

II. If she travelled 50 mph faster then she actually did it would have

taken her half the time.

8. What is the speed of the train?

I. A motorcycle takes an hour more then the train to cover the same

distance of 120 kms.

II. The train moves 6 km faster than the motorcycle.

9.By selling an article what is the profit percent gained?

I. 5% discount is given on list price.

II. If discount is not given, 20% profit is gained.

III. The cost price of the articles is Rs. 5000.

A. Only I and II B. Only II and II

C. Only I and III D. All I, II and III

E. None of these

10. A man mixes two types of rice (X and Y) and sells the mixture at the

rate of Rs. 17 per kg. Find his profit percentage.

I. The rate of X is Rs. 20 per kg.

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II. The rate of Y is Rs. 13 per kg.

( Same options from first problem)

11. A shopkeeper sells some toys at Rs. 250 each. What percent profit

does he make? To find the answer, which of the following information

given in Statements I and II is/are necessary?

I. Number of toys sold.

II. Cost price of each toy.

A. Only I is necessary

B. Only II is necessary

C. Both I and II are necessary

D. Either I or II is necessary

E. None of these

12. A and B together can complete a task in 7 days. B alone can do it in

20 days. What part of the work was carried out by A?

I. A completed the job alone after A and B worked together for 5 days.

II. Part of the work done by A could have been done by B and C together

in 6 days.

( Same options from first problem)

13. How long will Machine Y, working alone, take to produce x candles?

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I. Machine X produces x candles in 5 minutes.

II. Machine X and Machine Y working at the same time produce x

candles in 2 minutes.

( Same options from first problem)

14. How many workers are required for completing the construction work

in 10 days?

I. 20% of the work can be completed by 8 workers in 8 days.

II. 20 workers can complete the work in 16 days.

III. One-eighth of the work can be completed by 8 workers in 5 days.

A. I only

B. II and III only

C. III only

D. I and III only

E. Any one of the three

15. The area of playground is 1600 m2. What is the perimeter?

I. It is a perfect square playground.

II. It costs Rs. 3200 to put a fence around the playground at the rate of

Rs. 20 per metre.

( Same options from first problem)

16. What is the area of a right-angled triangle?

I. The perimeter of the triangle is 30 cm.

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II. The ratio between the base and the height of the triangle is 5 : 12.

III. The area of the triangle is equal to the area of a rectangle of length 10

cm.

A. I and II only

B. II and III only

C. I and III only

D. III, and either I or II only

E. None of these

17. The area of a rectangle is equal to the area of right-angles triangle.

What is the length of the rectangle?

I. The base of the triangle is 40 cm.

II. The height of the triangle is 50 cm.

( Same options from first problem)

DATA INTERPRETATION - WORKSHEET

Problem 1:

Study the following table and answer the questions based on it

Expenditures of a Company (in Lakh Rupees) per Annum Over the given

Years.

Year

Item of Expenditure

Salary Fuel and

Transport Bonus

Interest on

Loans Taxes

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1998 288 98 3.00 23.4 83

1999 342 112 2.52 32.5 108

2000 324 101 3.84 41.6 74

2001 336 133 3.68 36.4 88

2002 420 142 3.96 49.4 98

1. What is the average amount of interest per year which the company

had to pay during this period?

A. Rs. 32.43 lakhs B. Rs. 33.72 lakhs

C. Rs. 34.18 lakhs D. Rs. 36.66 lakhs

2. The total amount of bonus paid by the company during the given

period is approximately what percent of the total amount of salary paid

during this period?

A. 0.1% B. 0.5% C. 1% D. 1.25%

3. Total expenditure on all these items in 1998 was approximately what

percent of the total expenditure in 2002?

A. 62% B. 66% C. 69% D. 71%

4. The total expenditure of the company over these items during the year

2000 is?

A. Rs. 544.44 lakhs B. Rs. 501.11 lakhs C. Rs. 446.46 lakhs D. Rs.

478.87 lakhs

5. The ratio between the total expenditure on Taxes for all the years and

the total expenditure on Fuel and Transport for all the years respectively

is approximately?

A. 4:7 B. 10:13 C. 15:18 D. 5:8

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Problem 2:

The bar graph given below shows the sales of books (in thousand

number) from six branches of a publishing company during two

consecutive years 2000 and 2001.

Sales of Books (in thousand numbers) from Six Branches - B1, B2,

B3, B4, B5 and B6 of a publishing Company in 2000 and 2001.

1. What is the ratio of the total sales of branch B2 for both years to the

total sales of branch B4 for both years?

A. 2:3 B. 3:5 C. 4:5 D. 7:9

2. Total sales of branch B6 for both the years is what percent of the total

sales of branches B3 for both the years?

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A. 68.54% B. 71.11% C. 73.17% D. 75.55%

3. What percent of the average sales of branches B1, B2 and B3 in 2001

is the average sales of branches B1, B3 and B6 in 2000?

A. 75% B. 77.5% C. 82.5% D. 87.5%

4. What is the average sales of all the branches (in thousand numbers) for

the year 2000?

A. 73 B. 80 C. 83 D. 88

5. Total sales of branches B1, B3 and B5 together for both the years (in

thousand numbers) is?

A. 250 B. 310 C. 435 D. 560

Problem 3:

The following pie-chart shows the percentage distribution of the

expenditure incurred in publishing a book. Study the pie-chart and the

answer the questions based on it.

Various Expenditures (in percentage) Incurred in Publishing a Book

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1. If for a certain quantity of books, the publisher has to pay Rs. 30,600

as printing cost, then what will be amount of royalty to be paid for these

books?

A. Rs. 19,450 B. Rs. 21,200 C. Rs. 22,950 D. Rs.

26,150

2. What is the central angle of the sector corresponding to the expenditure

incurred on Royalty?

A. 15º B. 24º C. 54º D. 48º

3. The price of the book is marked 20% above the C.P. If the marked

price of the book is Rs. 180, then what is the cost of the paper used in a

single copy of the book?

A. Rs. 36 B. Rs. 37.50 C.Rs. 42 D.Rs. 44.25

4. If 5500 copies are published and the transportation cost on them

amounts to Rs. 82500, then what should be the selling price of the book

so that the publisher can earn a profit of 25%?

A. Rs. 187.50 B. Rs. 191.50 C.Rs. 175 D.Rs. 180

5. Royalty on the book is less than the printing cost by:

A. 5% B.33 1/5% C.20% D.25%

Problem 4:

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The following line graph gives the ratio of the amounts of imports by a

company to the amount of exports from that company over the period

from 1995 to 2001.

Ratio of Value of Imports to Exports by a Company Over the Years.

1. If the imports in 1998 was Rs. 250 crores and the total exports in the

years 1998 and 1999 together was Rs. 500 crores, then the imports in

1999 was ?

A. Rs. 250 crores B. Rs. 300 crores C. Rs. 357 crores D. Rs. 420

crores

2. The imports were minimum proportionate to the exports of the

company in the year ?

A. 1995 B. 1996 C. 1997 D.2000

3. What was the percentage increase in imports from 1997 to 1998 ?

A. 72 B. 56 C. 28 D. Data inadequate

4. If the imports of the company in 1996 was Rs. 272 crores, the exports

from the company in 1996 was?

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A. Rs. 370 crores B. Rs. 320 crores C. Rs. 280 crores D. Rs. 275

crores

5. In how many of the given years were the exports more than the imports

?

A. 1 B. 2 C. 3 D. 4

CONTENTS

SL.NO. TOPIC PAGE

NO.

1 ACKNOWLEDGEMENT 3

2 BASICS OF COMPUTER 4-9

3 CPU 10-14

4 NETWORKING AND

INTERNET

15-17

5 COMPUTER DATA STORAGE 18-26

6 BASIC HARDWARE

COMPONENTS

27-29

7 COMPUTER NETWORKS 30

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8 OSI LAYERS 30-36

9 INTERFACES 37

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ACKNOWLEDGEMENT

We owe a great many thanks to a great many people who helped and

supported us during the writing of this report.

Our deep sense of gratitude to the trainees of HCL Technologies for their

support and guidance.

We would also thank our Institution and our faculty members without

whom this industrial training in HCL would have been a distant reality.

We also extend my heartfelt thanks to my family and well wishers.

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BASICS OF COMPUTERS.

A computer is a programmable machine designed to sequentially

and automatically carry out a sequence of arithmetic or logical

operations. The particular sequence of operations can be changed

readily, allowing the computer to solve more than one kind of

problem. An important class of computer operations on some

computing platforms is the accepting of input from human

operators and the output of results formatted for human

consumption. The interface between the computer and the human

operator is known as the user interface.

Conventionally a computer consists of some form of memory for

data storage, at least one element that carries out arithmetic and

logic operations, and a sequencing and control element that can

change the order of operations based on the information that is

stored. Peripheral devices allow information to be entered from an

external source, and allow the results of operations to be sent out.

A computer's processing unit executes series of instructions that

make it read, manipulate and then store data. Conditional

instructions change the sequence of instructions as a function of

the current state of the machine or its environment.

The first electronic digital computers were developed in the mid-

20th century (1940–1945). Originally, they were the size of a large

room, consuming as much power as several hundred modern

personal computers (PCs). In this era mechanical analog

computers were used for military applications.

Modern computers based on integrated circuits are millions to

billions of times more capable than the early machines, and

occupy a fraction of the space. Simple computers are small

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enough to fit into mobile devices, and mobile computers can be

powered by small batteries. Personal computers in their various

forms are icons of the Information Age and are what most people

think of as "computers". However, the embedded computers found

in many devices from mp3 players to fighter aircraft and

from toys to industrial robots are the most numerous.

First general-purpose computers

In 1801, Joseph Marie Jacquard made an improvement to

the textile loom by introducing a series of punched paper cards as

a template which allowed his loom to weave intricate patterns

automatically. The resulting Jacquard loom was an important step

in the development of computers because the use of punched

cards to define woven patterns can be viewed as an early, albeit

limited, form of programmability.

It was the fusion of automatic calculation with programmability that

produced the first recognizable computers. In 1837, Charles

Babbage was the first to conceptualize and design a fully

programmable mechanical computer, his analytical engine. Limited

finances and Babbage's inability to resist tinkering with the design

meant that the device was never completed ; nevertheless his son,

Henry Babbage, completed a simplified version of the analytical

engine's computing unit (the mill) in 1888. He gave a successful

demonstration of its use in computing tables in 1906. This machine

was given to the Science museum in South Kensington in 1910.

In the late 1880s, Herman Hollerith invented the recording of data

on a machine readable medium. Prior uses of machine readable

media, above, had been for control, not data. "After some initial

trials with paper tape, he settled on punched cards. To process

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these punched cards he invented the tabulator, and

the keypunch machines. These three inventions were the

foundation of the modern information processing industry. Large-

scale automated data processing of punched cards was performed

for the 1890 United States Census by Hollerith's company, which

later became the core of IBM. By the end of the 19th century a

number of ideas and technologies, that would later prove useful in

the realization of practical computers, had begun to

appear: Boolean algebra, the vacuum tube (thermionic valve),

punched cards and tape, and the teleprinter.

During the first half of the 20th century, many

scientific computing needs were met by increasingly

sophisticated analog computers, which used a direct mechanical

or electrical model of the problem as a basis for computation.

However, these were not programmable and generally lacked the

versatility and accuracy of modern digital computers.

Atanasoff–Berry Computer (ABC) is the world's first electronic

digital computer Atanasoff is considered the father of the

computer . Conceived in 1937 by Iowa State College physics

professor John Atanasoff, and built with the assistance of graduate

student Clifford Berry the machine was not programmable, being

designed only to solve systems of linear equations. The computer

did employ parallel computation. A 1973 court ruling in a patent

dispute found that the patent for the 1946 ENIAC computer

derived from the Atanasoff–Berry Computer.

The inventor of the program-controlled computer was Konrad

Zuse, who built the first working computer in 1941 and later in

1955 the first computer based on magnetic storage

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George Stibitz is internationally recognized as a father of the

modern digital computer. While working at Bell Labs in November

1937, Stibitz invented and built a relay-based calculator he dubbed

the "Model K" (for "kitchen table", on which he had assembled it),

which was the first to use binary circuits to perform an arithmetic

operation. Later models added greater sophistication including

complex arithmetic and programmability.

A succession of steadily more powerful and

flexible computing devices were constructed in the 1930s and

1940s, gradually adding the key features that are seen in modern

computers. The use of digital electronics (largely invented

by Claude Shannon in 1937) and more flexible programmability

were vitally important steps, but defining one point along this road

as "the first digital electronic computer" is difficult. Shannon

194 Notable achievements include.

Stored-program architecture

Several developers of ENIAC, recognizing its flaws, came up with

a far more flexible and elegant design, which came to be known as

the "stored program architecture" or von Neumann architecture.

This design was first formally described by John von Neumann in

the paper First Draft of a Report on the EDVAC, distributed in

1945. A number of projects to develop computers based on the

stored-program architecture commenced around this time, the first

of these being completed in Great Britain. The first working

prototype to be demonstrated was the Manchester Small-Scale

Experimental Machine (SSEM or "Baby") in 1948. The Electronic

Delay Storage Automatic Calculator (EDSAC), completed a year

after the SSEM at Cambridge University, was the first practical,

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non-experimental implementation of the stored program design

and was put to use immediately for research work at the university.

Shortly thereafter, the machine originally described by von

Neumann's paper—EDVAC—was completed but did not see full-

time use for an additional two years.

Nearly all modern computers implement some form of the stored-

program architecture, making it the single trait by which the word

"computer" is now defined. While the technologies used in

computers have changed dramatically since the first electronic,

general-purpose computers of the 1940s, most still use the von

Neumann architecture.

Beginning in the 1950s, Soviet scientists Sergei

Sobolev and Nikolay Brusentsov conducted research on ternary

computers, devices that operated on a base three numbering

system of −1, 0, and 1 rather than the conventional binary

numbering system upon which most computers are based. They

designed theSetun, a functional ternary computer, at Moscow

State University. The device was put into limited production in the

Soviet Union, but supplanted by the more common binary

architecture.

Semiconductors and microprocessors

Computers using vacuum tubes as their electronic elements were

in use throughout the 1950s, but by the 1960s had been largely

replaced by transistor-based machines, which were smaller, faster,

cheaper to produce, required less power, and were more reliable.

The first transistorised computer was demonstrated at

the University of Manchester in 1953. In the 1970s, integrated

circuit technology and the subsequent creation

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of microprocessors, such as the Intel 4004, further decreased size

and cost and further increased speed and reliability of computers.

By the late 1970s, many products such as video

recorders contained dedicated computers called microcontrollers,

and they started to appear as a replacement to mechanical

controls in domestic appliances such as washing machines. The

1980s witnessed home computers and the now

ubiquitous personal computer. With the evolution of the Internet,

personal computers are becoming as common as

the television and the telephone in the house.

Modern smartphones are fully programmable computers in their

own right, and as of 2009 may well be the most common form of

such computers in existence.

Programs

The defining feature of modern computers which distinguishes

them from all other machines is that they can be programmed.

That is to say that some type of instructions (the program) can be

given to the computer, and it will carry process them. While some

computers may have strange concepts "instructions" and "output"

(see quantum computing), modern computers based on the von

Neumann architecture often have machine code in the form of

an imperative programming language.

In practical terms, a computer program may be just a few

instructions or extend to many millions of instructions, as do the

programs for word processors and web browsers for example. A

typical modern computer can execute billions of instructions per

second (gigaflops) and rarely makes a mistake over many years of

operation. Large computer programs consisting of several million

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instructions may take teams of programmers years to write, and

due to the complexity of the task almost certainly contain errors.

Stored program architecture

A 1970s punched card containing one line from a

FORTRAN program. The card reads: "Z(1) = Y + W(1)" and is

labelled "PROJ039" for identification purposes.

This section applies to most common RAM machine-based

computers.

In most cases, computer instructions are simple: add one number

to another, move some data from one location to another, send a

message to some external device, etc. These instructions are read

from the computer's memory and are generally carried out

(executed) in the order they were given. However, there are

usually specialized instructions to tell the computer to jump ahead

or backwards to some other place in the program and to carry on

executing from there. These are called "jump" instructions

(or branches). Furthermore, jump instructions may be made to

happen conditionally so that different sequences of instructions

may be used depending on the result of some previous calculation

or some external event. Many computers directly

support subroutines by providing a type of jump that "remembers"

the location it jumped from and another instruction to return to the

instruction following that jump instruction.

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Program execution might be likened to reading a book. While a

person will normally read each word and line in sequence, they

may at times jump back to an earlier place in the text or skip

sections that are not of interest. Similarly, a computer may

sometimes go back and repeat the instructions in some section of

the program over and over again until some internal condition is

met. This is called the flow of control within the program and it is

what allows the computer to perform tasks repeatedly without

human intervention.

Comparatively, a person using a pocket calculator can perform a

basic arithmetic operation such as adding two numbers with just a

few button presses. But to add together all of the numbers from 1

to 1,000 would take thousands of button presses and a lot of

time—with a near certainty of making a mistake. On the other

hand, a computer may be programmed to do this with just a few

simple instructions. For example:

mov #0, sum ; set sum to 0

mov #1, num ; set num to 1

loop: add num, sum ; add num to sum

add #1, num ; add 1 to num

cmp num, #1000 ; compare num to 1000

ble loop ; if num <= 1000, go back

to 'loop'

halt ; end of program. stop

running

Once told to run this program, the computer will perform the

repetitive addition task without further human intervention. It will

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almost never make a mistake and a modern PC can complete the

task in about a millionth of a second.

Machine code

In most computers, individual instructions are stored as machine

code with each instruction being given a unique number (its

operation code or opcode for short). The command to add two

numbers together would have one opcode, the command to

multiply them would have a different opcode and so on. The

simplest computers are able to perform any of a handful of

different instructions; the more complex computers have several

hundred to choose from—each with a unique numerical code.

Since the computer's memory is able to store numbers, it can also

store the instruction codes. This leads to the important fact that

entire programs (which are just lists of these instructions) can be

represented as lists of numbers and can themselves be

manipulated inside the computer in the same way as numeric

data. The fundamental concept of storing programs in the

computer's memory alongside the data they operate on is the crux

of the von Neumann, or stored program, architecture. In some

cases, a computer might store some or all of its program in

memory that is kept separate from the data it operates on. This is

called the Harvard architecture after theHarvard Mark I computer.

Modern von Neumann computers display some traits of the

Harvard architecture in their designs, such as in CPU caches.

While it is possible to write computer programs as long lists of

numbers (machine language) and while this technique was used

with many early computers, it is extremely tedious and potentially

error-prone to do so in practice, especially for complicated

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programs. Instead, each basic instruction can be given a short

name that is indicative of its function and easy to remember—

a mnemonic such as ADD, SUB, MULT or JUMP. These

mnemonics are collectively known as a computer's assembly

language. Converting programs written in assembly language into

something the computer can actually understand (machine

language) is usually done by a computer program called an

assembler. Machine languages and the assembly languages that

represent them (collectively termed low-level programming

languages) tend to be unique to a particular type of computer. For

instance, an ARM architecture computer (such as may be found in

a PDA or a hand-held videogame) cannot understand the machine

language of an Intel Pentium or the AMD Athlon 64 computer that

might be in a PC

Higher-level languages and program design

Though considerably easier than in machine language, writing

long programs in assembly language is often difficult and is also

error prone. Therefore, most practical programs are written in

more abstract high-level programming languages that are able to

express the needs of the programmer more conveniently (and

thereby help reduce programmer error). High level languages are

usually "compiled" into machine language (or sometimes into

assembly language and then into machine language) using

another computer program called a compiler. High level languages

are less related to the workings of the target computer than

assembly language, and more related to the language and

structure of the problem(s) to be solved by the final program. It is

therefore often possible to use different compilers to translate the

same high level language program into the machine language of

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many different types of computer. This is part of the means by

which software like video games may be made available for

different computer architectures such as personal computers and

various video game consoles.

The task of developing large software systems presents a

significant intellectual challenge. Producing software with an

acceptably high reliability within a predictable schedule and budget

has historically been difficult; the academic and professional

discipline of software engineering concentrates specifically on this

challenge.

Function

A general purpose computer has four main components:

the arithmetic logic unit (ALU), the control unit, the memory, and

the input and output devices (collectively termed I/O). These parts

are interconnected by busses, often made of groups of wires.

Inside each of these parts are thousands to trillions of

small electrical circuits which can be turned off or on by means of

an electronic switch. Each circuit represents a bit (binary digit) of

information so that when the circuit is on it represents a "1", and

when off it represents a "0" (in positive logic representation). The

circuits are arranged in logic gates so that one or more of the

circuits may control the state of one or more of the other circuits.

The control unit, ALU, registers, and basic I/O (and often other

hardware closely linked with these) are collectively known as

a central processing unit (CPU). Early CPUs were composed of

many separate components but since the mid-1970s CPUs have

typically been constructed on a single integrated circuit called

a microprocessor.

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Control unit

Diagram showing how a particular MIPS architecture instruction

would be decoded by the control system.

The control unit (often called a control system or central controller)

manages the computer's various components; it reads and

interprets (decodes) the program instructions, transforming them

into a series of control signals which activate other parts of the

computer. Control systems in advanced computers may change

the order of some instructions so as to improve performance.

A key component common to all CPUs is the program counter, a

special memory cell (a register) that keeps track of which location

in memory the next instruction is to be read from.

Since the program counter is (conceptually) just another set of

memory cells, it can be changed by calculations done in the ALU.

Adding 100 to the program counter would cause the next

instruction to be read from a place 100 locations further down the

program. Instructions that modify the program counter are often

known as "jumps" and allow for loops (instructions that are

repeated by the computer) and often conditional instruction

execution (both examples of control flow).

It is noticeable that the sequence of operations that the control unit

goes through to process an instruction is in itself like a short

computer program—and indeed, in some more complex CPU

designs, there is another yet smaller computer called

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a microsequencer that runs a microcode program that causes all

of these events to happen.

Arithmetic/logic unit (ALU)

The ALU is capable of performing two classes of operations:

arithmetic and logic.

The set of arithmetic operations that a particular ALU supports

may be limited to adding and subtracting or might include

multiplying or dividing, trigonometry functions (sine, cosine, etc.)

and square roots. Some can only operate on whole numbers

(integers) whilst others use floating point to represent real

numbers—albeit with limited precision. However, any computer

that is capable of performing just the simplest operations can be

programmed to break down the more complex operations into

simple steps that it can perform. Therefore, any computer can be

programmed to perform any arithmetic operation—although it will

take more time to do so if its ALU does not directly support the

operation. An ALU may also compare numbers and return boolean

truth values (true or false) depending on whether one is equal to,

greater than or less than the other ("is 64 greater than 65?").

Logic operations involve Boolean logic: AND, OR, XOR and NOT.

These can be useful both for creating complicated conditional

statements and processing boolean logic.

Superscalar computers may contain multiple ALUs so that they

can process several instructions at the same time. Graphics

processors and computers with SIMD and MIMD features often

provide ALUs that can perform arithmetic on vectors and matrices.

Memory

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Magnetic core memory was the computer memory of choice

throughout the 1960s, until it was replaced by semiconductor

memory.

A computer's memory can be viewed as a list of cells into which

numbers can be placed or read. Each cell has a numbered

"address" and can store a single number. The computer can be

instructed to "put the number 123 into the cell numbered 1357" or

to "add the number that is in cell 1357 to the number that is in cell

2468 and put the answer into cell 1595". The information stored in

memory may represent practically anything. Letters, numbers,

even computer instructions can be placed into memory with equal

ease. Since the CPU does not differentiate between different types

of information, it is the software's responsibility to give significance

to what the memory sees as nothing but a series of numbers.

In almost all modern computers, each memory cell is set up to

store binary numbers in groups of eight bits (called a byte). Each

byte is able to represent 256 different numbers (2^8 = 256); either

from 0 to 255 or −128 to +127. To store larger numbers, several

consecutive bytes may be used (typically, two, four or eight).

When negative numbers are required, they are usually stored

in two's complement notation. Other arrangements are possible,

but are usually not seen outside of specialized applications or

historical contexts. A computer can store any kind of information in

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memory if it can be represented numerically. Modern computers

have billions or even trillions of bytes of memory.

The CPU contains a special set of memory cells

called registers that can be read and written to much more rapidly

than the main memory area. There are typically between two and

one hundred registers depending on the type of CPU. Registers

are used for the most frequently needed data items to avoid

having to access main memory every time data is needed. As data

is constantly being worked on, reducing the need to access main

memory (which is often slow compared to the ALU and control

units) greatly increases the computer's speed.

Computer main memory comes in two principal varieties: random-

access memory or RAM and read-only memory or ROM. RAM can

be read and written to anytime the CPU commands it, but ROM is

pre-loaded with data and software that never changes, so the CPU

can only read from it. ROM is typically used to store the

computer's initial start-up instructions. In general, the contents of

RAM are erased when the power to the computer is turned off, but

ROM retains its data indefinitely. In a PC, the ROM contains a

specialized program called the BIOS that orchestrates loading the

computer'soperating system from the hard disk drive into RAM

whenever the computer is turned on or reset. In embedded

computers, which frequently do not have disk drives, all of the

required software may be stored in ROM. Software stored in ROM

is often called firmware, because it is notionally more like

hardware than software. Flash memory blurs the distinction

between ROM and RAM, as it retains its data when turned off but

is also rewritable. It is typically much slower than conventional

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ROM and RAM however, so its use is restricted to applications

where high speed is unnecessary.

In more sophisticated computers there may be one or more

RAM cache memories which are slower than registers but faster

than main memory. Generally computers with this sort of cache

are designed to move frequently needed data into the cache

automatically, often without the need for any intervention on the

programmer's part.

Input/output (I/O)

Hard disk drives are common storage devices used with

computers.

I/O is the means by which a computer exchanges information with

the outside world. Devices that provide input or output to the

computer are called peripherals. On a typical personal computer,

peripherals include input devices like the keyboard and mouse,

and output devices such as the display and printer. Hard disk

drives, floppy disk drives and optical disc drives serve as both

input and output devices. Computer networking is another form of

I/O.

Often, I/O devices are complex computers in their own right with

their own CPU and memory. A graphics processing unit might

contain fifty or more tiny computers that perform the calculations

necessary to display 3Dgraphics. Modern desktop

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computers contain many smaller computers that assist the main

CPU in performing I/O.

Multitasking

While a computer may be viewed as running one gigantic program

stored in its main memory, in some systems it is necessary to give

the appearance of running several programs simultaneously. This

is achieved by multitasking i.e. having the computer switch rapidly

between running each program in turn.

One means by which this is done is with a special signal called

an interrupt which can periodically cause the computer to stop

executing instructions where it was and do something else instead.

By remembering where it was executing prior to the interrupt, the

computer can return to that task later. If several programs are

running "at the same time", then the interrupt generator might be

causing several hundred interrupts per second, causing a program

switch each time. Since modern computers typically execute

instructions several orders of magnitude faster than human

perception, it may appear that many programs are running at the

same time even though only one is ever executing in any given

instant. This method of multitasking is sometimes termed "time-

sharing" since each program is allocated a "slice" of time in turn.

Before the era of cheap computers, the principal use for

multitasking was to allow many people to share the same

computer.

Seemingly, multitasking would cause a computer that is switching

between several programs to run more slowly — in direct

proportion to the number of programs it is running. However, most

programs spend much of their time waiting for slow input/output

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devices to complete their tasks. If a program is waiting for the user

to click on the mouse or press a key on the keyboard, then it will

not take a "time slice" until the event it is waiting for has occurred.

This frees up time for other programs to execute so that many

programs may be run at the same time without unacceptable

speed loss.

Multiprocessing

Cray designed many supercomputers that used multiprocessing

heavily.

Some computers are designed to distribute their work across

several CPUs in a multiprocessing configuration, a technique once

employed only in large and powerful machines such

as supercomputers, mainframe computers and servers.

Multiprocessor and multi-core (multiple CPUs on a single

integrated circuit) personal and laptop computers are now widely

available, and are being increasingly used in lower-end markets as

a result.

Supercomputers in particular often have highly unique

architectures that differ significantly from the basic stored-program

architecture and from general purpose computers. They often

feature thousands of CPUs, customized high-speed interconnects,

and specialized computing hardware. Such designs tend to be

useful only for specialized tasks due to the large scale of program

organization required to successfully utilize most of the available

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resources at once. Supercomputers usually see usage in large-

scale simulation, graphics rendering,

and cryptography applications, as well as with other so-called

"embarrassingly parallel" tasks.

Networking and the Internet

Computers have been used to coordinate information between

multiple locations since the 1950s. The U.S.

military's SAGE system was the first large-scale example of such a

system, which led to a number of special-purpose commercial

systems like Sabre.

In the 1970s, computer engineers at research institutions

throughout the United States began to link their computers

together using telecommunications technology. This effort was

funded by ARPA (now DARPA), and the computer network that it

produced was called the ARPANET. The technologies that made

the Arpanet possible spread and evolved.

In time, the network spread beyond academic and military

institutions and became known as the Internet. The emergence of

networking involved a redefinition of the nature and boundaries of

the computer. Computer operating systems and applications were

modified to include the ability to define and access the resources

of other computers on the network, such as peripheral devices,

stored information, and the like, as extensions of the resources of

an individual computer. Initially these facilities were available

primarily to people working in high-tech environments, but in the

1990s the spread of applications like e-mail and the World Wide

Web, combined with the development of cheap, fast networking

technologies like Ethernet and ADSL saw computer networking

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become almost ubiquitous. In fact, the number of computers that

are networked is growing phenomenally. A very large proportion

of personal computers regularly connect to the Internet to

communicate and receive information. "Wireless" networking,

often utilizing mobile phone networks, has meant networking is

becoming increasingly ubiquitous even in mobile computing

environments.

Required technology

Computational systems as flexible as a personal computer can be

built out of almost anything. For example, a computer can be

made out of billiard balls (billiard ball computer); this is an

unintuitive and pedagogical example that a computer can be made

out of almost anything. More realistically, modern computers are

made out of transistors made

of photolithographed semiconductors.

Historically, computers evolved from mechanical computers and

eventually from vacuum tubes to transistors.

There is active research to make computers out of many promising

new types of technology, such as optical computing, DNA

computers, neural computers, and quantum computers. Some of

these can easily tackle problems that modern computers cannot

(such as how quantum computers can break some modern

encryption algorithms by quantum factoring).

Logic gates are a common abstraction which can apply to most of

the above digital or analog paradigms.

The ability to store and execute lists of instructions

called programs makes computers extremely versatile,

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distinguishing them from calculators. The Church–Turing thesis is

a mathematical statement of this versatility: any computer with

a minimum capability (being Turing-complete) is, in principle,

capable of performing the same tasks that any other computer can

perform. Therefore any type of computer

(netbook, supercomputer, cellular automaton, etc.) is able to

perform the same computational tasks, given enough time and

storage capacity.

Limited-function computers

Conversely, a computer which is limited in function (one that is not

"Turing-complete") cannot simulate arbitrary things. For example,

simple four-function calculators cannot simulate a real computer

without human intervention. As a more complicated example,

without the ability to program a gaming console, it can never

accomplish what a programmable calculator from the 1990s could

(given enough time); the system as a whole is not Turing-

complete, even though it contains a Turing-complete component

(the microprocessor). Living organisms (the body, not the brain)

are also limited-function computers designed to make copies of

themselves; they cannot be reprogrammed without genetic

engineering.

Virtual computers

A "computer" is commonly considered to be a physical device.

However, one can create a computer program which describes

how to run a different computer, i.e. "simulating a computer in a

computer". Not only is this a constructive proof of the Church-

Turing thesis, but is also extremely common in all modern

computers. For example, some programming languages use

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something called an interpreter, which is a simulated computer

built using software that runs on a real, physical computer; this

allows programmers to write code (computer input) in a different

language than the one understood by the base computer (the

alternative is to use a compiler). Additionally, virtual machines are

simulated computers which virtually replicate a physical computer

in software, and are very commonly used by IT. Virtual

machines are also a common technique used to create emulators,

such game console emulators.

Artificial intelligence

A computer will solve problems in exactly the way they are

programmed to, without regard to efficiency nor alternative

solutions nor possible shortcuts nor possible errors in the code.

Computer programs which learn and adapt are part of the

emerging field of artificial intelligence and machine learning.

Hardware

The term hardware covers all of those parts of a computer that

are tangible objects. Circuits, displays, power supplies, cables,

keyboards, printers and mice are all hardware.

Software

Software refers to parts of the computer which do not have a

material form, such as programs, data, protocols, etc. When

software is stored in hardware that cannot easily be modified (such

as BIOSROM in an IBM PC compatible), it is sometimes called

"firmware" to indicate that it falls into an uncertain area somewhere

between hardware and software.

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Programming languages

Programming languages provide various ways of specifying

programs for computers to run. Unlike natural languages,

programming languages are designed to permit no ambiguity and

to be concise. They are purely written languages and are often

difficult to read aloud. They are generally either translated

into machine code by a compiler or an assembler before being

run, or translated directly at run time by an interpreter. Sometimes

programs are executed by a hybrid method of the two techniques.

There are thousands of different programming languages—some

intended to be general purpose, others useful only for highly

specialized applications.

Computer data storage

1 GB of SDRAM mounted in a personal computer. An example

of primary storage.

40 GB PATA hard disk drive (HDD); when connected to a

computer it serves assecondary storage.

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160 GB SDLT tape cartridge, an example of off-line storage. When

used within a robotic tape library, it is classified as tertiarystorage

instead.

Computer data storage, often called storage or memory, refers

to computer components and recording media that retain

digital data. Data storage is one of the core functions and

fundamental components of computers.

In contemporary usage, memory usually refers

to semiconductor storage random-access memory,

typically DRAM (Dynamic-RAM). Memory can refer to other forms

of fast but temporary storage. Storage refers to storage devices

and their media not directly accessible by the CPU,

(secondary or tertiary storage) — typically hard disk drives, optical

disc drives, and other devices slower than RAM but more

permanent. Historically, memory has been calledmain

memory, real storage or internal memory while storage devices

have been referred to as secondary storage, external

memory or auxiliary/peripheral storage.

The distinctions are fundamental to the architecture of computers.

The distinctions also reflect an important and significant technical

difference between memory and mass storage devices, which has

been blurred by the historical usage of the term storage.

Nevertheless, this article uses the traditional nomenclature.

Many different forms of storage, based on various natural

phenomena, have been invented. So far, no practical universal

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storage medium exists, and all forms of storage have some

drawbacks. Therefore a computer system usually contains several

kinds of storage, each with an individual purpose.

A digital computer represents data using the binary numeral

system. Text, numbers, pictures, audio, and nearly any other form

of information can be converted into a string of bits, or binary

digits, each of which has a value of 1 or 0. The most common unit

of storage is the byte, equal to 8 bits. A piece of information can be

handled by any computer whose storage space is large enough to

accommodate the binary representation of the piece of

information, or simply data. For example, using eight million bits,

or about one megabyte, a typical computer could store a short

novel.

Traditionally the most important part of every computer is

the central processing unit (CPU, or simply a processor), because

it actually operates on data, performs any calculations, and

controls all the other components. The CPU consists of two (2)

main parts: Control Unit and Arithmetic Logic Unit (ALU). The

former controls the flow of data between the CPU and memory

whilst the latter is used for performing arithmetic and logical

operations on data.

Without a significant amount of memory, a computer would merely

be able to perform fixed operations and immediately output the

result. It would have to be reconfigured to change its behavior.

This is acceptable for devices such as desk calculators or

simple digital signal processors. Von Neumann machines differ in

having a memory in which they store their

operating instructions and data. Such computers are more

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versatile in that they do not need to have their hardware

reconfigured for each new program, but can simply

be reprogrammed with new in-memory instructions; they also tend

to be simpler to design, in that a relatively simple processor may

keep state between successive computations to build up complex

procedural results. Most modern computers are von Neumann

machines.

In practice, almost all computers use a variety of memory types,

organized in a storage hierarchy around the CPU, as a trade-off

between performance and cost. Generally, the lower a storage is

in the hierarchy, the lesser its bandwidth and the greater its

access latency is from the CPU. This traditional division of storage

to primary, secondary, tertiary and off-line storage is also guided

by cost per bit.

Hierarchy of storage

Various forms of storage, divided according to their distance from

the central processing unit. The fundamental components of a

general-purpose computer are arithmetic and logic unit, control

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circuitry, storage space, and input/output devices. Technology and

capacity as in common home computers around 2005

Primary storage

Primary storage (or main memory or internal memory), often

referred to simply as memory, is the only one directly accessible

to the CPU. The CPU continuously reads instructions stored there

and executes them as required. Any data actively operated on is

also stored there in uniform manner.

Historically, early computers used delay lines, Williams tubes, or

rotating magnetic drums as primary storage. By 1954, those

unreliable methods were mostly replaced by magnetic core

memory. Core memory remained dominant until the 1970s, when

advances in integrated circuit technology allowed semiconductor

memory to become economically competitive.

This led to modern random-access memory (RAM). It is small-

sized, light, but quite expensive at the same time. (The particular

types of RAM used for primary storage are also volatile, i.e. they

lose the information when not powered).

As shown in the diagram, traditionally there are two more sub-

layers of the primary storage, besides main large-capacity RAM:

Processor registers are located inside the processor. Each register

typically holds a word of data (often 32 or 64 bits). CPU

instructions instruct the arithmetic and logic unit to perform various

calculations or other operations on this data (or with the help of it).

Registers are the fastest of all forms of computer data storage.

Processor cache is an intermediate stage between ultra-fast

registers and much slower main memory. It's introduced solely to

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increase performance of the computer. Most actively used

information in the main memory is just duplicated in the cache

memory, which is faster, but of much lesser capacity. On the other

hand, main memory is much slower, but has a much greater

storage capacity than processor registers. Multi-level hierarchical

cache setup is also commonly used—primary cache being

smallest, fastest and located inside the processor; secondary

cache being somewhat larger and slower.

Main memory is directly or indirectly connected to the central

processing unit via a memory bus. It is actually two buses (not on

the diagram): an address bus and a data bus. The CPU firstly

sends a number through an address bus, a number called memory

address, that indicates the desired location of data. Then it reads

or writes the data itself using the data bus. Additionally, a memory

management unit (MMU) is a small device between CPU and RAM

recalculating the actual memory address, for example to provide

an abstraction of virtual memory or other tasks.

As the RAM types used for primary storage are volatile (cleared at

start up), a computer containing only such storage would not have

a source to read instructions from, in order to start the computer.

Hence, non-volatile primary storage containing a small startup

program (BIOS) is used to bootstrap the computer, that is, to read

a larger program from non-volatile secondary storage to RAM and

start to execute it. A non-volatile technology used for this purpose

is called ROM, for read-only memory (the terminology may be

somewhat confusing as most ROM types are also capable

of random access).

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Many types of "ROM" are not literally read only, as updates are

possible; however it is slow and memory must be erased in large

portions before it can be re-written. Some embedded systems run

programs directly from ROM (or similar), because such programs

are rarely changed. Standard computers do not store non-

rudimentary programs in ROM, rather use large capacities of

secondary storage, which is non-volatile as well, and not as costly.

Secondary storage

A hard disk drive with protective cover removed.

Secondary storage (also known as external memory or auxiliary

storage), differs from primary storage in that it is not directly

accessible by the CPU. The computer usually uses

its input/output channels to access secondary storage and

transfers the desired data using intermediate area in primary

storage. Secondary storage does not lose the data when the

device is powered down—it is non-volatile. Per unit, it is typically

also two orders of magnitude less expensive than primary storage.

Consequently, modern computer systems typically have two

orders of magnitude more secondary storage than primary storage

and data is kept for a longer time there.

In modern computers, hard disk drives are usually used as

secondary storage. The time taken to access a given byte of

information stored on a hard disk is typically a few thousandths of

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a second, or milliseconds. By contrast, the time taken to access a

given byte of information stored in random access memory is

measured in billionths of a second, or nanoseconds. This

illustrates the significant access-time difference which

distinguishes solid-state memory from rotating magnetic storage

devices: hard disks are typically about a million times slower than

memory. Rotating optical storage devices, such

as CD and DVDdrives, have even longer access times. With disk

drives, once the disk read/write head reaches the proper

placement and the data of interest rotates under it, subsequent

data on the track are very fast to access. As a result, in order to

hide the initial seek time and rotational latency, data is transferred

to and from disks in large contiguous blocks.

When data reside on disk, block access to hide latency offers a ray

of hope in designing efficient external memory algorithms.

Sequential or block access on disks is orders of magnitude faster

than random access, and many sophisticated paradigms have

been developed to design efficient algorithms based upon

sequential and block access. Another way to reduce the I/O

bottleneck is to use multiple disks in parallel in order to increase

the bandwidth between primary and secondary memory.

Some other examples of secondary storage technologies

are: flash memory (e.g. USB flash drives or keys), floppy

disks, magnetic tape, paper tape, punched cards, standalone RAM

disks, andIomega Zip drives.

The secondary storage is often formatted according to a file

system format, which provides the abstraction necessary to

organize data into files and directories, providing also additional

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information (called metadata) describing the owner of a certain file,

the access time, the access permissions, and other information.

Most computer operating systems use the concept of virtual

memory, allowing utilization of more primary storage capacity than

is physically available in the system. As the primary memory fills

up, the system moves the least-used chunks (pages) to secondary

storage devices (to a swap file or page file), retrieving them later

when they are needed. As more of these retrievals from slower

secondary storage are necessary, the more the overall system

performance is degraded.

Tertiary storage

Large tape library. Tape cartridges placed on shelves in the front,

robotic arm moving in the back. Visible height of the library is

about 180 cm.

Tertiary storage or tertiary memory, provides a third level of

storage. Typically it involves a robotic mechanism which

will mount (insert) and dismountremovable mass storage media

into a storage device according to the system's demands; this data

is often copied to secondary storage before use. It is primarily

used for archiving rarely accessed information since it is much

slower than secondary storage (e.g. 5–60 seconds vs. 1–10

milliseconds). This is primarily useful for extraordinarily large data

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stores, accessed without human operators. Typical examples

include tape libraries and optical jukeboxes.

When a computer needs to read information from the tertiary

storage, it will first consult a catalog database to determine which

tape or disc contains the information. Next, the computer will

instruct a robotic arm to fetch the medium and place it in a drive.

When the computer has finished reading the information, the

robotic arm will return the medium to its place in the library.

Off-line storage

Off-line storage is a computer data storage on a medium or a

device that is not under the control of a processing unit. The

medium is recorded, usually in a secondary or tertiary storage

device, and then physically removed or disconnected. It must be

inserted or connected by a human operator before a computer can

access it again. Unlike tertiary storage, it cannot be accessed

without human interaction.

Off-line storage is used to transfer information, since the detached

medium can be easily physically transported. Additionally, in case

a disaster, for example a fire, destroys the original data, a medium

in a remote location will probably be unaffected, enabling disaster

recovery. Off-line storage increases generalinformation security,

since it is physically inaccessible from a computer, and data

confidentiality or integrity cannot be affected by computer-based

attack techniques. Also, if the information stored for archival

purposes is accessed seldom or never, off-line storage is less

expensive than tertiary storage.

In modern personal computers, most secondary and tertiary

storage media are also used for off-line storage. Optical discs and

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flash memory devices are most popular, and to much lesser extent

removable hard disk drives. In enterprise uses, magnetic tape is

predominant. Older examples are floppy disks, Zip disks, or

punched cards.

Characteristics of storage

A 1GB DDR RAM module (detail)

Storage technologies at all levels of the storage hierarchy can be

differentiated by evaluating certain core characteristics as well as

measuring characteristics specific to a particular implementation.

These core characteristics are volatility, mutability, accessibility,

and addressibility. For any particular implementation of any

storage technology, the characteristics worth measuring are

capacity and performance.

Volatility

Non-volatile memory

Will retain the stored information even if it is not constantly

supplied with electric power. It is suitable for long-term storage of

information.

Volatile memory

Requires constant power to maintain the stored information. The

fastest memory technologies of today are volatile ones (not a

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universal rule). Since primary storage is required to be very fast, it

predominantly uses volatile memory.

Dynamic random-access memory

A form of volatile memory which also requires the stored

information to be periodically re-read and re-written, or refreshed,

otherwise it would vanish.

Static random-access memory

A form of volatile memory similar to DRAM with the exception that

it never needs to be refreshed as long as power is applied. (It

loses its content if power is removed).

Mutability

Read/write storage or mutable storage

Allows information to be overwritten at any time. A computer

without some amount of read/write storage for primary storage

purposes would be useless for many tasks. Modern computers

typically use read/write storage also for secondary storage.

Read only storage

Retains the information stored at the time of manufacture,

and write once storage (Write Once Read Many) allows the

information to be written only once at some point after

manufacture. These are called immutable storage. Immutable

storage is used for tertiary and off-line storage. Examples

include CD-ROM and CD-R.

Slow write, fast read storage

Read/write storage which allows information to be overwritten

multiple times, but with the write operation being much slower than

the read operation. Examples include CD-RW and flash memory.

Accessibility

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Random access

Any location in storage can be accessed at any moment in

approximately the same amount of time. Such characteristic is well

suited for primary and secondary storage.

Sequential access

The accessing of pieces of information will be in a serial order, one

after the other; therefore the time to access a particular piece of

information depends upon which piece of information was last

accessed. Such characteristic is typical of off-line storage.

Addressability

Location-addressable

Each individually accessible unit of information in storage is

selected with its numerical memory address. In modern

computers, location-addressable storage usually limits to primary

storage, accessed internally by computer programs, since

location-addressability is very efficient, but burdensome for

humans.

File addressable

Information is divided into files of variable length, and a particular

file is selected with human-readable directory and file names. The

underlying device is still location-addressable, but the operating

system of a computer provides the file system abstraction to make

the operation more understandable. In modern computers,

secondary, tertiary and off-line storage use file systems.

Content-addressable

Each individually accessible unit of information is selected based

on the basis of (part of) the contents stored there. Content-

addressable storage can be implemented

using software (computer program) or hardware (computer

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device), with hardware being faster but more expensive option.

Hardware content addressable memory is often used in a

computer's CPU cache.

Capacity

Raw capacity

The total amount of stored information that a storage device or

medium can hold. It is expressed as a quantity

of bits or bytes (e.g. 10.4 megabytes).

Memory storage density

The compactness of stored information. It is the storage capacity

of a medium divided with a unit of length, area or volume (e.g. 1.2

megabytes per square inch).

Performance

Latency

The time it takes to access a particular location in storage. The

relevant unit of measurement is typically nanosecond for primary

storage, millisecond for secondary storage, and second for tertiary

storage. It may make sense to separate read latency and write

latency, and in case of sequential access storage, minimum,

maximum and average latency.

Throughput

The rate at which information can be read from or written to the

storage. In computer data storage, throughput is usually

expressed in terms of megabytes per second or MB/s, though bit

rate may also be used. As with latency, read rate and write rate

may need to be differentiated. Also accessing media sequentially,

as opposed to randomly, typically yields maximum throughput.

Energy use

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Storage devices that reduce fan usage, automatically shut-down

during inactivity, and low power hard drives can reduce energy

consumption 90 percent.

2.5 inch hard disk drives often consume less power than larger

ones. Low capacity solid-state drives have no moving parts and

consume less power than hard disks. Also, memory may use more

power than hard disks. computer utilizingain to the primary

storage, which is shared between multiple processors in a much

lesser degreeDirect-attached storage (DAS) is a traditional mass

storage, that does not use any network. This is still a most popular

approach. This retronym was coined recently, together with NAS

and SAN.Network-attached storage (NAS) is mass storage

attached to a computer which another computer can access at file

level over a local area network, a private wide area network, or in

the case ofonline file storage, over the Internet. NAS is commonly

associated with the NFS and CIFS/SMB protocolsSAN) is a

specialized network, that provides other computers with storage

capacity. The crucial difference between NAS and SAN is the

former presents and manages file systems to client computers,

whilst the latter provides access at block-addressing (raw) level,

leaving it to attaching systems to manage data or file systems

within the provided capacity. SAN is commonly associated

with Fibre Channel networks.

Basic hardware components

Apart from the physical communications media themselves as

described above, networks comprise additional basic hardware

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building blocks interconnecting their terminals, such as network

interface cards (NICs), hubs, bridges, switches, and routers.

Network interface cards

A network card, network adapter, or NIC (network interface card)

is a piece of computer hardware designed to allow computers to

physically access a networking medium. It provides a low-level

addressing system through the use of MAC addresses.

Each Ethernet network interface has a unique MAC address which

is usually stored in a small memory device on the card, allowing

any device to connect to the network without creating an address

conflict. Ethernet MAC addresses are composed of six octets.

Uniqueness is maintained by the IEEE, which manages the

Ethernet address space by assigning 3-octet prefixes to equipment

manufacturers. The list of prefixes is publicly available.

Repeaters and hubs

A repeater is an electronic device that receives a signal, cleans it

of unnecessary noise, regenerates it, and retransmits it at a higher

power level, or to the other side of an obstruction, so that the

signal can cover longer distances without degradation. In most

twisted pair Ethernet configurations, repeaters are required for

cable that runs longer than 100 meters. A repeater with multiple

ports is known as a hub. Repeaters work on the Physical Layer of

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the OSI model. Repeaters require a small amount of time to

regenerate the signal. This can cause a propagation delay which

can affect network communication when there are several

repeaters in a row. Many network architectures limit the number of

repeaters that can be used in a row (e.g. Ethernet's 5-4-3 rule).

Today, repeaters and hubs have been made mostly obsolete by

switches (see below).

Repeaters Hub

Bridges

A network bridge connects multiple network segments at the data

link layer (layer 2) of the OSI model. Bridges broadcast to all ports

except the port on which the broadcast was received. However,

bridges do not promiscuously copy traffic to all ports, as hubs do,

but learn which MAC addresses are reachable through specific

ports. Once the bridge associates a port and an address, it will

send traffic for that address to that port only.

Bridges learn the association of ports and addresses by examining

the source address of frames that it sees on various ports. Once a

frame arrives through a port, its source address is stored and the

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bridge assumes that MAC address is associated with that port.

Bridges come in three basic types:

Local bridges: Directly connect LANs

Remote bridges: Can be used to create a wide area network

(WAN) link between LANs. Remote bridges, where the

connecting link is slower than the end networks, largely have

been replaced with routers.

Wireless bridges: Can be used to join LANs or connect remote

stations to LANs.

Switches

A network switch is a device that forwards and filters OSI layer

2 datagrams (chunks of data communication) between ports

(connected cables) based on the MAC addresses in the

packets.[15]

A switch is distinct from a hub in that it only forwards

the frames to the ports involved in the communication rather than

all ports connected. A switch breaks the collision domain but

represents itself as a broadcast domain. Switches make

forwarding decisions of frames on the basis of MAC addresses. A

switch normally has numerous ports, facilitating a star topology for

devices, and cascading additional switches.[16]

Some switches are

capable of routing based on Layer 3 addressing or additional

logical levels; these are called multi-layer switches. The

term switch is used loosely in marketing to encompass devices

including routers and bridges, as well as devices that may

distribute traffic on load or by application content (e.g., a

Web URL identifier).

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Routers

A router is an internetworking device that

forwards packets between networks by processing information

found in the datagram or packet (Internet protocol information

from Layer 3 of the OSI Model). In many situations, this

information is processed in conjunction with the routing table (also

known as forwarding table). Routers use routing tables to

determine what interface to forward packets (this can include the

"null" also known as the "black hole" interface because data can

go into it, however, no further processing is done for said data).

Firewalls

A firewall is an important aspect of a network with respect to

security. It typically rejects access requests from unsafe sources

while allowing actions from recognized ones. The vital role

firewalls play in network security grows in parallel with the constant

increase in 'cyber' attacks for the purpose of stealing/corrupting

data, planting viruses, etc.

A computer network, often simply referred to as a network, is a

collection of hardware components and computers interconnected

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by communication channels that allow sharing of resources and

information.

Networks may be classified according to a wide variety of

characteristics such as the medium used to transport the

data, communications protocol used, scale, topology, and

organizational scope.

The rules and data formats for exchanging information in a

computer network are defined by communications protocols. Well-

known communications protocols are Ethernet, a hardware

and Link Layer standard that is ubiquitous in local area networks,

and the Internet Protocol Suite, which defines a set of protocols for

internetworking, i.e. for data communication between multiple

networks, as well as host-to-host data transfer, and application-

specific data transmission formats.

NETWORKS

A local area network (LAN) is a computer network that

interconnects computers in a limited area such as home, school,

computer laboratory or office building. The defining characteristics

of LANs, in contrast to wide area networks (WANs), include their

usually higher data-transfer rates, smaller geographic area, and

lack of a need for leased telecommunication lines.

ARCNET, Token Ring and other technology standards have been

used in the past, but Ethernet over twisted pair cabling, and Wi-

Fi are the two most common technologies currently used to build

LANs.

A wide area network (WAN) is a telecommunication network that

covers a broad area (i.e., any network that links across

metropolitan, regional, or national boundaries). Business and

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government entities utilize WANs to relay data among employees,

clients, buyers, and suppliers from various geographical locations.

In essence this mode of telecommunication allows a business to

effectively carry out its daily function regardless of location.

This is in contrast with personal area networks (PANs), local area

networks (LANs), campus area networks (CANs), or metropolitan

area networks(MANs) which are usually limited to a room, building,

campus or specific metropolitan area (e.g., a city) respectively.

A metropolitan area network (MAN) is a computer network that

usually spans a city or a large campus. A MAN usually

interconnects a number of local area networks (LANs) using a

high-capacity backbone technology, such as fiber-optical links, and

provides up-link services to wide area networks (or WAN) and the

Internet.

Description of OSI layers

According to recommendation X.200, there are seven layers, each

generically known as an N layer. An N+1 entity requests services

from the layer-N entity.

At each level, two entities (N-entity peers) interact by means of the

N protocol by transmitting protocol data units (PDU).

A Service Data Unit (SDU) is a specific unit of data that has been

passed down from an OSI layer to a lower layer, and which the

lower layer has not yet encapsulated into a protocol data unit

(PDU). An SDU is a set of data that is sent by a user of the

services of a given layer, and is transmitted semantically

unchanged to a peer service user.

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The PDU at any given layer, layer N, is the SDU of the layer

below, layer N-1. In effect the SDU is the 'payload' of a given PDU.

That is, the process of changing a SDU to a PDU, consists of an

encapsulation process, performed by the lower layer. All the data

contained in the SDU becomes encapsulated within the PDU. The

layer N-1 adds headers or footers, or both, to the SDU,

transforming it into a PDU of layer N-1. The added headers or

footers are part of the process used to make it possible to get data

from a source to a destination.

OSI Model

Data unit Layer Function

Host

layers

Data

7. Application Network process to application

6. Presentation

Data representation, encryption

and decryption, convert

machine dependent data to

machine independent data

5. Session Interhost communication

Segments 4. Transport End-to-end connections,

reliability and flow control

Media

layers

Packet/Datagram 3. Network Path determination andlogical

addressing

Frame 2. Data link Physical addressing

Bit 1. Physical Media, signal and binary

transmission

Some orthogonal aspects, such as management and security,

involve every layer.

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Security services are not related to a specific layer: they can be

related by a number of layers, as defined by ITU-T X.800

Recommendation.

These services are aimed to improve the CIA

triad(confidentiality, integrity, and availability) of transmitted data.

Actually the availability of communication service is determined

by network design and/or network management protocols.

Appropriate choices for these are needed to protect againstdenial

of service.

Layer 1: physical layer

The physical layer defines electrical and physical specifications for

devices. In particular, it defines the relationship between a device

and a transmission medium, such as a copper or optical cable.

This includes the layout

ofpins, voltages, cable specifications, hubs, repeaters, network

adapters, host bus adapters (HBA used in storage area networks)

and more.

The major functions and services performed by the physical layer

are:

Establishment and termination of a connection to

a communications medium.

Participation in the process whereby the communication

resources are effectively shared among multiple users. For

example, contention resolution and flow control.

Modulation, or conversion between the representation of digital

data in user equipment and the corresponding signals

transmitted over a communications channel. These are signals

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operating over the physical cabling (such as copper and optical

fiber) or over a radio link.

Parallel SCSI buses operate in this layer, although it must be

remembered that the logical SCSI protocol is a transport layer

protocol that runs over this bus. Various physical-layer Ethernet

standards are also in this layer; Ethernet incorporates both this

layer and the data link layer. The same applies to other local-area

networks, such as token ring, FDDI, ITU-T G.hn and IEEE 802.11,

as well as personal area networks such as Bluetooth and IEEE

802.15.4.

Layer 2: data link layer

The data link layer provides the functional and procedural means

to transfer data between network entities and to detect and

possibly correct errors that may occur in the physical layer.

Originally, this layer was intended for point-to-point and point-to-

multipoint media, characteristic of wide area media in the

telephone system. Local area network architecture, which included

broadcast-capable multiaccess media, was developed

independently of the ISO work in IEEE Project 802. IEEE work

assumed sublayering and management functions not required for

WAN use. In modern practice, only error detection, not flow control

using sliding window, is present in data link protocols such

as Point-to-Point Protocol (PPP), and, on local area networks, the

IEEE 802.2 LLC layer is not used for most protocols on the

Ethernet, and on other local area networks, its flow control and

acknowledgment mechanisms are rarely used. Sliding window flow

control and acknowledgment is used at the transport layer by

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protocols such as TCP, but is still used in niches where X.25 offers

performance advantages.

The ITU-T G.hn standard, which provides high-speed local area

networking over existing wires (power lines, phone lines and

coaxial cables), includes a complete data link layer which provides

both error correction and flow control by means of a selective

repeat Sliding Window Protocol.

Both WAN and LAN service arrange bits, from the physical layer,

into logical sequences called frames. Not all physical layer bits

necessarily go into frames, as some of these bits are purely

intended for physical layer functions. For example, every fifth bit of

the FDDI bit stream is not used by the layer.

Layer 3: network layer

The network layer provides the functional and procedural means of

transferring variable length data sequences from a source host on

one network to a destination host on a different network, while

maintaining the quality of service requested by the transport layer

(in contrast to the data link layer which connects hosts within the

same network). The network layer performs

network routingfunctions, and might also perform fragmentation

and reassembly, and report delivery errors. Routers operate at this

layer, sending data throughout the extended network and making

the Internet possible. This is a logical addressing scheme – values

are chosen by the network engineer. The addressing scheme is

not hierarchical.

The network layer may be divided into three sublayers:

Subnetwork access – that considers protocols that deal with the

interface to networks, such as X.25;

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Subnetwork-dependent convergence – when it is necessary to

bring the level of a transit network up to the level of networks on

either side

Subnetwork-independent convergence – handles transfer across

multiple networks.

Feature Name TP0 TP1 TP2 TP3 TP4

Connection oriented network Yes Yes Yes Yes Yes

Connectionless network No No No No Yes

Concatenation and separation No Yes Yes Yes Yes

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Layer 4:

transport

layer

The transport

layer provides

transparent

transfer of

data between

end users,

providing

reliable data

transfer

services to the

upper layers.

The transport

layer controls the reliability of a given link through flow control,

segmentation/desegmentation, and error control. Some protocols

are state- and connection-oriented. This means that the transport

layer can keep track of the segments and retransmit those that fail.

The transport layer also provides the acknowledgement of the

successful data transmission and sends the next data if no errors

occurred.

OSI defines five classes of connection-mode transport protocols

ranging from class 0 (which is also known as TP0 and provides the

least features) to class 4 (TP4, designed for less reliable networks,

similar to the Internet). Class 0 contains no error recovery, and

was designed for use on network layers that provide error-free

connections. Class 4 is closest to TCP, although TCP contains

Segmentation and reassembly Yes Yes Yes Yes Yes

Error Recovery No Yes Yes Yes Yes

Reinitiate connection (if an

excessive number of PDUs are

unacknowledged)

No Yes No Yes No

Multiplexing and

demultiplexing over a

single virtual circuit

No No Yes Yes Yes

Explicit flow control No No Yes Yes Yes

Retransmission on timeout No No No No Yes

Reliable Transport Service No Yes No Yes Yes

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functions, such as the graceful close, which OSI assigns to the

session layer. Also, all OSI TP connection-mode protocol classes

provide expedited data and preservation of record boundaries.

Detailed characteristics of TP0-4 classes are shown in the

following table:

Perhaps an easy way to visualize the transport layer is to compare

it with a Post Office, which deals with the dispatch and

classification of mail and parcels sent. Do remember, however,

that a post office manages the outer envelope of mail. Higher

layers may have the equivalent of double envelopes, such as

cryptographic presentation services that can be read by the

addressee only. Roughly speaking, tunneling protocols operate at

the transport layer, such as carrying non-IP protocols such

as IBM's SNA or Novell's IPX over an IP network, or end-to-end

encryption with IPsec. While Generic Routing

Encapsulation (GRE) might seem to be a network-layer protocol, if

the encapsulation of the payload takes place only at endpoint,

GRE becomes closer to a transport protocol that uses IP headers

but contains complete frames or packets to deliver to an

endpoint. L2TP carries PPP frames inside transport packet.

Although not developed under the OSI Reference Model and not

strictly conforming to the OSI definition of the transport layer,

the Transmission Control Protocol (TCP) and the User Datagram

Protocol(UDP) of the Internet Protocol Suite are commonly

categorized as layer-4 protocols within OSI.

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Layer 5: session layer

The session layer controls the dialogues (connections) between

computers. It establishes, manages and terminates the

connections between the local and remote application. It provides

for full-duplex,half-duplex, or simplex operation, and establishes

checkpointing, adjournment, termination, and restart procedures.

The OSI model made this layer responsible for graceful close of

sessions, which is a property of the Transmission Control Protocol,

and also for session checkpointing and recovery, which is not

usually used in the Internet Protocol Suite. The session layer is

commonly implemented explicitly in application environments that

use remote procedure calls.

Layer 6: presentation layer

The presentation layer establishes context between application-

layer entities, in which the higher-layer entities may use different

syntax and semantics if the presentation service provides a

mapping between them. If a mapping is available, presentation

service data units are encapsulated into session protocol data

units, and passed down the stack.

This layer provides independence from data representation

(e.g., encryption) by translating between application and network

formats. The presentation layer transforms data into the form that

the application accepts. This layer formats and encrypts data to be

sent across a network. It is sometimes called the syntax layer.

The original presentation structure used the basic encoding rules

of Abstract Syntax Notation One (ASN.1), with capabilities such as

converting an EBCDIC-coded text file to an ASCII-coded file,

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orserialization of objects and other data structures from and

to XML.

Layer 7: application layer

The application layer is the OSI layer closest to the end user,

which means that both the OSI application layer and the user

interact directly with the software application. This layer interacts

with software applications that implement a communicating

component. Such application programs fall outside the scope of

the OSI model. Application-layer functions typically include

identifying communication partners, determining resource

availability, and synchronizing communication. When identifying

communication partners, the application layer determines the

identity and availability of communication partners for an

application with data to transmit. When determining resource

availability, the application layer must decide whether sufficient

network or the requested communication exist. In synchronizing

communication, all communication between applications requires

cooperation that is managed by the application layer.

Cross-layer functions

There are some functions or services that are not tied to a given

layer, but they can affect more than one layer. Examples are

security service (telecommunication)[3]

as defined by ITU-

T X.800 Recommendation.

management functions, i.e. functions that permit to configure,

instantiate, monitor, terminate the communications of two or

more entities: there is a specific application layer

protocol, common management information protocol (CMIP) and

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its corresponding service, common management information

service (CMIS), they need to interact with every layer in order to

deal with their instances.

MPLS operates at an OSI-model layer that is generally

considered to lie between traditional definitions of layer 2 (data

link layer) and layer 3 (network layer), and thus is often referred

to as a "layer-2.5" protocol. It was designed to provide a unified

data-carrying service for both circuit-based clients and packet-

switching clients which provide a datagram service model. It can

be used to carry many different kinds of traffic, including IP

packets, as well as native ATM, SONET, and Ethernet frames.

ARP is used to translate IPv4 addresses (OSI layer 3) into

Ethernet MAC addresses (OSI layer 2)

Interfaces

Neither the OSI Reference Model nor OSI protocols specify any

programming interfaces, other than as deliberately abstract service

specifications. Protocol specifications precisely define the

interfaces between different computers, but the software interfaces

inside computers are implementation-specific.

For example Microsoft Windows' Winsock, and Unix's Berkeley

sockets and System V Transport Layer Interface, are interfaces

between applications (layer 5 and above) and the transport (layer

4).NDIS and ODI are interfaces between the media (layer 2) and

the network protocol (layer 3).

A set of techniques where by a sequence of information-carrying

quantities occurring at discrete instances of time is encoded into

a corresponding regular sequence of electromagnetic carrier

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pulses. Varying the amplitude, polarity, presence or absence,

duration, or occurrence in time of the pulses gives rise to the

four basic forms of pulse modulation: pulse-amplitude

modulation (PAM), pulse-code modulation (PCM), pulse-width

modulation (PWM, also known as pulse-duration

modulation, PDM), and pulse-position modulation (PPM).

Analog-to-digital conversion

An important concept in pulse modulation is analog-to-digital

(A/D) conversion, in which an original analog (time- and

amplitude-continuous) information signal s(t) is changed at the

transmitter into a series of regularly occurring discrete pulses

whose amplitudes are restricted to a fixed and finite number of

values. An inverse digital-to-analog (D/A) process is used at the

receiver to reconstruct an approximation of the original form

of s(t). Conceptually, analog-to-digital conversion involves two

steps. First, the range of amplitudes of s(t) is divided or

quantized into a finite number of predetermined levels, and each

such level is represented by a pulse of fixed amplitude. Second,

the amplitude of s(t) is periodically measured or sampled and

replaced by the pulse representing the level that corresponds to

the measurement. See also Analog-to-digital converter; Digital-

to-analog converter.

According to the Nyquist sampling theorem, if sampling occurs at

a rate at least twice that of the bandwidth of s(t), the latter can

be unambiguously reconstructed from its amplitude values at the

sampling instants by applying them to an ideal low-pass filter

whose bandwidth matches that of s(t).

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Quantization, however, introduces an irreversible error, the so-

called quantization error, since the pulse representing a sample

measurement determines only the quantization level in which

the measurement falls and not its exact value. Consequently, the

process of reconstructing s(t) from the sequence of pulses yields

only an approximate version of s(t).

Pulse-amplitude modulation

In PAM the successive sample values of the analog signal s(t) are

used to effect the amplitudes of a corresponding sequence of

pulses of constant duration occurring at the sampling rate. No

quantization of the samples normally occurs (Fig. 1a, b). In

principle the pulses may occupy the entire time between

samples, but in most practical systems the pulse duration, known

as the duty cycle, is limited to a fraction of the sampling

interval. Such a restriction creates the possibility of interleaving

during one sample interval one or more pulses derived from

other PAM systems in a process known as time-

division multiplexing (TDM). See also Multiplexing and multiple

access. sine wave. (a) Analog signal, s(t). (b) Pulse-

amplitude modulation. (c) Pulse-width modulation. (d) Pulse-

position modulation.">

Forms of pulse modulation for the case where the analog

signal, s(t), is a sine wave. (a) Analog signal, s(t). (b) Pulse-

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amplitude modulation. (c) Pulse-width modulation. (d) Pulse-

position modulation.

Pulse-width modulation

In PWM the pulses representing successive sample values of s(t)

have constant amplitudes but vary in time duration in direct

proportion to the sample value. The pulse duration can be

changed relative to fixed leading or trailing time edges or a fixed

pulse center. To allow for time-division multiplexing, the

maximum pulse duration may be limited to a fraction of the time

between samples (Fig. 1c).

Pulse-position modulation

PPM encodes the sample values of s(t) by varying the position of

a pulse of constant duration relative to its nominal time of

occurrence. As in PAM and PWM, the duration of the pulses is

typically a fraction of the sampling interval. In addition, the

maximum time excursion of the pulses may be limited (Fig. 1d).

Pulse-code modulation

Many modern communication systems are designed to transmit

and receive only pulses of two distinct amplitudes. In these so-

called binary digital systems, the analog-to-digital conversion

process is extended by the additional step of coding, in which

the amplitude of each pulse representing a quantized sample

of s(t) is converted into a unique sequence of one or more pulses

with just two possible amplitudes. The complete conversion

process is known as pulse-code modulation.

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Figure 2a shows the example of three successive quantized

samples of an analog signal s(t), in which sampling occurs

every T seconds and the pulse representing the sample is limited

to T/2 seconds. Assuming that the number of quantization levels

is limited to 8, each level can be represented by a unique

sequence of three two-valued pulses. In Fig. 2b these pulses are

of amplitude V or 0, whereas in Fig. 2c the amplitudes are V and

−V.

Pulse-code modulation. (a) Three successive quantized samples

of an analog signal. (b) With pulses of amplitude V or 0. (c) With

pulses of amplitude V or −V.

PCM enjoys many important advantages over other forms of pulse

modulation due to the fact that information is represented by a

two-state variable. First, the design parameters of a PCM

transmission system depend critically on the bandwidth of the

original signal s(t) and the degree of fidelity required at the

point of reconstruction, but are otherwise largely independent of

the information content of s(t). This fact creates the possibility

of deploying generic transmission systems suitable for many

types of information. Second, the detection of the state of a

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two-state variable in a noisy environment is inherently simpler

than the precise measurement of the amplitude, duration, or

position of a pulse in which these quantities are not constrained.

Third, the binary pulses propagating along a medium can be

intercepted and decoded at a point where the

accumulated distortion and attenuation are sufficiently low to

assure high detection accuracy. New pulses can then be

generated and transmitted to the next such decoding point. This

so-called process of repeatering significantly reduces the

propagation of distortion and leads to a quality of transmission

that is largely independent of distance.

Time-division multiplexing

An advantage inherent in all pulse modulation systems is their

ability to transmit signals from multiple sources over a common

transmission system through the process of time-division

multiplexing. By restricting the time duration of a pulse

representing a sample value from a particular analog signal to a

fraction of the time between successive samples, pulses derived

from other sampled analog signals can be accommodated on the

transmission system.

One important application of this principle occurs in the

transmission of PCM telephone voice signals over a digital

transmission system known as a T1 carrier. In standard T1 coding,

an original analog voice signal is band-limited to 4000 hertz by

passing it through a low-pass filter, and is then sampled at the

Nyquist rate of 8000 samples per second, so that the time

between successive samples is 125 microseconds. The samples

are quantized to 256 levels, with each of them being represented

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by a sequence of 8 binary pulses. By limiting the duration of a

single pulse to 0.65 microsecond, a total of 193 pulses can be

accommodated in the time span of 125 microseconds between

samples. One of these serves as asynchronization marker that

indicates the beginning of such a sequence of 193 pulses, while

the other 192 pulses are the composite of 8 pulses from each of

24 voice signals, with each 8-pulse sequence occupying a

specified position. T1 carriers and similar types of digital carrier

systems are in widespread use in the world's telephone networks.

Bandwidth requirements

Pulse modulation systems may incur a significant bandwidth

penalty compared to the transmission of a signal in its analog

form. An example is the standard PCM transmission of an analog

voice signal band-limited to 4000 hertz over a T1 carrier. Since

the sampling, quantizing, and coding process produces 8 binary

pulses 8000 times per second for a total of 64,000 binary pulses

per second, the pulses occur every 15.625 microseconds.

Depending on the shape of the pulses and the amount

of intersymbol interference, the required transmission bandwidth

will fall in the range of 32,000 to 64,000 hertz. This compares to

a bandwidth of only 4000 hertz for the transmission of the signal

in analog mode. See alsoBandwidth requirements

(communications).

Applications

PAM, PWM, and PPM found significant application early in the

development of digital communications, largely in the domain of

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radio telemetry for remote monitoring and sensing. They have

since fallen into disuse in favor of PCM.

Since the early 1960s, many of the world's telephone network

providers have gradually, and by now almost completely,

converted their transmission facilities to PCM technology. The

bulk of these transmission systems use some form of time-

division multiplexing, as exemplified by the 24-voice channel T1

carrier structure. These carrier systems are implemented over

many types of transmission media, including twisted pairs of

telephone wiring, coaxial cables, fiber-optic cables, and

microwave. See also Coaxial cable; Communications

cable; Microwave; Optical communications; Optical

fibers; Switching systems (communications).

The deployment of high-speed networks such as the Integrated

Service Digital Network (ISDN) in many parts of the world has also

relied heavily on PCM technology. PCM and various modified

forms such as delta modulation (DM) and adaptive differential

pulse-code modulation (ADPCM) have also found significant

application in satellite transmission systems. See

also Communications satellite; Data communications; Electrical

communications; Integrated services digital network

(ISDN); Modulation.

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Sampled signal (discrete signal): discrete time, continuous values.

Quantized signal: continuous time, discrete values.

Digital signal (sampled, quantized): discrete time, discrete values.

Quantization, in mathematics and digital signal processing, is the

process of mapping a large set of input values to a smaller set –

such as rounding values to some unit of precision. A device

oralgorithmic function that performs quantization is called

a quantizer. The error introduced by quantization is referred to

as quantization error or round-off error. Quantization is involved

to some degree in nearly all digital signal processing, as the

process of representing a signal in digital form ordinarily involves

rounding. Quantization also forms the core of essentially all lossy

compression algorithms.

Because quantization is a many-to-few mapping, it is an

inherently non-linear and irreversible process (i.e., because the

same output value is shared by multiple input values, it is

impossible in general to recover the exact input value when given

only the output value).

The set of possible input values may be infinitely large, and may

possibly be continuous and therefore uncountable (such as the set

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of all real numbers, or all real numbers within some limited range).

The set of possible output values may be finite or countably

infinite. The input and output sets involved in quantization can be

defined in rather general way. For example, vector quantization is

the application of quantization to multi-dimensional (vector-valued)

input data.[1]

There are two substantially different classes of applications where

quantization is used:

The first type, which may simply be called rounding quantization,

is the one employed for many applications, to enable the use of

a simple approximate representation for some quantity that is to

be measured and used in other calculations. This category

includes the simple rounding approximations used in everyday

arithmetic. This category also includes analog-to-digital

conversion of a signal for a digital signal processing system

(e.g., using a sound card of a personal computer to capture an

audio signal) and the calculations performed within most digital

filtering processes. Here the purpose is primarily to retain as

much signal fidelity as possible while eliminating unnecessary

precision and keeping the dynamic range of the signal within

practical limits (to avoid signal clipping or arithmetic overflow). In

such uses, substantial loss of signal fidelity is often

unacceptable, and the design often centers around managing

the approximation error to ensure that very little distortion is

introduced.

The second type, which can be called rate–distortion

optimized quantization, is encountered insource coding for

"lossy" data compression algorithms, where the purpose is to

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manage distortion within the limits of the bit rate supported by a

communication channel or storage medium. In this second

setting, the amount of introduced distortion may be managed

carefully by sophisticated techniques, and introducing some

significant amount of distortion may be unavoidable. A quantizer

designed for this purpose may be quite different and more

elaborate in design than an ordinary rounding operation. It is in

this domain that substantial rate–distortion theory analysis is

likely to be applied. However, the same concepts actually apply

in both use cases.

The analysis of quantization involves studying the amount of data

(typically measured in digits or bits or bit rate) that is used to

represent the output of the quantizer, and studying the loss of

precision that is introduced by the quantization process (which is

referred to as thedistortion). The general field of such study of rate

and distortion is known as rate–distortion theory.

Module

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4

Signal Representation

and Baseband

Processing

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Lesson

21

Nyquist Filtering and

Inter Symbol

Interference

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After reading this lesson, you will learn about:

Power spectrum of a random binary

sequence; Inter symbol interference

(ISI); Nyquist filter for avoiding

ISI; Practical improvisation of ideal

Nyquist filter; Raised Cosine (RC) filter and Root–Raised Cosine

(RRC) filtering; Nyquist’s sampling theorem plays a significant role in the

design of pulse- shaping filters, which enable us to restrict the bandwidth of

information-bearing pulses. In this lesson, we start with a short discussion on the

spectrum of a random sequence and then focus on the concepts of Nyquist Filtering.

Specifically, we develop an idea about the narrowest (possible) frequency band that will be needed

for transmission of information at a given symbol

rate. Consider a random binary sequence shown in Fig.4.21.1

(a) following a common style (NRZ: Non-Return-to-Zero Pulses) for its

representation. Note that a binary random sequence may be represented in several

ways. For example, consider Fig.4.21.1 (b). The impulse sequence of Fig.4.21.1 (b) is

an instantaneously sampled version of the NRZ sequence in Fig.4.21.1 (a) with a

sampling rate of one sample/pulse. The information embedded in the random binary sequence

of Fig.4.21.1 (a) is fully preserved in the impulse sequence of Fig.4.21.1 (b). Verify

that you can easily read out the logical sequence only by looking at the impulses. So,

the two waveforms are equivalent so far as their information content is concerned.

The obvious difference, from practical standpoint, is the energy carried

by a pulse. 1 0 0 1 0 1

1 +a

0 Time

-a T = t Fig.4.21.1(a) Sketch of a NRZ ( Non-Return-to-Zero)

waveform for a random binary sequence 10 0 1011 +1 0 -1 Fig.4.21.1 (b): Instantaneously sampled version of the

NRZ sequence in Fig.4.21.1 (a) Version 2 ECE IIT,

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Power Spectrum of Random Binary Sequence

Let us consider the NRZ representation of Fig.4.21.2(a)

where in a pulse is of height ‘a’ and duration Tb. We wish to get an idea about the spectrum

of such sequences. The sequence may be viewed as a sample function of a random

process, say X(t). So, our approach is to find the ACF of X(t) first and then take its

Fourier Transform. Now, the starting instant of observation need not synchronize with

the start time of a pulse. So, we assume that the starting time of the first pulse (i.e. the initial

delay)’td’ is equally likely to lie anywhere between 0 and

Tb, i.e., 1 ,0 tT = = () T pt = 4.21.1 d b b n 0, elsewhere

Next, the ’0’-s and ‘1’-s are equally likely. So, we can

readily note that, E[X(t)] = 0. +a t -a td Tb Fig. 4.21.2(a): A random NRZ sequence representing

an information sequence ACF of X(t) Let Rx (tk,ti) denote the ACF

of X(t). Rx (tk,ti) = E[X(tk) · X(ti)]

4.21.2 Two distinct cases are to be considered: a) when the shift,

i.e. ¦tk - ti¦ is greater than the bit duration Tb and b) when ¦tk -

ti¦= Tb. Case-I: Let, ¦tk - ti¦>

Tb. In this case, X (tk) and X (ti) occur in different bit

intervals and hence they are independent of each other. This

implies, E [X (tk) · X (ti)] = E [X (ti)] · E [X

(tk)] = 0; 4.21.3 Case-II: ¦tk - ti¦<Tb . For simplicity, let us set tk

= 0 and ti <tk = 0. In this case, the random variables X(tk) and X(ti) occur

in the same pulse interval iff td < Tb - ¦tk - ti¦. Further, both X(tk) and X(ti) are of

same magnitude ‘a’ and same polarity.

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Thus, we get a conditional

expectation: at T t t < -- i 4.21.4 ·= dbk EXt Xt elsewhere () () 2 , k it 0, d Averaging this result over all possible values of

td, we get, - - Ttt ·= () () () EXt Xt apt dt b ki 2 d k i d 0 - - - tt a dt a Ttt == - - 1, t t = b ki

2

2 k i k T TT b i d b b 0 4.21.5 By similar argument, it can be shown that for other values

of tk , the ACF of a binary waveform is a function of the time shift t = tk - ti. So, the

autocorrelation function Rx(t) can be expressed as

[Fig.4.21.2(b)]: t t -< aT T t = R 4.21.6 b () 2 1, b x = 0, T t b Rx(t) a2

t Tb -Tb 0 Fig. 4.21.2(b): Auto Correlation Function Rx(t) for a

random binary waveform Now the power spectral density of the random process X(t)

can be obtained by taking the Fourier Transform of

Rx(t ): () ()

t p t T 2 1 exp

2

=- - Sf a j f dt b T x b T - b ( = 4.21.7 22 sin aT c fT) b b A rough sketch of Sx(f) is shown in Fig. 4.21.2(c). Note

that the spectrum has a peak 1 . A value of ‘a2.Tb’. The spectrum stretches towards ±8 and it

has nulls at ± T n. b normalized version of the spectrum is shown in Fig.

4.21.2(d) where the amplitude is normalized with respect to the peak amplitude and the

frequency axis is expressed in terms of ‘fTb’.

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Sx(f) a2 Tb f -3/Tb -2/Tb -1/Tb

1/Tb 2/Tb 3/Tb Fig. 4.21.2(c) : A sketch of the power spectral density,

Sx(f) for a random binary NRZ pulse sequence

Fig. 4.21.2(d) : Normalized power spectral density, Sx(f)

for a random binary NRZ pulse sequence The wide stretch of the spectrum is understandable as the

time pulses are sharply limited within a bit duration. But then, if such a

random NRZ sequence is used to modulate a carrier sinusoid, one can easily imagine that the

modulated signal spectrum will also have an infinite width and hence, such a

modulated signal cannot be transmitted without distortion though a wireless channel, which is

band-limited. If the modulated signal is forced through a band limited channel without

appropriate spectral shaping, the spectrum of the modulated signal at the receiver will be

truncated due to the band pass characteristic of the channel. From the principle of time-

frequency duality, one can now guess that the energy of a transmitted pulse will no more be

limited within its typical duration ‘Tb’.

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Alternatively, the energy of one pulse will spill over the

time slot of one or more subsequent pulses, causing Inter Symbol Interference

(ISI). So, over a specific pulse duration ‘Tb’, the receiver will collect energy due to one

desired and multiple undesired pulses. A typically vulnerable situation is when a negative

pulse appears in a string of positive pulses or vice versa. In general, the received signal

becomes more vulnerable to noise and upon demodulation; the information sequence

may be erroneous. The extent of degradation in the quality of received information depends

on the time spread of energy of a transmitted pulse and how this effect of ISI is addresses

in the receiver. Another reason why sharp rectangular pulses, even though

designed following Gramm-Scmidt orthogonalization procedure, are

not good for band-limited channels is that, one is simply not allowed to use the full

bandwidth that may be presented by a physical channel. Specifically, most

wireless transmissions must have a priori approval from concerned regulatory authority of a

country. It is mandatory that signal transmission is done precisely over the narrow

portion of the allocated bandwidth so that the adjacent bands can be allocated for other

transmission schemes. Further, to conserve bandwidth for various transmission

applications, the narrowest feasible frequency slot only may be allocated. So,

it is necessary to address the issue of ISI in general and the issue of small transmission

bandwidth by shaping pulses. There are several equalization techniques available for

addressing the issue of ISI. Many of these techniques probe the physical channel and use the

channel state information in devising powerful adaptive equalizers. In this course, we

skip further discussion on equalization and focus only on the issue of pulse shaping

for reduction in transmission bandwidth. Nyquist Filter for avoiding ISI Let us recollect the second part of Nyquist’s sampling

theorem for low pass signals which says that a signal band limited to B Hz can be

recovered from a sequence of uniformly spaced and instantaneous samples of the signal

taken at least at the rate of 2B samples per second. Following this theorem, we may

now observe that the impulse sequence of Fig.4.21.1 (b), which contains information

‘1001011’, can be equivalently described by an analog signal, band limited to ½ x

1sample/pulse. That is, if the bit rate is 1 bit/sec and we sample it at 1 sample/sec, the minimum

bandwidth necessary is ½ Hz! An ideal low-pass filter with brick-wall type frequency

response and having a cutoff of 0.5 Hz will generate an equivalent analog waveform (pulse

sequence) when fed with the random impulse

sequence. Recollect that the impulse response of an ideal low-pass

filter is a sinc function [y = sinc(x), Fig. 4. 21.3]. We note that a) sinc(0)

= 1.0 and the impulse response has nulls at x = ±1, ±2, …. sec, b) null-to-null

time-width of the prominent pulse is 2 sec, c) the pulse is symmetric around t = 0 and d)

the peak amplitude of the pulse is of the same polarity as that of the input impulse.

As the filter is a linear network, the output analog waveform is simply superposition of sinc

pulses, where the peaks of adjacent sinc pulses are separated in time by 1 sec, which is

the sampling interval. Version 2 ECE IIT,

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Fig. 4. 21.3(a) Plot of y =

sinc(x) t R t R s 2 s 2 Fig. 4. 21.3(b) Sketch of output of Nyquist filter for

positive and negative impulses In general, if the information symbol rate is Rs

symbols/sec, i.e. the symbol 1 second, the single-sided bandwidth of the low-

pass filter, known interval is Ts = R

s R Hz. popularly as the equivalent Nyquist Bandwidth,

is BN = 2 s

A simple extension of the above observations implies that

for instantaneous samples of random information-bearing pulse (or symbol)

sequence (@ 1 sample per symbol) will exhibit nulls at ±nTs seconds at the filter

output. Now, assuming an ideal noise less baseband channel of bandwidth BN and zero (or

fixed but known) delay, we can comment that the same filter output waveform will

appear at the input of the receiver. Though the waveform will appear much differ ent from the

initial symbol sequence, the information (embedded in the polarity and magnitude) can

be retrieved without any effect of the other pulses by sampling the received baseband

signal exactly at peak position of each shaped pulse as at those time instants all other pulses

have nulls. This ideal brick- wall type filter is known as the Nyquist Filter for zero-ISI

condition. Fig. 4.21.4 highlights some important features of

Nyquist Filter.

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Fig. 4.21.4 Typical impulse response of a Nyquist filter

[5 samples /symbol duration have been used to obtain the responses using a digital computer.

±32 sample duration around the normalized peak value of 1.0 is

shown] Mathematical Explanation Let us consider an ideal lowpass filter with single-sided

bandwidth W. It is usually convenient to deal with the normalized impulse response

h(t) of the filter where in, Wt p ) 2 sin 4.21.8 h(t)=sinc (2Wt) = ( ) Wt p 2

Now if an impulse of strength ‘A’ is applied at the input of

the filter at t = t1, the filter o/p may be expressed as, y(t) = Ah(t- t1)=A.sinc 2W (t- t1) 4.21.9 An information-carrying symbol sequence may be

represented as ( - . t A d ), where Ai = ± A and ti = i.Ts 4.21.10 8 i t i = 0 i The response of the low-pass filter to this

sequence is, 4.21.11 - = )} ( 2 { sin . ) ( t W c A t y 8 i t i = 0 i 1 = Now, if we set /2 R = s 2.T W and sample the output of the filter at t =

m.Ts, it is easy s to see that,

m m () 4.21.12 = = = = A i) - .sinc(m A i - m .sinc2W A ) m.T ( T t y m i i s s = = 0 i 0 i So, we obtain the peak of the m-th pulse clean and devoid

of any interference from previous pulses.

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Practical improvisations: The above discussion on Nyquist bandwidth is ideal

because neither a low-pass filter (LPF) with brick-wall type response can be realized

physically nor can an ideal impulse sequence be generated to represent a discrete information

sequence. Further, note that if there is any error in the sampling instant (which, for a

practical system is very likely to occur from time to time due to the effects of thermal noise

and other disturbances), contribution from the other adjacent pulses will creep into

the sample value and will cause Inter Symbol Interference. For example, the second

lobe peak is only about 13 dB lower compared to the main lobe peak and the decay of the

side-peaks of a sin x/x function is not very rapid with increase in x. So, the

contribution of these peaks from adjacent symbols may be significant. Fortunately, the

desired features of pulse epoch and zero crossings are not unique to a ‘sinc’ pulse. Other pulse

shapes are possible to design with similar features, though the bandwidth requirement for

transmitting a discrete information sequence will be more compared to the

corresponding Nyquist bandwidth (BN). Two such relevant constructs are known as a) Raised

Cosine (RC) filter and b) Root-Raised Cosine (RRC)

filter. Raised Cosine Filter

Let us denote a normalized pulse shape which avoids ISI

as x(t). then, = 1 x(0) and 0 n 0, ) t x 4.21.13 ( = ± nT t = s

Some of the practical requirements on x(t) are

the following: (a) Energy in the main pulse is as much as possible

compared to the total energy distributed beyond the first nulls around the main peak.

This ensures better immunity against noise at the receiver for a given signal

transmission power. So, it is desired that the magnitde of the local maxima of the i-th pulse of

x(t) between iT= t < (i+1)T decreases monotonically and rapidly

with time. (b) The pulse shape x(t) should be so chosen that some

error in instants of sampling at the receiver does not result in appreciable ISI. These two

requirements are usually depicted in the form of a mask in technical

standards. Out of several mathematical possibilities, the following

amplitude mask is very useful for application on the ideal Nyquist filter

impulse response h(t): t cos ßp T ) s t m ß = 4.21.14 ( 2 ) / (4 - 1 2 2 T t s The resulting pulse shape is known as a Raised Cosine

pulse with a roll-off of ‘ß’( 0 ß =1). he Raised Cosine pulse is

described as: t ßp cos T pRC (t) =

sinc(t/Ts). )

2 s ß / (4 - 1 T t 2 2 s 4.21.15 The normalized spectrum of Raised Cosine

pulse is:

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1 ß - H(f) = 1, for () = , f 2 T s ß ß ß p T 1 1 1 + - - = () ( ) ( ) s = = - f f for , cos 2 ß T T T 2 2 2 2 s s s + 1 = 0, for () f 2 ß > 4.21.16 T s

Figs. 4.21.5 (a)-(c) highlight features of a Raised Cosine

(RC) filter. The roll off factor ‘ß’ is used to find a trade off between the absolute

bandwidth that is to be used and the difficulty (in terms of the order of the filter and the

associated delay and inaccuracy) in implementing the filter. The minimum usable bandwidth

is obviously the Nyquist bandwidth, BN (for ß =0) for which the filter is unrealizable

in practice while a maximum absolute bandwidth of 2BN (for ß = 1.0) makes it much

easier to design the filter. ß lies between 0.2 and 0.5 for most of the practical systems where

transmission bandwidth is not a luxury. Use of analog components was dominant

earlier though the modern trend is to use digital filters. While digital FIR structure usually

ensures better accuracy and performance, IIR structure is also used to reduce the

necessary hardware.

Fig. 4.21.5 (a) Typical impulse response of a Raised

Cosine filter with a roll off factor ß = 0.5[5 samples /symbol duration have been used to obtain

the responses using a digital computer. ±32 sample duration around the normalized

peak value of 1.0 is shown]

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Fig. 4.21.5 (b) Typical amplitude spectrum of a Raised

Cosine filter with a roll off factor ß = 0.5. The magnitude is not normalized. Multiply the

normalized frequency values shown above by a factor of 5 to read the frequency

normalized to symbol rate. For example, i) 0.1(from the above figure)x5 = 0.5 = (1 -ß).Rs,

where ß =0.5 and Rs = 1. The transition band of the filter starts here and ii) 0.3(from the

above figure)x5 = 1.5 = (1 +ß).Rs. The stop band of the filter

starts here.

Fig. 4.21.5 (c) Typical phase spectrum of a Raised

Cosine filter with a roll off factor ß = 0.5. Multiply the normalized frequency values shown above

by a factor of 5 to read the frequency normalized to symbol

rate. The side lobe peaks in the impulse response of a Raised

Cosine filter decreases faster with time and hence results in less ISI compared to

the ideal Nyquist filter in case of sampling error in the

receiver. There is another interesting issue in the design of pulse

shaping filters when it comes to applying the concepts in a practical

communication transceiver. From our discussion so far, it may be apparent that the pulse-shaping

filter is for use in the transmitter only, before the signal is modulated by a carrier

or launched in the physical channel. However, there is always a need for an equivalent

lowpass filter in the receiver to eliminate out-of-band noise before demodulation and

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out. This purpose is accomplished in a practical receiver by

splitting a Raised Cosine filter in two parts. Each part is known as a Root Raised

Cosine (RRC) filter. One RRC filter is placed in the transmitter while the other part is

placed in the receiver. The transmit RRC filter does the job of pulse shaping and

bandwidth-restriction fully while not ensuring the zero-ISI condition completely. In case of a

linear time invariant physical channel, the receiver RRC filter, in tandem with the

transmit RRC filter, fully ensures zero-ISI condition. Additionally, it filters out undesired out-

of-band thermal noise. On the whole, this approach ensures zero-ISI condition in the

demodulator where it is necessary and it also effectively ensures that the equivalent noise-

bandwidth of the received signal is equal to the Nyquist bandwidth BN. The overall

complexity of the transceiver is reduced without any degradation in performance compared

to a system employing a RC filter in the transmitter and a different out-of-band noise

eliminating filter in the receiver. Figs. 4.21.6 (a)-(c) summarize some features of a Root-

Raised Cosine filter.

Fig. 4.21.6 (a) Typical impulse response of a Root Raised

Cosine filter with a roll off factor ß = 0.5[5 samples /symbol duration have been used

to obtain the responses using a digital computer. ±32 sample duration around the

normalized peak value of 0.5 is shown]

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Fig. 4.21.6 (b) Typical amplitude spectrum of a Raised

Cosine filter with a roll off factor ß = 0.5. Multiply the normalized frequency values shown

above by a factor of 5 to read the frequency normalized to

symbol rate.

Fig. 4.21.6 (c) Typical phase spectrum of a Raised

Cosine filter with a roll off factor ß = 0.5. Multiply the normalized frequency values shown above

by a factor of 5 to read the frequency normalized to symbol

rate. We will mention another practical issue related to the

design and implementation of pulse-shaping filters. Usually, an information sequence is

presented at the input of a pulse shaping filter in the form of pulses (.e.g. bipolar

NRZ) and the width of such a pulse is not negligibly small (as in that case the energy per

pulse would be near zero and hence some pulses may even go unrecognized). This finite

non-zero width of the pulses causes a distortion in the RC or RRC pulse shape. The

situation is somewhat analogous to what happens when we go for flat-top sampling of a band-

limited analog signal instead of instantaneous sampling. The finite pulse-width results in

amplitude fluctuation and again introduces ISI. It needs a relatively simple pulse magnitude

correction which is x commonly referred as

‘ x sin amplitude compensation’ to restore the

quality of shaped Version 2 ECE IIT,

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% Sample the signal 100 times per

second, for 2 seconds.

Fs = 100;

t = [0:2*Fs+1]'/Fs;

Fc = 10; % Carrier frequency

pulses. The transfer function of the RRC shaping filter in

the transmitter is scaled appropriately to incorporate the amplitude

correction. Problems Q4.21.1) Mention possible causes of Inter Symbol

Interference (ISI) in a digital communication receiver. Q4.21.2) Sketch the power spectrum of a long random

binary sequence when the two logic levels are represented by +5V

and 0V. Q4.21.3) Sketch the frequency response characteristics of

an ideal Nyquist low pass filter. Q4.21.4) What are the practical difficulties in

implementing an ideal Nyquist low pass filter?

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x = sin(2*pi*t); % Sinusoidal

signal

% Modulate x using single- and

double-sideband AM.

ydouble = ammod(x,Fc,Fs);

ysingle = ssbmod(x,Fc,Fs);

% Compute spectra of both

modulated signals.

zdouble = fft(ydouble);

zdouble =

abs(zdouble(1:length(zdouble)/2+1)

);

frqdouble = [0:length(zdouble)-

1]*Fs/length(zdouble)/2;

zsingle = fft(ysingle);

zsingle =

abs(zsingle(1:length(zsingle)/2+1)

);

frqsingle = [0:length(zsingle)-

1]*Fs/length(zsingle)/2;

% Plot spectra of both modulated

signals.

figure;

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subplot(2,1,1);

plot(frqdouble,zdouble);

title('Spectrum of double-sideband

signal');

subplot(2,1,2);

plot(frqsingle,zsingle);

title('Spectrum of single-sideband

signal');

FM MODULATION

[r, c] = size(x);

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if r*c == 0 y = []; return; end; if (r == 1) x = x(:); len = c; else len = r; end; Df = 2*pi*beta*100; x = interp(x,20); x = x*Df; Fs = 20*8000; Fc = 40000; t = (0:1/Fs:((size(x,1)-1)/Fs))'; t = t(:, ones(1, size(x, 2))); x = 2 / Fs * pi * x; x = [zeros(1, size(x, 2)); cumsum(x(1:size(x,1)-1, :))]; y = cos(2 * pi * Fc * t + x ); plot(y)

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FM DEMODULATION

function x = FMdemod(y,beta) [r, c] = size(y); if r*c == 0 y = []; return; end; if (r == 1) y = y(:); len = c; else len = r; end; Fc = 40000; Fs = 8000*20; pi2 = 2*pi; [num, den] = butter(5, Fc * 2 / Fs); sen = 2*pi*beta*100; %pre-process the filter. if abs(den(1)) < eps

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error('First denominator filter coefficient must be non-zero.'); else num = num/den(1); if (length(den) > 1) den = - den(2:length(den)) / den(1); else den = 0; end; num = num(:)'; den = den(:)'; end; len_den = length(den); len_num = length(num); x = y; y = 2 * y; ini_phase = pi/2; for ii = 1 : size(y, 2) z1 = zeros(length(den), 1); s1 = zeros(len_num, 1); intgl = 0; memo = 0; for i = 1:size(y, 1)

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%start with the zero-initial condition integer. vco_out = cos(pi2 * intgl+ini_phase); if len_num > 1 s1 = [y(i, ii) * vco_out; s1(1:len_num-1)]; else s1 = y(i, ii); end tmp = num * s1 + den * z1; if len_den > 1 z1 = [tmp; z1(1:len_den-1)]; else z1 = tmp; end; intgl = rem(((tmp*sen + Fc)/ Fs + intgl), 1); x(i, ii) = tmp; end; end; x = x; x = decimate(x,20);

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KARPAGA VINAGA COLLEGE OF ENGINERING AND TECHNOLOGY

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

EC1351- DIGITAL COMMUNICATION

QUESTION BANK

SEM/YEAR/SEC: VI/III/B STAFF I/C: A.JOTHIMANI

UNIT-I PULSE MODULATION

Part A

1. Give the mathematical form of sampling process

2. List out uses of sampling theorem

3. Define instantaneous sampling

4. What is anti aliasing effect

5. What is PWM

6. List out the types of Quantization

7. What is

8. Transfer 01101001 in to Manchester code

9. What is noise consideration in PCM

10. Define processing gain.

11. What are the two limitations of Delta Modulation? (Apr/May-04)

12. What you understand by the term ‘aliasing’ ? (Nov/Dec05)

13. A band pass signal has the spectral range that extends from 20 KHz to82

KHz. Find the acceptable range of sampling frequency fs . (Nov/Dec05)

14. State Sampling Theroem ( May/June -07)

15. Define Quantization error ( May/June -07)

16. What is SNR of PCM system if the number of quantization levels? (Apr/May-06)

17. State band pass sampling theorem. (Apr/May-06)

18. Plot the magnitude spectrum of the ideally sampled version of the signal

M(t)=2cos(200t)+40sin(290t).Assuming that the sampling rate is 1khz.

(Apr/May - 08) 19.Define position modulation scheme with a suitable diagram (Apr/May -08)

20.State band pass sampling theorem. (Apr/May-04)

Part B

1. Drive the expression for the sampling process in time domain. (16)

2. What are all the types of sampling technique and explain about any two. (16)

3. a. Explain the generation of PPM and PWM with neat circuit diagram. (8)

b. Explain the quantization process with PCM block diagram. (8)

4. Write brief notes on 1) TDMA (8 )

2) FDMA (8)

5. Compare DM with ADM and explain linear prediction filter. (16)

What is meant by a compander?What are the two types of compression?(8)

(ii) Explain the frame format and signaling scheme used in T1 carrier system?(8)

(Apr/May-04)

6..(i) Derive expressions quantization noise and signal-to-noise in a PCM system using a

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Uniform quantiser.(10).

(ii) A sinusoidal signal is transmitted using PCM.An output SNR of 55.8dB is required

Find the number of representation levels required to achieve this performance.

( Nov/Dec-05)

7. (i) Explain with a neat diagram, the direct sequence spread binary PSK system?(8)

(ii) Obtain the expression for the processing gain(PG) in this system.

(8) (Apr/May-06)

8.. (i) Derive the expression for SNR in PCM system and compare it with

Delta modulation. Explain how can the SNR be improved in a PCM system.(10)

(ii) Show that prediction error variance is less than the variance of the predictor

input Of the predictor of order one. (Apr/May-06) 9.

(i) With supporting derivation, prove that if a signal contains no frequencies higher than W

Hertz,it may be reconstructed from its samples at a sequence of points spaced ½ W seconds

apart. (8)

(ii) Explain the principle of Delta modulation and derive an expression for average output

noise power in delta modulation.(8) (Apr/May-06)

10. Explain the process of quantization, encoding and decoding in pulse code

Modulation?In what way differential PCM is better than PCM?(16) ( May/June 07)

11. Describe the following systems by presenting appropriate diagrams.

(i) Time Division Multiplexing(8)

(ii) Delta Modulation.(8) ( May/June -07)

12.(a) (i) Give the block diagram of differential pulse code modulation scheme and

explain the principles in detail(9)

(ii) Obtain an expression for the processing gain of a DPCM system.(4)

(iii)Suppose an existing standard PCM system for the voice signal is replaced by

a DPCM processing gain 6dB,while maintaining the (SNR)Q. What will be the

reduction in the bit rate achieved by DPCM?(3) (Apr/May -08)

13.(i) For a Uniform quantizer,discuss the way in which the number of quantization

Levels (L) influence the bandwidth and the Quantization noise.(3)

(ii) Discuss the need for Non-uniform Quantization of speech signal (2).

(iii) Outline the principles of compander used for speech signal.(4)

(iv) Give the bit rate of a Delta modulator fed with samples at a rate of

40ksamples/sec.(2)

(v)How does adaptive delta modulation help in alleviating the problems associated with

Delta modulation scheme.(5) (Apr/May -08)

14. Explain the principle,operation ,signals ,constellation diagram transmitter and

receiver of a MSK system.(16) (May/June-07)

UNIT -2

BASEBAND PULSE TRANSMISSION

Part A

1. Define matched filter

2. What is decision device

3. Draw the block diagram of base band binary data transmission system

4. What is Nyquist channel

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5. Draw the frequency and time response of raised cosine spectrum

6. What you mean by correlative level coding

7. Define LMSA

8. What is DFE

9. List out the use of eye patterns

10. Define zero crossing effect

11.. How is the eye pattern obtained on the CRO? (Apr/May-04)

12. What is the condition for zeros Inter symbol Interference? (Apr/May-04)

13.What is the purpose of using an eye pattern? (Nov/Dec05)

14. Why do you need adaptive equalization in switched telephone network?

(Nov/Dec05)

15.What is an ideal Nyquist channel? (Apr/May-06)

16. What is meant by intersymbol interference? (Apr/May-06)

17. Why do we need equalization in base band pulse transmission? ( May/June -07)

18. A message has zero mean value and a peak value of 10V. It is to be quantized using a

Step size of 0.1V with one level coinciding to 0 V. Find number of bits required

for encoding the quantized signal. ( May/June -07)

19. Give the Nyquist criterion for zero ISI.Plot the impulse response of an ideal Nyquist

channel. (Apr/May -08)

20.Draw an eye pattern and respresent the ways in which it could be used to evaluate the

performance a baseband pulse transmission system. (Apr/May -08)

Part B

1. Derive the expression for the matched filter and prove any one property. (16)

2. Prove the effect ISI using necessary block diagram and design. (16)

3. Write brief notes on Nyquist’s criterion for distortion less base

band binary transmission. (16)

4. Explain the six array PAL transmission. (16)

5. a) (i) Explain the cross talk produced due to high frequency cutoff of the channel in

PAM. What is the effect of pulse width on the cross talk factor?(8)

(ii) Compare the power spectra of different binary formats.(8) (Apr/May-06)

6. (i) With neat sketches, explain the duo binary signaling scheme.(8)

(ii) Write briefly about eye pattern and adaptive equalization for data transmission.(8)

(Apr/May-06)

7. a) (i) Generate the code words for (7,4) Hamming code.(8)

(ii) State and prove the properties of syndrome decoding. (8) (Apr/May-06)

8. (i) Explain briefly about linear block codes.(4)

(ii) Evaluate the syndrome S for all five probable single error patterns in (5,1)

Repetition code.(6)

(iii) Briefly explain the Viterbi decoding algorithm .(6) (Apr/May-06)

9. Why do we adoptive equalizer and design any one type of adoptive equalizer. (16)

10..Find the optimum filter frequency response How that maximizes the output SNR

When the input noise is not a white noise.(16) ( May/June -07)

11. Discuss the following:

(i) Adaptive Equalization(8)

(ii) Base band M-ary PAM transmission. (8) ( May/June -07)

12. (a) (i) What are the special features of correlative coding?(2)

(ii) Draw the block diagram of a Duobinary encoder and obtain an expression for

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the frequency response of the same.(6)

(iii) What is the need for precoder.(1)

(iv) For the binary sequence 011010100110 give the output of a Duobinary

encoder (in the absence of a precoder).(3)

(v) For the binary sequence 011010100110 give the output of a Duobinary

Encoder with a precoder. (4) (Apr/May-08)

(vi) (i) Obtain the expression for the average probability of symbol error assuming

NRZ signaling,if the noise is modeled as AWGN.(10)

(ii) Compare a baseband binary PAM system with that of M-ary PAM system.(6)

(Apr/May -08)

13. (i) Sketch the time response and frequency response of signal with raised cosine pulse

spectrum.(8)

(ii) Why is the precoding used with duobinary signaling scheme?Draw the block diagram

of precoderand explain its operation (8) (Apr/May-04)

14. (i) Draw the block diagram of an adaptive filter and explain the LMS algorithm.(10)

(ii) Explain how eye pattern is used to study the performance of a data transmission

system.

(6) (Apr/May-04).

15. What do you understand by the term Inter –Symbol Interference (ISI)?Discuss in detail

the Nyquist criterion for minimizing ISI.Explain the difficulties in implementing it in a

practical system.(16). (Nov/Dec05)

16.(i)Discuss the merits and demerits of Duobinary Signalling.(6)

(ii)The binary data(0 1 1 1 0 0 1 0 1 ) are applied to the input of a duobinary

encoder.Construct the duobinary encoder output and corresponding receiver output,without a

precoder .Suppose due to error during transmission ,the level produced by the third digit is

reduced to zero,construct the new receiver output(10). (Nov/Dec05)

UNIT - 3

PASSBAND DATA TRANSMISSION

Part - A

1. Define FSK

2. Draw the Basic block diagram of FSK

3. Define MFSK

4. Draw the wave form of the MPSK

5. What is MSK

6. Define Non-coherent FSK

7. Differentiate coherent FSK from Non-coherent FSK

8. What do you mean by Probability error.

9. Define Passband transmission.

10 Draw the baseband signal.

11. How is the transfer function of the Matched Filter related to the spectrum of the signal?

(Apr/May-04)

12. In minimum shift keying what is the relation between the signal frequencies and bit rate?

(Apr/May-04)

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13. Draw an illustrative figure to show the operation of a correlation receiver?

(Nov/Dec05)

14.Highlight the major difference between a QPSK signal and aMSK signal. (Nov/Dec05)

15. Compare the probability of error of PSK with that of FSK. (Apr/May-06)

16. State the difference between coherent and non coherent binary modulation schemes.

(Apr/May-06)

17. Differentiate coherent and non coherent receivers. ( May/June -07)

18. What do you understand by continuous phase frequency shift keying?

( May/June -07)

19.Plot the power spectrum of a BPSK signal operated with a carrier frequency of

140MHz,modulated by data bits at a rate of 2400 bits/sec.What is the bandwidth

requirement? (Apr/May -08)

20.Give the signal space representations of QPSK.How is the performance of the system

related to the distances between the symbols in the signal space ? (Apr/May -08)

Part B

1. Derive the expression for the BPSK technique. (16)

2. Explain the MFSK and derive the pe. (16)

3. Draw the space diagram of MSK and explain about the same. (16)

4. Derive the error probability expression for the MPSK. (16)

5. Compare the all types of Passband transmission. (16)

6. (i) Discuss briefly about Minimum Shift Keying for a Continuous Phase Frequency Shift

Keying(CPFSK) signal.(8)

(ii)With necessary equations and signal space diagram, explain briefly about FSK

system.(8) (Apr/May-06)

7 (i) Explain the Quadriphase Shift keying (QPSK) modulation(8)

(ii) Obtain probability of error interms of Eb and No (Apr/May-06)

8.(i) In a PSK system, the received waveforms S1(t)=A Coswt,S2(t)=-A Coswt are

Coherently detected with a matched filter. The value of A is 20 mV, and the bit

Rate is 1 Mbps. Assume that the noise power spectral density h/2=10-11 W/Hz.

Find the proability of error Pe.(6)

(ii) Enumerate on carrier and symbol synchronization(5+5) ( May/June -07)

9. (i) Give the block diagram for the generation and detection of BFSK signal and

Give a brief explanation of the same.(8)

(ii)What are the special properties of MSK scheme?(4)

(iii) Compare the performance of BPSK with that of BFSK.(4) (Apr/May -08)

10. (i) Discuss the generation and detection of QPSK with suitable block

Diagrams (10) .

Write a note on DPSK.(6) (Apr/May -08)

11. Explain the direct sequence spread spectrum modulation with coherent binary

Phase shift keying.(16) (May/June-07)

12. Describe the following:

(i) Signal space dimensionality and processing gain. (8)

(ii). Maximum length and Gold codes. (8) (May/June-07)

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13. (i) Explain the technique used for generating N orthonormal basis functions.(8)

(ii) What do you understand by coherent detection?(2)

(iii)Discuss a method for bit synchronization.(6) (Apr/May-04)

14.(i) Draw the block diagram of QPISIC transmitter and coherent QPSK receriver and

explain their operation.(10)

(ii)Compare the BER of coherent PSK coherent QPSK and coherent FSK. (6) (Apr/May-

04)

Unit- 4

ERROR CONTROL CODING

Part A

1. Define the Channel coding

2. List out the uses of the Channel Coding

3. Differentiate Sourced coding from the Channel coding.

4. Prove any two properties of Block Codes.

5. What is syndrome?

6. Define Code rate.

7. What is BSC?

8. Draw the channel diagram of the BPSK system

9. What is turbo codes

10. Define CRC.

11. What is meant by syndrome of a linear block code? (Apr/May-04)

12. The minimum transmitter power requirement in the absence of channel coding of a

communication System is 2W. If a channel coding scheme of 3 dB is incorporated in the

system give the minimum required transmitter power. (Apr/May-04)

13.Explain the fundamental difference between block codes and convolutional codes.

(Nov/Dec05)

14.How will you define coding gain with reference to error control codes ? (Nov/Dec05)

15. List t7. What is meant by BCH code? (Apr/May-06)

16. What is a convolution code? (Apr/May-06) he advantages of Turbo codes

( May/June -07)

17. Define Minimum distance. ( May/June -07)

Part B

1. Explain the channel coding technique used in the Block codes. (16)

2. What are all the message vectors can be extracted from the

code vector that was generated by 1+X+X3. (16)

3. Explain the Viterbi algorithm for the detection of any two

message vectors (One Word) from the code vectors that

were generated by (111), (101) and (011). (16)

4. Explain the construction of Block Code and

Explain how error syndrome is calculated. (16)

5. Explain in detail about Orthogonal codes, Biorthogonal codes and

Transorthogonal codes. (16)

6. Consider a (7,4) linear block code with the parity-check matrix give by

(i)Construct code words for this (7,4) code

(ii)Show that this code is Hamming code.

(iii)Illustrate the relation between the minimum distance and the structure of the

parity-check matrix H by considering the word 0101100. (May/June-07)

7.Let g(x) be the generator polynomial of a cyclic code C.Find a scheme in

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Encoding the data sequence (d0,d1…………dk-1) into an (n,k) system code C.

(16) (May/June-07)

8.(a) (i) Give the properties of cyclic codes.(3)

(ii) Give the details of implementation of cyclic encoder and decoder based on

Linear feedback shift registers.(9)

(iii) A communication system operates at a bit rate of 10kbps with a transmitter

power of 50Watts with a BER of 10^-7 using BPSK.If a channel coder of

coding gain 3 dB is incorporated into the system,calculate the transmitter power required to

maintain the same BER.(assume the modulation scheme remains the same).State the

disadvantage of including the channel coder,if there is any.(4) (Apr/May -08)

9. (i) Draw the block diagram of a convolutional encoder of constraint length 3

And code rate ½.(3)

(ii) Draw the state diagram for the encoder you have given in.(5)

(iii) Discuss the decoder algorithm in detail for the encoder given by you.(8)

(iv) Define Hamming distance between linear block codes.Give the error

correction and error detection capability in terms of Hamming distance.

(Apr/May -08) (v) Give the special features of Trellis codes. (Apr/May -08)

10.(i)Write the generator matrix and parity check matrix of a (7,4) Hamming code.(6)

(ii)Describe a decoding procedure for linear block code.(6)

(iii)Explain the features of RS code (4) (Apr/May-04)

11.(i)Draw the diagram of a rate -1/2 convolutional encoder with constraint length 12.What is

the generator polynomial of the encoder? Find the encoded sequence you have drawn

,corresponding to the message sequence(10011).(12)

(ii)obtain the Trellis diagram of the encoder that you have drawn.(4). (Apr/May-04)

UNIT- 5

SPREAD SPECTRUM MODULATION

Part A

1 What are all the types of SSM

2 What is the use of PN sequence

3 List out the prosperities of PN sequence

4 What is the processing gain of SSM.

5 Define jamming margin

6 Define FHSSM

7 Define SFHSS.

8 Define FFHSM.

9 Draw the PN sequence generator for [6,4,2,1]

10 What you mean by runs in PN sequence

11 State the Gold Theorem

12. When is the PN sequence called as maximal length sequence?

13. What is meant by processing gain of DS spread spectrum system? (Apr/May-04)

14. Write down the properties of PN sequences. (Nov/Dec05)

15How do you define processing gain for aDS-CDMA system? (Nov/Dec05)

16.fine frequency hopping? April/may-06)

17 What are the applications of Spread spectrum modulation? April/may-06

18 What do you mean by Jamming? ( May/June -07)

19List any two applications of spread spectrum modulation. ( May/June -07)

Part B 1. Generate the PN sequence for [5, 4, 3, 1] and prove the all properties of the same. (16)

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2. List out and prove the Properties of the PN sequence. (16)

3. Explain the concept of DS/BPSK. (16)

4. List out the uses of the FH. Explain the FH techniques with the frequency spectrum. (16)

5. Write shorts notes on

1. Probability Error. (4)

2. Jamming Margin (4)

3. (S/N) of the DS System. (8)

6.(i). Discuss the configuration of a Direct Sequence spread spectrum system with

BPSK in detail.(13)

(ii) In a direct sequence spread spectrum system with BPSK,the information bit

Duration is 4.095 ms and the PN chip duration is 1us.Calculate the processing

Gain of the system and the feedback shift length used for the PN sequence.

(3) (Apr/May -08) 7.

With suitable block diagrams ,discuss the structure of the transmitter and

Receiver of frequency hopping spread spectrum systems.(10).

(ii)What is the difference between slow FH and fast FH systems?(2)

Assume that a slow FH system employs M-FSK data modulation with N hop bands and

operates with a binary input data rate of R bits/sec.What is the minimum bandwidth

requirement of the system?(4) (Apr/May -08)

8 give the properties of maximal length sequence(PN sequence). (Apr/May -08)

9List any two applications of spread spectrum. (Apr/May -08)

10State and explain the properties of maximal length sequences.(6)

(ii)Draw the block diagram of DS- spread spectrum system transmitter and receiver and

explain the function performed by each block in brief.(10) (Apr/May-04)

11Explain the principle of operation of frequency hopped M-ary FSK spread spectrum

system(10)

(ii)Discuss the ways in which fast hopping scheme and slow frequency hopping (spread

spectrum) schemes could be used to mitigate multipath effect.(6) (Apr/May-04)


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