Topic 14 Digital Technology

Analogue and Digital Signals

Base 10

5 0 3 7

Base 10

103 102 101 100

5 0 3 7

Base 10

5037 = 5 x 103 + 0 x 102 + 3 x 101 + 7 x 100

103 102 101 100

5 0 3 7

Base 2 (binary numbers)

1 x 24 + 0 x 23 + 1 x 22 + 0 x 21 + 1 x 20 = 21

26 25 24 23 22 21 20

1 0 1 0 1

Base 2 (binary numbers)

?

26 25 24 23 22 21 20

1 0 0 0 0 1

Base 2 (binary numbers)

1 x 25 + 1 x 20 = 33

26 25 24 23 22 21 20

1 0 0 0 0 1

Four-bit words

1 x 23 + 1 x 20 = 9

26 25 24 23 22 21 20

1 0 0 1

Four bit words

1001, 1110, etc.

A three bit word can be made into a four bit word by adding a zero to the front

4 = 100 (3 bit word) = 0100 (4 bit word)

Four bit words

With 4 bit words we can represent a total of 16 numbers in binary numbers (0 to 15)

Total number of numbers = 24

Total number of numbers = 2(# of bits)

Example question

How many numbers can be represented with 6 bit words?

Example question

How many numbers can be represented with 6 bit words?

# numbers = 26 = 64

Another example question

Write the decimal number 65 using eight bits.

Another example question

Write the decimal number 65 using eight bits.

65 = 1x26 + 0x25 + 0x24 + 0x23 + 0x22 + 0x21 + 1x20

65 = 1000001 = 01000001 (eight bits)

Another example question

Write the decimal number 65 using eight bits.

65 = 1x26 + 0x25 + 0x24 + 0x23 + 0x22 + 0x21 + 1x20

65 = 1000001 = 01000001 (eight bits)

Least significent bit (LSB)

Most significent bit (MSB)

Analogue signals

Analogue signals are continuous signals that vary in proportion to the physical mechanism that created the signal.

Digital signal

is a coded form of signal that takes the discrete values of 0 or 1 only

Converting from analogue to digital

Another example

t/ms

V/V

2

4

6

8

1 2 3 4

Another example

t/ms

V/V

2

4

6

8

1 2 3 4

Take a sample of the signal every 1 ms (sampling frequency or rate of 1000 Hz)

Another example

t/ms

V/V

2

4

6

8

1 2 3 4

This is called a pulse amplitude modulated signal (PAM)

Time/ms PAM signal /V

0 0

1 2

2 4

3 6

4 8

Another example

t/ms

V/V

2

4

6

8

1 2 3 4

Note the duration of each sample is is very short (1 μs or less)

Time/ms PAM signal /V

0 0

1 2

2 4

3 6

4 8

Converting into 2 bit words

t/ms

V/V

2

4

6

8

1 2 3 4

Time/ms PAM signal /V

0 0

1 2

2 4

3 6

4 8

Converting into 2 bit words

t/ms

V/V

2

4

6

8

1 2 3 4

Binary code

PAM signal /V

00 0 ≤ V < 2

01 2 ≤ V < 4

10 4 ≤ V < 6

11 6 ≤ V < 8

The maximum number of words is 22 = 4, so we have to split the voltages into 4 levels (0-2,2-4,4-6,6-8)

Converting into 2 bit words

t/ms

V/V

2

4

6

8

1 2 3 4

So in 2-bit words we can transmit the signal as 00011011

Binary code

PAM signal /V

00 0 ≤ V < 2

01 2 ≤ V < 4

10 4 ≤ V < 6

11 6 ≤ V < 8

Converting into 2 bit words

t/ms

V/V

2

4

6

8

1 2 3 4

Binary code

PAM signal /V

000 0 ≤ V < 1

001 1 ≤ V < 2

010 2 ≤ V < 3

011 3 ≤ V < 4

100 4 ≤ V < 5

101 5 ≤ V < 6

110 6 ≤ V < 7

111 7 ≤ V < 8

We can reduce the loss of information by sampling every 0.5ms and using 3 bit words (maximum of 23 = 8)

Quantization

Binary code

PAM signal /V

000 0 ≤ V < 1

001 1 ≤ V < 2

010 2 ≤ V < 3

011 3 ≤ V < 4

100 4 ≤ V < 5

101 5 ≤ V < 6

110 6 ≤ V < 7

111 7 ≤ V < 8

The process of dividing the range of the analogue system into levels is called quantization.

The number of quantization levels used depends on the number of bits used. Number of levels = 2n

Quantization error

Binary code

PAM signal /V

000 0 ≤ V < 1

001 1 ≤ V < 2

010 2 ≤ V < 3

011 3 ≤ V < 4

100 4 ≤ V < 5

101 5 ≤ V < 6

110 6 ≤ V < 7

111 7 ≤ V < 8

In this case the quantization error is 1 V. Two analogue signals that differ than less than the quantization error will be assigned the same binary number.

                                                                                                             

                                                                                     

Compact disks

Paths (1.6μm apart)

pits

land

A spiral of unconnected pits are sensed with a laser beam. The pattern of pits on a CD store information as a series of

binary '1's and '0's. As the disc rotates a laser beam is used to produce a '1' every time it finds a pit edge.

Destructive interference when path difference is λ/2 (depth is

λ/4)

This corresponds to a binary ‘1’

                                                                                                                                                                                                                                                              

Example question

Calculate the pit depth of a CD being read by a laser of λ = 600 nm.

                                                                

Example question

Calculate the pit depth of a CD being read by a laser of λ = 600 nm.

Pit depth = λ/4 = 150 nm                                                                 

Another example

Information is imprinted on a CD at a rate of 44 100 words per second. The information is in 32-bit words. A CD lasts for 74 minutes. Calculate the storage capacity of the CD

Another example

Information is imprinted on a CD at a rate of 44 100 words per second. The information is in 32-bit words. A CD lasts for 74 minutes. Calculate the storage capacity of the CD

# of bits = 44100 x 32 x 60 x 74 = 6.27 x 109 bits

1 byte = 8 bits

6.27 x 109 bits = (6.27 x 109)/8 bytes = 780 Mb

Questions!