eng241 digital design week #5 arithmetic circuits
TRANSCRIPT
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ENG241 ENG241 Digital DesignDigital Design
Week #5 Arithmetic Circuits
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TopicsTopics
Binary Adders Binary Ripple Carry Adder 1’s and 2’s Complement Binary Subtraction Binary Adder-Subtractors Binary Multipliers BCD Arithmetic
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ResourcesResources
Chapter #5, Mano Sections 5.2 Binary Adders 5.3 Binary Subtraction 5.4 Binary Adders-Subtractors 5.5 Binary Multiplications 5.7 HDL Representations -- VHDL
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Half AdderHalf Adder
S = XY’ + X’Y = X Y C = X.Y
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Recall: Arithmetic -- additionRecall: Arithmetic -- addition
Binary similar to decimal arithmetic
+ 10001
00110
1 1 1 0 1
No carries 1 0 1 1 0 0
1 0 1 1 0
+ 1 0 1 1 1
1 0 1 1 0 1
Carries
Remember: 1+1 is 2 (or (10)2), which results in a carry1+1+1 is 3 (or (11)2) which also results in a carry
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Full AdderFull Adder
Three inputs: X Y Third is Cin Z
Two outputs: Sum Cout
S
Full Adder
x y
ZCout
Implementation?
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Straight Forward Implementation:
What is this?
Z
S
K Map for S
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YZXZXYC
Y
X
Z
Y
Z
X
C
Straight Forward Implementation:
K Map for C
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Implementation IssuesImplementation Issues
If we try to implement the Optimized Boolean functions directly we will need how many gateshow many gates? Seven AND gates and two OR Gates!!
Can we do better? Share Logic Hierarchical Design.
YZXZXYC
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Any Alternatives?Any Alternatives?
Try to make use of hierarchy to design a 1-bit full adder from two half adders.
Also, try to share logic between the Sum output and Carry output.
Half Adder S = X Y C = XY
Full Adder S = X Y Z C = XY + XZ + YZ
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A Different Way to Represent CA Different Way to Represent C
1
1 1 1
X
YZ
0
1
00 01 11 10
XY
XYZ
XYZ
C = XY + XYZ + XYZ
C = XY + Z (XY + XY)
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Two Half Adders (and an OR)Two Half Adders (and an OR)
How many Gates do we need?
Full Adder
x y
ZC
S
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Binary Ripple-Carry AdderBinary Ripple-Carry Adder
A Parallel binary adder is a digital circuit that produces the arithmetic sum of two binary numbers using only combinational logic.
The parallel adder uses “n” full adders in parallel, with all input bits applied simultaneously to produce the sum.
The full adders are connected in cascade, with the carry output from one full adder connected to the carry input of the next full adder.
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Binary Ripple-Carry AdderBinary Ripple-Carry Adder
Straightforward – connect full adders Carry-out to carry-in chain
C0 in case this is part of larger chain, maybe just set to zero
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Hierarchical 4-Bit AdderHierarchical 4-Bit Adder
We can easily use hierarchy here1. Design half adder
2. Use TWO half adders to create full adder
3. Use FOUR full adders to create 4-bit adder
VHDL CODE?
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VHDL Half Adder (DATA FLOW)VHDL Half Adder (DATA FLOW)
entity half_adder is
port (x,y: in std_logic;
s,c: out std_logic);
end half_adder;
architecture dataflow of half_adder is
begin
s <= x xor y;
c <= x and y;
end dataflow
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VHDL Full Adder (Structural)VHDL Full Adder (Structural)
entity full_adder is
port (x, y, z: in std_logic;
s, c: out std_logic);
end full_adder;
architecture struc_dataflow of full_adder is
hs
hc
tc
component half_adder
port (x, y : in std_logic;
s, c : out std_logic);
end component;
signal hs, hc, tc: std_logic;begin
HA1: half_adder
port map (x, y, hs, hc);
HA2: half_adder
port map (hs, z, s, tc);
c <= tc or hc;
end struc_dataflow
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Any Problems with this Design?Any Problems with this Design?
Delay Approx how much?
Imagine a 64-bit adder Look at carry chain
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Carry Propagation & DelayCarry Propagation & Delay
One problem with the addition of binary numbers is the length of time to propagate the ripple carry from the least significant bit to the most significant bit.
The gate-level propagation path for a 4-bit ripple carry adder of the last example:
Note: The "long path" is from A0 or B0 through the circuit to S3.
A3B3
S3
B2
S2
B1
S1 S0
B0
A2 A1 A0
C4
C3 C2 C1 C0
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SubtractionSubtraction
We managed to design an Adder easily. For subtraction, we will also need to design a
Subtractor!! Can we perform subtraction using the Adder
Circuit we designed earlier? YES, we can use the concept of Complements.
X = Y – Z X = Y + complement(Z)
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Complements?Complements?
There are two types of complements two types of complements for each base-r system The radix complement, the (r’s) complement. The diminished radix complement, (r-1)’s comp.
For Decimal System 10’s complement 9’s complement
For Binary Systems 2’s complement 1’s complement
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Complements of Decimal SystemComplements of Decimal System
The 9’s complement of a decimal number is obtained by subtracting each digit from 9. Example: The 9’s complement of 546700 is 999999 – 546700 = 453299
The 10’s complement is obtained by adding 1 to the 9’s complement: Example: The 10’s complement of 546700 is 999999 – 546700 = 453299 + 1 = 453300 Or, 1000000 – 546700 = 453300 Or, leave all least significant 0’s unchanged, subtract
the first nonzero LSD from 10, and subtract all higher significant digits from 9.
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UnsignedUnsigned Decimal Subtraction Decimal Subtraction
Using 10’s complement, perform the subtraction 72532 – 3250. No complements: 72532 – 3250 = 69282
M = 72532 (5-digits), N = 3250 (4-digits) Since N has only 4 digits append a zero N=03250 10’s complement of N (03250)
99999 – 03250 = 96749 + 1 = 96750 Now add Now add M to the 10’s comp of N
72532 + 96750 = 169282 (carry occurred) The occurrence of the end carry indicates that M > N Discard end carry (169282 – 100000 = 69282)
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Example #1
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Using 10’s complement, perform the subtraction 3250 - 72532. No complements: 3250 - 72532 = - 69282
M = 3250 (4-digits), N = 72532 (5-digits) Since M has only 4 digits append a zero M=03250 10’s complement of N (72532)
99999 – 72532 = 27467 + 1 = 27468 Now add Now add M to the 10’s comp of N
03250 + 27468 = 30718 (There is no end carry!) No end carry indicates that M < N (make correction) Answer: -(10’s complement of 30718) = -69282
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UnsignedUnsigned Decimal Subtraction Decimal Subtraction
Example #2
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Binary SubtractionBinary Subtraction
We’ll use unsigned subtraction to motivate use of complemented representation
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1’s Complement1’s Complement
1’s Complement (Diminished Radix Complement)
● All ‘0’s become ‘1’s
● All ‘1’s become ‘0’s
Example (10110000)2
(01001111)2
If you add a number and its 1’s complement …
1 0 1 1 0 0 0 0
+ 0 1 0 0 1 1 1 1
1 1 1 1 1 1 1 1
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1’s Complement: Example1’s Complement: Example
Notice that the 1’s complement of the number 1011001 can be obtained by complementing each bit
2n - 1 1 1 1 1 1 1 1
- N 1 0 1 1 0 0 1
1’s Compl. 0 1 0 0 1 1 0
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2’s Complement2’s Complement
2’s Complement (Radix Complement)
● Take 1’s complement then add 1
● Toggle all bits to the left of the first ‘1’ from the right
Example:
Number:
1’s Comp.:
0 1 0 1 0 0 0 0
1 0 1 1 0 0 0 0
0 1 0 0 1 1 1 1
+ 1
OR
1 0 1 1 0 0 0 0
00001010
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2’s Complement: Example2’s Complement: Example
Notice that the 2’s complement of the number 011001 can be obtained by complementing each bit and adding 1.
2n 1 0 0 0 0 0 0
- N 0 1 1 0 0 1
1’s Comp 1 0 0 1 1 0
2’s Compl. 1 0 0 1 1 1
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ExampleExample
Borrow 1 1 1 0 0
(M) Minuend 1 0 0 1 1
(N) Subtrahend - 1 1 1 1 0
Difference 1 0 1 0 1
19 – 30 = 21 !!!!!
Incorrect Result!!
Minuend is smaller than Subtrahend
How can we know if the result is incorrect? How to fix the problem?
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ExampleExample
Borrow 1 1 1 0 0
(M) Minuend 1 0 0 1 1
(N) Subtrahend - 1 1 1 1 0
Difference 1 0 1 0 1
Correct Diff - 0 1 0 1 1
If no borrow, then result is non-negative (minuend >= subtrahend).
Since there is borrow, result must be negative.
The result must be corrected to a negative number.
19 – 30 = -11
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Algorithm: Subtraction of two n-digit Numbers M-N can be done as follows
1. Subtract N from M If no borrow, then M N and result is OK! Otherwise, N > M so result must be
subtracted from 2n and a minus sign should be appended
2. NOTE: Subtraction of a binary number from 2n to obtain an n-digit result is called 2’s complement
3. Circuit?
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Adder/Subtractor Circuit!!Adder/Subtractor Circuit!!
Binary Adder Binary Subtractor
EXPENSIVE!!
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How to get rid of Subtraction Operation?How to get rid of Subtraction Operation?
Use complements of numbers to replace the subtraction operation with addition only.
Any Idea?
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Subtraction of Subtraction of Unsigned NumbersUnsigned Numbers Using ComplementsUsing Complements
1. M – N Equivalent to M + (2’s complement of N)
2. Add (2’s complement of N) to M This is M + (2n – N) = M – N + 2n
Notice we are using addition to achieve subtraction.
3. If M N, will generate will generate carry! • Discard carry• Result is positive M - N
4. If M < N, no end carry no end carry will be generated! • Take 2’s complement of result• Place minus sign in front
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ExampleExample
X = 1010100 minus Y = 1000011 Notice that X > Y The 2’s complement of Y=1000011
is obtained first by getting the 1’s complement 0111100 and then adding 1 (0111101)
X 1 0 1 0 1 0 0
+ 2’s comp Y 0 1 1 1 1 0 1
Sum 1 0 0 1 0 0 0 1
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Example 2Example 2
Y = 1000011 minus X = 1010100 Notice Y < X
No end carry Answer: - (2’s complement of Sum) - 0010001
Y 1 0 0 0 0 1 1
+ 2’s comp X 0 1 0 1 1 0 0
Sum 1 1 0 1 1 1 1
We said numbers are unsigned. What does this mean?
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Adder-SubtractorAdder-Subtractor
I. By using 2’s complement approach we were able to get rid of the design of a subtractor.
II. Need only adder and complementer for input to subtract
III. Need selective complementer to make negative output back from 2’s complement
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Selective 1’s Complementer?Selective 1’s Complementer?
ControlWhen X = 0 we transfer Y to output
When X = 1 we complement Y
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DesignDesign
Inverts each bit of B if S is 1
Adds 1 to make 2’s complement
S low for add,high for subtract
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Negative NumbersNegative Numbers
Computers Represent Information in ‘0’s and ‘1’s
● ‘+’ and ‘−’ signs have to be represented in ‘0’s and ‘1’s
3 Systems
● Signed Magnitude
● 1’s Complement
● 2’s Complement
All three use the left-most bit to represent the sign:
♦ ‘0’ positive
♦ ‘1’ negative
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Signed Binary NumbersSigned Binary Numbers
First review signed representations Signed magnitude
Left bit is sign, 0 positive, 1 negative Other bits are number 0 0001001 +9 1 0001001 -9
2’s complement 1’s complement
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Signed Magnitude RepresentationSigned Magnitude Representation
Magnitude is magnitude, does not change with sign
(+3)10 ( 0 0 1 1 )2
(−3)10 ( 1 0 1 1 )2
Sign Magnitude
S Magnitude (Binary)
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1’s Complement Representation1’s Complement Representation
Positive numbers are represented in “Binary”
Negative numbers are represented in “1’s Comp.”
(+3)10 (0 011)2
(−3)10 (1 100)2
There are 2 representations for ‘0’!!!!!!
(+0)10 (0 000)2
(−0)10 (1 111)2
0 Magnitude (Binary)
1 Code (1’s Comp.)
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1’s Complement Range1’s Complement Range
4-Bit Representation
24 = 16 Combinations
− 7 ≤ Number ≤ + 7
−23+1 ≤ Number ≤ +23 − 1
n-Bit Representation
−2n−1+1 ≤ Number ≤ +2n−1 − 1
Decimal 1’s Comp.+ 7 0 1 1 1+ 6 0 1 1 0+ 5 0 1 0 1+ 4 0 1 0 0+ 3 0 0 1 1+ 2 0 0 1 0+ 1 0 0 0 1+ 0 0 0 0 0− 0 1 1 1 1− 1 1 1 1 0− 2 1 1 0 1− 3 1 1 0 0− 4 1 0 1 1− 5 1 0 1 0− 6 1 0 0 1− 7 1 0 0 0
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2’s Complement Representation2’s Complement Representation
Positive numbers are represented in “Binary”
Negative numbers are represented in “2’s Comp.”
(+3)10 (0 011)2
(−3)10 (1 101)2
There is 1 representation for ‘0’
(+0)10 (0 000)2
(−0)10 (0 000)2
0 Magnitude (Binary)
1 Code (2’s Comp.)
1’s Comp. 1 1 1 1
+ 1
1 0 0 0 0
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2’s Complement Range2’s Complement Range
4-Bit Representation
24 = 16 Combinations
− 8 ≤ Number ≤ + 7
−23 ≤ Number ≤ + 23 − 1
n-Bit Representation
−2n−1 ≤ Number ≤ + 2n−1 − 1
Decimal 2’s Comp.+ 7 0 1 1 1+ 6 0 1 1 0+ 5 0 1 0 1+ 4 0 1 0 0+ 3 0 0 1 1+ 2 0 0 1 0+ 1 0 0 0 1+ 0 0 0 0 0− 1 1 1 1 1− 2 1 1 1 0− 3 1 1 0 1− 4 1 1 0 0− 5 1 0 1 1− 6 1 0 1 0− 7 1 0 0 1− 8 1 0 0 0
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Convert 2’s Complement to DecimalConvert 2’s Complement to Decimal
bit index 7 6 5 4 3 2 1 0
bit weighting -27 26 25 24 23 22 21 20
Example 0 1 0 1 0 0 1 0
Decimal 0x-27 1x26 0x25 1x24 0x23 0x22 1x21 0x20
64 + 16 + 2 = 82
bit index 7 6 5 4 3 2 1 0
bit weighting -27 26 25 24 23 22 21 20
Example 1 0 1 0 1 1 1 0
Decimal 1x-27 0x26 1x25 0x24 1x23 1x22 1x21 0x20
-128 + 32 + 8 + 4 + 2 = -82
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Number RepresentationsNumber Representations
4-Bit Example
Unsigned Binary
Signed Magnitude
1’s Comp. 2’s Comp.
Range 0 ≤ N ≤ 15 -7 ≤ N ≤ +7 -7 ≤ N ≤ +7 -8 ≤ N ≤ +7
PositiveBinary Binary Binary Binary
Negative XBinary 1’s Comp. 2’s Comp.
0 0 0
1 1 1
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Example in Example in 8-bit8-bit byte byte
Represent +9 in different ways Signed magnitude 00001001
1’s Complement 000010012’s Complement 00001001
Represent -9 in different waysSigned magnitude 100010011’s Complement 111101102’s Complement 11110111
The
Same!
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ObservationsObservations
All positive numbers are the same
1’s Comp and Signed Mag have two zeros
2’s Comp has more negative than positive
All negative numbers have 1 in high-order bit
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Advantages/DisadvantagesAdvantages/Disadvantages
Signed magnitude has problem that we need to correct after subtraction
One’s complement has a positive and negative zero
Two’s complement is most popular i.e arithmetic operations are easy
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Signed Magnitude RepresentationSigned Magnitude Representation
Magnitude is magnitude, does not change with sign
(+3)10 ( 0 0 1 1 )2
(−3)10 ( 1 0 1 1 )2
Can’t include the sign bit in ‘Addition’0 0 1 1 (+3)10
+ 1 0 1 1 (−3)10
1 1 1 0 (−6)10
Sign Magnitude
S Magnitude (Binary)
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Signed Magnitude RepresentationSigned Magnitude Representation
The signed-magnitude system is used in ordinary arithmetic, but is awkward when employed in computer arithmetic (Why?)
1. We have to separately handle the sign 2. Perform the correction.
Therefore the signed complement is normally used.
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Binary Subtraction Using 1’s Comp. AdditionBinary Subtraction Using 1’s Comp. Addition
Change “Subtraction” to “Addition”
If “Carry” = 1then add it to theLSB, and the resultis positive(in Binary)
If “Carry” = 0then the resultis negative(in 1’s Comp.)
0 1 0 1
+ 1 1 1 0
(5)10 – (1)10
(+5)10 + (-1)10
0 0 1 1+
0 1 0 0
0 1 0 1
+ 1 0 0 1
(5)10 – (6)10
(+5)10 + (-6)10
0 1 1 1 0
1 1 1 0
+ 4 − 1
1
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Two’s ComplementTwo’s Complement
To Add: easy on any combination of positive
and negative numbers To subtract
Take 2’s complement of subtrahend Add This performs A + ( -B), same as A – B
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Binary Subtraction Using 2’s Comp. AdditionBinary Subtraction Using 2’s Comp. Addition
Change “Subtraction” to “Addition”
If “Carry” = 1ignore it, and the result is positive(in Binary)
If “Carry” = 0then the resultis negative(in 2’s Comp.)
0 1 0 1
+ 1 1 1 1
(5)10 – (1)10
(+5)10 + (-1)10
1 0 1 0 0
0 1 0 1
+ 1 0 1 0
(5)10 – (6)10
(+5)10 + (-6)10
0 1 1 1 1
+ 4 − 1
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Examples from BookExamples from Book
Addition (+6) + 13 (-6) + 13 (+6) + (- 13) (-6) + (-13)
Subtraction (-6) - (-13) (+6) - (-13)
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Addition of Two Positive NumbersAddition of Two Positive Numbers
Addition (+6) + 13 = +19 00000110 +6+00001101 +13-------------- 00010011 +19
If a carry out appears it should be discarded.
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Addition of :Addition of : a Positive and Negative Numbers a Positive and Negative Numbers
Addition (-6) + 13 = +7 11111010 (this is 2’s comp of +6)+00001101--------------1 00000111
The carry out was discarded
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Subtraction of Two NumbersSubtraction of Two Numbers
The subtraction of two signed binary numbers (when negative numbers are in 2’s complement form) can be accomplished as follows:1. Take the 2’s complement of the
subtrahend (including the sign bit)2. Add it to the minuend.3. A Carry out of the sign bit position is
discarded.
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Subtraction of Two NumbersSubtraction of Two Numbers
Subtraction (+6) – (+13) = -7 00000110 00000110- 00001101 + 11110011 (2’s comp)-------------- ----------- 11111001
What is 11111001? Take its 2’s complement=> 00000111 The magnitude is 7 So it must be -7
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Circuit for 2’s complement NumbersCircuit for 2’s complement Numbers
No Correction is needed if the signed numbers are in 2’s complement representation
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Sign ExtensionSign Extension
Sign extension is the operation, in computer arithmetic, of increasing the number of bits of a binary number while preserving the number’s sign (positive/negative) and value.
This is done by appending digits to the most significant side of the number
Examples: 2’s complement (6-bits 8-bits)
00 1010 0000 1010 2’s complement (5-bits 8-bits):
10001 1111 0001
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OverflowOverflow
In order to obtain a correct answer when adding and subtracting, we must ensure that the result has a sufficient number of bits to accommodate the sum.
If we start with two n-bit numbers and we end up with a number that is n+1 bits, we say an overflow has occurred.
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OverflowOverflow
Two cases of overflow for addition of signed numbers Two large positive numbers overflow
into sign bit Not enough room for result
Two large negative numbers added Same – not enough bits
Carry out can be OK
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ExamplesExamples
Two signed numbers +70 and +80 are stored in 8-bit registers8-bit registers.
The range The range of binary numbers, expressed in decimal, that each register can accommodate is from +127 to -128.
Since the sum of the two stored numbers is 150, it exceeds the capacity of an 8-bit register.
The same applies for -70 and -80.
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Overflow DetectionOverflow Detection
Carries: 0 1 Carries: 1 0 +70 0 1000110 -70 1 0111010 +80 0 1010000 -80 1 0110000------ ------------- ---- ------------- +150 1 0010110 -150 0 11010101. The addition of +70 and +80 resulted in a negative
number!2. The addition of -70 and -80 also resulted in an
incorrect value which is positive number!3. An overflow condition can be detected by
observing the carry into the sign bit position and the carry out of the sign bit position.
4. If the the carry in and carry out of the sign bit are not equal an overflow has occurred.
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Circuit for Overflow DetectionCircuit for Overflow Detection
Condition is that either Cn-1 or Cn is high, but not both
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BCD AdditionBCD Addition
One decimal digit + one decimal digit
● If the result is 1 decimal digit ( ≤ 9 ), then it is a simple binary addition
Example:
● If the result is two decimal digits ( ≥ 10 ), then binary addition gives invalid combinations
Example:
5
+ 3
8
0 1 0 1
+ 0 0 1 1
1 0 0 0
5
+ 5
1 0
0 1 0 1
+ 0 1 0 1
1 0 1 00 0 0 1 0 0 0 0
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BCD AdditionBCD AdditionIf the binary result
is greater than 9,correct the result byadding 6
5
+ 5
1 0
0 1 0 1
+ 0 1 0 1
1 0 1 0
+ 0 1 1 0
0 0 0 1 0 0 0 0
Two Decimal Digits
Multiple Decimal Digits
3 5 1
0 0 1 1 0 1 0 1 0 0 0 1
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ENG241/Digital Design 72
BCD ArithmeticBCD Arithmetic
8 1000 Eight+5 +0101 Plus Five 13 1101 is 13 (> 9)
Note that the result is MORE THAN 9, so must be represented by two digits!
To correct the digit, add 6 1000 Eight 8
+5 +0101 Plus 5 13 1101 is 13 (> 9)
+0110 so add 6 carry = 1 0011 leaving 3 + cy
0001 | 0011 Final answer (two digits)
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ENG241/Digital Design 73
BCD AdditionBCD Addition
4-bit binary adder
Addend Augend
Input
Carry
4-bit binary adder
BCD Sum
0 or 6
DetectionCircuit for
Invalid BCDOutput
Carry
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BCD AdditionBCD Addition
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Recall: Arithmetic -- multiplicationRecall: Arithmetic -- multiplication
1 0 1 1
0 0 0 0
1 0 1 1
1 1 0 1 1 1
1 0 1 1
X 1 0 1
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MultiplierMultiplier
Multiply by doing single-bit multiplies and shifts Combinational circuit to accomplish this?
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Combinational MultiplierCombinational Multiplier
AND computes A0 B0
Half adder computes sum. Will need FA for larger multiplier.
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Larger Multiplier: ResourcesLarger Multiplier: Resources
For J multiplier bits and K multiplicand bits we need J x K AND gates (J-1) K-bit adders to produce a
product of J+K bits.
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Larger MultiplierLarger Multiplier
A k=4-bit by j=3-bit Binary Multiplier.
J = 3
K = 4
Jxk = 12 AND Gates
(J-1) Adders
Of k bits each
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Carry Look ahead Adder
Note that add itself just 2 level
Idea is to separate carry from adder function Then make carry approx 2-level all way
across larger adder
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One-bit Subtractor
Inputs: Borrow in, minuend subtrahend
Outputs: Difference, borrow out Truth Table?
1-bit subM S
Bout
D
Bin
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Correcting Result
Borrow 1 1 1 0 0
Minuend 1 0 0 1 1
Subtrahend - 1 1 1 1 0
Difference 1 0 1 0 1 0 1 0 1 1
1 0 0 0 0 0
- 1 0 1 0 1
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Correcting Result
If M is minuend and N subtrahend of numbers length n, difference was (2n + M) – N
What we want is magnitude of N-M with minus sign in front
We can get the correct result by subtracting previous result from 2n
N - M = 2n – (M – N + 2n)
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Interpretation of the incorrect result
Borrow 1 1 1 0 0
Minuend 1 0 0 1 1
Subtrahend - 1 1 1 1 0
Difference 1 0 1 0 1
1 0 0 0 0 0
+ 1 0 0 1 1
- 1 1 1 1 0
1 0 1 0 1
M
N
25
(2n + M) – N
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Correcting Result
What, mathematically, does it mean to borrow?
It means that M < N If borrowing at digit i-1 you are
adding 2i
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Designs Aren’t Like This
That’s why people use complemented interpretation for signed numbers2’s complement1’s complement
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2’s Complement
The 2’s complement of 10101 is 01011
1 0 0 0 0 0
- 1 0 1 0 1
0 1 0 1 1
The circuit that performs this is a complementer
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1’s Complement
Given: binary number N with n digits 1’s complement defined as
(2n – 1) - N Note that (2n – 1) is number with n
digits, all of them 1For n = 4, (2n – 1) = 1111
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2’s Complement
Given: binary number N with n digits 2’s complement defined as
2n – N for N 00 for N = 0
Exception is so that the result will always have n bits
2’s complement is just a 1 added to 1’s complement
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Still Remember? Unsigned Arithmetic Subtraction
1 0 1 1 0
- 1 0 0 1 0
0 0 1 0 0
No borrows
1 1 1 1 0
- 1 0 0 1 1
Borrows
0 - 1 results in a borrow
11
11
0 0
0
1
0
Subtrahend
Minuend
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Four-bit Carry Look Ahead
Adder functionseparated fromcarry
Notice adder has A, B, C inand S out, as well as G,P out.
Reference
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ENG241/Digital Design 93
BCD Addition
F A F A F A F A
F A F A
Cin
A 3 A 2 A 1 A 0 B 3 B 2 B 1 B 0
Cout S 3 S 2 S 1 S 0
0
CO CI
S
CO CI
S
CO CI
S
CO CI
S
CO CI
S
CO CI
S
1 1XX A1
A2 1X1X
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