Arithmetic Circuits I

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Iterative Combinational Circuits

• Like a hierachy, except functional blocks per bit

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Adders

• Great example of Iterative design• Design a 1-bit adder circuit, then

expand to n-bit adder• Look at

♦ Half adder – which is a 2-bit adder♦ Inputs are bits to be added

• Outputs: result and possible carry

♦ Full adder – includes carry in, really a 3-bit adder

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Half Adder

• S = X Y• C = XY

Fig. 4-2

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Full Adder

• Three inputs. Two are operand bits, third is Cin

• Two outputs: sum and carry

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K Map for S

• What is this?

In a half adder the sum bit:

S = X Y

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K Map for C

In a half adder the carry bit:

C = XY

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Using two half Adders to Build a Full Adder

• Full adder functions

• Half adder functionsS = X YC = XY

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Two Half Adders (and an OR)

XY

X Y

( )Z X Y

( )XY Z X Y

X Y Z

Fig. 4-4

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Ripple-Carry Adder

• Straightforward – connect full adders

• C4 : Chain carry-out to carry-in of FA of bits A4 & B4

♦ C0 in case this is part of larger chain

♦ otherwise just set to zero

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Hierarchical 4-Bit Adder

• We can easily use hierarchy here

• Design half adder• Use in full adder• Use full adder in 4-bit adder

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Binary Subtraction

• Example: ♦ Let M = (19)10 ; N = (30)10 ; How is M-N

computed?• In decimal notation: (19)10 – (30)10 = - (11)10

• In binary: (10011)2 - (11110)2 = - (01011)2

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Using 2’s & 1’s Complement Representation of Signed Numbers

• People use complemented interpretation for signed numbers♦ 1’s complement♦ 2’s complement

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1’s Complement

• Given: binary number N with n digits1’s complement is defined as(2n – 1) - N

• Note that (2n – 1) is a number with n bits, all of them 1♦ For n = 4, (2n – 1) = (15)10 = 1111

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Example: Find 1’s Complement of N =101 1001

• Notice that 1’s complement is complement of each bit

n= 7; 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

•1’s Complement of N is (2n – 1) - N

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2’s Complement

• Given: binary number N with n digits2’s complement defined as

2n – N for N 00 for N = 0

• Note that since 1’s complement is

(2n – 1) – N

♦ 2’s complement is just a 1 added to 1’s complement

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Example: Find 2’s Complement of N =101 1001

• To find 2’s complement of N♦ Find is 1’s complement of N then add 1

N 1 0 1 1 0 0 1

1’s Compl. 0 1 0 0 1 1 0

+1 0 0 0 0 0 0 1

2’s Compl. 0 1 0 0 1 1 1

• 2’s Complement of N is 2n - N

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Important Property

• Complement of a complement generates original number

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Subtraction Using 2s Complement:Compute M-N

1. Add 2’s complement of N to M♦ M + (2n – N) = { M – N + 2n } = F

2. If M N, this addition will generate a carry, 2n

• Discard carry; the result gives the answer of M-N which is positive

3. If M < N, no carry; M-N is negative; the answer is –(N-M). To compute the answer:

• Take 2’s complement of F• 2n – F = 2n - { M – N + 2n } = N-M

• Place minus sign in front { - (N-M) }; This is the answer

4. Hence to compute (M-N): do 1 & 2 if M N; OR 1 & 3 If M < N

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Compute M-N;Example 1: M > N• M = (84)10= (101 0100)2

• N = (67)10 = (100 0011)2;

♦ 2’s comp N = ( 011 1100 ) + 1 = 011 1101M 1 0 1 0 1 0 0

+ 2’s comp N 0 1 1 1 1 0 1

Sum 1 0 0 1 0 0 0 1

M > N; Carry generated; Discard carry; Result is positive M - N

1710

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Compute M-N; Example 2: M < N;M = 6710 = 100 0011; N = 8410 = 101 0100

• M = 100 0011 minus N = 101 0100

• No end carry♦ Answer: - (2’s complement of Sum)

• Answer: - 0010001

M 1 0 0 0 0 1 1

+ 2’s comp N 0 1 0 1 1 0 0

Sum 1 1 0 1 1 1 1

We said numbers are unsigned. What does this mean?How is -1710 represented?

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Subtractor

• Compute M-N♦ Add 2’s complement of N to M:

M + (2n – N) • If M >= N, need only “one adder”

and a “complementer of N” to do the subtraction.

• If M < N, we also need to take 2’s complement of adder’s output to produce magnitude of result2n -{ M + (2n – N)} = N-M

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Adder-Subtractor Design: (A+B)OR (A-B)

• Output is result if A >= B; Discard carry• Output is 2’s complement of result if B > A

S is low for add,high for subtract

Inverts each bit of B if S is 1

Fig. 4-7

Add 1 tomake 2’s complement

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Signed Binary

1. Signed magnitude♦ Left bit is sign, 0 positive, 1 negative♦ Other bits are number

2. 2’s complement3. 1’s complement

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Example in 8-bit byte

• Represent -910 = -(0000 1001) in different ways

“Signed magnitude” of - (0000 1001) is1000 1001

“Signed 1’s Complement” of - (0000 1001) is 1111 0110

“Signed 2’s Complement” of - (0000 1001) is1111 0111

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Observations (assume 4-bit numbers)

• 1’s C and Signed Magnitude have two zeros

• 2’s C has more negative numbers than positive

• All negative numbers have 1 in highest-order bit

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Advantages/Disadvantages

• Signed magnitude & One’s complement has a positive and negative zero

• Two’s complement is most popular♦ Arithmetic operations easy

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Two’s Complement

• Addition easy on any combination of positive and negative numbers

• To compute A - B♦ Take 2’s complement of subtrahend B

to produce -B♦ Add to A

•This performs A + ( -B), same as A – B

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Examples• Assume we use 2’s complement

representation of signed numbers• Store each number in 8 bits 6 = (0000 0110) ; 13 = (0000 1101)-6 = (1111 1010) ; -13 = (1111 0011)

• Addition♦ 6 + 13♦ -6 + 13♦ 6 + (- 13)♦ (-6) + (-13)

• Subtraction♦ -6 - (-13)♦ 6 - (- 13)

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Overflow

• Overflow means that result can not be represented with the number of bits used

• Two cases of overflow for addition of signed numbers♦ Two large positive numbers

overflow into sign bit•Not enough bits for result

♦ Two large negative numbers added•Same – Not enough bits for result

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Examples:Work on Board

Assume

1- We use “2’s complement representation”

of signed numbers

2- Store each number in 4 bits : b3 b2 b1 b0

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Example 1

• 7 + 7

• Overflow: can not represent 14 using only 4 bits

• Generates NO CARRY, C4 = 0

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Example 2

• -7 - 7

• Overflow: can not represent -14 using only 4 bits

• Generates CARRY, C4 = 1

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Example 3

• 4 + 4

• Overflow: can not represent 8 using only 4 bits

• Generates NO CARRY, C4 = 0

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Example 4

• 7 – 7

•NO Overflow

• Generates CARRY, C4 = 1

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Overflow Detection• Condition for overflow:

♦ either Cn-1 or Cn is high, but not both

♦ 7 + 7; only Cn-1 (C3) is high ; overflow♦ -7 - 7; only Cn (C4) is high ; overflow♦ 4 + 4; only Cn-1 (C3) is high ; overflow♦ 7 – 7; BOTH Cn-1 & Cn (C4 & C3) are high;

no overflow

---------------------------------------------------------Fig. 5-9 Overflow Detection for Addition and SubtractionFig. 4-8