Question

QuestionB.28 [10] <§B.6> Now calculate the relative performance of adders. Assume that hardware corresponding to any equation containing only OR or AND terms, such as the equations for pi and gi on page B-40, takes one time unit T. Equations that consist of the OR of several AND terms, such as the equations for c1, c2, c3, and c4 on page B-40, would thus take two time units, 2T. The reason is it would take T to produce the AND terms and then an additional T to produce the result of the OR. Calculate the numbers and performance ratio for 4-bit adders for both ripple carry and carry lookahead. If the terms in equations are further defined by other equations, then add the appropriate delays for those intermediate equations, and continue recursively until the actual input bits of the adder are used in an equation. Include a drawing of each adder labeled with the calculated delays and the path of the worst-case delay highlighted.

Short answer

The relative performance of this adder:

Sum	Delay	Carry	Delay
S3	14	C4	19

Step-by-step solution

Define the concept.

The 4- bit adderhas the feature of carry propagating and carry generating.

For the carry propagating, the carry propagator (P) is used for propagating to the following state of this.

For carry generating, the carry generator (G) is used for generating the result query by not considering the carry of the input.

The computed sum of every 4-bit adder is represented by “S”.

T time is taken for producing the AND terms.

2T time is taken for each bit in the previous adder.

Hence the total required time is 8T, where “T” denotes the unit of time.

The relative performance of this adder:

Sum	${\text{S}}_{\text{n - 1}}\text{[n represents the number of bit in adder]}$
Delay	$\text{4n - 2[n represents the number of bit in adder]}$
Carry	${\text{C}}_{\text{n}}\text{[n represents the number of bit in adder]}$
Delay	$\text{4n + 3[n represents the number of bit in adder]}$

Determine the calculation.

Relative performance of this adder:

Sum	Delay	Carry	Delay
${\text{S}}_{\text{n - 1}}{\text{=S}}_{\text{4 - 1}}{\text{=S}}_3$	$\text{4n - 2 =(4×4)-2=16 - 2 = 14}$	$C_n=C_4$	$\text{4n + 3 =}\left(4\times 4\right)\text{+3 = 16 + 3 = 19}$

In the below mentioned figure-

• A0 and B0 are the two inputs for the first 1 bit adder.
• A1 and B1 are the two inputs for the second 1 bit adder.
• A2 and B2 are the two inputs for the third 1-bit adder.
• A3 and B3 are the two inputs for the fourth 1-bit adder.
• C0 is the “carry” for the first 1-bit adder.
• C1 is the “carry” for the second1-bit adder.
• C2 is the “carry” for the third 1-bit adder.
• C3 is the “carry” for the fourth 1-bit adder.
• The “super” propagates are P3, P2, P1, and P0.
• The “super” generator are G3, G2,G1, and G0.

The diagram of 4-bit adders with carry look ahead:

In the below mentioned figure-

• C0 is the “carry” for the first 1-bit adder.
• C1 is the “carry” for the second 1-bit adder.
• C2 is the “carry” for the third 1-bit adder.
• C3 is the “carry” for the fourth 1-bit adder.
• A0 and B0 are the two inputs for the first 1-bit adder.
• A1 and B1 are the two inputs for the second 1-bit adder.
• A2 and B2 are the two inputs for the third 1-bit adder.
• A3 and B3 are the two inputs for the fourth 1-bit adder.
• The “super” propagates are P3, P2, P1, and P0.
• The “super” generator are G3, G2,G1, and G0.

The diagram of 4-bit adders with ripple carry:

T time is taken for producing the AND terms.

2T time is taken for the each bit in the previous adder.

Hence the total required time is 8T, where “T” denotes the unit of time.

And the total required time for the adder by using “And” terms (4 x 1T) ==4T.

Ratio is (4 x 2T) : (4 x 1T) = 8T : 4T = 2: 1