HWK #8 Solution
Problem 8.0 (a)
In a 8-bit ripple adder, C
is calculated after 3 units of time. After that every C
is calculated in C
+ 2 units of time.
is calculated in 3+(2*6)=15 units of time. To calculate S
, it takes 15+1 =
of time. (The 1 unit is the delay for the last FA)
In the CSA shown in the diagram, S
is calculated after C
is calculated (we have to add the MUX delay)
is calculated in 3+(2*3) = 9 units of time. Adding the MUX delay, we calculate S
in 9+3 =
Problem 8.0 (b)
The P/G is calculated in 1 unit of time, carries in 2 units and the FA delay is 1 unit. The total latency is 1+2+1 =
of time for a 8-bit CLA.
If we use Carry Select as shown above, each of the 4-bit CLA will have a latency of 4 units of time (1+2+1 as explained above). The MUX will have a latency of 3 units. So the total latency is 4+3 =
Problem 8.0 (c)
MUX delay does not change.
Carry Select is faster as N increases.
Problem 8.0 (d)
Boolean property used
C8,0 * 1 + C4 * C8,1
C8,0*(C8,1 + C8,1') + C4*C8,1
C8,0*C8,1 + C8,0*C8,1' + C4*C8,1
C8,0*C8,1 + 0 + C4*C8,1
Apply C8,0' + C8,1 = 1
C8,0*C8,1 + C4*C8,1
Problem 8.2 (a)
The next-state table is shown below, and the state diagram to the right.