CSA
1.1 Aim of this Assignment
To show how the inputs to an adder circuit may be modified to make the resulting circuit perform other useful functions.
1.2Outline Circuit of the Device
The figure below shows a multi-bit adder having inputs A( a number), B(a number), and C0( a single bit). The adder forms an output, F, where F=A plus B plus C0. T is also a number, and Cin, M1 and M0 are each single bits. Two logic circuits generate outputs B and C0 as shown.
|
Adder |
|
A B C0 F |
|
Logic |
Logic |
M1 |
M2 |
|
A |
|
T |
|
Cin |
(figure 1)
2.1 Task
Obtain Boolean expressions for the two logic circuits such that the output F is according to the following table. Here, Bi(Ti, M1, M0) and C0(Cin, M1, M0).
| M1 M0 |
F |
| 0 0 |
A plus T |
| 0 1 |
A minus T |
| 1 0 |
A - 1 |
| 1 1 |
A+1 |
(table 1.1)
Here, F = A plus B plus C0. While Bi = (Ti, M1, M0) and C0(Cin, M1, M0), we can finish table 1.1 like below. F=A plus Bi and C.
| M1 M0 |
F |
Bi |
C0 |
| 0 0 |
A plus T |
Ti |
0 |
| 0 1 |
A minus T |
Not Ti |
1 |
| 1 0 |
A - 1 |
1 |
0 |
| 1 1 |
A+1 |
0 |
1 |
(table 1.2)
i) A logic expression that relates bit Bi to bit Ti and M1 and M0.
From figure 1.1,we can see:
Here,
M1 M0 Ti Bi
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 0
(table 1.3)
From the table1.3, we can draw a map for function Bi
.
|
Bi |
|
M1 |
|
Ti |
|
M0 |
| 0 |
1 |
0 |
1 |
| 1 |
0 |
0 |
1 |
(table 1.4)
From the table 1.4, here Boolean Expression for Bi.
Bi = M1M0’ + TiM0’ + Ti’M1’M0
( Ti’ = not Ti,
M1’ = not M1,
M0’ = not M0.)
Here, we can prove it:
| M1M0’ |
TiM0’ |
Ti’M1’M0 |
M1 M0 |
F |
Bi |
| 0 1 |
Ti 1 |
Ti’ 1 0 |
0 0 |
A plus T |
Ti |
| 0 1 |
Ti 0 |
Ti’ 1 1 |
0 1 |
A minus T |
Not Ti |
| 1 1 |
Ti 1 |
Ti’ 0 0 |
1 0 |
A - 1 |
1 |
| 1 0 |
Ti 0 |
Ti’ 0 1 |
1 1 |
A+1 |
0 |
(Table 1.5)
ii) A logic expression that relates C0 to Cin and M1 and M0.
While C0(Cin, M1, M0), we know that C0 depends on Cin and M1 and M0. From the figure 1.1, we can see:
M1 M0 Cin C0
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
(table 1.7)
From the table 1.7, we can draw a map for function C0.
|
C0 |
|
M1 |
|
Cin |
|
M0 |
| 0 |
1 |
1 |
0 |
| 0 |
1 |
1 |
0 |
(table 1.8)
From the table 1.8, we find the Boolean Expression for C0:
C0 = M0.
Here we can prove it :
| M1 M0 |
F |
C0 |
| 0 0 |
A plus T |
0 |
| 0 1 |
A minus T |
1 |
| 1 0 |
A - 1 |
0 |
| 1 1 |
A+1 |
1 |
3.1 Conclusion:
In this assignment, we mainly aim on how an adder circuit maybe modified to obtain different function. By finding out the logical expression of Bi and C0, we can see how F relates to the states of inputs M1, M0, Ti and C0. F= A +( M1M0’ + TiM0’ + Ti’M1’M0) + M0.