Question

3.1 for the circuit in figure 3 below, assume the inputs are, $overline{add/subt}=1$, a = 1001 and b = 1100.
3.1.1 write out the binary manipulation.
3.1.2 show the circuit with the binary numbers at each stage of manipulation.

Explanation:

Step1: Determine operation

Since $\overline{Add/Subt}=1$, the operation is subtraction. In binary subtraction using 2 - 's complement, we first find the 2 - 's complement of the subtrahend.

Step2: Find 2 - 's complement of B

The binary number $B = 1100$. First, find the 1 - 's complement: flip all the bits. The 1 - 's complement of $1100$ is $0011$. Then add 1 to get the 2 - 's complement. $0011+1 = 0100$.

Step3: Binary addition for subtraction

We now add $A = 1001$ and the 2 - 's complement of $B$.

• For the least - significant bit: $1 + 0=1$, $C_{in}=0$, so $\Sigma_0 = 1$.
• For the next bit: $0+0 = 0$, $C_{in}=0$, so $\Sigma_1 = 0$.
• For the third bit: $0 + 1=1$, $C_{in}=0$, so $\Sigma_2 = 1$.
• For the most - significant bit: $1+0 = 1$, $C_{in}=0$, so $\Sigma_3 = 1$.

The result of the binary manipulation is $1101$.

Step4: Show circuit with binary numbers

At the XOR gates (due to subtraction operation), when $\overline{Add/Subt}=1$:

• For the first XOR gate with inputs $A_0 = 1$ and $B_0 = 0$ (after 2 - 's complement), the output to the adder is $1$.
• For the second XOR gate with inputs $A_1 = 0$ and $B_1 = 0$, the output to the adder is $0$.
• For the third XOR gate with inputs $A_2 = 0$ and $B_2 = 1$, the output to the adder is $1$.
• For the fourth XOR gate with inputs $A_3 = 1$ and $B_3 = 0$, the output to the adder is $1$.

The adders then perform the addition as described above with $C_{in}$ starting from 0 for the least - significant bit adder and propagating the $C_{out}$ to the next higher - order adder.

Answer:

3.1.1: The binary manipulation is $1001 - 1100=1001+0100 = 1101$ (using 2 - 's complement for subtraction).
3.1.2: The circuit has XOR gates with inputs $A_i$ and the 2 - 's complement bits of $B_i$ ($i = 0,1,2,3$) giving outputs to the adders. The adders perform binary addition with $C_{in}$ starting at 0 for the least - significant bit adder and propagating $C_{out}$ to the next higher - order adder, resulting in $\Sigma_3\Sigma_2\Sigma_1\Sigma_0=1101$.