Unfortunately, I don't have access to specific figures or pages from Block 1, so I can't refer to Figure 3.15 on page 76. However, I can provide a general explanation of how an Adder-Subtractor circuit performs subtraction.
In a binary Adder-Subtractor circuit, subtraction is achieved by adding the two's complement of the number to be subtracted (B) to the other number (A). The two's complement of a binary number is obtained by inverting all bits and adding 1.
Let's consider the values A = 1011 and B = 0011 for subtraction (A - B):
1. Calculate Two's Complement of B:
- Invert all bits of B: \( \overline{B} = 1100 \)
- Add 1 to \( \overline{B} \): \( \overline{B} + 1 = 1101 \)
2. Perform Addition (A + \( \overline{B} + \text{Cin} \)):
- A: 1011
- \( \overline{B} \): 1101
- Cin: 0 (initial carry)
- Sum: 10110
3. Evaluate Overflow:
- Overflow occurs if the carry into the sign bit (MSB) and the carry out of the sign bit are different. In this case, there is no overflow.
4. Result:
- The sum is 10110, but since we are subtracting, we consider only the lower 4 bits (1010).
- The result of \( A - B \) is 1010.
In summary, the Adder-Subtractor circuit performs subtraction by adding the two's complement of the subtrahend (B) to the minuend (A), taking into account the initial carry (Cin) and evaluating for overflow. The result is the correct difference (A - B).
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