Consider the Adder-Subtractor circuit given in Unit 3 of Block 1. Explain how this circuit will perform subtraction (A-B) if the value of A is 1010 and B is 1011. You must list all the bit values, including Cin and Cout and overflow condition

In an Adder-Subtractor circuit, the subtraction operation is typically achieved by adding the two’s complement of the subtrahend (B) to the minuend (A).

Here’s how the subtraction ( A – B ) with ( A = 1010 ) and ( B = 1011 ) would be performed:

1. Convert B to its Two’s complement:

• B is ( 1011 ).
• To get the two’s complement of B, we first invert all the bits and then add 1:
  • Inverting B: ( 0100 ).
  • Adding 1: ( 0101 ).
• So, the two’s complement of B is ( 0101 ).

1. Add A and the Two’s complement of B:

• A is ( 1010 ).
• Two’s complement of B is ( 0101 ).
• Adding A and the two’s complement of B: [ \begin{array}{cccccc} & 1 & 0 & 1 & 0 & \text{(A)} \
  • & 0 & 1 & 0 & 1 & \text{(Two’s complement of B)} \
    \hline
    & 1 & 1 & 1 & 1 & \
    \end{array}
    ]
• This gives the result ( 1111 ), which is ( -1 ) in decimal.

1. Interpreting the Result:

• Since the result is ( 1111 ), which is ( -1 ) in two’s complement representation, it means that ( A – B = -1 ).
• In binary, ( -1 ) is represented as ( 1111 ).

1. Cin and Cout:

• In this case, ( \text{Cin} = 1 ) because subtraction operation requires a borrow.
• Cout doesn’t carry any significant meaning in the context of subtraction, but it’s generally the carry out of the most significant bit (MSB).
• In this case, Cout would be ( 1 ) since the result is negative.

1. Overflow Condition:

• Overflow occurs when the result of an operation exceeds the capacity of the register or the data type used to store it.
• In this case, since we are dealing with 4-bit numbers, the range of representable numbers is from ( -8 ) to ( 7 ).
• Since ( A – B = -1 ), it falls within this range and thus there is no overflow.