In binary addition using 1’s complement;
A. Addition of a positive and a negative binary number
We discuss the following cases under this.
Case I: When the positive number has greater magnitude.
In this case addition of numbers is performed after taking 1’s complement of the negative number and the end-around carry of the sum is added to the least significant bit.
The following examples will illustrate this method in binary addition using 1’s complement:
1. Find the sum of the following binary numbers:
(i) + 1110 and - 1101
Solution:
+ 1 1 1 0 ⇒ 0 1 1 1 0
Hence the required sum is + 0001.
(ii) + 1101 and - 1011
(Assume that the representation is in a signed 5-bit register).
Solution:
+ 1 1 0 1 ⇒ 0 1 1 0 1Hence the required sum is + 0010.
Case II: When the negative number has greater magnitude.
In this case the addition is carried in the same way as in case 1 but there will be non end-around carry. The sum is obtained by taking 1’s complement of the magnitude bits of the result and it will be negative.
The following examples will illustrate this method in binary addition using 1’s complement:
Find the sum of the following binary numbers represented in a sign-plus-magnitude 5-bit register:
(i) + 1010 and - 1100
Solution:
+ 1 0 1 0 ⇒ 0 1 0 1 0
Hence the required sum is – 0010.
(ii) + 0011 and - 1101.
Solution:
+ 0 0 1 1 ⇒ 0 0 0 1 1
Hence the required sum is – 1010.
B. When the two numbers are negative
For the addition of two negative numbers 1’s complements of both the numbers are to be taken and then added. In this case an end-around carry will always appear. This along with a carry from the MSB (i.e. the 4th bit in the case of sign-plus-magnitude 5-bit register) will generate a 1 in the sign bit. 1’s complement of the magnitude bits of the result of addition will give the final sum.
The following examples will illustrate this method in binary addition using 1’s complement:
Find the sum of the following negative numbers represented in a sign-plus-magnitude 5-bit register:
(i) -1010 and -0101
Solution:
- 1 0 1 0 ⇒ 1 0 1 0 1 (1’s complement)
1’s complement of the magnitude bits of sum is 1111 and the sign bit is 1.
Hence the required sum is -1111.
(ii) -0110 and -0111.
Solution:
- 0 1 1 0 ⇒ 1 1 0 0 1 (1’s complement)
1’s complement of 0010 is 1101 and the sign bit is 1.
Hence the required sum is - 1101.
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