Question

question 5 (15 points)
7 minutes perform 2s complement binary multiplication using booths algorithm on a 4 - bit multiplicand of -5 (base - 10) and 7 - bit multiplier of -34 (base - 10). what would be the appropriate 2s complement binary representations for the multiplicand?
a what would be the appropriate 2s complement binary representations for the multiplier?
a perform booths multiplication algorithm and enter the 11 bit answer.

Explanation:

Step1: Find 2's complement of multiplicand

The decimal number - 5 needs to be represented in 4 - bit 2's complement. First, find the binary of 5 which is 0101. Then, take 1's complement (1010) and add 1 to get 1011.

Step2: Find 2's complement of multiplier

The decimal number - 34 needs to be represented in 7 - bit 2's complement. First, find the binary of 34 which is 0100010. Then, take 1's complement (1011101) and add 1 to get 1011110.

Step3: Perform Booth's algorithm

1. Initialize:

• Let the multiplicand \(M = 1011\) (4 - bit) and multiplier \(Q = 1011110\) (7 - bit). Add an extra 0 bit at the end of \(Q\) for the algorithm, so \(Q = 10111100\). Also, initialize an accumulator \(A=00000000\) (8 - bit to match the eventual product size considering 4 - bit multiplicand and 7 - bit multiplier).

1. Iterative steps:

• Look at the last two bits of \(Q\) (initially \(Q_0 = 0\) and \(Q_{- 1}=0\)). Since \(Q_0 - Q_{-1}=0\), shift \(A\) and \(Q\) right (arithmetic shift for \(A\) and logical shift for \(Q\)). \(A = 00000000\), \(Q = 01011110\).
• Next iteration, \(Q_0 = 0\) and \(Q_{-1}=0\), so shift again. \(A = 00000000\), \(Q = 00101111\).
• Next, \(Q_0 = 1\) and \(Q_{-1}=0\), so \(A=A - M\). \(A = 00000000-1011=11110101\) (using 2's complement subtraction). Then shift \(A\) and \(Q\) right. \(A = 11111010\), \(Q = 00010111\).
• Next, \(Q_0 = 1\) and \(Q_{-1}=1\), so shift. \(A = 11111101\), \(Q = 00001011\).
• Next, \(Q_0 = 1\) and \(Q_{-1}=1\), so shift. \(A = 11111110\), \(Q = 00000101\).
• Next, \(Q_0 = 0\) and \(Q_{-1}=1\), so \(A = A+M\). \(A=11111110 + 1011=00001001\) (using 2's complement addition). Then shift \(A\) and \(Q\) right. \(A = 00000100\), \(Q = 00000010\).
• Next, \(Q_0 = 0\) and \(Q_{-1}=0\), so shift. \(A = 00000010\), \(Q = 00000001\).
• Next, \(Q_0 = 1\) and \(Q_{-1}=0\), so \(A = A - M\). \(A=00000010-1011 = 11111011\). Then shift \(A\) and \(Q\) right. \(A = 11111101\), \(Q = 00000000\).

1. The final product in \(A\) and \(Q\) (combined) is \(11110100010\) (11 - bit).

Answer:

• For the multiplicand: 1011
• For the multiplier: 1011110
• For the 11 - bit answer: 11110100010