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admin
asked
in Digital Logic
Sep 17, 2015

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Best answer

let the three bit multiplicand be $111$ and two bit multiplier be $11$

$1 1 1$

$ * $ $1 1$

${\color{Yellow} 1}$ ${\color{Green} 1}$ $1$

$1$ ${\color{Yellow} 1}$ ${\color{Green} 1}$ here we need 6 AND gates TO MULTIPLY

${\color{Red} 1}$ $0$ $1$ $0$ $1 $

TO ADD GREEN BITS WE NEED 1 HALF ADDER AND FOR YELLOW BIT WE NEED 1 FULL ADDER AND ONE MORE HALF ADDER FOR LAST BIT

AS WE KNOW HALF ADDER NEEDS 1 XOR(FOR SUM) AND 1 AND GATE(FOR CARRY BIT)

FULL ADDER NEED 2 XOR (FOR SUM)AND 2 AND GATE AND 1 OR GATE(FOR CARRY BIT)

THEREFORE 2 HALF ADDER =2-XOR GATE +2 AND GATE

1 FULL ADDER =2-XOR GATE+2-AND GATE+1-OR GATE

THEREFORE TOTAL 10 AND GATES 4 XOR GATES AND 1 OR GATE

THEN IT WOULD BE 10 AND GATE 4 XOR GATE AND 1 OR GATE.(OPTION C)

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