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If a 3 bit multiplicand is multiplied to a 2 bit multiplier, minimum number of two input AND, XOR, and OR gates, needed in the design are respectively?

A)    16,14,1
B)    7, 3, 1
C)    10, 4, 1
D)    8, 4, 1

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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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4 Comments

CAN U FURTHER TELL THE TOTAL DELAY?
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IT SHOULD BE 10 AND GATES RIGHT??

CAN SOMEONE PLZ TELL THE TOTAL DELAY TOO?? ASSUMING EVERY GATE HAS 1 UNIT OF DELAY.??
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can someone elaborate more why do we need here 2 half adder and 1 full adder  instead of 2 full adder and one half adder i am not getting it
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