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I am trying to build a logic circuit that multiplies any 4 digit binary number by 3.
I know that if I multiply/divide number by 2 it moves left/right the digits, but what I'm doing with multiply by 3?

  1. How to extract the equations that multiply the digits?

  2. I need to use a full adder?

I would like to get some advice. Thanks!

EDIT
I would like to get comments about my circuit of Multiplier 4 digits binary number by 3.

multiplier

JRN
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Ofir Attia
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    I cannot understand your circuit at all. What are these rectangular boxes? Aren't you supposed to have 6 bits for the output? – Tunococ Aug 01 '14 at 11:08

1 Answers1

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The easiest way I see is to note that $3n=2n+n$, so copy $n$, shift it one to the left, and add back to $n$

Ross Millikan
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  • ok, what I need to do now? I dont know how to implement it on a logic circuit? – Ofir Attia Jun 10 '13 at 10:46
  • That is electrical engineering design, which I can't help with. There is an electrical engineering stack exchange site. Hope that helps. – Ross Millikan Jun 10 '13 at 12:55
  • Here is the most related question from Electrial Engineering Stack Exchange: http://electronics.stackexchange.com/questions/72306/multiply-4-digit-binary-number-by-3-circuit. Good luck! – Xoque55 Jan 16 '14 at 22:19