two-bit comparator circuit using a 8-to-1 multiplexer

Discussion in 'Homework Help' started by lgolestan, Mar 14, 2010.

  1. lgolestan

    Thread Starter New Member

    Mar 14, 2010
    Hi everyone, :cool::)

    I should design a two bit comparator with a 8:1 multiplexer. that's all the question says.

    so I wrote the equations of a two-bit comparator like this:
    We've got 2 two-bit binaries like A and B , A= a1 a0 , B= b1 b0
    A and B should be compared with each other and tell us which of the following situations they have:
    E:Equal: A=B if ( a1=b1 and a0=b0) : (a1' a0' b1' b0') + (a1' a0 b1' b0 )+ (a1 a0' b1 b0' )+ (a1 a0 b1 b0 )
    G:Greater: A>B if (a1>b1 ) + (a1=b1) and (a0 > b0)
    L:Less: A<B if (a1<b1) + (a1=b1) and (a0 < b0 )

    now a 8:1 multiplexer has 3 selector, 8 inputs and 1 output.
    the only output of multiplexer should show if it's E,G, or L, but how can 1 output show that??????

    ** I can design a magnitude comparator, this question should be designed with one 8:1 multiplexer.

    Please tell me if you have any ideas about this. :rolleyes:
  2. lgolestan

    Thread Starter New Member

    Mar 14, 2010

    How about just making the A>B part with the 8:1 Multiplexer? anyone got any ideas?

    A=a1a0 , B=b1b0 are the inputs and if the (output of MUX)=1 it means A>B