# A > B

Discussion in 'Homework Help' started by transistortransistor, May 1, 2014.

1. ### transistortransistor Thread Starter New Member

Oct 9, 2013
11
0
Hi guys,

I have to implement a circuit that checks if A > B. So I have the following table:

And I have to implement it using FOUR mux 2:1 and ONE mux 4:1.

Here what I tried:

As it is possible to see from the table:
A1 = 0 and A0 = 0 => f(A1, A0, B1, B0) => 0;
A1 = 0 and A1 = 1 => f(A1, A0, B1, B0) => B1 = 0 and B0 = 0;
A1 = 1 and A0 = 0 => f(A1, A0, B1, B0) => B1 = 0;
A1 = 1 and A0 = 1 => f(A1, A0, B1, B0) => NAND (B1,B0);

I tried that scheme to implement the function, but i don't know how will I get the last function (A1=1 and A0=1) using this specification.

PS: Sorry my english.

2. ### WBahn Moderator

Mar 31, 2012
23,847
7,378
Look at your leftmost MUX circuits. What's the difference between them? Why are you using two circuits that do the exact same thing?

3. ### transistortransistor Thread Starter New Member

Oct 9, 2013
11
0
Well, they arent exactly the same. If I get the output of the bottom left-most MUX and use it as input to function 1 (A1=0, A0=1), then I can get a misfunction: if B1 = 0 and B1 = 1 it will also have an output of 1. What I need for function 1 is B1=0 and B1=0.

Anyway, the question also says I can use other gates if necessary. But it specifies 4 MUX 2:1 and one MUX 4:1

4. ### WBahn Moderator

Mar 31, 2012
23,847
7,378

So what, pray tell, is the difference?

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5. ### transistortransistor Thread Starter New Member

Oct 9, 2013
11
0
Yeah, you're right. Now with ur observation, it is possible to implement the function: