# Build a XOR with a MUX and a Invertor?

Discussion in 'Electronic Basics' started by Al Borowski, Oct 16, 2005.

1. ### Al BorowskiGuest

Input A of MUX = S1
Input B of MUX = NOT S1
Input S of MUX = S2

Working out the Truth Table....

If S1 and S2 are 0, the MUX output is S1, which is 0

If S1 and S2 are 1, the MUX output is NOT S1, which is 0

If S1 is 1 and S2 is 0, the MUX output is S1, which is 1

If S1 is 0 and S2 is 1, the MUX output is NOT S1, which is 1

Do I get the job?

Al

2. ### DavyGuest

Hi all,

How to Build a XOR with a MUX(2 to 1) and a invertor?

Any suggestions will be appreciated!
Best regards,
Davy

3. ### Frank BemelmanGuest

There aren't that many ways how you can wire them together.

Try each possible way, and figure out the truth table, and
compare that with an XOR truth table.

Good luck with your home work

4. ### DavyGuest

Hi Frank,

It's a interview problem and I just cannot answer it.
Anyone give an answer will be appreciated!

Best regards,
Davy

6. ### Fred BloggsGuest

An XOR on A and B is /A*B + A*/B using obvious notation. So you would
for example use A to drive the MUX bit select , then route B to the MUX
0 input and /B to the MUX 1 input:
View in a fixed-width font such as Courier.

7. ### Fred BloggsGuest

Not with that reasoning- better luck next time....

9. ### Anthony FremontGuest

I guess I'm just dense, but how is this different from Al's solution?
Granted that you hooked it up a bit different than Al, but the outcome
is the same AFAICT.

10. ### Jim ThompsonGuest

Yep. An XOR is just a 4-quadrant multiplier. And that architecture,
in high speed circuits, has less glitch issues (if implemented with
all inputs differential, as in ECL/PECL).

...Jim Thompson

11. ### Fred BloggsGuest

Fastest possible binary multiplier known
View in a fixed-width font such as Courier.

13. ### Fred BloggsGuest

The MUX transfer function is OUT= SEL*IN1+/SEL*IN0, the XOR is
OUT=A*/B+/A*B, the substitution is clear, in contrast to this gibberish:
Input A of MUX = S1
Input B of MUX = NOT S1
Input S of MUX = S2
with which only the perpetually confused could not be aggravated.

14. ### John LarkinGuest

Which is just a balanced modulator.

John

15. ### John LarkinGuest

Or is it that the perpetually aggravated are confused?

John

16. ### Fred BloggsGuest

The perpetually confused have no sense of comprehension and are
therefore never aggravated.

17. ### Rich GriseGuest

What percentage of your pay do we get for getting you this
job?

Thanks,
Rich

18. ### Rich GriseGuest

Well, I must be one of those "perpetually confused" then, because it
was perfectly clear to me. ;-)

Thanks,
Rich

19. ### Jim ThompsonGuest

Yep.

...Jim Thompson

20. ### Al BorowskiGuest

Even if you didn't know what nomenclature I used, I don't think it's
terribly hard to work out.

"Input A of MUX...Input B of MUX" implies I am calling the 2 inputs of
the MUX A & B.

"If S1 is 1 and S2 is 0" implies that S1 and S2 are the inputs to the
system.

"Input B of MUX = NOT S1 " implies that I am using the NOT gate to
invert an input, and attach it to a MUX input.

Admittedly a picture would have been clearer, but it would have taken me
longer to draw then it took me to solve the problem

I didn't bother consciously making the substitution. I just mentally
combined the parts in my head for twenty seconds until I found something
that worked.

cheers,

Al