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Build a XOR with a MUX and a Invertor?

A

Al Borowski

Jan 1, 1970
0
Davy said:
Hi all,

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

Any suggestions will be appreciated!
Best regards,
Davy


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
 
D

Davy

Jan 1, 1970
0
Hi all,

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

Any suggestions will be appreciated!
Best regards,
Davy
 
F

Frank Bemelman

Jan 1, 1970
0
Davy said:
Hi all,

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

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 ;)
 
D

Davy

Jan 1, 1970
0
Hi Frank,

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

Best regards,
Davy
 
F

Frank Bemelman

Jan 1, 1970
0
Davy said:
Hi Frank,

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

Well, there's your answer then -> "I cannot answer it".
 
F

Fred Bloggs

Jan 1, 1970
0
Davy said:
Hi Frank,

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

Best regards,
Davy

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.
 
F

Fred Bloggs

Jan 1, 1970
0
Al said:
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

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

Anthony Fremont

Jan 1, 1970
0
Fred Bloggs said:
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.


.
.
.
. MUX
. +-----------+
. |\ | |
. +-| o---|1 |
. | |/ | out|----> A XOR B
. B -+-------|0 |
. | sel |
. +-----------+
. |
. |
. |
. |
. A --------------+
.
.
.
. |
. A B | OUT
. ---------+-----
. 0 0 | 0
. |
. 0 1 | 1
. |
. 1 0 | 1
. |
. 1 1 | 0
. |

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.
 
J

Jim Thompson

Jan 1, 1970
0
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.


.
.
.
. MUX
. +-----------+
. |\ | |
. +-| o---|1 |
. | |/ | out|----> A XOR B
. B -+-------|0 |
. | sel |
. +-----------+
. |
. |
. |
. |
. A --------------+
.
.
.
. |
. A B | OUT
. ---------+-----
. 0 0 | 0
. |
. 0 1 | 1
. |
. 1 0 | 1
. |
. 1 1 | 0
. |

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
 
F

Fred Bloggs

Jan 1, 1970
0
Jim said:
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

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

John Fields

Jan 1, 1970
0
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.
 
F

Fred Bloggs

Jan 1, 1970
0
Anthony said:
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.

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.
 
J

John Larkin

Jan 1, 1970
0
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.

Or is it that the perpetually aggravated are confused?

John
 
F

Fred Bloggs

Jan 1, 1970
0
John said:
Or is it that the perpetually aggravated are confused?

John

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

Rich Grise

Jan 1, 1970
0
Hi Frank,

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

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

Thanks,
Rich
 
R

Rich Grise

Jan 1, 1970
0
Anthony said:
Fred Bloggs said:
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.
[circuit snipped]
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.
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.

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

Thanks,
Rich
 
A

Al Borowski

Jan 1, 1970
0
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.


What's so confusing about that?

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
 
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