Mux

[daveduggan]

I have and still am struggling with the Mux chip.

I built the truth table then or'd the lines where output was 1.

Then I used algebra to simplify the circuit.

I have done this a few times but testing fails.

I am not sure how to proceed to sort this out.

Any and all help is appreciated.

I do not understand the karnaugh concept as of yet.

may I also post my Xcode with my attempt solution?

[daveduggan]

[daveduggan]

[daveduggan]

[ybakos]

Your overall method is right but I don't believe you've simplified the boolean algebraic expression quite right.

I recommend first trying to implement the Mux without simplifying the big 4-part OR expression.

Another thing to think about is that "when sel is 0 it doesn't matter what b is"; and "when sel is 1 it doesn't matter what a is." If you think about it, then there are really only two cases where the out has a meaningful 1 output.

Give it a shot. If you get the non-simplified version to work, I'll post the simplified algebraic expression.

[daveduggan]

I would like to understand and learn how to do the algebra.

I started with:

!a*b*sel + a*!b*!sel + a*b*!sel + a*b*sel

then simplified to:

!a * b * sel + a(!b * !sel + b * !sel + b * sel)
 
                                        b( !sel + sel)  

!a * b * sel +  a ( !b * ! sel + b)

I wish i could learn how to do the algebra correctly.

Drawing the non simplified version seems like a pain but I will try and do it.

[daveduggan]

ok, I just completed mux without simplification.

It ran the script ok.

Should I submit it?

I still do not know how to simplify it using algebra.

[cadet1620]

daveduggan wrote

I still do not know how to simplify it using algebra.

Begin by thinking about the operation of the Mux. sel changes its behavior, a and b are just data. This means that sel is probably more important than a or b, so start the reduction with sel.

~a b sel + a ~b ~sel	+	a b ~sel + a b sel
(~a b sel + a b sel)	+	(a ~b ~sel + a b ~sel)		gather terms with sel and ~sel
(~a b + a b) sel	+	(a ~b + a b) ~sel		factor out sel and ~sel
[(~a + a)b] sel	+	[a (~b + b)] ~sel		factor out b and a
b sel	+	a ~sel		x + ~x = true

Are you talking about submitting it to Coursera? If so, then submit your version that passes the test. There are no grading points deducted for lack of simplification.

--Mark

[ccrcmxfx]

[Edited by admin to remove correct solution.]