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Question about implementing (boolean) function using multiplexer?

s3a

Jun 16, 2018
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Jun 16, 2018
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Hello, everyone.

I'm studying a past exam, and I'm stuck on something, but I'm thinking that the solution of the exam for the question I'm doing (Question 1 -- 3. a.k.a. Question 1 -- c.) has a (minor) error in it.

Question 1 -- 3. a.k.a. Question 1 -- c.:
https://www.docdroid.net/IeyzXy3/question-1-3-aka-question-1-c.pdf

Its provided solution:
https://www.docdroid.net/vNlE9vP/thesolution.pdf

My question:
Should the part with the (4-1) MUX drawing where there is a C be a C', instead? (I know this may seem like a small question, but I just want to make sure that I'm not misunderstanding anything.)

Any input would be greatly appreciated!
 

Laplace

Apr 4, 2010
1,252
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Apr 4, 2010
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1,252
I used Logisim to verify the K-map. The output for the selection A=1, B=1, certainly seems to match C' so I believe you are correct.
 

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Ratch

Mar 10, 2013
1,099
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Hello, everyone.

I'm studying a past exam, and I'm stuck on something, but I'm thinking that the solution of the exam for the question I'm doing (Question 1 -- 3. a.k.a. Question 1 -- c.) has a (minor) error in it.

Question 1 -- 3. a.k.a. Question 1 -- c.:
https://www.docdroid.net/IeyzXy3/question-1-3-aka-question-1-c.pdf

Its provided solution:
https://www.docdroid.net/vNlE9vP/thesolution.pdf

My question:
Should the part with the (4-1) MUX drawing where there is a C be a C', instead? (I know this may seem like a small question, but I just want to make sure that I'm not misunderstanding anything.)

Any input would be greatly appreciated!

Boolean expression not fully reduced.

A'B'+ABC'+AB'D=A'B'+ABC'+B'D

because

A'B'+ABC'+AB'D =
A'B'(1+D)+ABC'+AB'D=
A'B'+A'B'D+ABC'+AB'D=
A'B'+ABC'+A'B'D+AB'D=
A'B'+ABC'+B'D(A+A')=
A'B'+ABC'+B'D

Ratch
 
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