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The question says draw F(A,B,C,D)=∑(3,7,11,12,13). I derived A'B'CD+ABC'. I am trying to draw it using two inverters and 5 2:1 multiplexers but i couldn't connect the output to the separate components i wrote. I know the correct answer but i just couldn't understand it.This is how far i have come.

Here's the correct solution This is the correct answer.

Why is the last mux connected to the 0 instead of 1 like we did all the other components? And why did they give 1 to 1 in mux in the answer?

zedyjy
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1 Answers1

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OK, then maybe this will help:

F(A,B,C,D)=∑(3,7,11,12,13).

w/ 2 nots; 5 2/1 muxs

truth table
ABCD R
0000 0
0001 0
0010 0
0011 1
0100 0
0101 0
0110 0
0111 1
1000 0
1001 0
1010 0
1011 1
1100 1
1101 1
1110 0
1111 0

kmap
  \ CD 00 01 11 10
AB \ 
00     0  0  1  0
01     0  0  1  0
11     1  1  0  0
10     0  0  1  0

expression
ABC'+A'CD+B'CD

simplifying
ABC'+(A'+B')CD
ABC'+(A'+B')''CD
ABC'+(AB)'CD
(AB)'CD + ABC'

aux truth table:

(AB)'CD  ABC'  ((AB)'CD + ABC')
 0       0              0             see note 2
 0       1              1             see note 1
 1       0              1             see note 2
 1       1              1             see note 1

note 1: If ABC' is true (mux select is 1) then output is true (mux's 1 input is set to 1) note 2: If ABC' is false (mux select is 0) then output is (AB)'CD (mux's 0 input is set to (AB)'CD), the "see note 2" outputs are true only when (AB)'CD is true

Andrew
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  • I understood where I did wrong. Thank you. I still don't understand why we connected the last mux where output is connected, like that (it gives 1 to one and (AB)'CD goes to 0). – zedyjy Nov 08 '22 at 11:58
  • I do not know what you mean by "it gives 1 to one and (AB)'CD goes to 0". To understand the last mux, use a truth table... – Andrew Nov 08 '22 at 12:17
  • thank you very much you explained very well even though I couldn't make the question clear. – zedyjy Nov 08 '22 at 22:31