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I'm studying for my exam in digital electronics in mid april. There is a question with a multiplexer where I should write the correct truth table but I don't understand how to do it. It should be written with 1:s and 0:s. Can someone explain how to do this?

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  • \$\begingroup\$ open any web browser, enter www.google.com, enter multiplexer truth table, click any of the houndreds of resources even including video tutorials. come on... \$\endgroup\$ – Piglet Mar 27 '19 at 22:33
  • \$\begingroup\$ It's not that easy. Ofc I have done an extensive amount of googleing before asking here. Nobody covers how to go from curcuit to truth table with 1:s and 0:s. \$\endgroup\$ – Jean Doe Mar 28 '19 at 12:35
  • \$\begingroup\$ I don't get your problem. I just clicked and watched a random youtube tutorial on 8-1 MUX and they thoroughly explained how to get the output. maybe it would help if you'd pick one of those tutorials and explain which part you don't understand or if you would put a bit more effort into explaining what your problem is. \$\endgroup\$ – Piglet Mar 28 '19 at 15:23
  • \$\begingroup\$ The datasheets for real multiplexers (and other logic IC's) almost always have a truth table. \$\endgroup\$ – mkeith Mar 30 '19 at 5:55
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You have 4 inputs A,B,C and D and you're looking for the output F(A,B,C,D). s0,s1,s2 decide which of the 8 inputs is mapped to the output. As we already know the the input values (0,1, D or ¬D) we can write the output function F(A,B,C,D).

F(A,B,C,D) = ¬A¬B¬CD + ¬A¬BC0 + ¬AB¬C¬D + ¬ABC0 + A¬B¬C1 + A¬BC¬D + AB¬CD + ABC0
F(A,B,C,D) =  ¬A¬B¬CD + ¬AB¬C¬D + A¬B¬C + A¬BC¬D + AB¬CD

Then you create a table with all possible combinations of A,B,C and D and enter the resulting Y = F(A,B,C,D)

A B C D | Y
0 0 0 0 | 0
0 0 0 1 | 1
0 0 1 0 | 0
0 0 1 1 | 0
0 1 0 0 | 1
0 1 0 1 | 0
...

I hope I didn't make any mistakes.

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A B C | Y
0 0 0 | D
0 0 1 | L0
0 1 0 | ~D
0 1 1 | L0
1 0 0 | B1
1 0 1 | ~D
1 1 0 | D
1 1 1 | L0
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  • \$\begingroup\$ This may answer the question, but the truth table by itself with no explanation of how to arrive at it given the circuit won't help anyone learn how to produce a truth table from a circuit. I'm also not really certain that it is correct - the output column is odd. \$\endgroup\$ – JRE Mar 30 '19 at 14:17
  • \$\begingroup\$ Try the solution and compare the result with my suggested truth table \$\endgroup\$ – Vaibhav Mar 31 '19 at 15:05
  • \$\begingroup\$ Hard to do since the output column doesn't make any sense. \$\endgroup\$ – JRE Mar 31 '19 at 15:17
  • \$\begingroup\$ you need an explanation for that. Its very simple. INPUT D has no effect on output as there are three select lines, handled by A,B,C since if all of them are zero then I0 input will be the output. I0 input is D so whatever the value of D it will be the output and goes same for all. \$\endgroup\$ – Vaibhav Apr 1 '19 at 5:27

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