For truth tables, i have no problem dealing with several inputs and one output but what if the truth table has two outputs instead? I want to write it in Boolean Algebra but i gave it a go and I'm not sure how it works for two outputs.

0|0|0|0| 1| 1
0|0|0|1| 0| 0
0|0|1|0| 1| 0

ABCD will be the input and Z1 and Z2 will be the two outputs. If it is only Z1 i can create the boolean algebra out of it but what if we have two outputs? Any help?

Z1 = A'B'C'D'
Z2 = A'B'C'D'??
  • $\begingroup$ Can you give an example of a truth table for a single expression with two outputs? $\endgroup$ – vrugtehagel Mar 26 '17 at 7:46
  • $\begingroup$ why not just write two different expressions: $z_1=\bar A\bar B\bar D$ and $z_2=\bar A\bar B\bar C\bar D$ ? $\endgroup$ – user400188 Mar 26 '17 at 7:54

You would treat this problem the same way you would treat one with only one output: create two separate Karnaugh Maps, one for each output. I quickly made two with your example data which resulted in the following expressions:

Z1 = A'B'D'

Z2 = A'B'C'D'

Here is a Wikipedia article on Karnaugh Maps if you are not familiar with them: https://en.wikipedia.org/wiki/Karnaugh_map

Sources: As an Electrical Engineering student I spent a lot of time creating logic circuitry with multiple outputs and this was the method I used.

Edit: I'm not sure if this was what you were trying to get at, but if you wanted Z1 to depend on Z2, you would simply add Z2 as an additional input to Z1's equation. I hope this helps!

  • $\begingroup$ well to be honest, the question isn't clear. we have to write the boolean functions as boolean algebra terms for each single row. In the end, we are required to combine the terms for single rows into larger terms. I'm confused too. $\endgroup$ – Electric Mar 26 '17 at 8:15
  • $\begingroup$ also, I will be using the Karnaugh Map for the later stage to create the logic gates. right now, it's just the two outputs that is bothersome. my first comment was generally what the question wants so if you could understand it, do inform me on what it exactly meant. $\endgroup$ – Electric Mar 26 '17 at 8:18
  • $\begingroup$ For the first row you would have Z1 = A'B'C'D' and Z2 = A'B'C'D', the second row would only equal zero, and the third would be Z1 = A'B'CD' and Z2 still equal to zero. Ultimately you would add these expressions together to get the final answer. $\endgroup$ – Jared Cannon Mar 26 '17 at 8:25
  • $\begingroup$ The Karnaugh Map makes that a one-step process, but if this is an introductory exercise I guess I can understand why you would have to do it the long way first. $\endgroup$ – Jared Cannon Mar 26 '17 at 8:26
  • $\begingroup$ So for this example, the two Z1 terms would be added to equal A'B'C'D' + A'B'CD'. You can simplify that expression down to A'B'D' because the C and C' cancel each other out. $\endgroup$ – Jared Cannon Mar 26 '17 at 8:29

For two outputs you need two expressions. If you were hoping/thinking that there would be one boolean expression that somehow captured that ... then stop thinking that: a single boolean expression captures one and only one truth-function, and that is what the 'outputs' are: truth-functions. So: different outputs, different expressions.


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