# SOP and logic expression for a circuit

I'm having trouble doing the Sum-or-product of this truth table and then simplifying the logic expression.

For the SOP i got Y= A’.B’.C’.D’+ A.B’.C’.D’+ A’.B.C’.D’+ A’.B’.C.D’+ A’.B’.C’.D+ A.B.C’.D’+ A’.B.C.D’+ A’.B.C.D’+ A’.B.C’.D+ A’.B’.C.D+ A.B.C’.D+ A’.B.C.D+ A’.B’.C.D’+ A.B.C.D’+ A.B’.C.D

but I don't know where to go from here

• have idea about Karnaugh map/logic simplification. en.wikipedia.org/wiki/Karnaugh_map Commented Nov 4, 2020 at 4:34
• Why do you need more than 2 input bits to define 4 output states? Commented Nov 4, 2020 at 4:52
• It seems , if I presume what you are doing, you only need 2 inputs. Do you want this to be a quadrature detector with wind direction but with a low pass filter to rotate the outputs in the shortest direction but averaged over time? That requires a state sequence not just combinational logic. i.sstatic.net/XUqrG.png Redefine your problem better Commented Nov 4, 2020 at 5:08
• @Anarchy You only show one Y equation. If you have two outputs then there should be two SOP equations. Yes?
– jonk
Commented Nov 4, 2020 at 5:26
• @Anarchy Yes. Of course. Each equation accounts for one of the outputs. Is that difficult to follow why? Treat each output column as a separate table. Doesn't that make sense?
– jonk
Commented Nov 4, 2020 at 6:28

$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} E-W&\overline{B}\:\overline{A}&\overline{B}\: A&B \:A&B \:\overline{A}\\ \hline \overline{D}\:\overline{C}&1&1&1&1\\ \overline{D}\:C &1&0&0&1\\ D\: C &1&0&0&1\\ D\:\overline{C} &1&1&1&1 \end{array}\end{smallmatrix} & \begin{smallmatrix}\begin{array}{r|cccc} N-S&\overline{B}\:\overline{A}&\overline{B}\: A&B \:A&B \:\overline{A}\\ \hline \overline{D}\:\overline{C}&0&1&1&0\\ \overline{D}\:C&1&1&1&1\\ D\: C&1&1&0&0\\ D\:\overline{C}&1&1&0&0 \end{array}\end{smallmatrix} \end{array}$$