# Boolean expression with MUX circuits

Let's say I have a 4:1 MUX circuit of

inputs:

C--->...............................output: F2

C'--->........4:1 MUX........

C' --->.............................

C --->..............................

select signals: A B

A = most significant bits

B = least significant bits

How would you construct a boolean expression in terms of A,B, and C?

I only learned how to do it with the numerical inputs. If the inputs are 0,1,1,0 respectively, I would make a simple 2 variable k-map and construct a boolean expression with it, but I'm kinda confused how to build one with the variable inputs.

$$f2=ABC+A^|BC^|+AB^|C^|+A^|B^|C$$
You know that if A,B=1 then C is selected so $$f2=ABC$$(Only with this expression you get C as output) when A=B=1, therefore f2=C. Similarly if A=1,B=0 then C0 is selected at output so $$f2=A^|BC^|$$ therefore f2=C0 (where C0 is the C complement). Similarly if A=0,B=1 then C0 is selected so at output $$f2=AB^|C^|$$ therefore f2=C0. Similarly if A=0,B=0 then C0 is selected so at output $$f2=A^|B^|C^|$$ therefore f2=C.
Combining all these possibilities we get $$f2=ABC+A^|BC^|+AB^|C^|+A^|B^|C$$
You could reason thinking about the 'decision' taken by the multiplexer under the command signal of selection, e.g. if both the selection signals are low then then the output must be the first signal: f = IN0 (when 00), IN1 (when 01), IN2 (when 10), IN3 (when 11) or with a compact logic expression: $$f = \bar{A}\bar{B}\cdot IN_0 + \bar{A}B\cdot IN_1 + A\bar{B}\cdot IN_2 + AB\cdot IN_3 = \bar{A}\bar{B}C + \bar{A}B\bar{C} + A\bar{B}\bar{C} + ABC$$ Now it can be simplified in many ways: $$f = \bar{A}(\bar{B}C + B\bar{C}) + A(\bar{B}\bar{C}+BC) = \bar{A}(B\oplus C) + A(\overline{B \oplus C}) = A\oplus(B \oplus C)$$ I hope this can help you.