Digital Logic Help

[InverseOfNeo]

Need a lil help here guys. I designed a circuit that has 4 input and an output that is HIGH only when an odd number of inputs are HIGH. The circuit has 3 2-input XOR gates. So the expression woud be (A XOR B) XOR (C XOR D). But I need the simplification from A'B'C'D+A'B'CD'+A'BC'D'+A'BCD+AB'C'D'+AB'CD+ABC'D+ABCD' (derived from truth table from original question) to what I have above. I can simplify it but I dont get the XOR in the middle, I just get an OR which wouldnt give me the right result. Help would be much appreciated. Thanks guys.

Schematic

 

[rival]

wow, boolean algebra, we just did some of this in class and were supposed to take the test tonight but didnt have them ready...

cant help ya though, we just went over the basics

 

[InverseOfNeo]

Yea, I did the schematic in lab today though I cant explain how I figured it out. Even the TA was like shocked at how simple my circuit was because he was able to simplify it down to only two ICs and I did it with one. Most of the other people in my class did it with 4 ICs.

 

[TuxDave]

A'B'C'D+A'B'CD'+A'BC'D'+A'BCD+AB'C'D'+AB'CD+ABC'D+ABCD'

Ok, let's get started. So A xor B = AB' or A'B, let's use that.

Rearranging

C'D'(A'B+AB')+CD(A'B+B'A)+CD'(AB+A'B')+C'D(A'B'+AB)
C'D'(A xor B)+CD(A xor B)+CD'(A xnor B) + C'D(A xnor B)
(A xor B)(C'D'+CD)+(A xnor B)(CD'+C'D)
(A xor B)(C xnor D)+(A xnor B)(C xor D)
(A xor B) xor (C xor D)

If you don't get the last step, substitute with the following.
A xor B = X
A xnor B = X'
C xor D = Y
C xnor D = Y'

So the last expression was:
XY' + YX' = X xor Y


<------ Was a TA, but for an intro to analog circuits class.