f = d'[b + (c xor a)'] + dcb'

Also, if only a and c are active high, and the rest are active low, what changes have to be made when building the circuit?

 

Looks like & Im not a maths expert, no way near,,

if look f=freq d< maybe decibel, c=capacitance and Im sure xor a is " xor a is a xor a Lgoic gate.. the dcb it looks like you have to add it..

sorry this is a bit of a flimsy answer... eek hope Im not totally wrong!!!
if you send the complete formular, it may throw more light on it" For the maths wizzes on here they are really good sgtwookie for one..
regards
redlight

 

All the letters are just variables. They don't stand for anything.

 

Looks like a logic expression to me.

d' is the same as (not d) the dash being used to indicated negation.

All the letters are boolean or logic variables.

 

Looks like a logic expression to me.

d' is the same as (not d) the dash being used to indicated negation.

All the letters are boolean or logic variables.

Yes, this is a boolean logic equation that I am trying to simplify.

 

So where have you got to with it?

We don't do homework for people here, but will help if you show some working up to where you are stuck.

 

So where have you got to with it?

We don't do homework for people here, but will help if you show some working up to where you are stuck.

The deadline has passed and I already solved it by myself.

d'[b + (c xor a)'] + dcb'
d'[b + c'a + ca'] + dcb'
d'b + d'c'a + d'ca' + dcb'

 

Well done that's always the best way.

 

Can someone at least verify it's right?

 

You should be able to verify it yourself. Make a truth table for each. Is the outcome of each truth table the same? If so, then you were successful.

Unfortunately, this is not the case...

Let's call your solution g
Try, for example, the case where:
a = false
b = false
c = false
d = false

In this case f = true and g = false, therefore f ≠ g

 

Careful: (C XOR A)=C'A+CA'. So there is no way that (C XOR A)'=CA'+C'A. It's just the same expression in reverse.

The correct answer is CA+C'A'. But please post the proof as an exercise for yourself. Are you familia with the deMorgan's theorem? It is a must if you want to understand boolean expression conversion.