# Boolean expression to circuit diagram

Discussion in 'Homework Help' started by kishanpatel50, May 20, 2009.

1. ### kishanpatel50 Thread Starter New Member

May 20, 2009
3
0
Hi, I came across an exam question today which asked me to draw a circuit diagram using only 1 NOT gate, 1 XOR gate and 1 OR gate of the following expression

f = A'B'C' + A'B'C + AB'C + A'BC
Which i've simplified to

f=A'B' + B'C + A'C

Ive been trying for hours but i cant seem to find an answer.
I don't think its possible
There was a hint that said that it may be appropriate to use P=A+B but the final answer must be in terms of A,B and C
The gates can have any number of inputs.

the problem could also be that i interpreted the question wrong and
f = A'B'C' + A'BC + ABC + AB'C which simplifies to
f = A'B'C' + AC + BC

Either way, i cant make the circuit.

2. ### PRS Well-Known Member

Aug 24, 2008
989
35
Dog gone it, kish, I used to know this stuff. Your question makes me feel like I have forgotten too much. But if you reasoned out your simplification right, could you not realize the first term, A'B' with the Or gate (A or B)? For the second and third terms C is required in either case but also you need this A' + B'. So you have the case C(A'B + A'B') The latter is a case of ANDing C to an XOR gate with A and B inputs. So, can you draw a logic gate diagram from that?

3. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Are you sure for the kind of gates you have to use?

4. ### kishanpatel50 Thread Starter New Member

May 20, 2009
3
0
@Mik3 Yeah, im sure about the gates

@PRS A'B' is the same as (A or B)' ie all notted. If you meant that then fair enough.
This would lead to an equation (A+B)' + C(A xor B).
I understand how you to get there but doesnt allow me to only use the gates specified. with the above equation, i still need an AND gate and an extra NOR gate.

Last edited: May 20, 2009
5. ### Ratch New Member

Mar 20, 2007
1,068
3
kishanpatel50,

Let's continue on with what you have so far.

F = (A'B')(C)'+(A+B)C
F = (A'B')(C)'+(A'B')'C by DeMorgan's theorem
F = (A'B') XOR C

So, take a 2-input OR gate and attach inputs A and B. Feed the output of the OR gate to the NOT gate and the output of the NOT gate to one input of a XOR gate. Attach the C input to the other input of the XOR gate and the output of the XOR gate should be what you want.

Ratch

6. ### mik3 Senior Member

Feb 4, 2008
4,846
63

A'B' is an AND gate and not an OR gate.

7. ### kishanpatel50 Thread Starter New Member

May 20, 2009
3
0
A'B' is the same as (A+B)' which uses the OR and NOT the output.
Thank you all for all your help =D