Logic gates and truth table problems

[moenste]

Homework Statement

2. The attempt at a solution
Circuit:

A B G D E F C
0 0 1 1 1 1 0
1 1 0 1 0 1 0
0 0 1 1 1 1 0
1 1 0 1 0 1 0Answer:

Is (a) right? And any ideas what could be (b)?

 

[jedishrfu]

Your A and B columns should show all possible values {00, 01, 10, 11}

The output of the D NAND gate is: A' NAND B = (A' & B)' = D

and the F NAND gate is: A NAND B' = (A & B')'

 

[moenste]

Your A and B columns should show all possible values {00, 01, 10, 11}

The output of the D NAND gate is: A' NAND B = (A' & B)' = D

and the F NAND gate is: A NAND B' = (A & B')'

You mean I should have four columns?

A₁ 0101
A₂ 1010
B₁ 0101
B₂ 1010

Like this?

 

[jedishrfu]

No, you still have an A and a B column:

A   B
--  --
0   0
0   1
1   0
1   1

 

[moenste]

No, you still have an A and a B column:

A   B
--  --
0   0
0   1
1   0
1   1

But how did you derive these numbers? My original 0101 in A and B I got from the inverter table. But I don't see how to get the ones you mention.

 

[jedishrfu]

Your A and B inputs can be any binary value so since you have two inputs you have a possibility of 4 input values to the circuit as shown in the table above.

 

[moenste]

Your A and B inputs can be any binary value so since you have two inputs you have a possibility of 4 input values to the circuit as shown in the table above.

I get it now.

So:

A   B   G   D   E   F   C

0   0   1   1   1   1   0
0   1   1   0   0   1   1
1   0   0   1   1   0   1
1   1   0   1   0   1   0

 

[jedishrfu]

That's what I got too. The circuit diagram has a symmetry about it which is seen in the truth table.

 

[moenste]

That's what I got too. The circuit diagram has a symmetry about it which is seen in the truth table.

But what could be said about (b)? No idea about it.

 

[jedishrfu]

Look at the A and B inputs and then the C output doesn't that look like some gate you're familiar with?

 

[insightful]

I would say the "final" truth table should include only A, B, and C.

 

[moenste]

Look at the A and B inputs and then the C output doesn't that look like some gate you're familiar with?

AND has 0001 at output
OR 0111
NOR 1000
NAND 1110
Inverter 10

And I have at output 0110. The input is the same for this circuit and for AND, OR, NOR and NAND.

 

[insightful]

There is one more basic gate (or two, if you include its negative):

http://www.ee.surrey.ac.uk/Projects...ved=0CBkQ9QEwAGoVChMI-o3N7ceVyQIVw6MeCh00GgMp

 

[moenste]

There is one more basic gate (or two, if you include its negative):

http://www.ee.surrey.ac.uk/Projects...ved=0CBkQ9QEwAGoVChMI-o3N7ceVyQIVw6MeCh00GgMp

EXOR gate fits the output.

Thank you for the link.

 

[baudrunner]

The truth table for this problem should have three columns - two for A and B inputs, and one for the result C.