Solve 3 Bit Grey Code Decoding: Output Logic

[bd411]

Homework Statement

Given that the variables A[2:0] form a three bit grey code to represent numerical values:

i) Write down all binary combinations for A[2:0], in ascending order of numerical value

ii) A three bit grey code to decimal decoder is to be designed, where the outputs X[0:7] are active low. Write down the boolean functions of each output terminal.

Homework Equations

The Attempt at a Solution

I think the first part is easy enough, it's just grey code (only change one bit at a time)

A2:A0

0 000
1 001
2 011
3 010
4 110
5 111
6 101
7 100

The second part is confusing me though, and it's to do with the "active low part". Initially I got what's below but how is it any different to if the outputs weren't active low ?!

X0 = A2'A1'A0'
X1 = A2'A1'A0
X2 = A2'A1A0
X3 = A2'A1A0'
X4 = A2A1A0'
X5 = A2A1A0
X6 = A2A1'A0
X7 = A2A1'A0'

 

[LCKurtz]

If, for example, 2 would be A2'A1A0 if it was active high, you want the complement for active low. Use DeMorgan's law. The complement of a product is not the product of the complements.

 

[bd411]

If, for example, 2 would be A2'A1A0 if it was active high, you want the complement for active low. Use DeMorgan's law. The complement of a product is not the product of the complements.

Hmm ok so for X0 I had A2' . A1' . A0' when I should have all that [A2' . A1' . A0']'

So that becomes A2 + A1 + A0 ?

 

[LCKurtz]

Hmm ok so for X0 I had A2' . A1' . A0' when I should have all that [A2' . A1' . A0']'

So that becomes A2 + A1 + A0 ?

You tell me. Is that low only when they are all 0?

 

[bd411]

Yes, that function is low only when they are all 0.