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Decoder output logic

  • Thread starter bd411
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  • #1
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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'
 

Answers and Replies

  • #2
LCKurtz
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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.
 
  • #3
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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 ?
 
  • #4
LCKurtz
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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?
 
  • #5
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Yes, that function is low only when they are all 0.
 

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