Implementing boolean functions with decoder and external gate

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Discussion Overview

The discussion revolves around designing a combinational circuit using a decoder and external gates based on specific boolean functions F1, F2, and F3. Participants explore the implications of using a decoder, particularly in relation to handling minterms with fewer than the expected number of variables.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about how to handle minterms in F1, F2, and F3 that only involve two variables, questioning whether to set the third variable to 0.
  • Another participant notes a discrepancy in their implementation of F1, stating they calculated F1 = Σm(0,5), while the provided answer includes an additional minterm, leading to a discussion about the notation of minterms.
  • Concerns are raised regarding the requirement for each output equation from a decoder to include all input variables, particularly when some minterms do not contain all variables.
  • A participant introduces the concept of 'Don't Care' conditions, suggesting that the missing variable does not affect the outcome of the minterm.
  • Further clarification is provided on how to interpret the 'Don't Care' condition, with examples of how F1 can be satisfied under different variable conditions.

Areas of Agreement / Disagreement

Participants generally agree on the concept of 'Don't Care' conditions, but there remains uncertainty regarding how to apply this understanding to the design of the circuit and the implications for the minterms involved.

Contextual Notes

Some participants express uncertainty about the handling of minterms with fewer variables and the implications of 'Don't Care' conditions, indicating a need for further clarification on these topics.

CoolDude420
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Homework Statement


Design an combinational circuit using a decoder and external gates defined by the boolean functions F1, F2, F3(see picture)

Homework Equations

The Attempt at a Solution



024208f1ce.jpg


I'm quite confused as to the exact method in doing this. I understand that a decoder takes n inputs and produces 2^n outputs. The combination of the n inputs correspond to binary numbers, whatever binary numbers the inputs make, the corresponding output line will be 1. For example if x=0, y=0, z=0, that repersents binary 0, so D0 will be 1.

My confusion arises when there are minterms in F1,F2,F3 that only have 2 variables. Do I just set the third, non present variable to 0? Any ideas?
 
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I went ahead and implemented F1 to my understanding.

c639130899.jpg


I got F1 = Σm(0,5).
The answers given to us is F1 = Σ(0,5,7)
Any ideas

Also, is it better to write Σm or is just Σ fine?
 
I was always told that with decoders that each output equation contains all of the input variables but since the one of the minterms in the F1 output equation doesn't contain all variables, how do I deal with that?
 
CoolDude420 said:
...one of the minterms in the F1 output equation doesn't contain all variables, how do I deal with that?

The missing variable is a 'Don't Care.' That means the result of the minterm is the same regardless of the value of the missing variable.
 
Tom.G said:
The missing variable is a 'Don't Care.' That means the result of the minterm is the same regardless of the value of the missing variable.

I ended up connecting D0, D5 and D7 to an OR gate which basically resulted in the y term being canceled out by simplification.
 
CoolDude420 said:
I was always told that with decoders that each output equation contains all of the input variables but since the one of the minterms in the F1 output equation doesn't contain all variables, how do I deal with that?
Also how did you figure out that it's a don't care term without simplification
 
Tom.G said:
The missing variable is a 'Don't Care.'

CoolDude420 said:
...how did you figure out that it's a don't care term without simplification

F1 = ## \bar X \bar Y \bar Z##
or
##X Z##​

The equation for F1 says there are two different conditions, either of which can satisify it:
  • X, Y, Z are all False
-or-
  • X and Z are both True
Restating it; For F1 to be satisified, it is sufficient that both X and Z are True, nothing else matters and there is no reason to include anything else in the minterm; all other variables are extraneous.

By the same token, F1 can also be satisified if X, Y, Z are all False.
 

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