MHB Applying Shannons Expansion Theorem

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SUMMARY

The discussion focuses on applying Shannon's Expansion Theorem to the function f(w,x,y,z) = wy + ~w~x~y + ~w~y~z + wy~z using a 2:1 multiplexer (MUX). The theorem is utilized to derive the minimum equations for f(w,x,0,z) and f(w,x,1,z). The correct expansion results in the expression ~y(~w~x + ~w~z) + y(w + w~z), confirming the proper application of the theorem and addressing the initial confusion regarding the inclusion of the term ~wz.

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  • Understanding of Shannon's Expansion Theorem
  • Familiarity with Boolean algebra
  • Knowledge of 2:1 multiplexer (MUX) functionality
  • Ability to manipulate logical expressions
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  • Explore Boolean algebra simplification techniques
  • Learn about the implementation of multiplexers in logic circuits
  • Investigate other applications of Shannon's theorem in computer science
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Students and professionals in electrical engineering, computer science, and digital logic design who are working with Boolean functions and multiplexers.

shamieh
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Wasn't exactly sure where to post this. Wanted to see if I did this correctly.Can someone check my work please?

Problem: Consider f defined below. Apply Shannon's expansion theorem (also given below) with respect to input y as if you were implementing this function using a 2:1 MUX. Find the minimum equations for f(w,x,0,z) and f(w,x,1,z).
Shannons Expansion Theorem
f(w_1, w_2,...,w_n) = ~w_1 * f(0,w_2,...,w_n) + w_1 * f(1,w_2,...,w_n)

The function to be expanded: f(w,x,y,z) = wy + ~w~x~y + ~w~y~z + wy~z
Here is what I got for my solution.

~y(~w~x + ~w~z + ~wz) + y(w + w~z)
 
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Why do you have the term ~wz in the first part, the one that corresponds to y = 0?
 
After re-doing the problem I obtained this: ~y(~w~x + ~w~z) + y(w + w~z)
 
Last edited:
That's correct.
 
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