Digital Design - decoders and multiplexer help

Click For Summary
SUMMARY

The discussion focuses on implementing a function f(x3, x2, x1, x0) using digital design components, specifically four 2-to-4 decoders and a 4-to-1 multiplexer. The function outputs y1, which indicates whether the number of 1's in the input is even, and y0, which indicates whether the number of 1's is odd. A truth table is provided to clarify the outputs for all possible input combinations, confirming that for the input (0, 0, 0, 0), both y1 and y0 are 0, indicating an even count of 1's. The discussion highlights the importance of understanding the relationship between binary inputs and their corresponding outputs in digital logic design.

PREREQUISITES
  • Understanding of digital logic design principles
  • Familiarity with 2-to-4 decoders and 4-to-1 multiplexers
  • Knowledge of truth tables and their construction
  • Basic concepts of even and odd numbers in binary representation
NEXT STEPS
  • Study the implementation of 2-to-4 decoders in digital circuits
  • Learn about the operation and design of 4-to-1 multiplexers
  • Explore more complex digital logic functions using combinational logic
  • Practice creating truth tables for various digital logic scenarios
USEFUL FOR

Students and professionals in electrical engineering, computer engineering, and anyone involved in digital circuit design or logic implementation.

Saterial
Messages
54
Reaction score
0

Homework Statement


Let f(x3; x2; x1; x0) = (y1; y0) such that y1 = 1 if the number of 1's in x3x2x1x0 is even, 0 otherwise, and, output y0 = 1 if the number of 1's in x3x2x1x0 is odd, 0 otherwise.

1. Implement f with four 2-to-4 decoders.
2. Implement f with a 4-to-1 multiplexer with selection inputs x0; x1 (in this order)

Homework Equations


The Attempt at a Solution



How do I even start solving this? This was a bonus as it was not really taught to me so I have to learn myself.

(IN THE CASE OF 0, 0, 0, 0, both y1 and y0 are 0 right? because there are no 1's at all)

I begin by making a truth table of:
Code:
x3 x2 x1 x0   y1 y0
 0  0  0  0    0  0
 0  0  0  1    0  1
 0  0  1  0    0  1
 0  0  1  1    1  0
 0  1  0  0    0  1
 0  1  0  1    1  0
 0  1  1  0    1  0
 0  1  1  1    0  1
 1  0  0  0    0  1
 1  0  0  1    1  0
 1  0  1  0    1  0
 1  0  1  1    0  1
 1  1  0  0    1  0
 1  1  0  1    0  1
 1  1  1  0    0  1
 1  1  1  1    1  0

Where do I go from here?
 
Last edited by a moderator:
Physics news on Phys.org
Saterial said:
(IN THE CASE OF 0, 0, 0, 0, both y1 and y0 are 0 right? because there are no 1's at all)

Zero is an even number.

If x3..x0 | y1,y0 = 0,0,0,0 | 0,0 then y1 would indicate: "number of ones is not even" y0 would indicate "number of ones is not odd" --a contradiction.
 
So that would mean the output of y1 and y0 in my case is 1, 0 for 0, 0 ,0 0 in my case because the number of 1's is even and not odd?
 
Yes. And it should make your job somewhat easier as well.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K