combinational circuit

Discussion in 'Homework Help' started by Vasilis Gkagkaris, Jun 30, 2018.

1. Vasilis Gkagkaris Thread Starter New Member

Jan 30, 2018
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The digital circuit has inputs of three unmarked words A [2: 0], B [2: 0], and C [2: 0]. If it is

a number of words that have an odd value outweigh the number of words they have

even value then for R [2: 0] it is R [2: 0] = max (A [2: 0], B [2: 0], C [2: 0] ] =

min (A [2: 0], B [2: 0], C [2: 0]). A circuit is required to be designed to perform the above

mode. Only comparators, multiplexers and gates of any crowd are available

inputs.

It is recommended that the total circuit be based on individual, less sophisticated circuits and

initially to briefly describe the logic of the proposed solution

Last edited: Jun 30, 2018

Mar 31, 2012
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3. WBahn Moderator

Mar 31, 2012
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And your question is...?

Also (and this is likely a translation to English issue), the description is very confusing.

What does it mean for "a number of words that have an unnecessary value outweigh the number of words"

Is R[2:0] the output of the circuit?

Your equation can only be true if all three inputs are the same value.

What is the output supposed to be if they are not the same value.

4. Vasilis Gkagkaris Thread Starter New Member

Jan 30, 2018
9
0
a number of words that have an odd value outweigh the number of words they have

even value
my problem is that i do not know how to design it
yes r[2:0] is the output

5. WBahn Moderator

Mar 31, 2012
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Okay, so it seems like you want to do something based on whether or not the three input values consist mostly of odd or even values.

So first describe the easiest way you can think of to answer this specific question either Y (there are more odd valued words) or N (there are more even valued words).

Assuming that you have that part done, what is it you want to do in each case?

6. Vasilis Gkagkaris Thread Starter New Member

Jan 30, 2018
9
0
I didn't notice but the description had a second mistake
a number of words that have an odd value outweigh the number of words they have

even value then for R [2: 0] it is R [2: 0] = max (A [2: 0], B [2: 0], C [2: 0] ] in different case

R[2:0]=min (A [2: 0], B [2: 0], C [2: 0]).

7. WBahn Moderator

Mar 31, 2012
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So let's see if I understand the problem now:

Given three 3-bit words, if there are more odd-values words than even-valued words, output the word with the maximum value, else output the word with the minimum value.

If that is correct, the can you break this down into smaller problems and solve them individually? Note that I already asked you about one of them. There are about three others. With each of the small ones solved, then it is just a matter of combining the solutions to the small problems together to solve the overall problem.

8. Vasilis Gkagkaris Thread Starter New Member

Jan 30, 2018
9
0
First of all thank you for your time !!
but let's take the first case with maximum value
will I need one or two 4:1 multiplexer and three 2- bit comparator?
how i will connect them?

Last edited: Jul 1, 2018
9. WBahn Moderator

Mar 31, 2012
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YOU need to show YOUR best efforts to work YOUR homework.

Don't expect others to spoonfeed you the solutions.

Make an attempt!

There are many ways to approach the problem. Consider the simpler problem of finding the maximum of two 3-bit numbers.

10. Vasilis Gkagkaris Thread Starter New Member

Jan 30, 2018
9
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How i will connect magnitude comparator with multiplexer circuit ?
I tried to create a circuit with seven 2 x1 multiplexer
( A0=S0,B0=S1,C0=S2) are least significant numbers

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11. WBahn Moderator

Mar 31, 2012
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Where's the magnitude comparator you mention?

I can't make heads or tales of your truth table on the second page. What information does the Y column represent?

What is it that you are hoping to accomplish by just looking at the least significant bits of the three values?

What part of the problem are you trying to address?

We are not mind readers.

12. Vasilis Gkagkaris Thread Starter New Member

Jan 30, 2018
9
0
OK , can you delete the whole post?

13. WBahn Moderator

Mar 31, 2012
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No need to do that. Just try to communicate what you are doing and your reasoning more clearly.

14. Vasilis Gkagkaris Thread Starter New Member

Jan 30, 2018
9
0
I created a truth table with least significant numbers because I wanted to distinguish even numbers from odd numbers
if odd numbers are more than even numbers Y=1 in different case Y=0 (that's why I created and seven 2x1 multiplexer circuit)
my problem is I do not know how I use comparator and seven 2x1 multiplexer together that's why I did not design the comparator (in this part of the problem I want to help me)

Last edited: Aug 8, 2018
15. WBahn Moderator

Mar 31, 2012
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Look at your truth table? Do you agree with it?

You have five cases in which the output is 0 and three cases in which the output is 1. Does this make sense? Are there really more ways to have more even values than odd values than there are to have more odd values than even values?

Let's say I give you some black boxes. One box, MIN, accepts three 3-bit inputs and outputs a 3-bit value that is equal to the smallest of the three input values. Another box, MAX, behaves similarly except it outputs the largest of the three values. I also give you a box, ODD, that takes three 3-bit inputs and produces a 1 if there are more odd values than even values and a 0 otherwise. Finally, I give you a box called MUX that accepts two 3-bit inputs (V0 and V1) and a 1-bit inputs (SEL) that has a single 3-bit output (Y) that is equal to Vo if SEL = 0 and equal to V1 if SEL = 1;

Given these building blocks, can you design a circuit that solves the original problem?

16. Vasilis Gkagkaris Thread Starter New Member

Jan 30, 2018
9
0
OK i think I can solve it now
thank you

17. WBahn Moderator

Mar 31, 2012
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What's far more important than solving this problem is understanding how to go about solving problems in general.

Do you see what I did here? I did nothing more than the very thing that I recommended that you do back on the very first day of the thread in June -- break the problem down into small pieces and then combine the solutions to the small pieces to solve the overall problem.