Help is needed (digital logic)

Thread Starter

ChoparD

Joined Sep 29, 2012
6
If you please I have midterm exam and i need someone help with this questions

1. Simplify the following using algebraic manipulation:
(AC’+AB’+B’C)(A’C+B’C’+BC’+C’)

2. Design a combinational circuit, that outputs 1 whenever four-bits number is greater than 3 and odd, or divisible by 4.


3. Show the internal structure of the four_data_inputs multiplexer.


4. What is the value of X in the following equation: 23(x)=51(8)


5. Simplify the function using K-map

Y= ab’c’d’ + a’bc’d + abc’d’ + a’b’cd’ + a’bcd’ +a’bc’d’

thank you every one

Best Regards,
 
Last edited by a moderator:

WBahn

Joined Mar 31, 2012
29,932
Okay, so let's take them one at a time.

For the first one, show us what you have done thus far. Remember, while you are told to use boolean algebra, you can always do a truth table or K-map in parallel to either check your answer or to give you clues about which path to follow with the algebra.
 

Thread Starter

ChoparD

Joined Sep 29, 2012
6
Truth table of a function A B C D
0 0 0 0 0
1 0 0 0 1
2 0 0 1 0
3 0 0 1 1
4 0 1 0 0
5 0 1 0 1
6 0 1 1 0
7 0 1 1 1
8 1 0 0 0
9 1 0 0 1
10 1 0 1 0
11 1 0 1 1
12 1 1 0 0
13 1 1 0 1
14 1 1 1 0
15 1 1 1 1
 

WBahn

Joined Mar 31, 2012
29,932
Uh... This isn't a truth table of anything. It is just an enumeration of the bit patterns for all of the four bit unsigned binary integers.

To be a truth table, you have to indicate the values (T or F) for the function you are working with.

You have:

X = (AC’+AB’+B’C)(A’C+B’C’+BC’+C’)

If you multiply this out, you will have twelve terms to deal with.

Break that into the two products:

Y = (AC’+AB’+B’C)
Z = (A’C+B’C’+BC’+C’)

X = YZ

Can you simplify Y and/or Z separately?
 

panic mode

Joined Oct 10, 2011
2,703
Think...

4-bit numbers are 0000...1111 which in decimal is 0-15

odd numbers greater than 3 are: 5, 7, 9, 11, 13, 15

numbers greater than 3 that are divisible by 4 are : 4, 8, 12

so you need to develop function that will produce 1 when any of those numbers is at inputs. Truth table would then look like:

Rich (BB code):
No A B C D  F
00 0 0 0 0  0 
01 0 0 0 1  0
02 0 0 1 0  0 
03 0 0 1 1  0  
04 0 1 0 0  1
05 0 1 0 1  1
06 0 1 1 0  0
07 0 1 1 1  1 
08 1 0 0 0  1
09 1 0 0 1  1
10 1 0 1 0  0
11 1 0 1 1  1
12 1 1 0 0  1
13 1 1 0 1  1
14 1 1 1 0  0
15 1 1 1 1  1
Note, NOW this represents truth table because it has function (additional column, in this case I labelled it F).

To realize this function you can use any number of ways. But if you are asked to get optimal solutions then you need to use Karnough Table (K-Map), fill it with info from this truth table and use what you learned to get the result.
 
Last edited by a moderator:

WBahn

Joined Mar 31, 2012
29,932
i don't know how i do it
Okay, let's take a step back and get on the same page regarding what you do know how to do.

Prior to this course, what other courses have you taken that have anything to do with digital logic?

What topics have you already covered in this course to date?

Do you understand the basic logic operations of NOT, AND, and OR?

Do you understand what a Truth Table is and how to construct one?

Do you understand Boolean algebra and the Boolean identities.

Do you understand how to manipulate Boolean equations?

Do you understand DeMorgan's Theorems?

Do you understand Karnaugh maps?
 

Thread Starter

ChoparD

Joined Sep 29, 2012
6
First of all i would like to thank every one tried to help me,
I finish my exam and i hope that i get high grade.

and for the guy who say
That is a test unto it self!
Very soon I will share with you every one the Midterm exam.
 

WBahn

Joined Mar 31, 2012
29,932
First of all i would like to thank every one tried to help me,
I finish my exam and i hope that i get high grade.

and for the guy who say

Very soon I will share with you every one the Midterm exam.
I'm pretty sure (not positive) that he was referring to the list of questions I asked you.

While I always like to see actual exams, before you post please ask your instructor if they have objections, first.
 
Top