Digital Computer Electronics

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  • Topic: Boolean algebra, Logic gate, Logic
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PRESTON UNIVERSITY LAHORE CAMPUS
1-A Ahmed Block New Garden Town Lahore

Assignment-I: Digital Computer Electronics (DCE)
Semester: Fall-2012
Task: K-map Simplification

Student Name: Shaukat Mahmood (03007411855)
Registration No.16D2-212022
Program:B-Tech Pass (Advance)
Quarter:5th

Submitted to:Professor Assad Umer Khan

Assignment Task:
USE OF KARNAUGH MAP FOR SIMPLIFICATION OF LOGIC CIRCUIT/ EXPRESSION Rules for Karnaugh Map Simplification:
Followings are the rules to use k-Map for logic circuit simplification 1. Enter a 1 on the Karnaugh map for each fundamental product term that corresponds to 1 output, in the truth table. Enter 0s elsewhere. 2. Encircle the octets, quads, and pairs. Remember to roll and overlap to get the largest groups possible. 3. If any isolated 1s remain, encircle them.

4. Eliminate redundant groups if any exist.
5. Indicate don’t condition, if any, with X instead of 1s or 0s. 6. Treat X as 0 or 1 to form octets, quads or pair, as per simplification requirement. 7. Write the Boolean equation by ORing the products corresponding to the encircled groups. 8. Draw the equivalent logic circuit.

Example-1: Simplify the following Boolean Equation using K-Map Y= A’B’C’D+ A’B’CD’+ A’B’CD+ A’BC’D’+ A’BCD’+ AB’C’D’+ AB’C’D+ AB’CD’+ ABC’D’+ ABC’D+ ABCD’ Solution:

Now for
Group-IGroup-IIGroup-IIi
ABCDABCDABCD
000111000010
001111010110
A’B’D10001110
10011010
AC’CD’
Simplified SOP is
Y=A’B’D+AC’+CD’

Example-2: Simplify the following Boolean Equation using K-Map Y=A’BC’D’+ABC’D’+ABC’D+ABCD+ABCD’+AB’C’D’_AB’C’D+AB’CD+AB’CD’

Now for, Group-IGroup-II
ABCDABCD
01011100
11011101
BC’D1111
1110
1000
1001
1011
1010
A
Simplified SOP equation is Y=A+BC’D

Example-3: Simplify the following Boolean Equation using K-Map Y=A’BC’D’+A’BCD’+ABC’D’+ABCD’

Now for, Quad Group of 1’s
ABCD
0100
0110
1100
1110
BD’
Simplified SOP equation is
Y=BD’
Simplified logic diagram

Example-4: Simplify the following Boolean Equation using K-Map Y=A’BC’D+ABC’D+ABCD+AB’CD
Soultion:
Truth Table for given SOP equationKarnaugh Map for Given SOP equation Ref No.| Input| Output| Basic PT| | | | Group-I| | Redundant Group| | A| B| C| D| Y| | | | CD| | | | | | |

| 23| 22| 21| 20| | | | | AB| C’D’| C’D| CD| CD’| | | 0| 0| 0| 0| 0| | | | | A’B’| 0| 0| 0| 0| | | 1| 0| 0| 0| 1| | | | | A’B| 0| 1| 0| 0| | | 2| 0| 0| 1| 0| | | | | AB| 0| 1| 1| 0| | | 3| 0| 0| 1| 1| | | | | AB’| 0| 0| 1| 0| | | 4| 0| 1| 0| 0| | | | | | | | | | | |

5| 0| 1| 0| 1| 1| A’BC’D| | | | | | Group-II| | 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| 1| AB’CD| | | | | | | | | | 12| 1| 1| 0| 0| | | | | | | | | | | |
13| 1| 1| 0| 1| 1| ABC’D| | | | | | | | | | 14| 1| 1| 1| 0| | | | | | | | | | | |
15| 1| 1| 1| 1| 1| ABCD| | | | | | | | | |

Now for, Group-IGroup-II
ABCDABCD
01011111
11011011
BC’DACD
Simplified SOP equation is
Y=BC’D+ACD

Example-5: Simplify the following Boolean Equation using K-Map (Redundant Group example) Y= A’BCD+ AB’C’D+ AB’CD’+ AB’CD+ ABC’D’+ ABC’D+ ABCD’+ ABCD

Soultion:
Truth Table for given SOP equationKarnaugh Map for Given SOP equation Ref No.| Input| Output| Basic PT| | | | Group-I| Group-II| | A| B| C| D|...
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