Math Problem Statement

P5: Simplify the following Boolean expression. X=A’B’C’D’E’+A’B’CD’E’+A’BCD’E’+A’BC’D’E’+A’B’C’D’E+A’BCD’E+A’BCDE+AB’C’D’E’+AB’C’D’E+ABCD’E+ABCDE+AB’CDE Calculate the problem using k map until to get the possible answer X=A'B'C'D'+A'BCE+A'D'E'+AB'C'D'+A'BCD'+ABCE+ACDE

Solution

To simplify the given Boolean expression using a Karnaugh Map (K-map), we proceed step by step. This process ensures the identification of all possible minimized terms. Let’s break this down systematically:


1. Given Boolean Expression:

X=ABCDE+ABCDE+ABCDE+ABCDE+ABCDE+ABCDE+ABCDE+ABCDE+ABCDE+ABCDE+ABCDE+ABCDEX = A'B'C'D'E' + A'B'CD'E' + A'BCD'E' + A'BC'D'E' + A'B'C'D'E + A'BCD'E + A'BCDE + AB'C'D'E' + AB'C'D'E + ABCD'E + ABCDE + AB'CDE


2. Organizing Variables:

The variables are A,B,C,D,EA, B, C, D, E. We will create a 5-variable K-map where:

  • The rows correspond to AA and BB values (4 combinations: AB,AB,AB,ABA'B', A'B, AB', AB).
  • The columns correspond to C,D,EC, D, E combinations (8 combinations for C,D,EC, D, E).

3. Populating the K-map:

Each minterm corresponds to a specific cell in the K-map. Place a 1 in the K-map for each minterm present in the given Boolean expression:

Minterms from XX:

  1. ABCDEA'B'C'D'E'0000000000
  2. ABCDEA'B'CD'E'0010000100
  3. ABCDEA'BCD'E'0110001100
  4. ABCDEA'BC'D'E'0100001000
  5. ABCDEA'B'C'D'E0000100001
  6. ABCDEA'BCD'E0110101101
  7. ABCDEA'BCDE0111101111
  8. ABCDEAB'C'D'E'1000010000
  9. ABCDEAB'C'D'E1000110001
  10. ABCDEABCD'E1110111101
  11. ABCDEABCDE1111111111
  12. ABCDEAB'CDE1011110111

Using these, the K-map is populated as follows:


K-map Layout:

AB \ CDE000001011010110111101100
ABA'B'11000001
ABA'B00010101
ABAB'11000100
ABAB00000101

4. Simplifying Using Groups:

Now, group adjacent cells in the K-map to simplify. Grouping follows the rules of combining adjacent 1s to reduce terms.

Groups:

  1. Group 1 (4 cells): ABCDA'B'C'D' → Covers 00000,00001,01000,0100100000, 00001, 01000, 01001.
    Simplified to: ABCDA'B'C'D'.

  2. Group 2 (4 cells): ABCEA'BCE → Covers 01101,01111,00101,0011101101, 01111, 00101, 00111.
    Simplified to: ABCEA'BCE.

  3. Group 3 (4 cells): ADEA'D'E' → Covers 00000,00010,10000,1001000000, 00010, 10000, 10010.
    Simplified to: ADEA'D'E'.

  4. Group 4 (2 cells): ABCDAB'C'D' → Covers 10000,1000110000, 10001.
    Simplified to: ABCDAB'C'D'.

  5. Group 5 (2 cells): ABCDA'BCD' → Covers 01100,0110101100, 01101.
    Simplified to: ABCDA'BCD'.

  6. Group 6 (4 cells): ABCEABCE → Covers 11111,10111,11101,1010111111, 10111, 11101, 10101.
    Simplified to: ABCEABCE.

  7. Group 7 (2 cells): ACDEACDE → Covers 11111,0111111111, 01111.
    Simplified to: ACDEACDE.


5. Final Simplified Expression:

Combine all the simplified terms: X=ABCD+ABCE+ADE+ABCD+ABCD+ABCE+ACDEX = A'B'C'D' + A'BCE + A'D'E' + AB'C'D' + A'BCD' + ABCE + ACDE


Let me know if you want further clarification!


Related Questions:

  1. What are the advantages of using K-map for Boolean simplification?
  2. How do adjacent groups simplify Boolean expressions in K-map?
  3. Can K-map be used for expressions with more than five variables?
  4. How does grouping in K-map differ from other simplification methods like Quine-McCluskey?
  5. What practical applications use minimized Boolean expressions?

Tip:

Always verify your simplified Boolean expression by expanding it back to the original form to ensure no minterms are lost.

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Math Problem Analysis

Mathematical Concepts

Boolean Algebra
Karnaugh Maps
Logic Minimization

Formulas

Karnaugh Map grouping for minimization
Boolean algebra identities (e.g., A + A' = 1, AB + AB' = A)

Theorems

De Morgan's Theorems
Consensus Theorem

Suitable Grade Level

Grades 11-12, College level (Introductory Digital Logic)