Question
Write the logic expression for the following Karnaugh
Map whose output is Z. Be sure to group in the most
advantageous way to simplify your SOP.
AB C
00 01 11 10 0 1 1 1 0 1
1 0 0
1
Map whose output is Z. Be sure to group in the most
advantageous way to simplify your SOP.
AB C
00 01 11 10 0 1 1 1 0 1
1 0 0
1
Answers
GPT-4o mini
To construct the logic expression for the given Karnaugh map, we first need to layout the values based on the information you provided.
### Karnaugh Map Setup
Let's assign variables for better readability:
- Let \(A\) be the row variable (top).
- Let \(B\) be the column variable (left).
- Let \(C\) be for the columns of the inner positions.
The arrangement provided in the format appears to be in the following form:
\[
\begin{array}{|c|c|c|c|c|}
\hline
AB \backslash C & 00 & 01 & 11 & 10 \\
\hline
00 & 1 & 1 & 1 & 0 \\
\hline
01 & 1 & 0 & 0 & 1 \\
\hline
\end{array}
\]
The respective values for \(A\) and \(B\) in binary in this arrangement are:
- **Row \(00\)** (i.e., \(A=0, B=0\)): Values of C are 00, 01, 11, 10, giving us \(1, 1, 1, 0\)
- **Row \(01\)** (i.e., \(A=0, B=1\)): Values of C are 00, 01, 11, 10, giving us \(1, 0, 0, 1\)
### Output from the Karnaugh Map
Based on your description, the K-Map values translate to:
- \( (A,B,C) = (0,0,0) \) : \( 1 \)
- \( (A,B,C) = (0,0,1) \) : \( 1 \)
- \( (A,B,C) = (0,1,0) \) : \( 1 \)
- \( (A,B,C) = (1,1,0) \) : \( 1 \)
- All other combinations yield \( 0 \).
Consolidating our positions gives us:
- \( m_0 = (0,0,0) \)
- \( m_1 = (0,0,1) \)
- \( m_2 = (0,1,0) \)
- \( m_5 = (1,1,0) \)
### Grouping the Ones
To simplify the Sum of Products (SOP) expression, we can group the ones in the K-map:
1. **Group 1**: Combine the 1s in the first row (00, 00), (00, 01), (01, 00):
- **This can be simplified to**: \( \overline{A} \overline{B} \) (for C = 0 or C=1, thus \( \overline{C} + C\))
2. **Group 2**: Only the position \( (1,0) \) i.e., (01, 01) stays as is:
- **This can be simplified to**: \( \overline{A}B\overline{C} \)
### Final Expression
Combining both simplified groups,
The minimal SOP expression for \( Z \) is:
\[
Z = \overline{A} \overline{B} + A \overline{C}
\]
This should be your final expression as per your request!
### Karnaugh Map Setup
Let's assign variables for better readability:
- Let \(A\) be the row variable (top).
- Let \(B\) be the column variable (left).
- Let \(C\) be for the columns of the inner positions.
The arrangement provided in the format appears to be in the following form:
\[
\begin{array}{|c|c|c|c|c|}
\hline
AB \backslash C & 00 & 01 & 11 & 10 \\
\hline
00 & 1 & 1 & 1 & 0 \\
\hline
01 & 1 & 0 & 0 & 1 \\
\hline
\end{array}
\]
The respective values for \(A\) and \(B\) in binary in this arrangement are:
- **Row \(00\)** (i.e., \(A=0, B=0\)): Values of C are 00, 01, 11, 10, giving us \(1, 1, 1, 0\)
- **Row \(01\)** (i.e., \(A=0, B=1\)): Values of C are 00, 01, 11, 10, giving us \(1, 0, 0, 1\)
### Output from the Karnaugh Map
Based on your description, the K-Map values translate to:
- \( (A,B,C) = (0,0,0) \) : \( 1 \)
- \( (A,B,C) = (0,0,1) \) : \( 1 \)
- \( (A,B,C) = (0,1,0) \) : \( 1 \)
- \( (A,B,C) = (1,1,0) \) : \( 1 \)
- All other combinations yield \( 0 \).
Consolidating our positions gives us:
- \( m_0 = (0,0,0) \)
- \( m_1 = (0,0,1) \)
- \( m_2 = (0,1,0) \)
- \( m_5 = (1,1,0) \)
### Grouping the Ones
To simplify the Sum of Products (SOP) expression, we can group the ones in the K-map:
1. **Group 1**: Combine the 1s in the first row (00, 00), (00, 01), (01, 00):
- **This can be simplified to**: \( \overline{A} \overline{B} \) (for C = 0 or C=1, thus \( \overline{C} + C\))
2. **Group 2**: Only the position \( (1,0) \) i.e., (01, 01) stays as is:
- **This can be simplified to**: \( \overline{A}B\overline{C} \)
### Final Expression
Combining both simplified groups,
The minimal SOP expression for \( Z \) is:
\[
Z = \overline{A} \overline{B} + A \overline{C}
\]
This should be your final expression as per your request!