VIPSolutionsGive step-by-step solution with explanation and final answer:a) Write the truth table for the three-input XOR (exclusive OR) logic function with inputs , y, z. b) Using the Karnaugh map simplification method, derive the minimal expression of the three-input XOR logic function in Sum of Products (SOP) form and draw its logic circuit. ©) Draw the circuit of the three-input XOR logic function using the minimum number of two-input NOR (universal) logic gates.
Question:
Give step-by-step solution with explanation and final answer:
a) Write the truth table for the three-input XOR (exclusive OR) logic function with inputs , y, z.
b) Using the Karnaugh map simplification method, derive the minimal expression of the three-input XOR
logic function in Sum of Products (SOP) form and draw its logic circuit.
©) Draw the circuit of the three-input XOR logic function using the minimum number of two-input NOR
(universal) logic gates.
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Asked by: sri
Created at: 2025-05-24 17:33:57
Asked by: sri
Created at: 2025-05-24 17:33:57
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Let's solve each part step by step:
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### **a) Truth Table for 3-input XOR (x, y, z):**
The XOR function outputs 1 when an odd number of inputs are 1.
| x | y | z | F (XOR) |
|---|---|---|---------|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
---
### **b) Karnaugh Map and Minimal SOP Expression**
**Step 1:** Write minterms from the truth table (where F = 1):
- Minterms: \( m_1, m_2, m_4, m_7 \)
- Binary: 001, 010, 100, 111
**Step 2:** SOP Expression:
\[
F(x, y, z) = x'y'z + x'yz' + xy'z' + xyz
\]
**K-map:**
| | yz = 00 | yz = 01 | yz = 11 | yz = 10 |
|------|---------|---------|---------|---------|
| x=0 | 0 | 1 | 0 | 1 |
| x=1 | 1 | 0 | 1 | 0 |
No further grouping can be done, so the minimal SOP is as above.
**Logic Circuit:**
Each term is an AND gate, and they are all OR'd together.
---
### **c) Minimum Number of Two-input NOR Gates for 3-input XOR**
First, recall:
- XOR for two variables: \( A \oplus B = (A \bar{B}) + (\bar{A} B) \)
- For three variables: \( x \oplus y \oplus z = (x \oplus y) \oplus z \)
**Express XOR using only NOR gates:**
1. **NOT using NOR:**
\( \overline{A} = (A \downarrow A) \)
2. **OR using NOR:**
\( A + B = \overline{(\overline{A} \downarrow \overline{B})} \)
3. **AND using NOR:**
\( A \cdot B = \overline{(\overline{A} + \overline{B})} \)
**Implement 2-input XOR using NOR:**
\[
A \oplus B = (A \downarrow (A \downarrow B)) \downarrow (B \downarrow (A \downarrow B))
\]
**For 3-input XOR:**
\[
F = x \oplus y \oplus z = (x \oplus y) \oplus z
\]
- Implement \( x \oplus y \) using NORs as above.
- Then, use the result and z in the same XOR NOR implementation.
**Minimum Number of NOR Gates:**
- Each 2-input XOR requires 5 NOR gates.
- So, for 3-input XOR: \( 5 + 5 = 10 \) NOR gates.
---
### **Final Answers**
#### a) **Truth Table:**
| x | y | z | F (XOR) |
|---|---|---|---------|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
#### b) **Minimal SOP Expression:**
\[
F(x, y, z) = x'y'z + x'yz' + xy'z' + xyz
\]
#### c) **3-input XOR using Two-input NOR Gates:**
- Implemented as two cascaded 2-input XOR circuits, each requiring 5 NOR gates.
- **Minimum required: 10 NOR gates.**
---
> If you need the logic circuit diagrams, let me know and I can provide schematics!
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