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Exercises
Exercises
Digital Fundamentals Exercise
Find the output of the following logic circuit if A = 1 and B = 0. The circuit consists of an AND gate followed by a NOT gate.
- AND gate output: 1⋅0=01 \cdot 0 = 0
- NOT gate output: 0‾=1\overline{0} = 1
Convert the decimal number 29 to binary.
- Divide the number by 2 and record the remainder:
- 29 ÷ 2 = 14, remainder 1
- 14 ÷ 2 = 7, remainder 0
- 7 ÷ 2 = 3, remainder 1
- 3 ÷ 2 = 1, remainder 1
- 1 ÷ 2 = 0, remainder 1
Binary: 11101
Convert the binary number 1101 to decimal.
Answer: 13 in decimal
Simplify the Boolean expression: 𝐴 ⋅ 𝐵 + 𝐴 ⋅ 𝐵 ′ A⋅B+A⋅B ′
Using the Distributive Law:
A⋅(B+B′)=A⋅1=AA \cdot (B + B’) = A \cdot 1 = A
In a 4-to-1 multiplexer, if S1 = 1, S0 = 0, and the input values are I0 = 0, I1 = 1, I2 = 0, I3 = 1, find the output.
The select lines (S1, S0) = (1, 0) select I2 as the output.
I2 = 0
Answer: 0
Add the BCD numbers 0101 (5) and 0110 (6).
- Binary sum: 0101 + 0110 = 1011 (invalid BCD)
- Add 0110 (6) for BCD correction:
1011 + 0110 = 0001 0001
Answer: 0001 0001 (11 in BCD)
What type of flip-flop is described by the following characteristic equation? 𝑄 𝑛 𝑒 𝑥 𝑡 = 𝐷 Q next =D
This is a D Flip-Flop, where the next state simply follows the input.
Find the 2's complement of 1101.
- Invert the digits: 1101 → 0010
- Add 1: 0010 + 1 = 0011
Answer: 0011
Write the truth table for a half adder.
| A | B | Sum | Carry |
|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
Simplify the following Boolean expression using a 2-variable K-Map: 𝑓 ( 𝐴 , 𝐵 ) = ∑ ( 1 , 2 ) f(A,B)=∑(1,2)
| A\B | 0 | 1 |
|---|---|---|
| 0 | 0 | 1 |
| 1 | 1 | 0 |
The simplified Boolean expression is:
f(A,B)=A⊕Bf(A, B) = A \oplus B
Answer: A XOR B
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