Given,
A = x (y′ + z′)
= xy′ + xz′
To express as a complete sum-of-products, expand using all variable combinations:
= xy′(z + z′) + xz′(y + y′)
= xy′z + xy′z′ + xyz′ + xy′z′
After simplification, duplicate terms are removed:
A = xyz′ + xy′z + xy′z′
UGC NET Computer Science Digital Logic Previous Year Questions (PYQs) – Page 1 of 2
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Showing 16 questions
The number of gate inputs, required to realize expression $ABC+A\overline{B}CD+E\overline{F}+AD$ is
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Solution
Given expression:
$ABC+A\overline{B}CD+E\overline{F}+AD$
Number of inputs in product terms:
$ABC=3$
$A\overline{B}CD=4$
$E\overline{F}=2$
$AD=2$
Total inputs for AND gates:
$3+4+2+2=11$
There are $4$ product terms, so OR gate needs $4$ inputs.
Total gate inputs:
$11+4=15$
Consider a logic gate circuit, with 8 input lines $(D_0,D_1,\ldots,D_7)$ and 3 output lines $(A_0,A_1,A_2)$ specified by following operations
$A_2=D_4+D_5+D_6+D_7$
$A_1=D_2+D_3+D_6+D_7$
$A_0=D_1+D_3+D_5+D_0$
Where $+$ indicates logical OR operation. This circuit is
$A_2=D_4+D_5+D_6+D_7$
$A_1=D_2+D_3+D_6+D_7$
$A_0=D_1+D_3+D_5+D_0$
Where $+$ indicates logical OR operation. This circuit is
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Solution
In an octal to binary encoder, there are $8$ input lines and $3$ output lines.
The outputs represent binary form of input lines $D_0$ to $D_7$.
Standard equations are:
$A_2=D_4+D_5+D_6+D_7$
$A_1=D_2+D_3+D_6+D_7$
$A_0=D_1+D_3+D_5+D_7$
So, the given circuit represents an Octal to Binary encoder.
The outputs represent binary form of input lines $D_0$ to $D_7$.
Standard equations are:
$A_2=D_4+D_5+D_6+D_7$
$A_1=D_2+D_3+D_6+D_7$
$A_0=D_1+D_3+D_5+D_7$
So, the given circuit represents an Octal to Binary encoder.
What will be the number of states when a $MOD-2$ counter is followed by a $MOD-5$ counter?
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Solution
When one counter is followed by another counter, the total number of states is the product of their mod values.
Here,
First counter is $MOD-2$
Second counter is $MOD-5$
So, total number of states is
$2 \times 5=10$
For line drawing, which of the following algorithms are used?
A. DDA algorithm
B. Bresenham algorithm
C. Mid-point algorithm
D. Curve algorithm
Choose the most appropriate answer from the options given below:
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Solution
Line drawing algorithms used in computer graphics are:
DDA algorithm
Bresenham algorithm
Mid-point algorithm is generally used for circle drawing.
Consider a $3$-bit counter, designed using $T$ flip flops as shown below.
Assume the initial state of the counter given by $PQR$ as $000$, what are the next three states?
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Solution
From the circuit,
$T_P=R$
$T_Q=P'$
$T_R=Q'$
For $T$ flip-flop:
If $T=0$, output remains same.
If $T=1$, output toggles.
Initial state:
$PQR=000$
Step 1:
$T_P=R=0,\ T_Q=P'=1,\ T_R=Q'=1$
So,
$PQR=011$
Step 2:
For $PQR=011$,
$T_P=R=1,\ T_Q=P'=1,\ T_R=Q'=0$
So,
$PQR=101$
Step 3:
For $PQR=101$,
$T_P=R=1,\ T_Q=P'=0,\ T_R=Q'=1$
So,
$PQR=000$
Therefore, next three states are:
$011, 101, 000$
Match List-I with List-II.
| List-I (computing systems) | List-II |
| (A) Half Adder | (I) Has $n$ input and $2^n$ output |
| (B) Decoder | (II) CPU Storage unit |
| (C) Register | (III) Used to store program at runtime |
| (D) Main memory | (IV) 2-bit addition circuit |
Choose the correct answer from the options given below :
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Solution
Half Adder → performs addition of two bits → 2-bit addition circuit → (IV)
Decoder → has $n$ inputs and $2^n$ outputs → (I)
Register → CPU internal storage → (II)
Main memory → stores program during execution → (III)
Thus
(A) → (IV)
(B) → (I)
(C) → (II)
(D) → (III)
Consider the following statements regarding combinational and sequential circuits.
(A) Output of combinational circuits depends on the only current input.
(B) Output of combinational circuit depends on the both current input and previous output.
(C) Output of sequential circuit depends on the current input.
(D) Output of sequential circuit depends on both current input and previous output.
Choose the correct answer from the options given below :
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Solution
Combinational circuit → output depends only on current input → (A) True
Sequential circuit → output depends current input + previous state → (D) True
(B) False
(C) Incomplete / incorrect
Find the Boolean expression for the logic circuit shown below:
$(1-\text{NAND gate},\ 2-\text{NOR gate},\ 3-\text{NOR gate})$
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Solution
Gate $1$ is a NAND gate with inputs $A$ and $B$.
So, output of gate $1$ is
$\overline{AB}$
Now, gate $2$ is a NOR gate.
Its inputs are $\overline{A}$ and $B$.
So, output of gate $2$ is
$\overline{\overline{A}+B}$
Using De Morgan's law,
$\overline{\overline{A}+B}=A\overline{B}$
Now, gate $3$ is also a NOR gate.
Its inputs are $\overline{AB}$ and $A\overline{B}$.
So,
$Y=\overline{\overline{AB}+A\overline{B}}$
Now,
$\overline{AB}=\overline{A}+\overline{B}$
Therefore,
$Y=\overline{\overline{A}+\overline{B}+A\overline{B}}$
Since $\overline{B}+A\overline{B}=\overline{B}$,
$Y=\overline{\overline{A}+\overline{B}}$
Using De Morgan's law,
$Y=AB$
Consider the Boolean expression A(x, y, z) = x (y′ + z′).
Which of the following is the complete sum-of-products form of the given Boolean expression?
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Solution
Out of following steps in the proper sequence for simplifying a Boolean function using a Karnaugh map (K-map).
(A) Identify and group the largest possible cluster of $1$'s
(B) Draw the K-map for the given Boolean function
(C) Write the simplified Boolean expression from the grouped clusters
(D) Transfer the truth table values to the K-map
Choose the correct answer from the options given below :
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Solution
Correct K-map procedure:
Step 1 → Draw the K-map
Step 2 → Fill truth table values
Step 3 → Group largest clusters of $1$'s
Step 4 → Write simplified expression
Sequence:
(B) → (D) → (A) → (C)
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