Question 1 |
A counter is constructed with three D flip-flops. The input-output pairs are named (D_{0},\:Q_{0}), (D_{1},\:Q_{1}) and (D_{2},\:Q_{2}), where the subscript 0 denotes the least significant bit. The output sequence is desired to be the Gray-code sequence 000, \:001, \:011,\: 010, \:110,\:111,\: 101 and 100, repeating periodically. Note that the bits are listed in the Q_{2}\:Q_{1}\:Q_{0} format. The combinational logic expression for D_{1} is
Q_{2}Q_{1}Q_{0} | |
Q_{2}Q_{0}+Q_{1}\overline{Q_{0}} | |
\overline{Q_{2}}Q_{0}+Q_{1}\overline{Q_{0}} | |
Q_{2}Q_{1}+\overline{Q_{2}}\:\overline{Q_{1}} |
Question 1 Explanation:
D_{1}=\bar{Q}_{2} Q_{0}+Q_{1} \bar{Q}_{0}
Question 2 |
Consider the following circuit which uses a 2-to-1 multiplexer as shown in the figure below. The Boolean expression for output F in terms of A and B is
A\oplus B | |
\overline{A+B} | |
A+B | |
\overline{A\oplus B} |
Question 2 Explanation:
In the given multiplexer,
I=0=\bar{A}, I_1=A
Select =B
F=\bar{B}I_0+BI_1
F=\bar{B}\bar{A}+AB=\overline{A\oplus B}
I=0=\bar{A}, I_1=A
Select =B
F=\bar{B}I_0+BI_1
F=\bar{B}\bar{A}+AB=\overline{A\oplus B}
Question 3 |
A Boolean function f(A,B,C,D)= \Pi(1,5,12,15) is to be implemented using an 8x1 multiplexer (A is MSB). The inputs ABC are connected to the select inputs S_2S_1S_0 of the multiplexer respectively.
Which one of the following options gives the correct inputs to pins 0,1,2,3,4,5,6,7 in order?
Which one of the following options gives the correct inputs to pins 0,1,2,3,4,5,6,7 in order?
D,0,D,0,0,0,\bar{D},D | |
\bar{D},1,\bar{D},1,1,1,D,\bar{D} | |
D,1,D,1,1,1,\bar{D},D | |
\bar{D},0,\bar{D},0,0,0,\bar{D},D |
Question 3 Explanation:
Boolean function f(A,B,C,D)=\Pi (1,5,12,15) is implemented with 8x1 MUX
Question 4 |
In the 4 x 1 multiplexer, the output F is given by F=A\oplus B. Find the required input 'I_{3}I_{2}I_{1}I_{0}'.
1010 | |
0110 | |
1000 | |
1110 |
Question 4 Explanation:
F=A\oplus B
\;\;=\bar{A}B+A\bar{B}
\;\;=\bar{A}\bar{B}I_0+\bar{A}BI_1+A\bar{B}I_2+ABI_3
\Rightarrow \; I_0=0, I_1=1, I_2=1, i_3=0
I_3I_2I_1I_0=0110
\;\;=\bar{A}B+A\bar{B}
\;\;=\bar{A}\bar{B}I_0+\bar{A}BI_1+A\bar{B}I_2+ABI_3
\Rightarrow \; I_0=0, I_1=1, I_2=1, i_3=0
I_3I_2I_1I_0=0110
Question 5 |
A 3-bit gray counter is used to control the output of the multiplexer as shown in
the figure. The initial state of the counter is 000_2. The output is pulled high. The
output of the circuit follows the sequence
I_{0}1,1,I_{1},I_{3},1,1,I_{2} | |
I_{0},1,I_{1},1,I_{2},1,I_{3},1 | |
1,I_{0},1,I_{1},I_{2},1,I_{3},1 | |
I_{0},I_{1},I_{2},I_{3},I_{0},I_{1},I_{2},I_{3} |
Question 5 Explanation:
For \begin{matrix} A_2 & A_1 & A_0 & S_0 & S_1\\ 0 & 0 & 0 & 0 & 0 \end{matrix}
MUX is enabled and output is I_0
For \begin{matrix} A_2 & A_1 & A_0 & S_0 & S_1\\ 0 & 0 & 1 & 0 & 0 \end{matrix}
MUX is disable and output is '1'
similarly, for
MUX is enabled and output is I_0
For \begin{matrix} A_2 & A_1 & A_0 & S_0 & S_1\\ 0 & 0 & 1 & 0 & 0 \end{matrix}
MUX is disable and output is '1'
similarly, for
There are 5 questions to complete.