Question 1 |
For the components in the sequential circuit shown below, t_{pd} is the propagation delay, t_{setup} is the setup time, and t_{hold} is the hold time. The maximum clock frequency (rounded
off to the nearest integer), at which the given circuit can operate reliably, is ____ MHz.
76.92 | |
46.23 | |
85.12 | |
121.23 |
Question 1 Explanation:
Total propagation delay =(t_{pd}+t_{set-up})_{max}=8ns+5ns=13ns
\therefore Frequency of operations =\frac{1000}{13}MHz=76.92MHz
\therefore Frequency of operations =\frac{1000}{13}MHz=76.92MHz
Question 2 |
The state diagram of a sequence detector is shown below. State S_0 is the initial state
of the sequence detector. If the output is 1, then
the sequence 01010 is detected | |
the sequence 01011 is detected | |
the sequence 01110 is detected | |
the sequence 01001 is detected |
Question 3 |
In the circuit shown, the clock frequency, i.e., the frequency of the Clk signal, is 12?kHz. The frequency of the signal at Q_2 is____ kHz.
2 | |
4 | |
6 | |
8 |
Question 3 Explanation:
\begin{aligned} M O D &=3 \\ f_{Q 2} &=\frac{f_{c \mid k}}{3}=\frac{12}{3} k H z=4 k H z \end{aligned}
Question 4 |
A 4-bit shift register circuit configured for right-shift operation, i.e, D_{in}\rightarrow A,\; A\rightarrow B , \; B\rightarrow C, \; C\rightarrow D, as shown. If the present state of the shift register is ABCD = 1101, the number of clock cycles required to reach the state ABCD = 1111 is _________.
5 | |
10 | |
15 | |
20 |
Question 4 Explanation:
So, 10 clock cycles are required.
Question 5 |
In the latch circuit shown, the NAND gates have non-zero, but unequal propagation delays. The present input condition is: P = Q = '0'. If the input condition is changed simultaneously to P = Q = '1', the outputs X and Y are
X = '1', Y = '1' | |
either X = '1', Y = '0' or X = '0', Y = '1' | |
either X = '1', Y = '1' or X = '0', Y = '0' | |
X = '0', Y = '0' |
Question 5 Explanation:
Present input condition : P=Q=0
\Rightarrow Corresponding outputs are X=Y=1
When input condition is changed to P=Q=1 from P=Q=0 :
Possibility - 1 :
Let gate-1 is faster than gate-2, then the possible outputs are X=0, Y=1
Possibility - 2 :
Let gate-2 is faster than gate-1, then the possible outputs are X=1, Y=0
Question 6 |
Consider the D-Latch shown in the figure, which is transparent when its clock input CK is high and has zero propagation delay. In the figure, the clock signal CLK1 has a 50% duty cycle and CLK2 is a one-fifth period delayed version of CLK1. The duty cycle at the output latch in percentage is___________.
20 | |
25 | |
30 | |
35 |
Question 6 Explanation:
Duty cycle of output
=\frac{\frac{T_{\mathrm{CLK}}}{2}-\frac{T_{\mathrm{CLK}}}{5}}{T_{\mathrm{CK}}} \times 100=\frac{3}{10} \times 100=30 \%
Question 7 |
For the circuit shown in the figure, the delay of the bubbled NAND gate is 2 ns and that of the counter is assumed to be zero.
If the clock (Clk) frequency is 1 GHz, then the counter behaves as a
If the clock (Clk) frequency is 1 GHz, then the counter behaves as a
mod-5 counter | |
mod-6 counter | |
mod-7 counter | |
mod-8 counter |
Question 7 Explanation:
Clock frequency= 1 GHz
\Rightarrow Clock time period = 1 ns
If the propagation delay of the NANO gate were 0 ns, the circuit would have behaved as MOD 6 counter.
However, the delay of NANO gate is 2 ns. During this time, two more clock pulses would reach the counter, and therefore it would count two more states.
Hence, it acts as MOD 8 counter.
\Rightarrow Clock time period = 1 ns
If the propagation delay of the NANO gate were 0 ns, the circuit would have behaved as MOD 6 counter.
However, the delay of NANO gate is 2 ns. During this time, two more clock pulses would reach the counter, and therefore it would count two more states.
Hence, it acts as MOD 8 counter.
Question 8 |
Assume that all the digital gates in the circuit shown in the figure are ideal, the resistor R= 10 k\Omega and the supply voltage is 5 V. The D flip-flops D1, D2, D3, D4 and D5 are initialized with logic values 0,1,0,1 and 0, respectively. The clock has a 30% duty cycle.
The average power dissipated (in mW) in the resistor R is ________
The average power dissipated (in mW) in the resistor R is ________
0.5 | |
1 | |
1.5 | |
2 |
Question 8 Explanation:
The waveform of the gate output
Average power dissipated
P=\frac{V^{2}}{R} \times \frac{T_{\mathrm{ON}}}{T}=\frac{5^{2}}{10 \mathrm{k}} \times \frac{3 T}{5 T}=1.5 \mathrm{mW}
Question 9 |
A three bit pseudo random number generator is shown. Initially the value of output
Y \equiv Y_{2} Y_{1} Y_{0} is set to 111. The value of output Y after three clock cycles is
000 | |
001 | |
010 | |
100 |
Question 9 Explanation:
The given circuit is similar to
After three clock Y is 100.
After three clock Y is 100.
Question 10 |
The circuit shown consists of J-K flip-flops, each with an active low asynchronous reset (\bar{R_{d}} input).The counter corresponding to this circuit is
(
a modulo-5 binary up counter | |
a modulo-6 binary down counter | |
a modulo-5 binary down counter | |
a modulo-6 binary up counter |
Question 10 Explanation:
Q is applied as negative edge triggering clock.
\Rightarrow Counter is up counter.
Reset signal =\overline{Q_{2} \cdot Q_{0}}
The counter resets at 101
\Rightarrow Counter is MOD-5
\Rightarrow Counter is up counter.
Reset signal =\overline{Q_{2} \cdot Q_{0}}
The counter resets at 101
\Rightarrow Counter is MOD-5
There are 10 questions to complete.