# Sequential Circuits

 Question 1
A state transition diagram with states A, B, and C, and transition probabilities $p_1,p_2,...,p_7$ is shown in the figure (e.g., $p_1$ denotes the probability of transition from state A to B). For this state diagram, select the statement(s) which is/are universally true

 A $p_2+p_3=p_5+p_6$ B $p_1+p_3=p_4+p_6$ C $p_1+p_4+p_7=1$ D $p_2+p_5+p_7=1$
GATE EC 2022   Digital Circuits
Question 1 Explanation:
$\left.\begin{matrix} A &\rightarrow &A &P_7 \\ A&\rightarrow &B &P_1 \\ A&\rightarrow &C &P_4 \end{matrix}\right\} \Rightarrow P_1+P_4+P_7=1$
$\left.\begin{matrix} B &\rightarrow &A &P_2 \\ B&\rightarrow &B &P_3 \end{matrix}\right\} \Rightarrow P_2+P_3=1$
$\left.\begin{matrix} C &\rightarrow &A &P_6 \\ C&\rightarrow &C &P_5 \end{matrix}\right\} \Rightarrow P_5+P_6=1$
$P_2+P_3=P_5+P_6\Rightarrow$ Option (A) is correct.
$P_1+P_4+P_7=1\Rightarrow$ Option (C) is correct.
 Question 2
For the circuit shown, the clock frequency is $f_o$ and the duty cycle is 25%. For the signal at the Q output of the Flip-Flop, _______.

 A frequency is $f_0/4$ and duty cycle is 50% B frequency is $f_0/4$ and duty cycle is 25% C frequency is $f_0/2$ and duty cycle is 50% D frequency is $f_0$ and duty cycle is 25%
GATE EC 2022   Digital Circuits
Question 2 Explanation:
2-bit counter
\begin{aligned} MSB&&LSB(J,K)\\ 0&&0\\ 0&&1\\ 1&&0\\ 1&&1 \end{aligned}

Duty cycle =50%
Output frequency =$f_0/4$
 Question 3
The propagation delay of the exclusive$-\text{OR} (\text{XOR})$ gate in the circuit in the figure is $3\:ns$. The propagation delay of all the flip-flops is assumed to be zero. The clock $(\text{Clk})$ frequency provided to the circuit is $500\: \text{MHz}$.

Starting from the initial value of the flip-flop outputs $Q_{2}Q_{1}Q_{0} = 1\;1\;1$ with $D_{2}=1$, the minimum number of triggering clock edges after which the flip-flop outputs $Q_{2}Q_{1}Q_{0}$ becomes $1\; 0\; 0$ (in integer) is
 A 3 B 4 C 5 D 6
GATE EC 2021   Digital Circuits
Question 3 Explanation:

$\therefore\quad$Total 5 clocks required.
 Question 4
The propagation delays of the XOR gate, AND gate and multiplexer (MUX) in the circuit shown in the figure are 4 ns, 2 ns and 1 ns, respectively.

If all the inputs P, Q, R, S and T are applied simultaneously and held constant, the maximum propagation delay of the circuit is
 A 3 ns B 5 ns C 6 ns D 7 ns
GATE EC 2021   Digital Circuits
Question 4 Explanation:
Case -1 : when T=0
Propogation delay $=t_{AND1}+t_{MUX2}=2+1=3ns$

Case -1 : when T=1
Propogation delay $=t_{AND2}+t_{MUX1}t_{AND3}+t_{MUX2}=2+1+2+1=6ns$
 Question 5
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.
 A 76.92 B 46.23 C 85.12 D 121.23
GATE EC 2020   Digital Circuits
Question 5 Explanation:
Total propagation delay $=(t_{pd}+t_{set-up})_{max}=8ns+5ns=13ns$
$\therefore$ Frequency of operations $=\frac{1000}{13}MHz=76.92MHz$
 Question 6
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
 A the sequence 01010 is detected B the sequence 01011 is detected C the sequence 01110 is detected D the sequence 01001 is detected
GATE EC 2020   Digital Circuits
 Question 7
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.
 A 2 B 4 C 6 D 8
GATE EC 2019   Digital Circuits
Question 7 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 8
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 _________.
 A 5 B 10 C 15 D 20
GATE EC 2017-SET-1   Digital Circuits
Question 8 Explanation:

So, 10 clock cycles are required.
 Question 9
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
 A X = '1', Y = '1' B either X = '1', Y = '0' or X = '0', Y = '1' C either X = '1', Y = '1' or X = '0', Y = '0' D X = '0', Y = '0'
GATE EC 2017-SET-1   Digital Circuits
Question 9 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 10
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___________.
 A 20 B 25 C 30 D 35
GATE EC 2017-SET-1   Digital Circuits
Question 10 Explanation:

Duty cycle of output
$=\frac{\frac{T_{\mathrm{CLK}}}{2}-\frac{T_{\mathrm{CLK}}}{5}}{T_{\mathrm{CK}}} \times 100=\frac{3}{10} \times 100=30 \%$
There are 10 questions to complete.