# Sequential Logic Circuits

 Question 1
A sequence detector is designed to detect precisely 3 digital inputs, with overlapping sequences detectable. For the sequence (1,0,1) and input data (1,1,0,1,0,0,1,1,0,1,0,1,1,0):

what is the output of this detector?
 A 1,1,0,0,0,0,1,1,0,1,0,0 B 0,1,0,0,0,0,0,1,0,1,0,0 C 0,1,0,0,0,0,0,1,0,1,1,0 D 0,1,0,0,0,0,0,0,1,0,0,0
GATE EE 2020   Digital Electronics
Question 1 Explanation:
Given overlapping sequence input data $= 11010011010110$
It will give output $1$ when $101$ is detected.
\begin{aligned} 110 - 0\\ 101 - 1\\ 010 - 0\\ 100 - 0\\ 001 - 0\\ 011 - 0\\ 110 - 0\\ 101 - 1\\ 010 - 0\\ 101 - 1\\ 011 - 0\\ 110 - 0\\ \end{aligned}
So, output $= 010000010100$
 Question 2
Which one of the following statements is true about the digital circuit shown in the figure
 A It can be used for dividing the input frequency by 3. B It can be used for dividing the input frequency by 5. C It can be used for dividing the input frequency by 7. D It cannot be reliably used as a frequency divider due to disjoint internal cycles.
GATE EE 2018   Digital Electronics
Question 2 Explanation:

So, frequency will be divided by 5.
 Question 3
For the synchronous sequential circuit shown below, the output Z is zero for the initial conditions $Q_{A}Q_{B}Q_{C}=Q_{A}'Q_{B}'Q'_{C}=100$

The minimum number of clock cycles after which the output Z would again become zero is ________
 A 4 B 5 C 6 D 7
GATE EE 2017-SET-2   Digital Electronics
Question 3 Explanation:

The output Z will again become zero after 6th clock cycle.
 Question 4
The current state $Q_{A}Q_{B}$ of a two JK flip-flop system is 00. Assume that the clock rise-time is much smaller than the delay of the JK flip-flop. The next state of the system is
 A 00 B 01 C 11 D 10
GATE EE 2016-SET-1   Digital Electronics
Question 4 Explanation:
From the figure we get
$J_A=K_A=1$
$J_B=K_B=\bar{Q_A}$

So, next state will be 11.
 Question 5
In the following sequential circuit, the initial state (before the first clock pulse) of the circuit is $Q_{1}Q_{0}$=00. The state ($Q_{1}Q_{0}$), immediately after the $333^{rd}$ clock pulse is
 A 00 B 01 C 10 D 11
GATE EE 2015-SET-2   Digital Electronics
Question 5 Explanation:
From the circuit, we can find out that
$J_0=Q'_1,\;K_0=Q_1$
$J_1=Q_0,\;K_1=Q'_0$

So state diagram is

at every 4th clock, the system is at 00
So, at 332 it will be at 00
So, at 333 clock it will be at 01.
 Question 6
The figure shows a digital circuit constructed using negative edge triggered J-K flip flops. Assume a starting state of $Q_{2}Q_{1}Q_{0}=000$. This state $Q_{2}Q_{1}Q_{0}=000$ will repeat after _______ number of cycles of the clock CLK
 A 4 B 5 C 6 D 7
GATE EE 2015-SET-1   Digital Electronics
Question 6 Explanation:
JK flip-flop1 and 2 from a syncgronous sequential circuits and they are synchronized with the output of 0th JK flip-flop.

Number of cycles = 3 i.e. equal to 6 clock cycles.
 Question 7
A state diagram of a logic gate which exhibits a delay in the output is shown in the figure, where X is the don't care condition, and Q is the output representing the state. The logic gate represented by the state diagram is
 A XOR B OR C AND D NAND
GATE EE 2014-SET-3   Digital Electronics
Question 7 Explanation:

If any one of the input is zero, output is 1 and for both inputs as '1' , output is '0', which represents the NAND gate.
 Question 8
A JK flip flop can be implemented by T flip-flops. Identify the correct implementation.
 A A B B C C D D
GATE EE 2014-SET-2   Digital Electronics
Question 8 Explanation:
To obtain a JK flip-flop from a T flip-flop, we first constuct the characteristic table of JK flip-flop, and then obtain the excitation values for the T flip-flop as shown below

Now, assuming T to be an output, we solve it in terms of J, K, $Q_n$ inputs. This gives the defination of the logic to be applied on the T input.
Also, observing the given options, we solve for T using a maxterms map instead of using a minterms map, as shown below

$T=(J+Q_n)\cdot (K+\bar{Q_n})$
The circuit corresponding to this expression is given in option (B).
 Question 9
A cascade of three identical modulo-5 counters has an overall modulus of
 A 5 B 25 C 125 D 625
GATE EE 2014-SET-1   Digital Electronics
Question 9 Explanation:
Overall modulus =5 x 5 x 5=125
 Question 10
The clock frequency applied to the digital circuit shown in the figure below is 1 kHz. If the initial state of the output Q of the flip-flop is '0', then the frequency of the output waveform Q in kHz is
 A 0.25 B 0.5 C 1 D 2
GATE EE 2013   Digital Electronics
Question 10 Explanation:
$x=\overline{(Q\oplus \bar{Q})\cdot (Q \odot \bar{Q})}$
$\;\;=\overline{1\cdot 0}=1$
$\because \;\; X=1=T \Rightarrow$ Q always toggle whenever clock triggers.

$\therefore \; f_Q=\frac{f_{clk}}{2}=\frac{1kHz}{2}=0.5kHz$
There are 10 questions to complete.