Previous Year Questions- Sequential Logic Circuits - 2 | Analog and Digital Electronics - Electrical Engineering (EE) PDF Download

Q16: The state transition diagram for the logic circuit shown is (2012)
(a) (b) (c)
(d) Ans: (d)
Sol: When A = 1, then Q will be selected by MUX and feedback to D flip-flop which gives output Q again.
So, at A = 1, it holds its state. When A = 0, then will be selected by MUX and feedback to D flip-flop and output will be inverted.
Hence, for A =1, it holds the state and for A = 0, it interchange the state i.e. if Q = 0 then it will go to Q = 1 and then it will go to Q = 0.
Therefore, option (D) is correct.

Q17: Consider the given circuit
In this circuit, the race around (2012)
(a) does not occur
(b) occur when CLK = 0
(c) occur when CLK = 1 and A = B = 1
(d) occur when CLK = 1 and A = B = 0
Ans: (a)
Sol: The above circuit is S-R flip-flop. The race around condition occurs in J-K flip-flop.

Q18: A two bit counter circuit is shown below
If the state Q_AQ_B of the counter at the clock time t_n is '10' then the state Q_AQ_B of the counter at t_n + 3 (after three clock cycles) will be (2011)
(a) 00
(b) 01
(c) 10
(d) 11
Ans: (c)
Sol: Q_AQ_B at t_n + 3 is '10'

Q19: In the figure, as long as X₁ = 1 and X₂= 1, the output Q remains  (2005)
(a) at 1
(b) at 0
(c) at its initial value
(d) unstable
Ans: (d)
Sol: As no combination of 'Q' with (X₁ and X₂) = 1, output is stable.
It always switches its stable state from '1' to '0' and from '0' to '1'.

Q20: Select the circuit, which will produce the given output Q for the input signals X₁ and X₂ given in the figure (2005)
(a) (b) (c) (d) Ans: (a)
Sol: From the above table we can write,

Q21: The digital circuit shown in the figure works as (2005)
(a) JK flip-flop
(b) Clocked RS flip-flop
(c) T flip-flop
(d) Ring counter
Ans: (c)
Sol: Truth table of the circuit,

Q22: The digital circuit shown in figure generates a modified clock pulse at the output. Choose the correct output waveform from the options given below. (2004)
(a) (b) (c) (d) Ans: (b)
Sol: 
Q23: A digit circuit which compares two numbers A₃A₂A₁A₀ and B₃B₂B₁B₀ is shown in figure. To get output Y = 0, choose one pair of correct input numbers. (2004)
(a) 1010, 1010
(b) 0101, 0101
(c) 0010, 0010
(d) 0010, 1011
Ans: (d)
Sol: For a 4-input X-NOR gate output will be zero if number of '1's will be odd.
We also knoew that output of XOR gate will be '1's if number of '1's will be odd.
If number of inputs will be same then output of XOR gate will be '0', so all input to XNOR will be zero. So, output Y will be '1'. Only in option (D) the inputs are different, so Y will be zero.

Q24: The digital circuit using two inverters shown in figure will act as (2004)
(a) a bistable multi-vibrator
(b) an astable multi-vibrator
(c) a monostable multi-vibrator
(d) an oscillator
Ans: (a)
Sol: For the both states (0, 1); our system is stable. Therefore, it is bistable multivibrator.

Q25: An X-Y flip-flop, whose Characteristic Table is given below is to be implemented using a J-K flip flop (2003)
(a)
(b)
(c)
(d)
Ans: (d)
Sol: To make (X-Y) FF using (J-K) FF, (J) should be  and (K) should be (X).

Q26: The shift register shown in figure is initially loaded with the bit pattern 1010. Subsequently the shift register is clocked, and with each clock pulse the pattern gets shifted by one bit position to the right. With each shift, the bit at the serial input is pushed to the left most position (msb). After how many clock pulses will the content of the shift register become 1010 again ? (2003)
 (a) 3
(b) 7
(c) 11
(d) 15
Ans: (b)

Q27: The frequency of the clock signal applied to the rising edge triggered D flip-flop shown in figure is 10 kHz. The frequency of the signal available at Q is (2002)
(a)10 kHz
(b) 2.5 kHz
(c) 20 kHz
(d) 5 kHz
Ans: (d)
Sol: In toggle mode