Digital Circuits (Test 1)

Gate Exam : Ec Electronics And Communication Engineering

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Digital Circuits

Digital Circuits
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Q.1
In the following sequential circuit, the initial state (before the first clock pulse) of the circuit is Q1Q0 = 00. The state (Q1Q0), immediately after the 333rd clock pulse is
A. 00
B. 01
C. 10
D. 11
Answer : Option B
Explaination / Solution:



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Q.2
In the following circuit, X is given by
A.
B.
C. X = AB + BC + AC
D.
Answer : Option A
Explaination / Solution:



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Q.3
In an 8085 microprocessor, the shift registers which store the result of an addition and the overflow bit are, respectively
A. B and F
B. A and F
C. H and F
D. A and C
Answer : Option B
Explaination / Solution:

The shift registers A and F store the result of an addition and the overflow bit.

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Q.4
An 8085 assembly language program is given below. 
Line 1:   MVI A, B5H 
        2:   MVI B, OEH 
        3:   XRI 69H 
        4:   ADD B 
        5:   ANI 9BH 
        6:   CPI 9FH 
        7:   STA 3010H 
        8:   HLT
The contents of the accumulator just execution of the ADD instruction in line 4 will be
A. C3H
B. EAH
C. DCH
D. 69H
Answer : Option B
Explaination / Solution:

Line 1 : MVI A, B5H ; Move B5H to A
        2 : MVI B, 0EH ; Move 0EH to B
        3 : XRI 69H      ; [A] XOR 69H and store in A
                                 ; Contents of A is CDH
        4 : ADDB          ; Add the contents of A to contents of B and
                                 ; store in A, contents of A is EAH
        5 : ANI 9BH      ; [a] AND 9BH, and store in A,
                                 ; Contents of A is 8 AH
        6 : CPI 9FH      ; Compare 9FH with the contents of A
                                 ; Since 8 AH < 9BH, CY = 1
       7 : STA 3010 H  ; Store the contents of A to location 3010 H 
       8 : HLT               ; Stop
Thus the contents of accumulator after execution of ADD instruction is EAH. 


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Q.5
The Boolean expression  converted into the canonical product of sum (POS) form is
A.
B.
C.
D.
Answer : Option A
Explaination / Solution:

We have the SOP Boolean form,


Hence, in POS form, we have


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Q.6
The Boolean expression  simplifies to
A. X
B. Y
C. XY
D. X + Y
Answer : Option C
Explaination / Solution:

Given the Boolean expression

We simplify the expression as


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Q.7
Consider the Boolean function, .  Which one of the following is the complete set of essential prime implicants ?
A.
B. w, y, xz
C.
D.
Answer : Option D
Explaination / Solution:

k -map for the Boolean function can be given as

Therefore, the simplified So P from for the Boolean function is


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Q.8
The state transition diagram for the logic circuit shown is
A.
B.
C.
D.
Answer : Option D
Explaination / Solution:



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Q.9
For the circuit shown in the following, I0 - I3 are inputs to the 4:1 multiplexers, R(MSB) and S are control bits. 
The output Z can be represented by
A.
B.
C.
D.
Answer : Option A
Explaination / Solution:




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Q.10
A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles
A. and AND gate
B. an OR gate
C. an XOR gate
D. a NAND gate
Answer : Option C
Explaination / Solution:

Let A denotes the position of switch at ground floor and B denotes the position of switch at upper floor. The switch can be either in up position or down position.Following are the truth table given for different combinations of A and B

When the switches A and B are both up or both down, output will be zero (i.e.Bulb will be OFF). Any of the switch changes its position leads to the ON state of bulb. Hence, from the truth table, we get

i.e., the XOR gate


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