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Test: Number Systems, Boolean Algebra & Sequential Logic Circuits

20 Questions MCQ Test RRB JE for Electrical Engineering | Test: Number Systems, Boolean Algebra & Sequential Logic Circuits

Description

Attempt Test: Number Systems, Boolean Algebra & Sequential Logic Circuits | 20 questions in 60 minutes | Mock test for Railways preparation | Free important questions MCQ to study RRB JE for Electrical Engineering for Railways Exam | Download free PDF with solutions

QUESTION: 1

To perform product of maxterms Boolean function must be brought into

A.
AND terms
B.
OR terms
C.
NOT terms
D.
NAND terms

Solution:

QUESTION: 2

If (211)_x = (152)₈ , then the value of base x is

A.
6
B.
5
C.
7
D.
9

Solution:

In general, since we are comfortable dealing with Decimal system, the best way to go about under the circumstances is to convert one Number into decimal Number and then with the base x, convert another Number into Decimal Number and equate them to find the value of x.

So, (152)8 is 64*1+8*5+2, which is 106 in Decimal System.

(211)x is 2x^2+ 1*x+1 in Decimal System.

Now, we solve 2x^2+x+1 = 106

i. e. 2x^2+x-105=0

Therefore 2x^2–14x+15x-105=0

Therefore 2x(x-7)+15(x-7)=0

Therefore (x-7)(2x+15)=0

Therefore either x = 7 or x = -15/2.

Obviously x = -15/2 isnot the answer. So, x = 7 is the answer, which can be very easily verified. 2*7^2+1*7+1 = 106 which issame as (152)8.

Thus the base for 211 is 7.

QUESTION: 3

11001, 1001 and 111001 correspond to the 2’s complement representation of the following set of numbers

A.
25, 9 and 57 respectively
B.
-6, -6 and -6 respectively
C.
-7, -7 and -7 respectively
D.
-25, -9 and -57 respectively

Solution:

All are 2’s complement of 7

QUESTION: 4

A signed integer has been stored in a byte using 2’s complement format. We wish to store the same integer in 16-bit word. We should copy the original byte to the less significant byte of the word and fill the more significant byte with

A.
0
B.
1
C.
equal to the MSB of the original byte
D.
complement of the MSB of the original byte.

Solution:

See a example

QUESTION: 5

A computer has the following negative numbers stored in binary form as shown. The wrongly stored number is

A.
-37 as 1101 1011
B.
-89 as 1010 0111
C.
-48 as 1110 1000
D.
-32 as 1110 0000

Solution:

QUESTION: 6

The circuit shown in fig. is

A.
a MOD–2 counter
B.
a MOD–3 counter
C.
generate sequence 00, 10, 01, 00.....
D.
generate sequence 00, 10, 00, 10, 00 ......

Solution:

From table it is clear that it is a MOD–3 counter.

QUESTION: 7

The counter shown in fig. is

A.
MOD–8 up counter
B.
MOD–8 down counter
C.
MOD–6 up counter
D.
MOD–6 down counter

Solution:

It is a down counter because 0 state of previous FFs change the state of next FF. You may trace the following sequence, let initial state be000

QUESTION: 8

The counter shown in fig. counts from

A.
0 0 0 to 1 1 1
B.
1 1 1 to 0 0 0
C.
1 0 0 to 0 0 0
D.
0 0 0 to 1 0 0

Solution:

It is a down counter because the inverted FF output drive the clock inputs. The NAND gate will clear FFs A and B when the count tries to recycle to 111. This will produce as result of 100. Thus the counting sequence will be 100, 011, 010, 001, 000, 100 etc.

QUESTION: 9

The mod-number of the asynchronous counter shown in fig. is

A.
24
B.
48
C.
25
D.
36

Solution:

It is a 5 bit ripple counter. At 11000 the output of NAND gate is LOW. This will clear all FF. So it is a Mod 24 counter. Note that when 11000 occur, the CLR input is activated and all FF are immediately cleared. So it is a MOD 24 counter not MOD 25.

QUESTION: 10

The frequency of the pulse at z in the network shown in fig. is

A.
10 Hz
B.
160 Hz
C.
40 Hz
D.
5 Hz

Solution:

10-bit ring counter is a MOD–10, so it divides the 160 kHz input by 10. therefore, w = 16 kHz. The four-bit parallel counter is a MOD–16. Thus, the frequency at x = 1 kHz. The MOD–25 ripple counter produces a frequency at y = 40 Hz. (1 kHz/25 = 40 Hz).
The four-bit Johnson Counter is a MOD-8. This, the frequency at z = 5 Hz.

QUESTION: 11

The three-stage Johnson counter as shown in fig. is clocked at a constant frequency of f_c from the starting state of Q₂ Q₁Q₀ = 101. The frequency of output Q₂ Q₁Q₀ will be

A.
f_c/8
B.
f_c/6
C.
f_c/3
D.
f_c/2

Solution:

We see that 1 0 1 repeat after every two cycles, hence frequency will be f_c/2

QUESTION: 12

The counter shown in the fig. has initially Q₂Q₁Q₀ = 000. The status of Q₂ Q₁Q₀ after the first pulse is

A.
0 0 1
B.
0 1 0
C.
1 0 0
D.
1 0 1

Solution:

At first cycle

QUESTION: 13

A 4 bit ripple counter and a 4 bit synchronous counter are made by flips flops having a propagation delay of 10 ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S respectively, then

A.
R = 10 ns, S = 40 ns
B.
R = 40 ns, S = 10 ns
C.
R = 10 ns, S = 30 ns
D.
R = 30 ns, S = 10 ns

Solution:

In ripple counter delay 4T_d = 40 ns.

The synchronous counter are clocked simultaneously, then its worst delay will be equal to 10 ns.

QUESTION: 14

A 4 bit modulo–6 ripple counter uses JK flip-flop. If the propagation delay of each FF is 50 ns, the maximum clock frequency that can be used is equal to

A.
5 MHz
B.
10 MHz
C.
4 MHz
D.
20 Mhz

Solution:

4 bit uses 4 FF

QUESTION: 15

The initial contents of the 4-bit serial-in-parallel-out right-shift, register shown in fig. is 0 1 1 0. After three clock pulses are applied, the contents of the shift register will be

A.
0 0 0 0
B.
0 1 0 1
C.
1 1 1 1
D.
1 0 1 0

Solution:

QUESTION: 16

Consider the signed binary number A = 01010110 and B = 1110 1100 where B is the 1’s complement and MSB is the sign bit. In list-I operation is given, and in list-II resultant binary number is given.