Previous year question papers of Electrical Sub Engineer of MAHADISCOM

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1. Find the Fourier sine transform of f(x), where
f(x) = f(x) 1,0<x<a
{ 0, x>a
(a) √2/p (cos st / s)
(b) √2/p (1 - cos as / s) (Ans)
(c) √2 (1 - cos as)
(d) None of these
2. A random variable X with uniform density in the interval 0 to 1 is Quantized as follows:
If 0 ≤ X ≤ 0.3, xq = 0
If 0.3 ≤ X ≤ 1, xq = 0.7
Where Xq is the quatized value of x
The root mean square value of the quantization noise is
(a) 0.573
(b) 0.198 (ans)
(c) 2.205
(d) 0.266
Solution : Since it is uniform as
xq = 0 in the range 0≤x≤0.3
xq = 0.7 in the range 0.3≤x≤1
The square mean value is
¥
s2 = ∫ (x - xq)2 f (x) dx
-¥
1
= ∫ (x - xq)2 f (x) dx
0
0.3 0.1
= ∫ (x - 0)2 f (x) dx + ∫ (x - 0.7)2 f (x) dx
0 0.3
0.3 1
= [x3/3] + [x3/3 + 0.49 x - 1.4]
0 0.3
or s2 = 0.039
The root mean square quantization noise
RMS = √s2
= √0.039 = 0.198

3. Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 with the most appropriate item in Group 2.
Group 1 Group 2
1 : FM P : Slope overload
2 : DM Q : m Law
3 : PSK R : Envelope detector
4 : PCM S : Capture effect
T : Hilbert transform
U : Matched filter
(a) 1 - T, 2 - P, 3 - U, 4 - S
(b) 1 - S, 2U, 3 - P, 4 - T
(c) 1 - S, 2 - P, 3U, 4 - Q (ans)
(d) 1 - U, 2 - R, 3 - S, 4 - Q
Solution : FM --- Capture effect --- Receives only strong signal
DM ---- Slop over load Noise
PSK --- Matched filter
PCM - m law - Non linear quantization by using Companding with a law
V = log (1 + m |M|)
log (1 +m )

4. There analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is
(a) 115.2 kbps
(b) 28.8 kbps
(c) 57.6 kbps (ans)
(d) 38.4 kbps
Solution : The three analog Signals having BW 1200 Hz, 600Hz and 600 Hz are sampled at their respective Nyquist rate i.e. at 2400, 1200, 1200 sample/sec respectively.
The total of (2400 + 1200 + 1200) = 4800 sample/sec
The Bit rate = n. fs = (4800 sample/sec) x 12 = 57.6 Kbps
Where n = number of bit in a symbol
5. Find the correct match between group 1 and group 2.
Ground I
P - [1 + km (t)] A sin (wct)
Q - km (t) A sin (wct)
R - A sin [ w'c + k]'-¥ m (t) dt
S - A sin [wct + k '∫-¥ m (t) dt]
Solution :
Group II
W - Phase modulation
X - Frequency modulation
Y - Amplitude modulation
Z - DSB-SC modulation

P Q R S
(a) Z Y X W
(b) W X Y Z
(c) X W Z Y
(d) Y Z W X (ans)
Solution : The correct match is given below
[1 + km (t)] A sin (wct) Amplitude modulation
km (t) A sin (wct) DB-SC modulation
A sin [w'c + k]'-¥ m (t) dt Phase modulation
A sin [wct + k '∫-¥ m (t) dt] Frequency modulation

6. Which of the following analog modulation scheme requires the minimum transmitted power and minimum channel bandwidth?
(a) VSB
(b) DSB-SC
(c) SSB (ans)
(d) AM
Solution : VSB → fm + fc
DBS - SC → 2 fm
SSB → fm
AM → 2 fm
Thus SSB has minimum bandwidth and it required minimum power i.e. 17% as compared to AM.
7. A device with input x(t) and output y(t) is characteristic by : y (t) = x2(t). An FM signal with frequency deviation of 90 KHz and modulation signal bandwidth of 5 KHz is applied to this device. The bandwidth of the output signal is
(a) 370 KHz (ans)
(b) 190 KHz
(c) 380 KHz
(d) 95 KHz
Solution : In present case
∆f = 90; fm = 5
β = [∆f / fm] = [90/5] = 18
FM equation
A cos [wct + β = sin wmt]
= A cos [wct + 18 sin wmt]
y(t) = x2 (t) = A2 cos2 [wct + 18 Sin wmt]
Note : Cos2 q = [1 + Cos2q] / 2
If there is change in frequency the modulation index also changes in same ratio
y(t) = A2 [(1/2) + (1/2) Cos {2wct + 36Sin wmt}]
y(t) = [(A2/2) + (A2/2) Cos {2wct + 36Sin wmt}]
After the device,
β(new) = 36 = [∆f(new) / fm]
∆f(new) = 36 x 5 = 180
By carson's rule
Bandwidth = 2(∆f + fm)
= 2 (180 + 5)
Bandwidth = 370 kHz
9. A carrier is phase modulated (PM) with frequency deviation of 10 KHz by a single tone frequency of 1 KHz. If the single tone frequency is increased to 2 KHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is
(a) 21 kHz
(b) 22 kHz
(c) 42 kHz
(d) 44 kHz (ans)
Solution : ∆f = 10 KHz fm(new) = 2 KHz
fm = 1 KHz
By carson's Rule
BW = 2 (∆f + fm) = 2 (10 + 1) = 22 KHz
∆ f(new) = 2 x 10 = 20
BW(new) = 2 (20 + 2) = 44 kHz
10 If A and B be the set and Ac and Bc denote the complements of the sets. A and B, then set (A - B) È (B - A) È (A Ç B) is equal to
(a) A È B (Ans)
(b) Ac È Bc
(c) A Ç B
(d) Ac Ç Bc

11 Let G = G(V, E) has five vertices, then the maximum number of m of edges in E, if G is a multigraph ?
(a) 5
(b) 2
(c) 10
(d) Finite or infinite (Ans)
12 How many straight line can be drawn through 10 points on a circle ?
(a) 10
(b) 20
(c) 45 (Ans)
(d) Infinite
13 . The Fourier transform of unit step function u(t) is
(a) 1
(b) pd(w)
(c) pd(w) - 1/jw (Ans)
(d) pd(w) + 1/jw

14. The value of the integral ∫ e-2(x - t) d(t - 2) dt is
-∞
(a) e-2(x - 2) (Ans)
(b) e2(x - 2)
(c) e-2(x + 2)
(d) e2(x + 2)

15. The uint of Ñ x H is
(a) A
(b) A/m
(c) A/m2 (Ans)
(d) A-m

1. SCR is the solid state equivalent of
a. Transistor
b. Thyratron
c. Vacuum diode
d. Crystal diode

Answer: B
2. With gate open, if the supply voltage exceeds the breakover voltage of SCR, then SCR will conduct
a. False
b. True
c. For DC
d. For AC

Answer: B
3. The SCR is turned off when the anode current falls below
a. Forward current rating
b. Break over voltage
c. Holding current
d. Latching current

Answer: C
4. In a SCR circuit, the angle of conduction can be changed by changing
a. Anode voltage
b. Anode current
c. Forward current rating
d. Gate current

Answer: D
5. The normal way to close a SCR is by appropriate
a. Gate current
b. Cathode current
c. Anode current
d. Forward current

Answer: A
6. An SCR has ......PN junctions
a. Two
b. Four
c. Three
d. One

Answer: C
7. If gate current is increased, the anode-cathode voltage at which SCR closes is.....
a. Increased
b. Decreased
c. Maximum
d. Least

Answer: B
8. A conducting SCR can be opened by reducing ..... to zero
a. Supply voltage
b. Grid voltage
c. Grid current
d. Anode current

Answer: A
9. With gate open, a SCR can be turned-on by making supply voltage....
a. Minimum
b. Reverse
c. Equal to cathode voltage
d. Equal to breakover voltage

Answer: D
10. If firing angle in a SCR rectifier is increased, output is....
a. Increased
b. Maximum
c. Decreased
d. Unaffected

Answer: C
11. A SCR is a ...switch
A. Two directional
B. Unidirectional
C. Three directional
D. Four directional

Answer: B
12. The anode of SCR is always maintained at........potential with respect to cathode
A. Positive
B. Zero
C. negative
D. Equipotential

Answer: A
13. IF the chopper switching frequency is 200Hz and Ton time is 2ms, the duty cycle is
A. 0.4
B. 0.8
C. 0.6
D. None of the above

Answer: A
14. Speed control of induction motor can be effected by varying
A. Flux
B. Voltage input to stator
C. Keeping rotor coil open
D. None of the above

Answer: B
15. In a Dc motor if the field coil gets opened, the speed of the motor will
A. Decrease
B. Come to stop
C. Increase
D. None of the above

Answer: C