C.1) A message signal
m(t)=cos(2000*pi*t)+4cos(4000*pi*t) modulates the carrier c(t)=cos(2*pi*fc*t)
where fc=1MHz to produce an AM signal. For demodulating the generated AM signal
using an envelope detector, the time constant RC of the detector circuit should
satisfy
(a) 0.5ms
< RC < 1ms
(b) 1μs <<
RC < 0.5ms
(c) RC
<< 1 μs
(d) RC
>> 0.5 ms
Answer is (b)
Explanation:
Desired condition for the time constant RC of the detector circuit is (1/fc) << TRC < (1/fm). Thus we have
(1/1M) << TRC < (1/2K) ; 2K from the signal 4cos(4000*pi*t) = 4cos(2*2000*pi*t)
=> 1us << TRC < 0.5ms
Explanation:
Desired condition for the time constant RC of the detector circuit is (1/fc) << TRC < (1/fm). Thus we have
(1/1M) << TRC < (1/2K) ; 2K from the signal 4cos(4000*pi*t) = 4cos(2*2000*pi*t)
=> 1us << TRC < 0.5ms
C.2) Consider an angle modulated signal x(t) = 6cos[2Π*106t +
2*sin(8000Πt) + 4*cos(8000Πt)] . The
average power of x(t) is
(a) 10W
(b) 18W
(c) 20W
(d) 28W
Answer is (b)
Solution:
Average power of the signal = (Ac)^2/2
= (6^2)/2
= 36/2
= 18 W
Solution:
Average power of the signal = (Ac)^2/2
= (6^2)/2
= 36/2
= 18 W
C.3) In delta modulation, the slope overload distortion
can be reduced by
(a) decreasing
step size
(b) decreasing
granular noise
(c) decreasing
sampling rate
(d)
increasing step size
Answer is (d)
C.4) If the number
of bits per sample in a Pulse Coded Modulation (PCM) system is increased from 5
bits to 6 bits,the improvement in signal to quantization noise ratio will be
(a) 3
dB
(b) 6 dB
(c) 2pi
dB
(d) 0 dB
Answer is (b)
Explanation:
SNRdb = 6.02n+1.76 ; where n is the number of bits
SNRdb(5) = 6.02 x 5 +1.76 = 31.86 dB
SNRdb(5) = 6.02 x 6 +1.76 = 37.88 dB
SNRdb(6) - SNRdb(5) = 37.88 - 31.86 = 6.02 dB
Thus we have 6db as the answer
Explanation:
SNRdb = 6.02n+1.76 ; where n is the number of bits
SNRdb(5) = 6.02 x 5 +1.76 = 31.86 dB
SNRdb(5) = 6.02 x 6 +1.76 = 37.88 dB
SNRdb(6) - SNRdb(5) = 37.88 - 31.86 = 6.02 dB
Thus we have 6db as the answer
C.5) For a message
signal m(t) = cos(2*pi*fm*t) and carrier of frequency fc, which of the
following represents a single side-band signal?
(a)
cos(2*pi*fm*t) cos(2*pi*fc*t)
(b)
cos(2*pi*fc*t)
(c)
cos[2*pi*(fm+fc)*t]
(d)
1+cos(2*pi*fm*t) cos(2*pi*fc*t)
Answer is (c)
C.6) If 'E' denotes Expectation, the variance of a random
variable X is given by:
(a) E[x^2]
- E^2[x]
(b) E[x^2]
+ E^2[x]
(c) E[x^2]
(d) E^2[x]
Answer is (a)
C.7) A device with input x(t) & output y(t) is
characterized by: y(t) =
x^2(t). An FM
signal with frequency deviation of 90kHz & modulating signal bandwidth of
5kHz is applied to this device. The
bandwidth of the output signal is
(a) 370
kHz
(b) 190
kHz
(c) 380
kHz
(d) 95 kHz
Answer is (a)
Concept : Carson's Rule
C.8) Four messages
band limited to W, W, 2W and 3W respectively are to be multiplexed using time
division multiplexing. The minimum bandwidth required for transmission of this
TDM signal is
(a) W
(b) 3W
(c) 6W
(d) 7W
Answer is (c)
Explanation:
Minimum bandwidth required for transmission = 2 x (maximum frequency of the message signal)
Thus, we have 6W = 2 x (3W)
Explanation:
Minimum bandwidth required for transmission = 2 x (maximum frequency of the message signal)
Thus, we have 6W = 2 x (3W)
C.9) An analog signal is band-limited to 4 kHz, sampled at Nyquist
rate & the samples are quantized into 4 levels. The
quantized levels are assumed to be independent & equally probable. If we
transmit two quantized samples per second, the information rate is:
(a) 1
bit/sec
(b) 2
bits/sec
(c) 3
bits/sec
(d) 4
bits/sec
Answer is (d)
C.10) A 1kHz
signal is sampled at the rate of 1.8kHz and the samples are applied to an ideal
rectangular LPF with a cut-off frequency of 1.1kHz, then the output of the
filter contains
(a) only
800 Hz component
(b) 800 Hz
and 900 Hz components
(c) 800 Hz
and 100 Hz components
(d) 800
Hz, 900Hz and 100 Hz components
Answer is (a)
Explanation
The frequency at which the aliased signal will be present is given by fm±kfs ;where fm and fs represent message and sampling frequencies respectively with k being an integer. In this example,
frequency at 1kHz will get aliased at -0.8(=1+1.8)kHz and 2.8(=1+1.8)kHz.
Similarly the signal at -1kHz will get aliased at 0.8(=-1+1.8)kHz and -2.8(=-1-1.8)kHz.
After passing through an ideal rectangular LPF with cutoff frequency 1.1kHz, only the component of 0.8kHz remains.
Explanation
The frequency at which the aliased signal will be present is given by fm±kfs ;where fm and fs represent message and sampling frequencies respectively with k being an integer. In this example,
frequency at 1kHz will get aliased at -0.8(=1+1.8)kHz and 2.8(=1+1.8)kHz.
Similarly the signal at -1kHz will get aliased at 0.8(=-1+1.8)kHz and -2.8(=-1-1.8)kHz.
After passing through an ideal rectangular LPF with cutoff frequency 1.1kHz, only the component of 0.8kHz remains.
C.11)Find the correct match between group 1 & group
2
✖- multiplication
Group 1
P.{1 +
km(t) A sin(Omega ✖t)}
Q. km(t) A
sin(Omega ✖t)
R. A
sin{Omega ✖t + k m(t)}
S. A
sin[Omegk ✖t + k ✖Integral (m(t))]
Group 2
W. Phase Modulation
X.
Frequency Modulation
Y. Amplitude Modulation
Z. DSB-SC Modulation
(a) P-Y , Q- Z , R- W, S-X
(b) P-Z , Q -Y, R - W, S- X
(c)P- X, Q-W,R-Y,S-Z
(d)none of above
Answer is (a)
C.12) Power in the signal s(t) =
8cos(20*pi-pi/2) + 4sin(15*pi*t) is
(a) 40 (b) 41
(c) 42 (d) 82
Answer is (a)
Solution:
Power in the signal = [(Ac1)^2+(Ac2)^2]/2
= (8^2+4^2)/2 = (64+16)/2
= 80/2 = 40
C.13) An angle
modulated signal is expressed by: F(t) = cos(2 ✖10^8 Πt + 75 sin 2✖10^3Πt). The peak frequency deviation of carrier is
then:
(a) 1 kHz
(b) 7.5 kHz
(c) 75 kHz
(d)100MHz
Answer is (c)
C.14) Which of the following analog
modulation scheme requires the minimum transmitted power and minimum channel
bandwidth?
(a) VSB
(b) SSB
(c) DSB-SC
(d) AM
(b) SSB
(c) DSB-SC
(d) AM
Answer is (b)
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