Communication Systems A. it reduces the bandwidth requirement to half A. mf B. 3mf
1. A cordless telephone using separate frequencies B. it avoids phase distortion at low frequencies C. mf/3 D. mf/9
for transmission in base and portable units is known
as C. it results in better reception Frequency multiplier increase the deviation,
A. duplex arrangement D. none of the above
.
B. half duplex arrangement VSB (vestigial side band) transmission transmits
one side band fully and the other side band 8. In colour TV receiver, varactor diode is used for
C. either (a) or (b) partially thus, reducing the bandwidth
requirement. A. detection B. rectification
D. neither (a) nor (b)
5. A woofer should be fed from the input through a C. tuning D. both (a) and (b)
Separate frequencies for transmission from base
and portable units allows two way transmission A. low pass filter B. high pass filter In varactor diode the applied reverse bias controls
and is called duplex arrangement. the width and therefore capacitance of depletion
C. band pass filter D. band stop filter layer.
2. For attenuation of high frequencies we should
use Woofer is a low frequency loud speaker covering This capacitance is used for tuning.
the range 16 Hz to 500 Hz.
A. shunt capacitance B. series capacitance 9. The number of noise sources in a BJT are
6. The colour subcarrier and sidebands produced by
C. inductance D. resistance its modulation with the chrominance signals are A. 3 B. 2 C. 1 D. 4
accommodated in the standard channel width by
Since Xc for high frequencies is low, the high the process of Shot noise, partition noise and thermal noise.
frequencies are shunted to ground and are not
transmitted. A. frequency adjustment 10. Energy content of atmospheric noise
3. A modem is classified as low speed if data rate B. frequency interleaving A. does not depend on frequency
handled is
C. frequency changing B. decreases as frequency is increased
A. upto 100 bps B. upto 250 bps
D. frequency amalgamation C. increases as frequency is increased
C. upto 400 bps D. upto 600 bps
In frequency interleaving the space between D. either (a) or (c) depending on the temperature
When data rate in bits per second is upto 600, bundles is used for colour signal.
modem is low speed. Atmospheric noise decreases as frequency is
7. In FM signal with a modulation index mf is increased.
4. VSB modulation is preferred in TV because passed through a frequency tripler. The wave in the
output of the tripler will have a modulation index of
, 11. A 400 W carrier is amplitude modulated with m 13. Non-coherently detection is not possible for C. Pulse Repetition Frequency
= 0.75. The total power in AM is
A. PSK B. ASK D. Pulse Response Factor
A. 400 W B. 512 W
C. FSK D. both (a) and (c) PRF in radar systems means pulse repetition
C. 588 W D. 650 W frequency.
14. Tracking of extra terrestrial objects requires
17. Which of the following cannot be the Fourier
A. high transmitting power series expansion of a periodic signal?
.
B. very sensitive receiver A. x(t) = 2 cos t + 3 cos 3t
12. c(t) and m(t) are used to generate an FM signal.
If the peak frequency deviation of the generated FM C. fully steerable antenna B. x(t) = 2 cos pit + 7 cos t
signal is three times the transmission bandwidth of
the AM signal, then the coefficient of the term cos
D. all of the above C. x(t) = cos t + 0.5
[2pi(1008 x 10^3t)] in the FM signal (in terms of the
Bessel Coefficients) is
15.Assertion (A): Free space does not interfere with D. x(t) = 2 cos 1.5pit + sin 3.5 pit
normal radiation and propagation of radio waves
A. 5 J4(3) B. J8(3)
x(t) = cos t + 0.5
Reason (R): Free space has no magnetic or
C. J8(4) D. 5 J4(6)
gravitational fields. not satisfies the Dirichlet condition.
As, we known FM, s(t)
A. Both A and R are correct and R is correct The integration of constant term is ∞.
explanation of A
18. Which of the following is the indirect way of FM
B. Both A and R are correct but R is not correct generation?
explanation of A
Frequency deviation = 3 . (AM Bandwidth) = 6 fm A. Reactance bipolar transistor modulator
C. A is correct but R is wrong
B. Armstrong modulator
D. A is wrong but R is correct
C. Varactor diode modulator
and β= 6 is given in (d) only. Since free space does not have magnetic and other
fields, waves can propagate without any D. Reactance FM modulator
ωc + n ωm = (1008 x 103)2p interference.
It generates FM through phase modulation.
2p x 106 + n. 4p x 103 = (1008 x 103).2p 16. In radar systems PRF stands for
19. A telephone exchange has 9000 subscribers. If
n=4 A. Power Return Factor the number of calls originating at peak time is 10,
000 in one hour, the calling rate is
So, β = 6, n = 4 hence answer is 5j4(6). B. Pulse Return Factor
1. A cordless telephone using separate frequencies B. it avoids phase distortion at low frequencies C. mf/3 D. mf/9
for transmission in base and portable units is known
as C. it results in better reception Frequency multiplier increase the deviation,
A. duplex arrangement D. none of the above
.
B. half duplex arrangement VSB (vestigial side band) transmission transmits
one side band fully and the other side band 8. In colour TV receiver, varactor diode is used for
C. either (a) or (b) partially thus, reducing the bandwidth
requirement. A. detection B. rectification
D. neither (a) nor (b)
5. A woofer should be fed from the input through a C. tuning D. both (a) and (b)
Separate frequencies for transmission from base
and portable units allows two way transmission A. low pass filter B. high pass filter In varactor diode the applied reverse bias controls
and is called duplex arrangement. the width and therefore capacitance of depletion
C. band pass filter D. band stop filter layer.
2. For attenuation of high frequencies we should
use Woofer is a low frequency loud speaker covering This capacitance is used for tuning.
the range 16 Hz to 500 Hz.
A. shunt capacitance B. series capacitance 9. The number of noise sources in a BJT are
6. The colour subcarrier and sidebands produced by
C. inductance D. resistance its modulation with the chrominance signals are A. 3 B. 2 C. 1 D. 4
accommodated in the standard channel width by
Since Xc for high frequencies is low, the high the process of Shot noise, partition noise and thermal noise.
frequencies are shunted to ground and are not
transmitted. A. frequency adjustment 10. Energy content of atmospheric noise
3. A modem is classified as low speed if data rate B. frequency interleaving A. does not depend on frequency
handled is
C. frequency changing B. decreases as frequency is increased
A. upto 100 bps B. upto 250 bps
D. frequency amalgamation C. increases as frequency is increased
C. upto 400 bps D. upto 600 bps
In frequency interleaving the space between D. either (a) or (c) depending on the temperature
When data rate in bits per second is upto 600, bundles is used for colour signal.
modem is low speed. Atmospheric noise decreases as frequency is
7. In FM signal with a modulation index mf is increased.
4. VSB modulation is preferred in TV because passed through a frequency tripler. The wave in the
output of the tripler will have a modulation index of
, 11. A 400 W carrier is amplitude modulated with m 13. Non-coherently detection is not possible for C. Pulse Repetition Frequency
= 0.75. The total power in AM is
A. PSK B. ASK D. Pulse Response Factor
A. 400 W B. 512 W
C. FSK D. both (a) and (c) PRF in radar systems means pulse repetition
C. 588 W D. 650 W frequency.
14. Tracking of extra terrestrial objects requires
17. Which of the following cannot be the Fourier
A. high transmitting power series expansion of a periodic signal?
.
B. very sensitive receiver A. x(t) = 2 cos t + 3 cos 3t
12. c(t) and m(t) are used to generate an FM signal.
If the peak frequency deviation of the generated FM C. fully steerable antenna B. x(t) = 2 cos pit + 7 cos t
signal is three times the transmission bandwidth of
the AM signal, then the coefficient of the term cos
D. all of the above C. x(t) = cos t + 0.5
[2pi(1008 x 10^3t)] in the FM signal (in terms of the
Bessel Coefficients) is
15.Assertion (A): Free space does not interfere with D. x(t) = 2 cos 1.5pit + sin 3.5 pit
normal radiation and propagation of radio waves
A. 5 J4(3) B. J8(3)
x(t) = cos t + 0.5
Reason (R): Free space has no magnetic or
C. J8(4) D. 5 J4(6)
gravitational fields. not satisfies the Dirichlet condition.
As, we known FM, s(t)
A. Both A and R are correct and R is correct The integration of constant term is ∞.
explanation of A
18. Which of the following is the indirect way of FM
B. Both A and R are correct but R is not correct generation?
explanation of A
Frequency deviation = 3 . (AM Bandwidth) = 6 fm A. Reactance bipolar transistor modulator
C. A is correct but R is wrong
B. Armstrong modulator
D. A is wrong but R is correct
C. Varactor diode modulator
and β= 6 is given in (d) only. Since free space does not have magnetic and other
fields, waves can propagate without any D. Reactance FM modulator
ωc + n ωm = (1008 x 103)2p interference.
It generates FM through phase modulation.
2p x 106 + n. 4p x 103 = (1008 x 103).2p 16. In radar systems PRF stands for
19. A telephone exchange has 9000 subscribers. If
n=4 A. Power Return Factor the number of calls originating at peak time is 10,
000 in one hour, the calling rate is
So, β = 6, n = 4 hence answer is 5j4(6). B. Pulse Return Factor
