[AM modulation]

Prévia do material em texto

MultipleMultiple
ChoiceChoice
QuestionsQuestionsaand And Answnsweersrs
on Amplon Amplitudeitude
ModulationModulation
1) 1) Amplitude Amplitude modulation modulation isis  
a.a. Change in amplitude of the car Change in amplitude of the carrier accordingrier according
to modulating signalto modulating signal
b.b. Change in frequency of the car Change in frequency of the carrier accordingrier according
to modulating signalto modulating signal
c.c. Change in amplitude of the modulating signal Change in amplitude of the modulating signal
according to carrier signalaccording to carrier signal
d.d. Change in amplitude of the carrier  Change in amplitude of the carrier accordingaccording
to modulating signal frequencyto modulating signal frequency
ANSWER: (a) Change in amplitude of theANSWER: (a) Change in amplitude of the
carrier according to modulating signalcarrier according to modulating signal  
2) 2) The ability The ability of the of the receiver receiver to seleto select thect the
wanted signals among the wanted signals among the various incomingvarious incoming
signals is termed assignals is termed as  
a.a. Sensitivity Sensitivity
b.b. Selectivity Selectivity
c.c. Stability Stability
d.d. None of the above None of the above
ANSWER: ANSWER: (b) (b) SelectivitySelectivity  
3) 3) Emitter Emitter modulator modulator amplifier amplifier for for AmplitudeAmplitude
ModulationModulation  
a.a. Operates in class A mode Operates in class A mode
b.b. Has a low efficiency Has a low efficiency
c.c. Output power is small Output power is small
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
4) 4) Super Super heterodyne heterodyne receiversreceivers  
a.a. Have better sensitivity Have better sensitivity
b.b. Have high selectivity Have high selectivity
c.c. Need extra circuitry for  Need extra circuitry for frequency conversionfrequency conversion
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
5) 5) The The AM AM spectrum spectrum consists consists of of   
a.a. Carrier frequency Carrier frequency
b.b. Upper side band frequency Upper side band frequency
c.c. Lower side band frequency Lower side band frequency
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
6) 6) Standard Standard intermediate intermediate frequency frequency used used forfor
AM receiver isAM receiver is  
a.a. 455 MHz 455 MHz
b.b. 455 KHz 455 KHz
c.c. 455 Hz 455 Hz
d.d. None of the above None of the above
ANSWER: (b) 455 KHzANSWER: (b) 455 KHz  
7) 7) In the In the TV receivers, TV receivers, the device the device used forused for
tuning the receiver to the tuning the receiver to the incoming signal isincoming signal is  
a.a. Varactor diode Varactor diode
b.b. High pass Filter High pass Filter
c.c. Zener diode Zener diode
d.d. Low pass filter Low pass filter
ANSWER: (a) Varactor diodeANSWER: (a) Varactor diode  
8) 8) The mThe modulation odulation technique technique that uses that uses thethe
minimum channel bandwidth and transmittedminimum channel bandwidth and transmitted
power ispower is  
a.a. FM FM
b.b. DSB-SC DSB-SC
c.c. VSB VSB
d.d. SSB SSB
ANSWER: (d) SSBANSWER: (d) SSB  
  
9) 9) Calculate the Calculate the bandwidth bandwidth occupied occupied by a by a DSBDSB
signal when the modulating frequency lies insignal when the modulating frequency lies in
the range from 100 Hz to 10KHz.the range from 100 Hz to 10KHz.  
a.a. 28 KHz 28 KHz
b.b. 24.5 KHz 24.5 KHz
c.c. 38.6 KHz 38.6 KHz
d.d. 19.8 KHz 19.8 KHz
ANSWER: (d) 19.8 KHzANSWER: (d) 19.8 KHz  
10) 10) In AIn Amplitude mplitude Demodulation, Demodulation, the the conditioncondition
which the load resistor R must satisfy twhich the load resistor R must satisfy too
discharge capacitor C slowly between thedischarge capacitor C slowly between the
positive peaks of the carrier wave so that thepositive peaks of the carrier wave so that the
capacitor voltage will not discharge at thecapacitor voltage will not discharge at the
maximum rate of change of the maximum rate of change of the modulatingmodulating
wave (W is message bandwidth and ω iswave (W is message bandwidth and ω is
carrier frequency, in rad/sec) iscarrier frequency, in rad/sec) is  
a.a. RC 1/W RC > 1/W
c.c.  RC 1/ωRC > 1/ω  
ANSWER: (a) RC12 kHz OR
= 2(6 kHz) = 12 kHz= 2(6 kHz) = 12 kHz
17) 17) The total The total power power in an in an AmplitudeAmplitude
Modulated signal if the carrier of an AMModulated signal if the carrier of an AM
transmitter is 800 W and it is modulated 50transmitter is 800 W and it is modulated 50
percent.percent.  
a.a. 850 W 850 W
b.b. 1000.8 KW 1000.8 KW
c.c. 750 W 750 W
d.d. 900 W 900 W
ANSWER: (d) 900 WANSWER: (d) 900 W  
Explanation:Explanation:
The total power in an Amplitude ModulatedThe total power in an Amplitude Modulated
wave is given bywave is given by
PPTT = P = PCC (1+ m (1+ m
22
2)2)
Here, PHere, PCC = 800W, = 800W,
m = 0.5m = 0.5
therefore,therefore,
PPTT = 800 (1+ (0.5) = 800 (1+ (0.5)22/2) = 900 W/2) = 900 W
18) 18) An unAn unmodulated modulated AM signAM signal prodal produces auces a
current of 5.4 A. If the modulation is 100current of 5.4 A. If the modulation is 100
percent,percent,
calculate:calculate:  
(a) the carrier power,(a) the carrier power,
(b) the total power,(b) the total power,
(c) the sideband power when it (c) the sideband power when it is transmittedis transmitted
through an antenna having an impedance ofthrough an antenna having an impedance of
50Ω.50Ω.  
a.a. 1458 W, 2187.5 W, 729.25 W 1458 W, 2187.5 W, 729.25 W
b.b. 278 W, 2187.5 W, 1917.25 W 278 W, 2187.5 W, 1917.25 W
c.c. 1438 W, 2187.5 W, 759.25 W 1438 W, 2187.5 W, 759.25 W
d.d. 280 W, 2187.5 W, 750.25 W 280 W, 2187.5 W, 750.25 W
ANSWER: (a) 1458 W, 2187.5 W, 729.25 WANSWER: (a) 1458 W, 2187.5 W, 729.25 W  
Explanation:Explanation:
a) Pa) PCC=I=I
22
R = (5.4)R = (5.4)
22
*50 = 1458W*50 = 1458W
b) Ib) ITT = I = Icc√(1+m√(1+m22/2) = 5.4√(1+1/2) = 5.4√(1+122/2)/2)
=6.614 A=6.614 A
PPTT = I = ITT
22RR
= (6.614)= (6.614)22 * 50 * 50
= 2187.25 W= 2187.25 W
c) Pc) PSBSB = P = PTT   – – P PCC  
= 2187.25= 2187.25 – – 1458 W 1458 W
= 729.25W (for two bands)= 729.25W (for two bands)
For single band, PFor single band, PSBSB = 729.25/2 = 729.25/2
= 364. 625 W= 364. 625 W
19) 19) Calculate the Calculate the depth odepth of modf modulation when ulation when aa
transmitter radiates a signal of 9.8KW aftertransmitter radiates a signal of 9.8KW after
modulation and 8KW without modulation ofmodulation and 8KW without modulation of
the signal.the signal.  
a.a. 80% 80%
b.b. 67% 67%
c.c. 50% 50%
d.d. 100% 100%
ANSWER: (b) 67%ANSWER: (b) 67%  
Explanation:Explanation:
PPtotaltotal = 9.8KW= 9.8KW
PPcc = 8KW = 8KW
Power of the signal (PPower of the signal (Ptotaltotal) transmitted by a) transmitted by a
transmitter after modulation is given bytransmitter after modulation is given by
PPtotaltotal = P = Pcc (1+ m (1+ m22/2)/2)
Where PWhere Pcc is the power of carrier is the power of carrier i.e., withouti.e., without
modulationmodulation
M is the modulation indexM is the modulation index
Therefore,Therefore,
9.8= 8 (1+ m9.8= 8 (1+ m22/2)/2)
9.8/8=1+ m9.8/8=1+ m22/2/2
m=0.67 = 67%m=0.67 = 67%
20) 20) When AM When AM signal is signal is of 25KHof 25KHz, calculate z, calculate thethe
number of channels required in Mediumnumber of channels required in Medium
Frequency (MF) band of 300KHz-3000KHz.Frequency (MF) band of 300KHz-3000KHz.  
  
a.a. 94 94
b.b. 69 69
c.c. 85 85
d.d. 54 54
ANSWER: (d) 54ANSWER: (d) 54  
Explanation:Explanation:
Medium Frequency (MF) is the band ofMedium Frequency (MF) is the band of
frequencies from 300 KHz to frequencies from 300 KHz to 3MHz. The lower3MHz. The lower
portion of the MF band (300to 500 portion of the MF band (300to 500 kilohertz) iskilohertz) is
used for ground-wave transmission forused for ground-wave transmission for
reasonably long distances. The upper and lowerreasonably long distances. The upper and lower
ends of the mf band are ends of the mf band are used for naval purpose.used for naval purpose.
Frequency available in MF band= 3000Frequency available in MF band= 3000 – – 300 = 300 =
2700 KHz2700 KHz
The bandwidth required by 25 KHz signal = 2 The bandwidth required by 25 KHz signal = 2 **
25= 50 KHz25= 50 KHz
Therefore the number of channels available =Therefore the number of channels available =
2700/ 50 = 542700/ 50 = 54
21) 21) Calculate the Calculate the power power in onin one of e of the sidethe side
band in SSBSC modulation when the carrierband in SSBSC modulation when the carrier
power is 124W and there is 80% modulationpower is 124W and there is 80% modulation
depth in the amplitude modulated signal.depth in the amplitude modulated signal.  
a.a. 89.33 W 89.33 W
b.b. 64.85 W 64.85 W
c.c. 79.36 W 79.36 W
d.d. 102 W 102 W
ANSWER: (c) 79.36 WANSWER: (c) 79.36 W  
Explanation:Explanation:
Modulation Index = 0.8Modulation Index = 0.8
PPcc = 124W= 124W
Power in sidebands may be calculated as =Power in sidebands may be calculated as =
mm22 P Pcc/4/4
= (0.8)= (0.8)22 * 124/4 * 124/4
= 79.36 W= 79.36 W
22) 22) Calculate Calculate the the total total modulation modulation IndexIndex
when a carrier wave when a carrier wave is being modulated byis being modulated by
two modulating signals with modulationtwo modulating signals with modulation
indices 0.8 and 0.3.indices 0.8 and 0.3.  
a.a. 0.8544 0.8544
b.b. 0.6788 0.6788
c.c. 0.9999 0.9999
d.d. 0.5545 0.5545
ANSWER: (a) 0.8544ANSWER: (a) 0.8544  
Explanation:Explanation:
Here, mHere, m11 = 0.8 = 0.8
mm22 = 0.3 = 0.3
total modulation index mtotal modulation index mtt  = √( m= √( m11
22 + m + m22
22 ) )
= √( 0.8= √( 0.822 + 0.3 + 0.322 ) )
= √ 0.73= √ 0.73  
= 0.8544= 0.8544
23) 23) Calculate the Calculate the frequencies frequencies available available in in thethe
frequency spectrum when a 2MHz carrier isfrequency spectrum when a 2MHz carrier is
modulated by two sinusoidal signals of 350Hzmodulated by two sinusoidal signals of 350Hz
and 600Hz.and 600Hz.  
a.a. 2000.35, 1999.65 and 2000.6, 1999.4 2000.35, 1999.65 and 2000.6, 1999.4
b.b. 1999.35, 1999.65 and 2000.6, 2000.4 1999.35, 1999.65 and 2000.6, 2000.4
c.c. 2000.35, 2000.65 and 2000.6, 2000.4 2000.35, 2000.65 and 2000.6, 2000.4
d.d. 1999.35, 1999.65 and 1999.6, 1999.4 1999.35, 1999.65 and 1999.6, 1999.4
ANSWER: (a) 2000.35, 1999.65 and 2000.6,ANSWER: (a) 2000.35, 1999.65 and 2000.6,
1999.41999.4  
Explanation:Explanation:
The frequencies obtained in the spectrum afterThe frequencies obtained in the spectrum after
the amplitude modulation arethe amplitude modulation are
f f cc + f  + f mm and f  and f cc + f  + f mm  
therefore,therefore,
the available frequencies after modulation bythe available frequencies after modulation by
0.350 KHz are0.350 KHz are
2000KHz + 0.350 KHz = 2000.35 2000KHz + 0.350 KHz = 2000.35 and 2000KHzand 2000KHz – –  
0.350 KHz = 1999.650.350 KHz = 1999.65
the available frequencies after modulation bythe available frequencies after modulation by
0.6 KHz are0.6 KHz are
2000KHz + 0 .6 KHz = 2000.6 and 2000KHz2000KHz + 0 .6 KHz = 2000.6 and 2000KHz – – 0.6 0.6
KHz = 1999.4KHz = 1999.4
24) 24) If an If an AM sAM signal is ignal is represented represented byby
v = ( 15 + 3 v = ( 15 + 3 Sin( 2Π * 5 * 10Sin( 2Π * 5 * 1033  t) ) * Sin( 2t) ) * Sin( 2Π * 0.5 *Π * 0.5 *
101066 t) volts t) volts  
  
i) Calculate the values of the frequencies ofi) Calculate the values of the frequencies of
carrier and modulating signals.carrier and modulating signals.
ii) Calculate the value of modulation index.ii) Calculate the value of modulation index.
iii) Calculate the value of bandwidth of thisiii) Calculate the value of bandwidth of this
signal.signal.
a.a. 1.6 MHz and 8 KHz, 0 1.6 MHz and 8 KHz, 0.6, 16 MHz.6, 16 MHz
b.b. 1.9 MHz and 18 KHz, 0.2,  1.9 MHz and 18 KHz, 0.2, 16 KHz16 KHz
c.c. 2.4 MHz and 18 KHz, 0.2, 16 KHz 2.4 MHz and 18 KHz, 0.2, 16 KHz
d.d. 1.6 MHz and 8 KHz, 0.2, 16 KHz 1.6 MHz and 8 KHz, 0.2, 16 KHz
ANSWER: (d) 1.6 MHz and 8 KHz, 0.2, 16 KHzANSWER: (d) 1.6 MHz and 8 KHz, 0.2, 16 KHz  
Explanation:Explanation:
The amplitude modulated wave equation isThe amplitude modulated wave equation is
v = ( 10 + 2 v = (10 + 2 Sin( 2Π * 8 * 10Sin( 2Π * 8 * 1033  t) ) * Sin (2Π * 1.6 t) ) * Sin (2Π * 1.6 **
101066 t) volts t) volts
Instantaneous value of AM signal is representedInstantaneous value of AM signal is represented
by the equationby the equation
v = {Vv = {Vcc + V + Vmm  Sin ( ωSin ( ωmm  t )} * Sin (ωt )} * Sin (ωcc t ) t )
comparing it with the given equation,comparing it with the given equation,
VVcc = 10 V = 10 V
VVmm = 2V = 2V
ωωcc  (= 2Π f (= 2Π f cc) = 2Π * 1.6 * 10) = 2Π * 1.6 * 1066  
ωωmm  (= 2Π f (= 2Π f mm) = 2Π * 8 * 10) = 2Π * 8 * 1033  
(i) The carrier frequency f (i) The carrier frequency f cc is = 1.6 * 10 is = 1.6 * 1066 = 1.6 = 1.6
MHzMHz
The modulating frequency f The modulating frequency f mm is = 8* 10 is = 8* 1033 = 8 kHz = 8 kHz
(ii) The modulation index m = V(ii) The modulation index m = Vmm/V/Vcc = 2/10 = 0.2 = 2/10 = 0.2
(iii) The bandwidth BW = 2 f (iii) The bandwidth BW = 2 f mm = 16 kHz = 16 kHz
25) 25) An AM An AM signal has signal has a total a total power of power of 4848
Watts with 45% modulation. Calculate theWatts with 45% modulation. Calculate the
power in the carrier power in the carrier and the sidebands.and the sidebands.  
a.a. 39.59 W, 4.505 W 39.59 W, 4.505 W
b.b. 40.59 W, 4.205 W 40.59 W, 4.205 W
c.c. 43.59 W, 2.205 W 43.59 W, 2.205 W
d.d. 31.59 W, 8.205 W 31.59 W, 8.205 W
ANSWER: (c) 43.59 W, 2.205 WANSWER: (c) 43.59 W, 2.205 W  
Explanation:Explanation:
Given that PGiven that Ptt = 48 W = 48 W
Modulation index m = 0.45Modulation index m = 0.45
The total power in an AM is giveThe total power in an AM is given byn by
PPtt = P = Pcc ( 1 + m( 1 + m22/2)/2)
= P= Pcc ( 1 +0.45 ( 1 +0.4522/2)/2)
48 = P48 = Pcc * 1.10125 * 1.10125
Therefore, PTherefore, Pcc = 48/ 1.10125 = 48/ 1.10125
= 43.59 W= 43.59 W
The total power in two sidebands is 4843.59 =The total power in two sidebands is 4843.59 =
4.41 W4.41 W
So the power in each So the power in each sideband is 4.41/2 = 2.205sideband is 4.41/2 = 2.205
WW
26) 26) Calculate the Calculate the power power saved saved in anin an
Amplitude Modulated wave when it isAmplitude Modulated wave when it is
transmitted with 45% modulationtransmitted with 45% modulation
 – – Without carrier Without carrier
 – – Without carrier and a sideband Without carrier and a sideband  
a.a. 90%, 95% 90%, 95%
b.b. 82%, 91% 82%, 91%
c.c. 82%, 18% 82%, 18%
d.d. 68%, 16% 68%, 16%
ANSWER: (a) 90%, 95%ANSWER: (a) 90%, 95%  
Explanation:Explanation:
i) The total power in an AM is i) The total power in an AM is given bygiven by
PPtt = P = Pcc ( 1 + m ( 1 + m22/2)/2)
Given: m = 0.45Given: m = 0.45
Therefore PTherefore Ptt = P = Pcc ( 1 + 0.45 ( 1 + 0.4522/2 )/2 )
PPtt= P= Pcc *1.10125 *1.10125
PPcc/ P/ Ptt = 1/1.10125 = 1/1.10125
= 0.908= 0.908
= 90%= 90%
This shows that the carrier occupies 90% ofThis shows that the carrier occupies 90% of
total power. So 90% of total power total power. So 90% of total power may bemay be
saved if carrier is suppressed in the saved if carrier is suppressed in the AM signal.AM signal.
(ii) If one of the sidebands is also suppressed,(ii) If one of the sidebands is also suppressed,
half of the remaining power will be savedhalf of the remaining power will be saved
i.e., 10/2 = 5 %. So a total of 95% (90% + 5% )i.e., 10/2 = 5 %. So a total of 95% (90% + 5% )
will be saved when carrier and a will be saved when carrier and a side band areside band are
suppressed.suppressed.
  
27) 27) What What is the is the carrier frecarrier frequency quency in an in an AMAM
wave when its highest frequency component iswave when its highest frequency component is
850Hz and the bandwidth of the 850Hz and the bandwidth of the signal is 50Hz?signal is 50Hz?  
a.a. 80 Hz 80 Hz
b.b. 695 Hz 695 Hz
c.c. 625 Hz 625 Hz
d.d. 825 Hz 825 Hz
ANSWER: (d) 825 HzANSWER: (d) 825 Hz  
Explanation:Explanation:
Upper frequency = 850HzUpper frequency = 850Hz
Bandwidth = 50HzBandwidth = 50Hz
Therefore lower Frequency = 850Therefore lower Frequency = 850 – – 50= 800 Hz 50= 800 Hz
Carrier Frequency = (850-800)/2Carrier Frequency = (850-800)/2
= 825 Hz= 825 Hz
28) 28) Noise Noise figure of figure of merit merit in SSin SSB modB modulatedulated
signal issignal is  
a.a. 1 1
b.b. Less than 1 Less than 1
c.c. Greater than 1 Greater than 1
d.d. None of the above None of the above
ANSWER: (a) 1ANSWER: (a) 1  
29) 29) For For low lelow level modvel modulation, ulation, amplifier amplifier usedused
isis  
a.a. Class A Class A
b.b. Class C Class C
c.c. Class A & C Class A & C
d.d. None of the above None of the above
ANSWER: (a) Class AANSWER: (a) Class A  
30) 30) The antennThe antenna current a current of the of the transmitter transmitter isis
10A. Find the percentage of modulation when10A. Find the percentage of modulation when
the antenna current increases to 10.4A.the antenna current increases to 10.4A.  
a.a. 32% 32%
b.b. 28.5% 28.5%
c.c. 64% 64%
d.d. 40% 40%
ANSWER:(b) 28.5%ANSWER:(b) 28.5%  
Explanation:Explanation:
IItt = I = Icc  √(1+ m√(1+ m22/2)/2)
10.4= 10 √(1+ m10.4= 10 √(1+ m22/2)/2)
√ (1+ m√ (1+ m22/2) = 1.04/2) = 1.04
Therefore m = 0.285Therefore m = 0.285
= 28.5%= 28.5%
31) 31) What is What is the change the change in the in the value ofvalue of
transmitted power when the modulation indextransmitted power when the modulation index
changes from 0 to 1?changes from 0 to 1?  
a.a. 100% 100%
b.b. Remains unchanged Remains unchanged
c.c. 50% 50%
d.d. 80% 80%
ANSWER: (c) 50%ANSWER: (c) 50%  
Explanation:Explanation:
PPtt = P = Pcc ( 1 + m ( 1 + m22/2)/2)
PPtt= P= Pcc ( 1 + 0 ( 1 + 022/2) = P/2) = Pcc ..(1) ..(1)
New total power PNew total power Pt1t1= P= Pcc ( 1 + 1 ( 1 + 122/2)/2)
= P= Pcc *3/2 ..(2)*3/2 ..(2)
(2) / (1),(2) / (1),
We get , PWe get , Pt1t1/ P/ Ptt= 3/2= 1.5= 3/2= 1.5
PPt1t1= 1.5 P= 1.5 Ptt  
i.e. there is increase in total power i.e. there is increase in total power by 50%by 50%
32) 32) Function Function of of RF RF mixer mixer isis  
a.a. Addition of two signals Addition of two signals
b.b. Multiplication of two signals Multiplication of two signals
c.c. Rejection of noise Rejection of noise
d.d. None of the above None of the above
ANSWER: (b) Multiplication of two signalsANSWER: (b) Multiplication of two signals  
33) 33) If a If a receiver has receiver has poor cappoor capacity of acity of blockingblocking
adjacent channel interference then theadjacent channel interference then the
receiver hasreceiver has  
a.a. Poor selectivity Poor selectivity
b.b. Poor Signal to noise ratio Poor Signal to noise ratio
c.c. Poor sensitivity Poor sensitivity
d.d. None of the above None of the above
ANSWER: (a) Poor selectivityANSWER: (a) Poor selectivity  
  
34) 34) Advantage Advantage of uof using a sing a high freqhigh frequencyuency
carrier wave iscarrier wave is  
a.a. Signal can be transmitted over very long Signal can be transmitted over very long
distancesdistances
b.b. Dissipates very small power Dissipates very small power
c.c. Antenna height of the transmitter  Antenna height of the transmitter is reducedis reduced
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
35) 35) Advantage Advantage of uof using Vsing VSB transSB transmission mission isis  
a.a. Higher bandwidth than SSB Higher bandwidth than SSB
b.b. Less power required as compared to DSBSC Less power required as compared to DSBSC
c.c. Both a and b Both a and b
d.d. None of the above None of the above
ANSWER: (c) Both a and bANSWER: (c) Both a and b  
36) 36) Modulation Modulation is is required required forfor  
a.a. Reducing noise while transmission Reducing noise while transmission
b.b. Multiplexing the signals Multiplexing the signals
c.c. Reduction of Antenna height Reduction of Antenna height
d.d. Reduction in the complexity of circuitry Reduction in the complexity of circuitry
e.e. All of the above All of the above
ANSWER: (e) All of the ANSWER: (e) All of the aboveabove  
37)37) Bandwidth Bandwidth required required in Sin SSB-SC siSB-SC signal isgnal is
(f (f mm is modulating frequency): is modulating frequency):  
a.a. 2f  2f mm  
b.b.  2f  > 2f mm  
d.d. f  f mm  
ANSWER: (d) f ANSWER: (d) f mm  
Explanation:Explanation:
In an amplitude modulated wave, totalIn an amplitude modulated wave, total
bandwidth required is from f bandwidth required is from f cc + f  + f mm to f  to f cc   – – f  f mm  
i.e. BW = 2f i.e. BW = 2f mm where f  where f cc is carrier frequency. is carrier frequency.
In SSB-SC transmission, as the carrier and one ofIn SSB-SC transmission, as the carrier and one of
the sidebands are suppressed, the bandwidththe sidebands are suppressed, the bandwidth
remains as f remains as f mm..
38) 38) For For over over modulation, modulation, the the value ofvalue of
modulation index m ismodulation index m is  
a.a. m 1 m > 1
d.d. Not predetermined Not predetermined
ANSWER: (c) m > 1ANSWER: (c) m > 1  
39) 39) Demodulation Demodulation is:is:  
a.a. Detection Detection
b.b. Recovering information from modulated Recovering information from modulated
signalsignal
c.c. Both a and b Both a and b
d.d. None of the above None of the above
ANSWER: (c)Both a and bANSWER: (c)Both a and b  
40) 40) Calculate the Calculate the side band side band power in power in an SSBSan SSBSCC
signal when there is 50% modulation and thesignal when there is 50% modulation and the
carrier power is 50W.carrier power is 50W.  
a.a. 50 W 50 W
b.b. 25 W 25 W
c.c. 6.25 W 6.25 W
d.d. 12.5 W 12.5 W
ANSWER: (c) 6.25 WANSWER: (c) 6.25 W  
Explanation:Explanation:
The side band power is given byThe side band power is given by
PPcc mm22/2/2
= 50 * (0.5)= 50 * (0.5) 22/2/2
= 6.25W= 6.25W
41) 41) TRF TRF receiver receiver and and super super heterodyneheterodyne
receiver are used forreceiver are used for  
a.a. Detection of modulating signal Detection of modulating signal
b.b. Removal of unwanted signal Removal of unwanted signal
c.c. Both a and b Both a and b
d.d. None of the above None of the above
ANSWER: ANSWER: (c) Both a (c) Both a and band b  
42) 42) Disadvantage Disadvantage of usinof using a g a DSB or DSB or SSB SSB signalsignal
modulation ismodulation is  
  
a.a. Difficult to recover information at the Difficult to recover information at the
receiverreceiver
b.b. Carrier has to be locally generated at receiver Carrier has to be locally generated at receiver
c.c. Both a and b are correct Both a and b are correct
d.d. None of the above None of the above
ANSWER: (c) Both a and b are correctANSWER: (c) Both a and b are correct  
43) 43) Calculate the Calculate the modulation modulation index index when when thethe
un modulated carrier power is 15KW, and afterun modulated carrier power is 15KW, and after
modulation, carrier power is 17KW.modulation, carrier power is 17KW.  
a.a. 68% 68%
b.b. 51.63% 51.63%
c.c. 82.58% 82.58%
d.d. 34.66% 34.66%
ANSWER: ANSWER: (b) (b) 51.63%51.63%  
Explanation:Explanation:
The total power in an AM is The total power in an AM is given bygiven by
PPtt = P = Pcc ( 1 + m( 1 + m22/2)/2)
17 = 15(1 + m17 = 15(1 + m22/2)/2)
mm22/2 = 0.134/2 = 0.134
m = 0.5163m = 0.5163
= 51.63%= 51.63%
44) 44) An An AM AM transmitter transmitter has has an an antennaantenna
current changing from 5 A un current changing from 5 A un modulated to 5.8modulated to 5.8
A. What is the A. What is the percentage of modulation?percentage of modulation?  
a.a. 38.8% 38.8%
b.b. 83.14% 83.14%
c.c. 46.8% 46.8%
d.d. 25.2% 25.2%
ANSWER: (b) 83.14%ANSWER: (b) 83.14%  
Explanation:Explanation:
Modulation index m is given byModulation index m is given by
m= √ (2{Im= √ (2{Itt/I/Icc}}22-1)-1)
= √ (2 (5.8/5)= √ (2 (5.8/5)22 -1)-1)
= √ (2 (5.8/5)= √ (2 (5.8/5)22 -1)-1)
= 0.8314= 0.8314
= 83.14%= 83.14%
45) 45) Calculate the Calculate the power power in a in a DSB DSB SC siSC signalgnal
when the modulation is 60% with a carrierwhen the modulation is 60% with a carrier
power of 600W.power of 600W.  
a.a. 600 W 600 W
b.b. 540 W 540 W
c.c. 108 W 108 W
d.d. 300 W 300 W
ANSWER: ANSWER: (c) 108 (c) 108 WW  
Explanation:Explanation:
The total power in an AM is giveThe total power in an AM is given byn by
PPtt = Pc (1 + m = Pc (1 + m22/2)/2)
Given: m = 0.6Given: m = 0.6
Therefore DSB power = (mTherefore DSB power = (m22/2)P/2)Pcc  
= 600* (0.6)= 600* (0.6)22/2/2
= 108 W= 108 W
46) 46) Analog Analog communication communication indicates:indicates:  
a.a. Continuous signal with varying amplitude  Continuous signal with varying amplitude oror
phasephase
b.b. No numerical coding No numerical coding
c.c. AM or FM signal AM or FM signal
d.d. All of the above All of the above
ANSWER:(d) All of the aboveANSWER:(d) All of the above  
47) 47) Types Types of of analog analog modulation modulation are:are:  
a.a. Phase modulation Phase modulation
b.b. Frequency modulation Frequency modulation
c.c. Amplitude modulation Amplitude modulation
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
48) 48) What is What is the effect the effect on on the transmthe transmitteditted
power of AM signal when the modulationpower of AM signal when the modulation
index changes from 0.8 to 1?index changes from 0.8 to 1?  
a.a. 0.1364 0.1364
b.b. 0.3856 0.3856
c.c. 1.088 1.088
d.d. 0.5 0.5
ANSWER: (a) 0.1364ANSWER: (a) 0.1364  
Explanation:Explanation:
The total power in an AM is giveThe total power in an AM is given byn by
PPtt = P = Pcc (1 + m (1 + m22/2)/2)
Where PWhere Pcc is the carrier power and m is the is the carrier power and m is the
modulation index.modulation index.
  
Therefore,Therefore,
PPt1t1 = P = Pcc (1 + 0.8 (1 + 0.822/2) = 1.32 P/2) = 1.32 Pcc  
PPt2t2 = P = Pcc ( 1 + 1 ( 1 + 122/2) = 1.5 P/2) = 1.5 Pcc  
Increase in power = (1.5 PIncrease in power = (1.5 Pcc   – – 1.32 P 1.32 Pcc)/ 1.32 P)/ 1.32 Pcc  
= 0.1364= 0.1364
49) 49) Synchronous Synchronous detection detection meansmeans  
a.a. Extracting week signal from noise Extracting week signal from noise
b.b. Need a reference signal with predetermined Need a reference signal with predetermined
frequency and phasefrequency and phase
c.c. Both a and b Both a and b
d.d. None of the above None of the above
ANSWER: ANSWER: (c) (c) Both a Both a and band b  
50) 50) Analog Analog signal signal may bmay be convee converted intorted into
digital signal bydigital signal by  
a.a. Sampling Sampling
b.b. Amplitude modulation Amplitude modulation
c.c. Filtering Filtering
d.d. Mixing Mixing
ANSWER: (a) SamplingANSWER: (a) Sampling  
51) 51) The minThe minimum imum antenna antenna height requheight required foired forr
transmission in reference to wavelength λ istransmission in reference to wavelength λ is  
a.a.  λλ  
b.b.  λ/4λ/4  
c.c.  λ/2λ/2  
d.d.  4 λ4 λ  
ANSWER:(b) λ/4ANSWER:(b) λ/4  
52) 52) Advantages Advantages of of analog analog communicationcommunication
over digital communication are:over digital communication are:  
a.a. Data rate is low Data rate is low
b.b. Less transmission bandwidth is required Less transmission bandwidth is required
c.c. Synchronization is not needed Synchronization is not needed
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
53) 53) Radio Radio waves waves travel travel throughthrough  
a.a. Electromagnetic waves Electromagnetic waves
b.b. Water Water
c.c. Wires Wires
d.d. Fiber optic cable Fiber optic cable
ANSWER: (a) Electromagnetic wavesANSWER: (a) Electromagnetic waves  
54) 54) AM wave AM wave may be may be represented represented as E(t) as E(t) coscos
ωωcct where E(t) ist where E(t) is  
a.a. Envelope of the AM wave Envelope of the AM wave
b.b. Carrier signal Carrier signal
c.c. Amplitude of modulating signal Amplitude of modulating signal
d.d. None of the above None of the above
ANSWER: (a) Envelope of the AM waveANSWER:(a) Envelope of the AM wave  
55) 55) USB (UppeUSB (Upper Side r Side Band) is Band) is the band the band ofof
frequencyfrequency  
a.a. Above the carrier frequency Above the carrier frequency
b.b. Includes the carrier frequency Includes the carrier frequency
c.c. That lies in AM spectrum That lies in AM spectrum
d.d. Both a and c are correct Both a and c are correct
ANSWER:(d) ANSWER:(d) Both a and Both a and c are correctc are correct  
56) 56) LSB (LoweLSB (Lower Side r Side Band) is Band) is the band the band ofof
frequencyfrequency  
a.a. Below the carrier frequency Below the carrier frequency
b.b. Includes the carrier frequency Includes the carrier frequency
c.c. That lies in AM spectrum That lies in AM spectrum
d.d. Both a and c are correct Both a and c are correct
ANSWER: (d) Both a and c are correctANSWER: (d) Both a and c are correct  
57) 57) Bandwidth Bandwidth (B) of an (B) of an AM signAM signal is al is given bygiven by  
a.a.  B = 2 ωB = 2 ωmm  
b.b.  B = (ωB = (ωcc  + ω+ ωmm)) – –  (ω(ωcc   – –  ωωmm))
c.c.  ωmωm  
d.d. None of the above None of the above
e.e. Both a and b are correct Both a and b are correct
ANSWER: (e) Both a and b are correctANSWER: (e) Both a and b are correct  
58) 58) An oscillator An oscillator for an for an AM transmAM transmitter has itter has aa
100μH coil and a 10nF capacitor. If a100μH coil and a 10nF capacitor. If a
modulating frequency of 10 KHz modulates themodulating frequency of 10 KHz modulates the
oscillator, find the frequency range of the sideoscillator, find the frequency range of the side
bands.bands.  
  
a.a. 149 KHz to 169 KHz 149 KHz to 169 KHz
b.b. 184 KHz to 296 KHz 184 KHz to 296 KHz
c.c. 238 KHz to 296 KHz 238 KHz to 296 KHz
d.d. 155 KHz to 166 KHz 155 KHz to 166 KHz
ANSWER: (a) 149 KHz to 169 KHzANSWER: (a) 149 KHz to 169 KHz  
Explanation:Explanation:
Carrier frequency f Carrier frequency f cc  = 1/2Π√LC= 1/2Π√LC  
= 1/ 2Π√100 * 10 –= 1/ 2Π√100 * 10 – 6 * 10 * 10 6 * 10 * 10-9-9  
= 1/2Π * 10= 1/2Π * 10-6-6  
= 1.59 * 10= 1.59 * 1055 Hz Hz
= 159 KHz= 159 KHz
The modulating frequency f The modulating frequency f mm is 10KHz is 10KHz
Therefore the range of AM spectrum is given byTherefore the range of AM spectrum is given by
(f (f cc f  f mm ) to (f  ) to (f cc + f  + f mm ) )
= (159= (159 – – 10) to (159 + 10) 10) to (159 + 10)
= 149 KHz to 169 KHz= 149 KHz to 169 KHz
59) 59) In LIn Low ow level level Amplitude Amplitude ModulationModulation  
a.a. Modulation is done at lower power of c Modulation is done at lower power of carrierarrier
and modulating signaland modulating signal
b.b. Output power is low Output power is low
c.c. Power amplifiers are required to boost  Power amplifiers are required to boost thethe
signalsignal
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
60) 60) In HIn High igh level Alevel Amplitude mplitude ModulationModulation  
a.a. Modulation is done at high power of carrie Modulation is done at high power of carrierr
and modulating signaland modulating signal
b.b. Collector modulation method is High level Collector modulation method is High level
Amplitude ModulationAmplitude Modulation
c.c. Power amplifiers are used to boost the carrier Power amplifiers are used to boost the carrier
and modulating signals before modulationand modulating signals before modulation
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
61) 61) Square Square law law modulatorsmodulators  
a.a. Have non linear current-voltage Have non linear current-voltage
characteristicscharacteristics
b.b. Are used for Amplitude Modulation Are used for Amplitude Modulation
c.c. Are used for frequency modulation Are used for frequency modulation
d.d. Both a and b are correct Both a and b are correct
ANSWER: (d) Both a and b are correctANSWER: (d) Both a and b are correct  
62) 62) AM AM demodulation demodulation techniques techniques areare  
a.a. Square law demodulator Square law demodulator
b.b. Envelope detector Envelope detector
c.c. PLL detector PLL detector
d.d. Both a and b are correct Both a and b are correct
ANSWER: (d) Both a and b are correctANSWER: (d) Both a and b are correct  
63) 63) The proThe process cess of of recovering recovering informationinformation
signal from received carrier is known assignal from received carrier is known as  
a.a. Detection Detection
b.b. Modulation Modulation
c.c. Demultiplexing Demultiplexing
d.d. Sampling Sampling
ANSWER: (a) DetectionANSWER: (a) Detection  
64) 64) Ring Ring modulator modulator isis  
a.a. Is used for DSB SC generation Is used for DSB SC generation
b.b. Consists of four diodes connected in the  Consists of four diodes connected in the formform
of ringof ring
c.c. Is a product modulator Is a product modulator
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
65) 65) What What is is the mthe maximum aximum transmissiontransmission
efficiency of an AM signal?efficiency of an AM signal?  
a.a. 64.44% 64.44%
b.b. 33.33% 33.33%
c.c. 56.66% 56.66%
d.d. 75.55% 75.55%
ANSWER: (b) 33.33%ANSWER: (b) 33.33%  
66) 66) In synIn synchronous chronous detection detection of AM of AM signalsignal  
a.a. Carrier is locally generated Carrier is locally generated
b.b. Passed through a low pass filter Passed through a low pass filter
c.c. The original signal is recovered The original signal is recovered
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
  
67) 67) Requirements Requirements of of synchronous synchronous detectiondetection
of AM signal are:of AM signal are:  
a.a. Local generation of carrier Local generation of carrier
b.b. The frequency of the locally generated carrier The frequency of the locally generated carrier
must be identical to that of transmitted carriermust be identical to that of transmitted carrier
c.c. The phase of the locally generated carrier The phase of the locally generated carrier
must be synchronized to that of transmittedmust be synchronized to that of transmitted
carriercarrier
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
68) 68) Disadvantages Disadvantages of of using using synchronoussynchronous
detection of AM signal are:detection of AM signal are:  
a.a. Needs additional system for generation of Needs additional system for generation of
carriercarrier
b.b. Needs additional system for synchronization Needs additional system for synchronization
of carrierof carrier
c.c. Receiver is complex and costly Receiver is complex and costlyd.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
69) 69) Quadrature Quadrature Amplitude Amplitude Modulation Modulation (QAM)(QAM)
isis  
a.a. Have same bandwidth used for two DSB-SC Have same bandwidth used for two DSB-SC
signalssignals
b.b. Is also known as Bandwidth Conservation Is also known as Bandwidth Conservation
schemescheme
c.c. Is used in color television Is used in color television
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
70) 70) Pilot Pilot carrier carrier isis  
a.a. Used with DSB-SC  Used with DSB-SC signalsignal
b.b. A small carrier transmitted with modulated A small carrier transmitted with modulated
signalsignal
c.c. Used for synchronization with local carrier at Used for synchronization with local carrier at
the receiverthe receiver
d.d. All of the above All of the above
ANSWER:(d) All of the aboveANSWER:(d) All of the above  
71) 71) Generation of Generation of SSB SC SSB SC signal is signal is done bydone by  
a.a. Phase discrimination method Phase discrimination method
b.b. Frequency discrimination method Frequency discrimination method
c.c. Product modulator Product modulator
d.d. Both a and b Both a and b
ANSWER: (d) Both a andbANSWER: (d) Both a and b  
72) 72) Limitations Limitations of of Frequency Frequency discriminationdiscriminationmethod are:method are:  
a.a. Cannot be used for video signals Cannot be used for video signals
b.b. Designing of band pass filter is difficult Designing of band pass filter is difficult
c.c. Both a and b Both a and b
d.d. None of the above None of the above
ANSWER:(c) Both a and bANSWER:(c) Both a and b  
73) 73) Phase Phase shift shift method method isis  
a.a. Includes two balanced modulators Includes two balanced modulators
b.b. Two phase shifting networks Two phase shifting networks
c.c. Avoids the use of filters Avoids the use of filters
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
74) 74) Vestigial Vestigial side bside band sigand signals are nals are detected detected byby  
a.a. Filters Filters
b.b. Synchronous detection Synchronous detection
c.c. Balanced modulator Balanced modulator
d.d. None of the above None of the above
ANSWER: (b) Synchronous detectionANSWER: (b) Synchronous detection  
75) 75) Automatic Automatic gain gain control control isis  
a.a. Provides controlled signal amplitude at the Provides controlled signal amplitude at the
outputoutput
b.b. Adjusts the input to output gain to  Adjusts the input to output gain to a suitablea suitable
valuevalue
c.c. Is used in AM radio rece Is used in AM radio receiveriver
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
76) 76) In In an an Amplitude Amplitude ModulationModulation  
a.a. Amplitude of the carrier varies Amplitude of the carrier varies
b.b. Frequency of the carrier remains constant Frequency of the carrier remains constant
  
c.c. Phase of the carrier remains constant Phase of the carrier remains constant
d.d. All of the above All of the above
ANSWER: )(d) All of the aANSWER: )(d) All of the abovebove  
77) 77) If moIf modulation indulation index is dex is greater than greater than 11  
a.a. The baseband signal is not preserved in the The baseband signal is not preserved in the
envelope of the AM signalenvelope of the AM signal
b.b. The recovered signal is distorted The recovered signal is distorted
c.c. It is called over modulation It is called over modulation
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
78) 78) Examples Examples of low of low level mlevel modulation odulation areare  
a.a. Square law diode modulation Square law diode modulation
b.b. Switching modulation Switching modulation
c.c. Frequency discrimination method Frequency discrimination method
d.d. Both a and b Both a and b
ANSWER: (d) Both a and bANSWER: (d) Both a and b  
79) 79) Frequency Frequency components components of an of an AM wave AM wave (m(m
= modulation index) are= modulation index) are  
a.a.  Carrier frequency (ωCarrier frequency (ωcc ) with amplitude A ) with amplitude A
b.b.  Upper side band (ωUpper side band (ωcc  + ω+ ωmm) having amplitude) having amplitude
mA/2mA/2
c.c.  Lower side band (ωLower side band (ωcc   – –  ωωmm) having amplitude) having amplitude
mA/2mA/2
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
80) 80) Squelch Squelch circuit circuit isis  
a.a. Suppresses output audio Suppresses output audio
b.b. Works when there  Works when there is insufficient desiredis insufficient desired
input signalinput signal
c.c. Is used to suppress the unwanted channel Is used to suppress the unwanted channel
noise when there is no reception by thenoise when there is no reception by the
receiverreceiver
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
81) 81) In AutoIn Automatic gamatic gain controin control of l of the Athe AMM
receiverreceiver  
a.a. Gain of the receiver is adjusted Gain of the receiver is adjusted
b.b. The gain adjustment depends upon the The gain adjustment depends upon the
strength of the received signalstrength of the received signal
c.c. The output provided is a DC voltage The output provided is a DC voltage
d.d. All of the above All of the above
ANSWER: All of the aboveANSWER: All of the above  
82) 82) The factors The factors that deterthat determine the mine the sensitivitysensitivity
of super heterodyne receiver areof super heterodyne receiver are  
a.a. Gain of the IF  Gain of the IF amplifieramplifier
b.b. Noise figure of the receiver Noise figure of the receiver
c.c. Gain of RF amplifier Gain of RF amplifier
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
83) 83) Selectivity Selectivity of of a a receiver:receiver:  
a.a. Changes with incoming signal frequency Changes with incoming signal frequency
b.b. Is poorer at  Is poorer at high frequencieshigh frequencies
c.c. Is the rejection of the adjacent channel at the Is the rejection of the adjacent channel at the
receiverreceiver
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
84) 84) Advantages Advantages of usof using an ing an RF amRF amplifier are:plifier are:  
a.a. Better selectivity Better selectivity
b.b. Better sensitivity Better sensitivity
c.c. Improved signal to noise ratio Improved signal to noise ratio
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
85) 85) Intermediate Intermediate frequency frequency (IF) (IF) should should bebe
carefully chosen ascarefully chosen as  
a.a. High IF results in poor  High IF results in poor selectivityselectivity
b.b. High IF results in problems in t High IF results in problems in tracking ofracking of
signalssignals
c.c. Image frequency rejection becomes poor at Image frequency rejection becomes poor at
low IFlow IF
d.d. All of the above All of the above
ANSWER: (d) All of the aboveANSWER: (d) All of the above  
86) 86) Example Example of of continuous continuous wave wave analoganalog
modulation ismodulation is  
  
a.a. PCM PCM
b.b. DM DM
c.c. AM AM
d.d. PAM PAM
ANSWER: ANSWER: (c) (c) AMAM  
87) 87) The The standard standard value value for for IntermediateIntermediatefrequency (IF) in AM receivers isfrequency (IF) in AM receivers is  
a.a. 455 KHz 455 KHz
b.b. 580 KHz 580 KHz
c.c. 10.7 MHz 10.7 MHz
d.d. 50 MHz 50 MHz
ANSWER: (a) 455 KHzANSWER: (a) 455 KHz  
88) 88) The fuThe functions nctions of radof radio io receiver receiver areare  
a.a. Receive the Incoming modulated carrier by Receive the Incoming modulated carrier by
antennaantenna
b.b. Select the wanted signal and reject the Select the wanted signal and reject the
unwanted signals and noiseunwanted signals and noise
c.c. Detection and amplification of the Detection and amplification of the
information signal from the carrierinformation signal from the carrier
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
89) 89) Function Function of freqof frequency uency mixer mixer in suin superper
heterodyne receiver isheterodyne receiver is  
a.a. Amplification Amplification
b.b. Filtering Filtering
c.c. Multiplication of incoming signal and the Multiplication of incoming signal and the
locally generated carrierlocally generated carrier
d.d. None of the above None of the above
ANSWER: ANSWER: (c) Multiplication of (c) Multiplication of incoming signaincoming signall
and the locally generated carrierand the locally generated carrier  
90) 90) The advanThe advantages of tages of using using an RF an RF amplifieramplifier
areare  
a.a. Better sensitivity Better sensitivity
b.b. Improved signal to noise ratio Improved signal to noise ratio
c.c. Better selectivity Better selectivity
d.d. All of the above All of the above
ANSWER: (d) All of the ANSWER: (d) All of the aboveabove  
91) 91) The coThe costas stas receiver receiver is is used used forfor  
a.a. FM signalFM signal
b.b. DSB-SC signal DSB-SC signal
c.c. PCM signal PCM signal
d.d. DM signal DM signal
ANSWER: ANSWER: (b) DSB-S(b) DSB-SC signalC signal  
92) 92) Cross Cross talk talk isis – –  
a.a. The disturbance caused in the nearby The disturbance caused in the nearby
channel or circuit due to transmitted signalchannel or circuit due to transmitted signal
b.b. Adjacent frequency rejection Adjacent frequency rejection
c.c. Generation of closely lying side bands Generation of closely lying side bands
d.d. None of the above None of the above
ANSWER: ANSWER: (a) The disturban(a) The disturbance caused in thece caused in the
nearby channel or circuit due to nearby channel or circuit due to transmittedtransmitted
signalsignal  
93) 93) In termIn terms s of siof signal fgnal frequency requency (f (f ss) and) and
intermediate frequency (f intermediate frequency (f ii), the image), the image
frequency is given byfrequency is given by  
a.a. f  f ss + f  + f ii  
b.b. f  f ss + 2f  + 2f ii  
c.c. 2f  2f ss + f + f ii  
d.d. 2( f  2( f ss + f  + f ii))
ANSWER: ANSWER: (b) (b) f f ss + 2f  + 2f ii