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Angle Modulated Systems
1. Consider an FM wave
π(π‘) = cos[2ππππ‘ + π½1 sin 2ππ1π‘ + π½22ππ2π‘]
The maximum deviation of the instantaneous frequency from the carrier
frequency fc is
(a) π½1π1 + π½2π2
(b) π½1π2 + π½2π1
(c) π½1 + π½2
(d) π1 + π2
[GATE 2014: 1 Mark]
Soln. The instantaneous value of the angular frequency
ππ = ππ +π
π π(π·π π¬π’π§ ππ πππ + π·π π¬π’π§ ππ πππ)
ππ + ππ π·πππ ππ¨π¬ ππ πππ + ππ π·πππ ππ¨π¬ ππ πππ
ππ = ππ + π·πππ ππ¨π¬ ππ πππ + π·πππ ππ¨π¬ ππ πππ
Frequency deviation (βπ)πππ
= π·πππ + π·πππ
Option (a)
2. A modulation signal is π¦(π‘) = π(π‘) cos(40000ππ‘), where the baseband
signal m(t) has frequency components less than 5 kHz only. The minimum
required rate (in kHz) at which y(t) should be sampled to recover m(t) is ____
[GATE 2014: 1 Mark]
Soln. The minimum sampling rate is twice the maximum frequency called
Nyquist rate
The minimum sampling rate (Nyquist rate) = 10K samples/sec
3. Consider an angle modulation signal π₯(π‘) = 6πππ [2π Γ 103 +
2 sin(8000ππ‘) + 4 cos(8000ππ‘)]π. The average power of π₯(π‘) is
(a) 10 W
(b) 18 W
(c) 20 W
(d) 28 W
[GATE 2010: 1 Mark]
Soln. The average power of an angle modulated signal is
π¨π
π
π=
ππ
π
=18 W
Option (b)
4. A modulation signal is given by π (π‘) = πβππ‘ cos[(ππ + βπ)π‘] π’(π‘),
where,ππ πππ βπ are positive constants, and ππ β« βπ. The complex
envelope of s(t) is given by
(a) exp(βππ‘) ππ₯π[π(ππ + βπ)]π’(π‘)
(b) exp(βππ‘) exp(πβππ‘) π’(π‘)
(c) ππ₯π(πβππ‘)π’(π‘)
(d) ππ₯π[(πππ + βπ)π‘]
[GATE 1999: 1 Mark]
Soln. π(π) = πβππ ππ¨π¬[(ππ + βπ)π]π(π)
Complex envelope π(π)Μ = π(π)πβππππ
= [πβππππ(ππ+βπ)π. π(π)]πβππππ
= πβππππβπππ(π)
Option (b)
5. A 10 MHz carrier is frequency modulated by a sinusoidal signal of 500 Hz,
the maximum frequency deviation being 50 KHz. The bandwidth required.
as given by the Carsonβs rule is ___________
[GATE 1994: 1 Mark]
Soln. By carsonβs rule
π©πΎ = π(βπ + ππ)
= π(ππ + π. π)
= πππ π²π―π
6. π£(π‘) = 5[cos(106ππ‘) β sin(103ππ‘) Γ sin(106ππ‘)] represents
(a) DSB suppressed carrier signal
(b) AM signal
(c) SSB upper sideband signal
(d) Narrow band FM signal
[GATE 1994: 1 Mark]
Soln. π(π) = π ππ¨π¬(ππππ π) βπ
πππ¨π¬(πππ β πππ) π π +
π
πππ¨π¬(πππ + πππ)π π
Carrier and upper side band are in phase and lower side band is out of
phase with carrier
The given signal is narrow band FM signal
Option (d)
7. The input to a coherent detector is DSB-SC signal plus noise. The noise at
the detector output is
(a) the in-phase component
(b) the quadrature-component
(c) zero
(d) the envelope
[GATE 2003: 1 Mark]
Soln. The coherent detector rejects the quadrature component of noise
therefore noise at the output has in phase component only.
Option (a)
8. An AM signal and a narrow-band FM signal with identical carriers,
modulating signals and modulation indices of 0.1 are added together. The
resultant signal can be closely approximated by
(a) Broadband FM
(b) SSB with carrier
(c) DSB-SC
(d) SSB without carrier
[GATE 2004: 1 Mark]
Soln. π½π¨π΄(π) = π¨ ππ¨π¬ ππ π +π.ππ¨
πππ¨π¬(ππ + ππ)π +
π.ππ¨
πππ¨π¬(ππ β ππ)π
π½ππ΄(π)(ππππππππππ )
= π¨ ππ¨π¬ πππ +π. ππ¨
πππ¨π¬(ππ + ππ)π β
π. ππ¨
πππ¨π¬(ππ β ππ)π
π½π¨π΄(π) + π½ππ΄(π) = ππ¨ ππ¨π¬ πππ + π. ππ¨ ππ¨π¬(ππ + ππ)π
The resulting signal is SSB with carrier
Option (b)
9. The List-I (lists the attributes) and the List-II (lists of the modulation
systems). Match the attribute to the modulation system that best meets it.
List-I
(A) Power efficient transmission of signals
(B) Most bandwidth efficient transmission of voice signals
(C) Simplest receiver structure
(D) Bandwidth efficient transmission of signals with significant dc
component
List-II
(1) Conventional AM
(2) FM
(3) VSB
(4) SSB-SC
A B C D
(a) 4 2 1 3
(b) 2 4 1 3
(c) 3 2 1 4
(d) 2 4 3 1
[GATE 2011: 1 Mark]
Soln. FM is the most power efficient transmission of signals AM has the
simplest receiver. Vestigial sideband is bandwidth efficient transmission
of signals with sufficient dc components. Single sideband, suppressed
carrier (SSB-SC) is the most bandwidth efficient transmission of voice
signals.
Option (b)
10. The signal m(t) as shown I applied both to a phase modulator (with kp as the
phase constant) and a frequency modulator with (kf as the frequency
constant) having the same carrier frequency
The ratio ππ/ππ (in rad/Hz) for the same maximum phase deviation is
-2 0 2 4 6 8 t(seconds)
m(t)
-2
(a) 8Ο
(b) 4Ο
(c) 2Ο
(d) Ο
[GATE 2012: 2 Marks]
Soln. For a phase modulator, the instantaneous value of the phase angle ππ is
equal to phase of an unmodulated carrier ππ(π) plus a time varying
component proportional to modulation signal m(t)
ππ·π΄(π) = ππ πππ + πππ(π)
Maximum phase deviation (ππ·π΄)πππ
= π²π π¦ππ± π(π) = ππ²π
For a frequency modulator, the instantaneous value of the angular
frequency
ππ = ππ + ππ π²ππ(π)
The total phase of the FM wave is
πππ΄ = β« πππ π
= πππ + ππ π²π β« π(π)π π
π
π
(πππ΄)πππ = ππ π²π β« ππ π
π
π
= ππ π²π
π²π·
π²π=
ππ
π
= ππ
Option (b)
11. Consider the frequency modulated signal
10[cos 2π Γ 105π‘ + 5 sin(2π Γ 1500π‘) + 7.5 sin(2π Γ 1000π‘)]
with carrier frequency of 105 Hz. The modulation index is
(a) 12.5
(b) 10
(c) 7.5
(d) 5
[GATE 2008: 2 Marks]
Soln. Frequency modulated signal
ππ ππ¨π¬[ππ Γ ππππ + π π¬π’π§(ππ Γ πππππ) + π. π π¬π’π§(ππ Γ
ππππππ)]
The instantaneous value of the angular frequency
ππ = ππ +π
π π[π π¬π’π§(ππ Γ πππππ) + π. π π¬π’π§(ππ Γ πππππ)]
Frequency deviation
βπ= π Γ ππ Γ ππππ ππ¨π¬(ππ Γ ππππ) + π. π Γ ππ Γ ππππ ππ¨π¬(ππ Γ πππππ)
(βπ)πππ = ππ (ππππ + ππππ)
Frequency deviation (πΉ) =(βπ)πππ
ππ = ππππππ―π
Modulation index ππ =πππππ
ππππ
= 10
Option (b)
12. A message signal with bandwidth 10 KHz is Lower-Side Band SSB
modulated with carrier frequency ππ1= 106π»π§. The resulting signal is then
passed through a narrow-band frequency Modulator with carrier frequency
ππ2 = 109π»π§.
The bandwidth of the output would be
(a) 4 Γ 104π»π§
(b) 2 Γ 106π»π§
(c) 2 Γ 109π»π§
(d) 2 Γ 1010π»π§
[GATE 2006: 2 Marks]
Soln. Lower side band frequency = πππ β ππ
= πππ π²π―π
Considering this as the baseband signal, the bandwidth of narrow band
FM
= π Γ πππ π²π―π
β ππ΄π―π
Option (b)
13. A device with input π₯(π‘) and output π¦(π‘) is characterized by: π¦(π‘) = π₯2(π‘).
An FM signal with frequency deviation of 90 KHz and modulating signal
bandwidth of 5 KHz is applied to this device. The bandwidth of the output
signal is
(a) 370 KHz
(b) 190 KHz
(c) 380 KHz
(d) 95 KHz
[GATE 2005: 2 Marks]
Soln. Frequency deviation βπ= πππ²π―π
Modulating signal bandwidth = 5 KMz
When FM signal is applied to doubler frequency deviation doubles.
π©. πΎ = π(βπ + ππ)
= π(πππ + π)
= πππ π²π―π
Option (a)
14. An angle-modulation signal is given by
π (π‘) = cos(2π Γ 2 Γ 106π‘ + 2π Γ 30 sin 150π‘ + 2π Γ 40 cos 150π‘)
The maximum frequency and phase deviations of s(t) are
(a) 10.5KHz, 140Ο rad
(b) 6 KHz, 80Ο rad
(c) 10.5 KHz, 100Ο rad
(d) 7.5 KHz, 100Ο rad
[GATE 2002: 2 Marks]
Soln. The total phase angle of the carrier
π = πππ + ππ
π = ππ Γ π Γ πππ + ππ Γ ππ π¬π’π§ πππ π + ππ Γ ππ ππ¨π¬ ππππ
Instantaneous value of angular frequency ππ
ππ = ππ +π
π π(ππ Γ ππ π¬π’π§ ππππ + ππ Γ ππ ππ¨π¬ ππππ)
= ππ + ππ Γ ππ Γ πππ ππ¨π¬ ππππ β ππ Γ ππ Γ πππ π¬π’π§ ππππ
= ππ + ππ Γ ππππ ππ¨π¬ ππππ β ππ Γ ππππ π¬π’π§ πππ π
Frequency deviation βπ= ππ Γ ππππ[π ππ¨π¬ ππππ β π π¬π’π§ ππππ]
= πππππ βππ + ππ πππ /πππ
= πππππ πππ /πππ
= ππ Γ π. ππ² πππ /πππ
βπ=βπ
ππ = π. ππ²π―π
Phase deviation βπ is proportional to π½π
βπ = ππ βπππ + πππ
= ππ Γ ππ = ππππ πππ
Option (d)
15. In a FM system, a carrier of 100 MHz is modulated by a sinusoidal signal of
5 KHz. The bandwidth by Carsonβs approximation is 1MHz. If y(t) =
(modulated waveform)3, then by using Carsonβs approximation, the
bandwidth of y(t) around 300 MHz and the spacing of spectral components
are, respectively.
(a) 3 MHz, 5 KHz
(b) 1 MHz, 15 KHz
(c) 3 MHz, 15 KHz
(d) 1 MHz, 5 KHz
[GATE 2000: 2 Marks]
Soln. In an FM signal, adjacent spectral components will get separated by
modulating frequency ππ = ππ²π―π
π©πΎ = π(βπ + ππ) = ππ΄π―π
βπ + ππ = πππ π²π―π
βπ= πππ π²π―π
The nth order non-linearity makes the carrier frequency and frequency
deviation increased by n-fold, with baseband frequency fm unchanged.
(βπ)πππ = π Γ πππ
= ππππ π²π―π
π΅ππ π©πΎ = π(ππππ + π) Γ πππ
= π. ππ π΄π―π
β π π΄π―π
Option (a)
16. An FM signal with a modulation index 9 is applied to a frequency tripler.
The modulation index in the output signal will be
(a) 0
(b) 3
(c) 9
(d) 27
[GATE 1996: 2 Marks]
Soln. The frequency modulation index Ξ² is multiplied by n in ntimes
frequency multiplier.
πΊπ, π·β² = π Γ π
= 27
Option (d)
17. A signal π₯(π‘) = 2 cos(π. 104π‘) volts is applied to an FM modulator with
the sensitivity constant of 10 KHz/volt. Then the modulation index of the
FM wave is
(a) 4
(b) 2
(c) 4/Ο
(d) 2/Ο
[GATE 1989: 2 Marks]
Soln. Modulation index
π· =π²ππ¨π
ππ
π²π = πππ²π―π/ππππ
Am is the amplitude of modulating signal
fm is the modulating frequency
π· =ππ Γ πππ Γ π
π Γπππ
ππ
= π
Option (a)
18. A carrier π΄πΆ cos ππ π‘ is frequency modulated by a signal πΈπ cos ππ π‘.The
modulation index is mf. The expression for the resulting FM signal is
(a) π΄π cos[πππ‘ + ππ sin πππ‘]
(b) π΄π cos[πππ‘ + ππ cos πππ‘]
(c) π΄π cos[πππ‘ + 2π ππ sin πππ‘]
(d) π΄πΆ cos [πππ‘ +2ππππΈπ
ππcos πππ‘]
[GATE 1989: 2 Marks]
Soln. The frequency modulated signal
π½ππ΄(π) = π¨π ππ¨π¬[πππ + π²π β« π(π)π π]
Kf is the frequency sensitivity of the modulator
β« π(π)π π = β« π¬π ππ¨π¬ πππ π π =π¬π π¬π’π§ πππ
ππ
π½ππ΄(π) = π¨π ππ¨π¬ [πππ +π²ππ¬π
πππ¬π’π§ πππ]
= π¨π ππ¨π¬[πππ + ππ π¬π’π§ πππ]where mf is the modulation index
Option (a)
19. A message m(t) bandlimited to the frequency fm has a power of Pm. The
power of the output signal in the figure is
multiplyOutput signal
Ideal low
Pass filter
Cut of f=fm
Pass band
gain=1
(a) ππ cos π
2
(b) ππ
4
(c) πππ ππ2π
4
(d) πππππ 2π
4
[GATE 2000: 2 Marks]
Soln. Output of the multiplier = π(π) ππ¨π¬ πππ πππ(πππ + π½)
=π(π)
π[ππ¨π¬(ππππ + π½) + ππ¨π¬ π½]
Output of LPF π½π(π) =π(π)
πππ¨π¬ π½
=π
πππ¨π¬ π½ π(π)
Power of output signal
= π₯π’π¦π»ββ
π
π»β« π½π
π(π)π π
π»
π
= π₯π’π¦π»ββ
π
π»β«
πͺππππ½
πππ(π)π π
=πππππ½
ππ₯π’π¦π»ββ
π
π»β« ππ(π)π π
π»
π
=πππππ½
ππ·π
Option (d)
20. c(t) and m(t) are used to generate an FM signal. If the peak frequency
deviation of the generated FM signal is three times the transmission
bandwidth of the AM signal , then the coefficient of the term
5cos[2π(1008 Γ 103π‘)] in the FM signal (in terms of the Bessel
coefficients) is
(a) 5π½4(3)
(b) 5
2π½8(3)
(c) 5
2π½8(4)
(d) 5π½4(6)
[GATE 2003: 2 Marks]
Soln.
π½ππ΄(π) = β π¨π±π(ππ) ππ¨π¬(ππ + πππ)π
β
π=ββ
Peak frequency deviation of FM signal is three times the bandwidth of
AM signal
πΉπ = π Γ πππ = πππ
Modulation index
ππ =πΉπ
ππ=
πππ
ππ= π
π ππ¨π¬[ππ (ππππ Γ ππππ)] = π ππ¨π¬[ππ (ππππ + π Γ π) Γ πππ]
π = π
The required coefficient is ππ±π(π)
Option (d)
21. Consider a system shown in the figure. Let π(π) πππ π(π) denote the
Fourier transforms of π₯(π‘)πππ π¦(π‘) respectively. The ideal HPF has the
cutoff frequency 10 KHz.
Balanced
ModulatorHPF
10KHzx(t) Balanced
Modulator
10 KHz 13 KHz
X(f)
f (KHz)
-3 31-1
The positive frequencies where Y(f) has spectral peaks are
(a) 1 KHz and 24 KHz
(b) 2 KHz and 24 KHz
(c) 1 KHz and 14 KHz
(d) 2 KHz and 14 KHz
[GATE 2004: 2 Marks]
Soln. Input signal x(f) has the peaks at 1KHz and -1Mhz.
The output of balanced modulator will have peaks at
ππ Β± π, ππ Β± (βπ)ππ Β± π = ππ Β± π
= ππ πππ π π²π―π
ππ Β± (βπ) = ππ Β± (βπ) = π π²π―π πππ ππ π²π―π
9 MHz will be filtered out by the HPF
After passing through 13 KHz balanced modulator, the signal will have
ππ Β± ππ frequencies.
π(π) = ππ π² πππ π π²
Option (b)
Common Data for Questions 22 & 23
Consider the following Amplitude Modulated (AM) signal,
Where ππ < π΅ ππ΄π(π‘) = 10(1 + 0.5 sin 2π πππ‘) cos 2ππππ‘
22. The average side-band power for the AM signal given above is
(a) 25
(b) 12.5
(c) 6.25
(d) 3.125
[GATE 2006: 2 Marks]
Soln. The average sideband power for the AM signal is
π·πΊπ© = π·π
πππ
π
π·π β πππππππ πππππ
ππ β πππ πππππππ πππ ππ
π·π =π¨ππ
π=
πππ
π
= πππ
ππ = π. π
So,
π·πΊπ© = ππ(π. π)π
π
=ππ Γ π. ππ
π
= π. ππ πππππ
Option (c)
23. The AM signal gets added to a noise with Power Spectral Density Sn(f)
given in the figure below. The ratio of average sideband power to mean
noise power would be:
(a) 25
8π0π΅
(b) 25
4π0π΅
(c) 25
2π0π΅
(d) 25
π0π΅
[GATE 2006: Marks]
Soln. The AM signal gets added to a noise with spectral density ππ(π)
The noise power
π·π» = β« ππ(π)π π
β
ββ
= π β« ππ(π)π π
β
π
Noise power = π [π
πΓ
π©
πΓ
π΅π
π]
= π΅ππ©
Power in sidebands π·πΊπ© =ππ
π πππππ
π·πΊπ©
πππππ πππππ=
ππ
ππ΅ππ© ππππππ (π)