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Angle-Modulated-Systems1

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Angle Modulated Systems
1. Consider an FM wave
𝑓(𝑑) = cos[2πœ‹π‘“π‘ 𝑑 + 𝛽1 sin 2πœ‹π‘“1 𝑑 + 𝛽2 2πœ‹π‘“2 𝑑]
The maximum deviation of the instantaneous frequency from the carrier
frequency fc is
(a) 𝛽1 𝑓1 + 𝛽2 𝑓2
(c) 𝛽1 + 𝛽2
(b) 𝛽1 𝑓2 + 𝛽2 𝑓1
(d) 𝑓1 + 𝑓2
[GATE 2014: 1 Mark]
Soln. The instantaneous value of the angular frequency
πŽπ’Š = πŽπ’„ +
𝒅
𝒅𝒕
(𝜷𝟏 𝐬𝐒𝐧 πŸπ…π’‡πŸ 𝒕 + 𝜷𝟐 𝐬𝐒𝐧 πŸπ…π’‡πŸ 𝒕)
πŽπ’„ + πŸπ…πœ·πŸ π’‡πŸ 𝐜𝐨𝐬 πŸπ…π’‡πŸ 𝒕 + πŸπ…πœ·πŸ π’‡πŸ 𝐜𝐨𝐬 πŸπ…π’‡πŸ 𝒕
π’‡π’Š = 𝒇𝒄 + 𝜷𝟏 π’‡πŸ 𝐜𝐨𝐬 πŸπ…π’‡πŸ 𝒕 + 𝜷𝟐 π’‡πŸ 𝐜𝐨𝐬 πŸπ…π’‡πŸ 𝒕
Frequency deviation (βˆ†π’‡ )
π’Žπ’‚π’™
= 𝜷𝟏 π’‡πŸ + 𝜷𝟐 π’‡πŸ
Option (a)
2. A modulation signal is 𝑦(𝑑) = π‘š(𝑑) cos(40000πœ‹π‘‘), 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‘₯(𝑑) = 6π‘π‘œπ‘ [2πœ‹ × 103 +
2 sin(8000πœ‹π‘‘) + 4 cos(8000πœ‹π‘‘)]𝑉. The average power of π‘₯(𝑑) is
(a) 10 W
(c) 20 W
(b) 18 W
(d) 28 W
[GATE 2010: 1 Mark]
Soln. The average power of an angle modulated signal is
π‘¨πŸπ’„ πŸ”πŸ
=
𝟐
𝟐
=18 W
Option (b)
4. A modulation signal is given by 𝑠(𝑑) = 𝑒 −π‘Žπ‘‘ cos[(πœ”π‘ + βˆ†πœ” )𝑑] 𝑒(𝑑),
where,πœ”π‘ π‘Žπ‘›π‘‘ βˆ†πœ” are positive constants, and πœ”π‘ ≫ βˆ†πœ”. The complex
envelope of s(t) is given by
(a) exp(−π‘Žπ‘‘) 𝑒π‘₯𝑝[𝑗(πœ”π‘ + βˆ†πœ”)]𝑒(𝑑)
(b) exp(−π‘Žπ‘‘) exp(π‘—βˆ†πœ”π‘‘) 𝑒(𝑑)
(c) 𝑒π‘₯𝑝(π‘—βˆ†πœ”π‘‘)𝑒(𝑑)
(d) 𝑒π‘₯𝑝[(π‘—πœ”π‘ + βˆ†πœ”)𝑑]
[GATE 1999: 1 Mark]
Soln. 𝒔(𝒕) = 𝒆−𝒂𝒕 𝐜𝐨𝐬[(πŽπ’„ + βˆ†πŽ )𝒕]𝒖(𝒕)
Μƒ = 𝒔(𝒕)𝒆−π’‹πŽπ’„π’•
Complex envelope 𝒔(𝒕)
= [𝒆−𝒂𝒕 𝒆𝒋(πŽπ’„ +βˆ†πŽ)𝒕 . 𝒖(𝒕)]𝒆−π’‹πŽπ’„π’•
= 𝒆−𝒂𝒕 π’†π’‹βˆ†πŽπ’• 𝒖(𝒕)
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 Carson’s rule is ___________
[GATE 1994: 1 Mark]
Soln.
By carson’s rule
𝑩𝑾 = 𝟐(βˆ†π’‡ + π’‡π’Ž )
= 𝟐(πŸ“πŸŽ + 𝟎. πŸ“)
= 𝟏𝟎𝟏 𝑲𝑯𝒛
6. 𝑣(𝑑) = 5[cos(106 πœ‹π‘‘) − sin(103 πœ‹π‘‘) × sin(106 πœ‹π‘‘)] represents
(a) DSB suppressed carrier signal
(b) AM signal
(c) SSB upper sideband signal
(d) Narrow band FM signal
[GATE 1994: 1 Mark]
πŸ“
πŸ“
𝟐
𝟐
Soln. 𝒗(𝒕) = πŸ“ 𝐜𝐨𝐬(πŸπŸŽπŸ” 𝝅𝒕) − 𝐜𝐨𝐬(πŸπŸŽπŸ” − πŸπŸŽπŸ‘ ) 𝝅𝒕 + 𝐜𝐨𝐬(πŸπŸŽπŸ” + πŸπŸŽπŸ‘ )𝝅𝒕
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
(c) zero
(b) the quadrature-component
(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
(c) DSB-SC
(b) SSB with carrier
(d) SSB without carrier
[GATE 2004: 1 Mark]
Soln. 𝑽𝑨𝑴 (𝒕) = 𝑨 𝐜𝐨𝐬 πŽπ’„ 𝒕 +
𝟎.πŸπ‘¨
𝟐
𝐜𝐨𝐬(πŽπ’„ + πŽπ’Ž )𝒕 +
𝟎.πŸπ‘¨
𝟐
𝐜𝐨𝐬(πŽπ’„ − πŽπ’Ž )𝒕
𝑽𝑭𝑴 (𝒕)(π’π’‚π’“π’“π’π’˜π’ƒπ’‚π’π’…)
= 𝑨 𝐜𝐨𝐬 πŽπ’„ 𝒕 +
𝟎. πŸπ‘¨
𝟎. πŸπ‘¨
𝐜𝐨𝐬(πŽπ’„ + πŽπ’Ž )𝒕 −
𝐜𝐨𝐬(πŽπ’„ − πŽπ’Ž )𝒕
𝟐
𝟐
𝑽𝑨𝑴 (𝒕) + 𝑽𝑭𝑴 (𝒕) = πŸπ‘¨ 𝐜𝐨𝐬 πŽπ’„ 𝒕 + 𝟎. πŸπ‘¨ 𝐜𝐨𝐬(πŽπ’„ + πŽπ’Ž )𝒕
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 π‘˜π‘ /π‘˜π‘“ (in rad/Hz) for the same maximum phase deviation is
m(t)
-2
0
-2
2
4
6
8 t(seconds)
(a) 8π
(b) 4π
(c) 2π
(d) π
[GATE 2012: 2 Marks]
Soln. For a phase modulator, the instantaneous value of the phase angle ππ’Š is
equal to phase of an unmodulated carrier πŽπ’„ (𝒕) plus a time varying
component proportional to modulation signal m(t)
𝛙𝑷𝑴 (𝒕) = πŸπ…π’‡π’„ 𝒕 + π’Œπ’‘ π’Ž(𝒕)
Maximum phase deviation (𝛙𝑷𝑴 )π’Žπ’‚π’™
= 𝑲𝒑 𝐦𝐚𝐱 π’Ž(𝒕) = πŸπ‘²π’‘
For a frequency modulator, the instantaneous value of the angular
frequency
πŽπ’Š = πŽπ’„ + πŸπ…π‘²π’‡ π’Ž(𝒕)
The total phase of the FM wave is
𝛙𝑭𝑴 = ∫ πŽπ’Š 𝒅𝒕
𝒕
= πŽπ’„ 𝒕 + πŸπ…π‘²π’‡ ∫ π’Ž(𝒕)𝒅𝒕
𝟎
𝟐
(𝛙𝑭𝑴 )π’Žπ’‚π’™ = πŸπ…π‘²π’‡ ∫ πŸπ’…π’•
𝟎
= πŸ–π…π‘²π’‡
𝑲𝑷 πŸ–π…
=
𝑲𝒇
𝟐
= πŸ’π…
Option (b)
11.Consider the frequency modulated signal
10[cos 2πœ‹ × 105 𝑑 + 5 sin(2πœ‹ × 1500𝑑) + 7.5 sin(2πœ‹ × 1000𝑑)]
with carrier frequency of 105 Hz. The modulation index is
(a) 12.5
(c) 7.5
(b) 10
(d) 5
[GATE 2008: 2 Marks]
Soln. Frequency modulated signal
𝟏𝟎 𝐜𝐨𝐬[πŸπ… × πŸπŸŽπŸ“ 𝒕 + πŸ“ 𝐬𝐒𝐧(πŸπ… × πŸπŸ“πŸŽπŸŽπ’•) + πŸ•. πŸ“ 𝐬𝐒𝐧(πŸπ… ×
πŸπŸŽπŸŽπŸŽπŸŽπ’•)]
The instantaneous value of the angular frequency
πŽπ’Š = 𝝎 𝒄 +
𝒅
[πŸ“ 𝐬𝐒𝐧(πŸπ… × πŸπŸ“πŸŽπŸŽπ’•) + πŸ•. πŸ“ 𝐬𝐒𝐧(πŸπ… × πŸπŸŽπŸŽπŸŽπ’•)]
𝒅𝒕
Frequency deviation
βˆ†πŽ = πŸ“ × πŸπ… × πŸπŸ“πŸŽπŸŽ 𝐜𝐨𝐬(πŸπ… × πŸπŸ“πŸŽπŸŽ) + πŸ•. πŸ“ × πŸπ… × πŸπŸŽπŸŽπŸŽ 𝐜𝐨𝐬(πŸπ… × πŸπŸŽπŸŽπŸŽπ’•)
(βˆ†πŽ )π’Žπ’‚π’™ = πŸπ…(πŸ•πŸ“πŸŽπŸŽ + πŸ•πŸ“πŸŽπŸŽ)
Frequency deviation (𝜹) =
Modulation index π’Žπ’‡ =
= 10
Option (b)
(βˆ†πŽ )π’Žπ’‚π’™
πŸπ…
πŸπŸ“πŸŽπŸŽπŸŽ
πŸπŸ“πŸŽπŸŽ
= πŸπŸ“πŸŽπŸŽπŸŽπ‘―π’›
12.A message signal with bandwidth 10 KHz is Lower-Side Band SSB
modulated with carrier frequency 𝑓𝑐1 = 106 𝐻𝑧. The resulting signal is then
passed through a narrow-band frequency Modulator with carrier frequency
𝑓𝑐2 = 109 𝐻𝑧.
The bandwidth of the output would be
(a) 4 × 104 𝐻𝑧
(c) 2 × 109 𝐻𝑧
(b) 2 × 106 𝐻𝑧
(d) 2 × 1010 𝐻𝑧
[GATE 2006: 2 Marks]
Soln. Lower side band frequency = πŸπŸŽπŸ‘ − 𝟏𝟎
= πŸ—πŸ—πŸŽ 𝑲𝑯𝒛
Considering this as the baseband signal, the bandwidth of narrow band
FM
= 𝟐 × πŸ—πŸ—πŸŽ 𝑲𝑯𝒛
≈ πŸπ‘΄π‘―π’›
Option (b)
13.A device with input π‘₯(𝑑) and output 𝑦(𝑑) is characterized by: 𝑦(𝑑) = π‘₯ 2 (𝑑).
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
(c) 380 KHz
(b) 190 KHz
(d) 95 KHz
[GATE 2005: 2 Marks]
Soln. Frequency deviation βˆ†π’‡ = πŸ—πŸŽπ‘²π‘―π’›
Modulating signal bandwidth = 5 KMz
When FM signal is applied to doubler frequency deviation doubles.
𝑩. 𝑾 = 𝟐(βˆ†π’‡ + π’‡π’Ž )
= 𝟐(πŸπŸ–πŸŽ + πŸ“)
= πŸ‘πŸ•πŸŽ 𝑲𝑯𝒛
Option (a)
14.An angle-modulation signal is given by
𝑠(𝑑) = cos(2πœ‹ × 2 × 106 𝑑 + 2πœ‹ × 30 sin 150𝑑 + 2πœ‹ × 40 cos 150𝑑)
The maximum frequency and phase deviations of s(t) are
(a) 10.5KHz, 140π rad
(c) 10.5 KHz, 100π rad
(b) 6 KHz, 80π rad
(d) 7.5 KHz, 100π rad
[GATE 2002: 2 Marks]
Soln. The total phase angle of the carrier
𝛙 = π›šπœ 𝐭 + π›‰πŸŽ
𝛙 = πŸπ›‘ × πŸ × πŸπŸŽπŸ” + πŸπ›‘ × πŸ‘πŸŽ 𝐬𝐒𝐧 πŸπŸ“πŸŽ 𝐭 + πŸπ›‘ × πŸ’πŸŽ 𝐜𝐨𝐬 πŸπŸ“πŸŽπ­
Instantaneous value of angular frequency πŽπ’Š
πŽπ’Š = πŽπ’„ +
𝒅
(πŸπ… × πŸ‘πŸŽ 𝐬𝐒𝐧 πŸπŸ“πŸŽπ’• + πŸπ… × πŸ’πŸŽ 𝐜𝐨𝐬 πŸπŸ“πŸŽπ’•)
𝒅𝒕
= πŽπ’„ + πŸπ… × πŸ‘πŸŽ × πŸπŸ“πŸŽ 𝐜𝐨𝐬 πŸπŸ“πŸŽπ’• − πŸπ… × πŸ’πŸŽ × πŸπŸ“πŸŽ 𝐬𝐒𝐧 πŸπŸ“πŸŽπ’•
= πŽπ’„ + πŸπ… × πŸ’πŸ“πŸŽπŸŽ 𝐜𝐨𝐬 πŸπŸ“πŸŽπ’• − πŸπ… × πŸ”πŸŽπŸŽπŸŽ 𝐬𝐒𝐧 πŸπŸ“πŸŽ 𝒕
Frequency deviation βˆ†πŽ = πŸπ… × πŸπŸ“πŸŽπŸŽ[πŸ‘ 𝐜𝐨𝐬 πŸπŸ“πŸŽπ’• − πŸ’ 𝐬𝐒𝐧 πŸπŸ“πŸŽπ’•]
= πŸ‘πŸŽπŸŽπŸŽπ…√πŸ‘πŸ + πŸ’πŸ 𝒓𝒂𝒅/𝒔𝒆𝒄
= πŸπŸ“πŸŽπŸŽπ… 𝒓𝒂𝒅/𝒔𝒆𝒄
= πŸπ… × πŸ•. πŸ“π‘² 𝒓𝒂𝒅/𝒔𝒆𝒄
βˆ†π’‡ =
βˆ†πŽ
= πŸ•. πŸ“π‘²π‘―π’›
πŸπ…
Phase deviation βˆ†π›™ is proportional to 𝜽𝟎
βˆ†π = πŸπ…√πŸ‘πŸŽπŸ + πŸ’πŸŽπŸ
= πŸπ… × πŸ“πŸŽ = πŸπŸŽπŸŽπ… 𝒓𝒂𝒅
Option (d)
15. In a FM system, a carrier of 100 MHz is modulated by a sinusoidal signal of
5 KHz. The bandwidth by Carson’s approximation is 1MHz. If y(t) =
(modulated waveform)3, then by using Carson’s approximation, the
bandwidth of y(t) around 300 MHz and the spacing of spectral components
are, respectively.
(a) 3 MHz, 5 KHz
(c) 3 MHz, 15 KHz
(b) 1 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 π’‡π’Ž = πŸ“π‘²π‘―π’›
𝑩𝑾 = 𝟐(βˆ†π’‡ + π’‡π’Ž ) = πŸπ‘΄π‘―π’›
βˆ†π’‡ + π’‡π’Ž = πŸ“πŸŽπŸŽ 𝑲𝑯𝒛
βˆ†π’‡ = πŸ’πŸ—πŸ“ 𝑲𝑯𝒛
The nth order non-linearity makes the carrier frequency and frequency
deviation increased by n-fold, with baseband frequency fm unchanged.
(βˆ†π’‡ )π’π’†π’˜ = πŸ‘ × πŸ’πŸ—πŸ“
= πŸπŸ’πŸ–πŸ“ 𝑲𝑯𝒛
π‘΅π’†π’˜ 𝑩𝑾 = 𝟐(πŸπŸ’πŸ–πŸ“ + πŸ“) × πŸπŸŽπŸ‘
= 𝟐. πŸ—πŸ– 𝑴𝑯𝒛
≈ πŸ‘ 𝑴𝑯𝒛
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
(c) 9
(b) 3
(d) 27
[GATE 1996: 2 Marks]
Soln. The frequency modulation index β is multiplied by n in ntimes
frequency multiplier.
𝑺𝒐, 𝜷′ = πŸ‘ × πŸ—
= 27
Option (d)
17. A signal π‘₯(𝑑) = 2 cos(πœ‹. 104 𝑑) 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
(c) 4/π
(b) 2
(d) 2/π
[GATE 1989: 2 Marks]
Soln. Modulation index
𝜷=
𝑲𝒇 π‘¨π’Ž
π’‡π’Ž
𝑲𝒇 = πŸπŸŽπ‘²π‘―π’›/𝒗𝒐𝒍𝒕
Am is the amplitude of modulating signal
fm is the modulating frequency
𝜷=
𝟏𝟎 × πŸπŸŽπŸ‘ × πŸ
𝝅×πŸπŸŽπŸ’
πŸπ…
Option (a)
=πŸ’
18. A carrier 𝐴𝐢 cos πœ”π‘ 𝑑 is frequency modulated by a signal πΈπ‘š cos πœ”π‘š 𝑑.The
modulation index is mf. The expression for the resulting FM signal is
(a) 𝐴𝑐 cos[πœ”π‘ 𝑑 + π‘šπ‘“ sin πœ”π‘š 𝑑]
(b) 𝐴𝑐 cos[πœ”π‘ 𝑑 + π‘šπ‘“ cos πœ”π‘š 𝑑]
(c) 𝐴𝑐 cos[πœ”π‘ 𝑑 + 2πœ‹ π‘šπ‘“ sin πœ”π‘š 𝑑]
(d) 𝐴𝐢 cos [πœ”π‘ 𝑑 +
2πœ‹π‘šπ‘“ πΈπ‘š
πœ”π‘š
cos πœ”π‘š 𝑑]
[GATE 1989: 2 Marks]
Soln. The frequency modulated signal
𝑽𝑭𝑴 (𝒕) = 𝑨𝒄 𝐜𝐨𝐬[πŽπ’„ 𝒕 + 𝑲𝒇 ∫ π’Ž(𝒕)𝒅𝒕]
Kf is the frequency sensitivity of the modulator
∫ π’Ž(𝒕)𝒅𝒕 = ∫ π‘¬π’Ž 𝐜𝐨𝐬 πŽπ’Ž 𝒕 𝒅𝒕 =
𝑽𝑭𝑴 (𝒕) = 𝑨𝒄 𝐜𝐨𝐬 [πŽπ’„ 𝒕 +
π‘¬π’Ž 𝐬𝐒𝐧 πŽπ’Ž 𝒕
πŽπ’Ž
𝑲𝒇 π‘¬π’Ž
𝐬𝐒𝐧 πŽπ’Ž 𝒕]
πŽπ’Ž
= 𝑨𝒄 𝐜𝐨𝐬[πŽπ’„ 𝒕 + π’Žπ’‡ 𝐬𝐒𝐧 πŽπ’Ž 𝒕]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
multiply
Ideal low
Pass filter
Cut of f=fm
Pass band
gain=1
Output signal
(a)
(b)
π‘ƒπ‘š cos πœƒ
(c)
2
π‘ƒπ‘š
(d)
4
π‘ƒπ‘š 𝑠𝑖𝑛2 πœƒ
4
π‘ƒπ‘š π‘π‘œπ‘  2 πœƒ
4
[GATE 2000: 2 Marks]
Soln. Output of the multiplier = π’Ž(𝒕) 𝐜𝐨𝐬 𝝎𝟎 𝒕 𝒄𝒐𝒔(𝝎𝟎 𝒕 + 𝜽)
=
π’Ž(𝒕)
[𝐜𝐨𝐬(𝟐𝝎𝟎 𝒕 + 𝜽) + 𝐜𝐨𝐬 𝜽]
𝟐
Output of LPF π‘½πŸŽ (𝒕) =
π’Ž(𝒕)
𝟐
𝐜𝐨𝐬 𝜽
𝟏
= 𝐜𝐨𝐬 𝜽 π’Ž(𝒕)
𝟐
Power of output signal
𝑻
𝟏
= π₯𝐒𝐦 ∫ π‘½πŸπŸŽ (𝒕)𝒅𝒕
𝑻→∞ 𝑻
𝟎
𝟏 π‘ͺπ’π’”πŸ 𝜽 𝟐
= π₯𝐒𝐦 ∫
π’Ž (𝒕)𝒅𝒕
𝑻→∞ 𝑻
πŸ’
𝑻
π’„π’π’”πŸ 𝜽
𝟏
=
π₯𝐒𝐦 ∫ π’ŽπŸ (𝒕)𝒅𝒕
πŸ’ 𝑻→∞ 𝑻
𝟎
π’„π’π’”πŸ 𝜽
=
π‘·π’Ž
πŸ’
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œ‹(1008 × 103 𝑑)] in the FM signal (in terms of the Bessel
coefficients) is
5
(a) 5𝐽4 (3)
(c) 𝐽8 (4)
2
5
(b) 𝐽8 (3)
(d) 5𝐽4 (6)
2
[GATE 2003: 2 Marks]
Soln.
∞
𝑽𝑭𝑴 (𝒕) = ∑ 𝑨𝑱𝒏 (π’Žπ’‡ ) 𝐜𝐨𝐬(πŽπ’„ + π’πŽπ’Ž )𝒕
𝒏=−∞
Peak frequency deviation of FM signal is three times the bandwidth of
AM signal
πœΉπ’‡ = πŸ‘ × πŸπ’‡π’Ž = πŸ”π’‡π’Ž
Modulation index
π’Žπ’‡ =
πœΉπ’‡ πŸ”π’‡π’Ž
=
=πŸ”
π’‡π’Ž
π’‡π’Ž
πŸ“ 𝐜𝐨𝐬[πŸπ…(πŸπŸŽπŸŽπŸ– × πŸπŸŽπŸ‘ 𝒕)] = πŸ“ 𝐜𝐨𝐬[πŸπ…(𝟏𝟎𝟎𝟎 + πŸ’ × πŸ) × πŸπŸŽπŸ‘ ]
𝒏=πŸ’
The required coefficient is πŸ“π‘±πŸ’ (πŸ”)
Option (d)
21. Consider a system shown in the figure. Let 𝑋(𝑓) π‘Žπ‘›π‘‘ π‘Œ(𝑓) denote the
Fourier transforms of π‘₯(𝑑)π‘Žπ‘›π‘‘ 𝑦(𝑑) respectively. The ideal HPF has the
cutoff frequency 10 KHz.
x(t)
Balanced
Modulator
Balanced
Modulator
HPF
10KHz
10 KHz
13 KHz
X(f)
f (KHz)
-3
-1
1
3
The positive frequencies where Y(f) has spectral peaks are
(a) 1 KHz and 24 KHz
(c) 1 KHz and 14 KHz
(b) 2 KHz and 24 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
𝒇𝒄 ± 𝟏, 𝒇𝒄 ± (−𝟏)𝒇𝒄 ± 𝟏 = 𝟏𝟎 ± 𝟏
= 𝟏𝟏 𝒂𝒏𝒅 πŸ— 𝑲𝑯𝒛
𝒇𝒄 ± (−𝟏) = 𝟏𝟎 ± (−𝟏) = πŸ— 𝑲𝑯𝒛 𝒂𝒏𝒅 𝟏𝟏 𝑲𝑯𝒛
9 MHz will be filtered out by the HPF
After passing through 13 KHz balanced modulator, the signal will have
πŸπŸ‘ ± 𝟏𝟏 frequencies.
π’š(𝒇) = πŸπŸ’ 𝑲 𝒂𝒏𝒅 𝟐 𝑲
Option (b)
Common Data for Questions 22 & 23
Consider the following Amplitude Modulated (AM) signal,
Where π‘“π‘š < 𝐡
𝑋𝐴𝑀 (𝑑) = 10(1 + 0.5 sin 2πœ‹ π‘“π‘š 𝑑) cos 2πœ‹π‘“π‘ 𝑑
22. The average side-band power for the AM signal given above is
(a) 25
(c) 6.25
(b) 12.5
(d) 3.125
[GATE 2006: 2 Marks]
Soln. The average sideband power for the AM signal is
𝑷𝑺𝑩
π’Žπ’‚πŸ
= 𝑷𝒄
𝟐
𝑷𝒄 → π’„π’‚π’“π’“π’Šπ’†π’“ π’‘π’π’˜π’†π’“
π’Žπ’‚ → π’Žπ’π’…π’–π’π’‚π’•π’Šπ’π’ π’Šπ’π’…π’†π’™
π‘¨π’„πŸ 𝟏𝟎𝟐
𝑷𝒄 =
=
𝟐
𝟐
= πŸ“πŸŽπŽ
π’Žπ’‚ = 𝟎. πŸ“
So,
𝑷𝑺𝑩
=
(𝟎. πŸ“)𝟐
= πŸ“πŸŽ
𝟐
πŸ“πŸŽ × πŸŽ. πŸπŸ“
𝟐
= πŸ”. πŸπŸ“ π’˜π’‚π’•π’•π’”
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)
(b)
25
(c)
8𝑁0 𝐡
25
(d)
4𝑁0 𝐡
25
2𝑁0 𝐡
25
𝑁0𝐡
[GATE 2006: Marks]
Soln. The AM signal gets added to a noise with spectral density 𝒔𝒏 (𝒇)
The noise power
∞
𝑷𝑻 = ∫ 𝒔𝒏 (𝒇)𝒅𝒇
−∞
∞
= 𝟐 ∫ 𝒔𝒏 (𝒇)𝒅𝒇
𝟎
𝟏
𝑩
π‘΅πŸŽ
𝟐
𝟏
𝟐
Noise power = πŸ’ [ × ×
]
= π‘΅πŸŽ 𝑩
Power in sidebands 𝑷𝑺𝑩 =
𝑷𝑺𝑩
π’π’π’Šπ’”π’† π’‘π’π’˜π’†π’“
=
πŸπŸ“
πŸπŸ“
πŸ’π‘΅πŸŽ 𝑩
πŸ’
π’˜π’‚π’•π’•π’”
π’π’‘π’•π’Šπ’π’ (𝒃)
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