Angle Modulated Systems
1. Consider an FM wave
푓(푡) = cos[2휋푓푐푡 + 훽1 sin 2휋푓1푡 + 훽22휋푓2푡] The maximum deviation of the instantaneous frequency from the carrier
frequency fc is
(a) 훽1푓1 + 훽2푓2 (c) 훽1 + 훽2 (b) 훽1푓2 + 훽2푓1 (d) 푓1 + 푓2 [GATE 2014: 1 Mark]
Soln. The instantaneous value of the angular frequency
풅 흎 = 흎 + (휷 퐬퐢퐧 ퟐ흅풇 풕 + 휷 퐬퐢퐧 ퟐ흅풇 풕) 풊 풄 풅풕 ퟏ ퟏ ퟐ ퟐ
흎풄 + ퟐ흅휷ퟏ풇ퟏ 퐜퐨퐬 ퟐ흅풇ퟏ풕 + ퟐ흅휷ퟐ풇ퟐ 퐜퐨퐬 ퟐ흅풇ퟐ풕
풇풊 = 풇풄 + 휷ퟏ풇ퟏ 퐜퐨퐬 ퟐ흅풇ퟏ풕 + 휷ퟐ풇ퟐ 퐜퐨퐬 ퟐ흅풇ퟐ풕
Frequency deviation (∆ ) = 휷 풇 + 휷 풇 풇 풎풂풙 ퟏ ퟏ ퟐ ퟐ Option (a)
2. A modulation signal is 푦(푡) = 푚(푡) cos(40000휋푡), 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 푥(푡) = 6푐표푠[2휋 × 103 + 2 sin(8000휋푡) + 4 cos(8000휋푡)]푉. The average power of 푥(푡) 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
푨ퟐ ퟔퟐ 풄 = ퟐ ퟐ =18 W
Option (b)
−푎푡 4. A modulation signal is given by 푠(푡) = 푒 cos[(휔푐 + ∆휔)푡] 푢(푡), where,휔푐 푎푛푑 ∆휔 are positive constants, and 휔푐 ≫ ∆휔. The complex envelope of s(t) is given by
(a) exp(−푎푡) 푒푥푝[푗(휔푐 + ∆휔)]푢(푡) (b) exp(−푎푡) exp(푗∆휔푡) 푢(푡) (c) 푒푥푝(푗∆휔푡)푢(푡)
(d) 푒푥푝[(푗휔푐 + ∆휔)푡] [GATE 1999: 1 Mark]
−풂풕 Soln. 풔(풕) = 풆 퐜퐨퐬[(흎풄 + ∆흎)풕]풖(풕)
Complex envelope 풔̃(풕) = 풔(풕)풆−풋흎풄풕
= [풆−풂풕풆풋(흎풄+∆흎)풕. 풖(풕)]풆−풋흎풄풕
= 풆−풂풕풆풋∆흎풕풖(풕)
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
푩푾 = ퟐ(∆풇 + 풇풎)
= ퟐ(ퟓퟎ + ퟎ. ퟓ)
= ퟏퟎퟏ 푲푯풛
6. 푣(푡) = 5[cos(106휋푡) − sin(103휋푡) × sin(106휋푡)] represents (a) DSB suppressed carrier signal (b) AM signal (c) SSB upper sideband signal (d) Narrow band FM signal [GATE 1994: 1 Mark]
ퟓ ퟓ Soln. 풗(풕) = ퟓ 퐜퐨퐬(ퟏퟎퟔ흅풕) − 퐜퐨퐬(ퟏퟎퟔ − ퟏퟎퟑ) 흅풕 + 퐜퐨퐬(ퟏퟎퟔ + ퟏퟎퟑ)흅풕 ퟐ ퟐ 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]
ퟎ.ퟏ푨 ퟎ.ퟏ푨 Soln. 푽 (풕) = 푨 퐜퐨퐬 흎 풕 + 퐜퐨퐬(흎 + 흎 )풕 + 퐜퐨퐬(흎 − 흎 )풕 푨푴 풄 ퟐ 풄 풎 ퟐ 풄 풎
푽푭푴(풕)(풏풂풓풓풐풘풃풂풏풅) ퟎ. ퟏ푨 ퟎ. ퟏ푨 = 푨 퐜퐨퐬 흎 풕 + 퐜퐨퐬(흎 + 흎 )풕 − 퐜퐨퐬(흎 − 흎 )풕 풄 ퟐ 풄 풎 ퟐ 풄 풎
푽푨푴(풕) + 푽푭푴(풕) = ퟐ푨 퐜퐨퐬 흎풄풕 + ퟎ. ퟏ푨 퐜퐨퐬(흎풄 + 흎풎)풕 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 푘푝/푘푓 (in rad/Hz) for the same maximum phase deviation is m(t)
-2 0 2 4 6 8 t(seconds)
-2
(a) 8π (c) 2π (b) 4π (d) π [GATE 2012: 2 Marks]
Soln. For a phase modulator, the instantaneous value of the phase angle 흍풊 is equal to phase of an unmodulated carrier 흎풄(풕) plus a time varying component proportional to modulation signal m(t)
훙푷푴(풕) = ퟐ흅풇풄풕 + 풌풑풎(풕)
Maximum phase deviation (훙푷푴)풎풂풙
= 푲풑 퐦퐚퐱 풎(풕) = ퟐ푲풑
For a frequency modulator, the instantaneous value of the angular frequency
흎풊 = 흎풄 + ퟐ흅푲풇풎(풕)
The total phase of the FM wave is
훙푭푴 = ∫ 흎풊풅풕
풕
= 흎풄풕 + ퟐ흅푲풇 ∫ 풎(풕)풅풕 ퟎ
ퟐ
(훙푭푴)풎풂풙 = ퟐ흅푲풇 ∫ ퟐ풅풕 ퟎ
= ퟖ흅푲풇
푲 ퟖ흅 푷 = 푲풇 ퟐ
= ퟒ흅
Option (b)
11. Consider the frequency modulated signal 10[cos 2휋 × 105푡 + 5 sin(2휋 × 1500푡) + 7.5 sin(2휋 × 1000푡)] 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
ퟏퟎ 퐜퐨퐬[ퟐ흅 × ퟏퟎퟓ풕 + ퟓ 퐬퐢퐧(ퟐ흅 × ퟏퟓퟎퟎ풕) + ퟕ. ퟓ 퐬퐢퐧(ퟐ흅 × ퟏퟎퟎퟎퟎ풕)]
The instantaneous value of the angular frequency 풅 흎 = 흎 + [ퟓ 퐬퐢퐧(ퟐ흅 × ퟏퟓퟎퟎ풕) + ퟕ. ퟓ 퐬퐢퐧(ퟐ흅 × ퟏퟎퟎퟎ풕)] 풊 풄 풅풕 Frequency deviation
∆흎= ퟓ × ퟐ흅 × ퟏퟓퟎퟎ 퐜퐨퐬(ퟐ흅 × ퟏퟓퟎퟎ) + ퟕ. ퟓ × ퟐ흅 × ퟏퟎퟎퟎ 퐜퐨퐬(ퟐ흅 × ퟏퟎퟎퟎ풕)
(∆흎)풎풂풙 = ퟐ흅(ퟕퟓퟎퟎ + ퟕퟓퟎퟎ)
(∆ ) Frequency deviation (휹) = 흎 풎풂풙 = ퟏퟓퟎퟎퟎ푯풛 ퟐ흅
ퟏퟓퟎퟎퟎ Modulation index 풎 = 풇 ퟏퟓퟎퟎ = 10
Option (b)
12. A message signal with bandwidth 10 KHz is Lower-Side Band SSB 6 modulated with carrier frequency 푓푐1 = 10 퐻푧. The resulting signal is then passed through a narrow-band frequency Modulator with carrier frequency 9 푓푐2 = 10 퐻푧. The bandwidth of the output would be (a) 4 × 104퐻푧 (c) 2 × 109퐻푧 (b) 2 × 106퐻푧 (d) 2 × 1010퐻푧 [GATE 2006: 2 Marks]
Soln. Lower side band frequency = ퟏퟎퟑ − ퟏퟎ
= ퟗퟗퟎ 푲푯풛
Considering this as the baseband signal, the bandwidth of narrow band FM
= ퟐ × ퟗퟗퟎ 푲푯풛
≈ ퟐ푴푯풛
Option (b)
13. A device with input 푥(푡) and output 푦(푡) is characterized by: 푦(푡) = 푥2(푡). 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 ∆풇= ퟗퟎ푲푯풛
Modulating signal bandwidth = 5 KMz
When FM signal is applied to doubler frequency deviation doubles.
푩. 푾 = ퟐ(∆풇 + 풇풎)
= ퟐ(ퟏퟖퟎ + ퟓ)
= ퟑퟕퟎ 푲푯풛
Option (a)
14. An angle-modulation signal is given by 푠(푡) = cos(2휋 × 2 × 106푡 + 2휋 × 30 sin 150푡 + 2휋 × 40 cos 150푡) 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
훙 = 훚퐜퐭 + 훉ퟎ 훙 = ퟐ훑 × ퟐ × ퟏퟎퟔ + ퟐ훑 × ퟑퟎ 퐬퐢퐧 ퟏퟓퟎ 퐭 + ퟐ훑 × ퟒퟎ 퐜퐨퐬 ퟏퟓퟎ퐭
Instantaneous value of angular frequency 흎풊 풅 흎 = 흎 + (ퟐ흅 × ퟑퟎ 퐬퐢퐧 ퟏퟓퟎ풕 + ퟐ흅 × ퟒퟎ 퐜퐨퐬 ퟏퟓퟎ풕) 풊 풄 풅풕
= 흎풄 + ퟐ흅 × ퟑퟎ × ퟏퟓퟎ 퐜퐨퐬 ퟏퟓퟎ풕 − ퟐ흅 × ퟒퟎ × ퟏퟓퟎ 퐬퐢퐧 ퟏퟓퟎ풕
= 흎풄 + ퟐ흅 × ퟒퟓퟎퟎ 퐜퐨퐬 ퟏퟓퟎ풕 − ퟐ흅 × ퟔퟎퟎퟎ 퐬퐢퐧 ퟏퟓퟎ 풕
Frequency deviation ∆흎= ퟐ흅 × ퟏퟓퟎퟎ[ퟑ 퐜퐨퐬 ퟏퟓퟎ풕 − ퟒ 퐬퐢퐧 ퟏퟓퟎ풕]
= ퟑퟎퟎퟎ흅√ퟑퟐ + ퟒퟐ 풓풂풅/풔풆풄
= ퟏퟓퟎퟎ흅 풓풂풅/풔풆풄
= ퟐ흅 × ퟕ. ퟓ푲 풓풂풅/풔풆풄 ∆ ∆ = 흎 = ퟕ. ퟓ푲푯풛 풇 ퟐ흅
Phase deviation ∆훙 is proportional to 휽ퟎ
∆흍 = ퟐ흅√ퟑퟎퟐ + ퟒퟎퟐ
= ퟐ흅 × ퟓퟎ = ퟏퟎퟎ흅 풓풂풅
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 풇풎 = ퟓ푲푯풛
푩푾 = ퟐ(∆풇 + 풇풎) = ퟏ푴푯풛
∆풇 + 풇풎 = ퟓퟎퟎ 푲푯풛
∆풇= ퟒퟗퟓ 푲푯풛
The nth order non-linearity makes the carrier frequency and frequency
deviation increased by n-fold, with baseband frequency fm unchanged.
(∆풇)풏풆풘 = ퟑ × ퟒퟗퟓ
= ퟏퟒퟖퟓ 푲푯풛
푵풆풘 푩푾 = ퟐ(ퟏퟒퟖퟓ + ퟓ) × ퟏퟎퟑ
= ퟐ. ퟗퟖ 푴푯풛
≈ ퟑ 푴푯풛
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.
푺풐, 휷′ = ퟑ × ퟗ
= 27
Option (d)
17. A signal 푥(푡) = 2 cos(휋. 104푡) 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 푲풇푨풎 휷 = 풇풎
푲풇 = ퟏퟎ푲푯풛/풗풐풍풕
Am is the amplitude of modulating signal
fm is the modulating frequency
ퟏퟎ × ퟏퟎퟑ × ퟐ 휷 = = ퟒ 흅×ퟏퟎퟒ ퟐ흅 Option (a)
18. A carrier 퐴퐶 cos 휔푐 푡 is frequency modulated by a signal 퐸푚 cos 휔푚 푡.The modulation index is mf. The expression for the resulting FM signal is
(a) 퐴푐 cos[휔푐푡 + 푚푓 sin 휔푚푡]
(b) 퐴푐 cos[휔푐푡 + 푚푓 cos 휔푚푡]
(c) 퐴푐 cos[휔푐푡 + 2휋 푚푓 sin 휔푚푡] 2휋푚푓퐸푚 (d) 퐴퐶 cos [휔푐푡 + cos 휔푚푡] 휔푚 [GATE 1989: 2 Marks]
Soln. The frequency modulated signal
푽푭푴(풕) = 푨풄 퐜퐨퐬[흎풄풕 + 푲풇 ∫ 풎(풕)풅풕]
Kf is the frequency sensitivity of the modulator
푬풎 퐬퐢퐧 흎풎풕 ∫ 풎(풕)풅풕 = ∫ 푬풎 퐜퐨퐬 흎풎풕 풅풕 = 흎풎
푲풇푬풎 푽푭푴(풕) = 푨풄 퐜퐨퐬 [흎풄풕 + 퐬퐢퐧 흎풎풕] 흎풎
= 푨풄 퐜퐨퐬[흎풄풕 + 풎풇 퐬퐢퐧 흎풎풕]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
Ideal low multiply Pass filter Output signal Cut of f=fm Pass band gain=1
푃 cos 휃 푃 푠푖푛2휃 (a) 푚 (c) 푚 2 4 푃푚 푃 푐표푠2휃 (b) (d) 푚 4 4 [GATE 2000: 2 Marks]
Soln. Output of the multiplier = 풎(풕) 퐜퐨퐬 흎ퟎ풕 풄풐풔(흎ퟎ풕 + 휽) 풎(풕) = [퐜퐨퐬(ퟐ흎 풕 + 휽) + 퐜퐨퐬 휽] ퟐ ퟎ
풎(풕) Output of LPF 푽 (풕) = 퐜퐨퐬 휽 ퟎ ퟐ
ퟏ = 퐜퐨퐬 휽 풎(풕) ퟐ Power of output signal 푻 ퟏ ퟐ = 퐥퐢퐦 ∫ 푽ퟎ(풕)풅풕 푻→∞ 푻 ퟎ
ퟏ 푪풐풔ퟐ휽 = 퐥퐢퐦 ∫ 풎ퟐ(풕)풅풕 푻→∞ 푻 ퟒ
푻 풄풐풔ퟐ휽 ퟏ = 퐥퐢퐦 ∫ 풎ퟐ(풕)풅풕 ퟒ 푻→∞ 푻 ퟎ
풄풐풔ퟐ휽 = 푷 ퟒ 풎 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[2휋(1008 × 103푡)] in the FM signal (in terms of the Bessel coefficients) is
(a) 5퐽 (3) 5 4 (c) 퐽8(4) 5 2 (b) 퐽 (3) 2 8 (d) 5퐽4(6) [GATE 2003: 2 Marks]
Soln. ∞
푽푭푴(풕) = ∑ 푨푱풏(풎풇) 퐜퐨퐬(흎풄 + 풏흎풎)풕 풏=−∞ Peak frequency deviation of FM signal is three times the bandwidth of AM signal
휹풇 = ퟑ × ퟐ풇풎 = ퟔ풇풎
Modulation index
휹풇 ퟔ풇풎 풎풇 = = = ퟔ 풇풎 풇풎 ퟓ 퐜퐨퐬[ퟐ흅(ퟏퟎퟎퟖ × ퟏퟎퟑ풕)] = ퟓ 퐜퐨퐬[ퟐ흅(ퟏퟎퟎퟎ + ퟒ × ퟐ) × ퟏퟎퟑ]
풏 = ퟒ
The required coefficient is ퟓ푱ퟒ(ퟔ) Option (d)
21. Consider a system shown in the figure. Let 푋(푓) 푎푛푑 푌(푓) denote the Fourier transforms of 푥(푡)푎푛푑 푦(푡) respectively. The ideal HPF has the cutoff frequency 10 KHz.
Balanced x(t) HPF Balanced Modulator 10KHz Modulator
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
풇풄 ± ퟏ, 풇풄 ± (−ퟏ)풇풄 ± ퟏ = ퟏퟎ ± ퟏ = ퟏퟏ 풂풏풅 ퟗ 푲푯풛
풇풄 ± (−ퟏ) = ퟏퟎ ± (−ퟏ) = ퟗ 푲푯풛 풂풏풅 ퟏퟏ 푲푯풛 9 MHz will be filtered out by the HPF
After passing through 13 KHz balanced modulator, the signal will have ퟏퟑ ± ퟏퟏ frequencies.
풚(풇) = ퟐퟒ 푲 풂풏풅 ퟐ 푲
Option (b)
Common Data for Questions 22 & 23
Consider the following Amplitude Modulated (AM) signal,
Where 푓푚 < 퐵 푋퐴푀(푡) = 10(1 + 0.5 sin 2휋 푓푚푡) cos 2휋푓푐푡 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
풎풂ퟐ 푷 = 푷 푺푩 풄 ퟐ
푷풄 → 풄풂풓풓풊풆풓 풑풐풘풆풓
풎풂 → 풎풐풅풖풍풂풕풊풐풏 풊풏풅풆풙
푨풄ퟐ ퟏퟎퟐ 푷 = = 풄 ퟐ ퟐ = ퟓퟎ흎
풎풂 = ퟎ. ퟓ So, (ퟎ. ퟓ)ퟐ 푷 = ퟓퟎ 푺푩 ퟐ
ퟓퟎ × ퟎ. ퟐퟓ = ퟐ = ퟔ. ퟐퟓ 풘풂풕풕풔
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:
25 25 (a) (c) 8푁0퐵 2푁0퐵 25 25 (b) (d) 4푁0퐵 푁0퐵 [GATE 2006: Marks]
Soln. The AM signal gets added to a noise with spectral density 풔풏(풇) The noise power ∞
푷푻 = ∫ 풔풏(풇)풅풇 −∞
∞
= ퟐ ∫ 풔풏(풇)풅풇 ퟎ
ퟏ 푩 푵 Noise power = ퟒ [ × × ퟎ] ퟐ ퟏ ퟐ
= 푵ퟎ푩
ퟐퟓ Power in sidebands 푷 = 풘풂풕풕풔 푺푩 ퟒ
푷 ퟐퟓ 푺푩 = 풐풑풕풊풐풏 (풃) 풏풐풊풔풆 풑풐풘풆풓 ퟒ푵ퟎ푩