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Angle Modulated Systems

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Angle Modulated Systems 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 푲푷 ퟖ흅 = 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 th The n 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.
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