angle modulated systemsย ยท 10.the signal m(t) as shown i applied both to a phase modulator (with k...
TRANSCRIPT
Angle Modulated Systems
1. Consider an FM wave
๐(๐ก) = cos[2๐๐๐๐ก + ๐ฝ1 sin 2๐๐1๐ก + ๐ฝ22๐๐2๐ก]
The maximum deviation of the instantaneous frequency from the carrier
frequency fc is
(a) ๐ฝ1๐1 + ๐ฝ2๐2
(b) ๐ฝ1๐2 + ๐ฝ2๐1
(c) ๐ฝ1 + ๐ฝ2
(d) ๐1 + ๐2
[GATE 2014: 1 Mark]
Soln. The instantaneous value of the angular frequency
๐๐ = ๐๐ +๐
๐ ๐(๐ท๐ ๐ฌ๐ข๐ง ๐๐ ๐๐๐ + ๐ท๐ ๐ฌ๐ข๐ง ๐๐ ๐๐๐)
๐๐ + ๐๐ ๐ท๐๐๐ ๐๐จ๐ฌ ๐๐ ๐๐๐ + ๐๐ ๐ท๐๐๐ ๐๐จ๐ฌ ๐๐ ๐๐๐
๐๐ = ๐๐ + ๐ท๐๐๐ ๐๐จ๐ฌ ๐๐ ๐๐๐ + ๐ท๐๐๐ ๐๐จ๐ฌ ๐๐ ๐๐๐
Frequency deviation (โ๐)๐๐๐
= ๐ท๐๐๐ + ๐ท๐๐๐
Option (a)
2. A modulation signal is ๐ฆ(๐ก) = ๐(๐ก) cos(40000๐๐ก), 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 ๐ฅ(๐ก) = 6๐๐๐ [2๐ ร 103 +
2 sin(8000๐๐ก) + 4 cos(8000๐๐ก)]๐. The average power of ๐ฅ(๐ก) 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
๐จ๐
๐
๐=
๐๐
๐
=18 W
Option (b)
4. A modulation signal is given by ๐ (๐ก) = ๐โ๐๐ก cos[(๐๐ + โ๐)๐ก] ๐ข(๐ก),
where,๐๐ ๐๐๐ โ๐ are positive constants, and ๐๐ โซ โ๐. The complex
envelope of s(t) is given by
(a) exp(โ๐๐ก) ๐๐ฅ๐[๐(๐๐ + โ๐)]๐ข(๐ก)
(b) exp(โ๐๐ก) exp(๐โ๐๐ก) ๐ข(๐ก)
(c) ๐๐ฅ๐(๐โ๐๐ก)๐ข(๐ก)
(d) ๐๐ฅ๐[(๐๐๐ + โ๐)๐ก]
[GATE 1999: 1 Mark]
Soln. ๐(๐) = ๐โ๐๐ ๐๐จ๐ฌ[(๐๐ + โ๐)๐]๐(๐)
Complex envelope ๐(๐)ฬ = ๐(๐)๐โ๐๐๐๐
= [๐โ๐๐๐๐(๐๐+โ๐)๐. ๐(๐)]๐โ๐๐๐๐
= ๐โ๐๐๐๐โ๐๐๐(๐)
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 Carsonโs rule is ___________
[GATE 1994: 1 Mark]
Soln. By carsonโs rule
๐ฉ๐พ = ๐(โ๐ + ๐๐)
= ๐(๐๐ + ๐. ๐)
= ๐๐๐ ๐ฒ๐ฏ๐
6. ๐ฃ(๐ก) = 5[cos(106๐๐ก) โ sin(103๐๐ก) ร sin(106๐๐ก)] represents
(a) DSB suppressed carrier signal
(b) AM signal
(c) SSB upper sideband signal
(d) Narrow band FM signal
[GATE 1994: 1 Mark]
Soln. ๐(๐) = ๐ ๐๐จ๐ฌ(๐๐๐๐ ๐) โ๐
๐๐๐จ๐ฌ(๐๐๐ โ ๐๐๐) ๐ ๐ +
๐
๐๐๐จ๐ฌ(๐๐๐ + ๐๐๐)๐ ๐
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. ๐ฝ๐จ๐ด(๐) = ๐จ ๐๐จ๐ฌ ๐๐ ๐ +๐.๐๐จ
๐๐๐จ๐ฌ(๐๐ + ๐๐)๐ +
๐.๐๐จ
๐๐๐จ๐ฌ(๐๐ โ ๐๐)๐
๐ฝ๐ญ๐ด(๐)(๐๐๐๐๐๐๐๐๐๐ )
= ๐จ ๐๐จ๐ฌ ๐๐๐ +๐. ๐๐จ
๐๐๐จ๐ฌ(๐๐ + ๐๐)๐ โ
๐. ๐๐จ
๐๐๐จ๐ฌ(๐๐ โ ๐๐)๐
๐ฝ๐จ๐ด(๐) + ๐ฝ๐ญ๐ด(๐) = ๐๐จ ๐๐จ๐ฌ ๐๐๐ + ๐. ๐๐จ ๐๐จ๐ฌ(๐๐ + ๐๐)๐
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 ๐๐/๐๐ (in rad/Hz) for the same maximum phase deviation is
-2 0 2 4 6 8 t(seconds)
m(t)
-2
(a) 8ฯ
(b) 4ฯ
(c) 2ฯ
(d) ฯ
[GATE 2012: 2 Marks]
Soln. For a phase modulator, the instantaneous value of the phase angle ๐๐ is
equal to phase of an unmodulated carrier ๐๐(๐) plus a time varying
component proportional to modulation signal m(t)
๐๐ท๐ด(๐) = ๐๐ ๐๐๐ + ๐๐๐(๐)
Maximum phase deviation (๐๐ท๐ด)๐๐๐
= ๐ฒ๐ ๐ฆ๐๐ฑ ๐(๐) = ๐๐ฒ๐
For a frequency modulator, the instantaneous value of the angular
frequency
๐๐ = ๐๐ + ๐๐ ๐ฒ๐๐(๐)
The total phase of the FM wave is
๐๐ญ๐ด = โซ ๐๐๐ ๐
= ๐๐๐ + ๐๐ ๐ฒ๐ โซ ๐(๐)๐ ๐
๐
๐
(๐๐ญ๐ด)๐๐๐ = ๐๐ ๐ฒ๐ โซ ๐๐ ๐
๐
๐
= ๐๐ ๐ฒ๐
๐ฒ๐ท
๐ฒ๐=
๐๐
๐
= ๐๐
Option (b)
11. Consider the frequency modulated signal
10[cos 2๐ ร 105๐ก + 5 sin(2๐ ร 1500๐ก) + 7.5 sin(2๐ ร 1000๐ก)]
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
๐๐ ๐๐จ๐ฌ[๐๐ ร ๐๐๐๐ + ๐ ๐ฌ๐ข๐ง(๐๐ ร ๐๐๐๐๐) + ๐. ๐ ๐ฌ๐ข๐ง(๐๐ ร
๐๐๐๐๐๐)]
The instantaneous value of the angular frequency
๐๐ = ๐๐ +๐
๐ ๐[๐ ๐ฌ๐ข๐ง(๐๐ ร ๐๐๐๐๐) + ๐. ๐ ๐ฌ๐ข๐ง(๐๐ ร ๐๐๐๐๐)]
Frequency deviation
โ๐= ๐ ร ๐๐ ร ๐๐๐๐ ๐๐จ๐ฌ(๐๐ ร ๐๐๐๐) + ๐. ๐ ร ๐๐ ร ๐๐๐๐ ๐๐จ๐ฌ(๐๐ ร ๐๐๐๐๐)
(โ๐)๐๐๐ = ๐๐ (๐๐๐๐ + ๐๐๐๐)
Frequency deviation (๐น) =(โ๐)๐๐๐
๐๐ = ๐๐๐๐๐๐ฏ๐
Modulation index ๐๐ =๐๐๐๐๐
๐๐๐๐
= 10
Option (b)
12. A message signal with bandwidth 10 KHz is Lower-Side Band SSB
modulated with carrier frequency ๐๐1= 106๐ป๐ง. The resulting signal is then
passed through a narrow-band frequency Modulator with carrier frequency
๐๐2 = 109๐ป๐ง.
The bandwidth of the output would be
(a) 4 ร 104๐ป๐ง
(b) 2 ร 106๐ป๐ง
(c) 2 ร 109๐ป๐ง
(d) 2 ร 1010๐ป๐ง
[GATE 2006: 2 Marks]
Soln. Lower side band frequency = ๐๐๐ โ ๐๐
= ๐๐๐ ๐ฒ๐ฏ๐
Considering this as the baseband signal, the bandwidth of narrow band
FM
= ๐ ร ๐๐๐ ๐ฒ๐ฏ๐
โ ๐๐ด๐ฏ๐
Option (b)
13. A device with input ๐ฅ(๐ก) and output ๐ฆ(๐ก) is characterized by: ๐ฆ(๐ก) = ๐ฅ2(๐ก).
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 โ๐= ๐๐๐ฒ๐ฏ๐
Modulating signal bandwidth = 5 KMz
When FM signal is applied to doubler frequency deviation doubles.
๐ฉ. ๐พ = ๐(โ๐ + ๐๐)
= ๐(๐๐๐ + ๐)
= ๐๐๐ ๐ฒ๐ฏ๐
Option (a)
14. An angle-modulation signal is given by
๐ (๐ก) = cos(2๐ ร 2 ร 106๐ก + 2๐ ร 30 sin 150๐ก + 2๐ ร 40 cos 150๐ก)
The maximum frequency and phase deviations of s(t) are
(a) 10.5KHz, 140ฯ rad
(b) 6 KHz, 80ฯ rad
(c) 10.5 KHz, 100ฯ rad
(d) 7.5 KHz, 100ฯ rad
[GATE 2002: 2 Marks]
Soln. The total phase angle of the carrier
๐ = ๐๐๐ญ + ๐๐
๐ = ๐๐ ร ๐ ร ๐๐๐ + ๐๐ ร ๐๐ ๐ฌ๐ข๐ง ๐๐๐ ๐ญ + ๐๐ ร ๐๐ ๐๐จ๐ฌ ๐๐๐๐ญ
Instantaneous value of angular frequency ๐๐
๐๐ = ๐๐ +๐
๐ ๐(๐๐ ร ๐๐ ๐ฌ๐ข๐ง ๐๐๐๐ + ๐๐ ร ๐๐ ๐๐จ๐ฌ ๐๐๐๐)
= ๐๐ + ๐๐ ร ๐๐ ร ๐๐๐ ๐๐จ๐ฌ ๐๐๐๐ โ ๐๐ ร ๐๐ ร ๐๐๐ ๐ฌ๐ข๐ง ๐๐๐๐
= ๐๐ + ๐๐ ร ๐๐๐๐ ๐๐จ๐ฌ ๐๐๐๐ โ ๐๐ ร ๐๐๐๐ ๐ฌ๐ข๐ง ๐๐๐ ๐
Frequency deviation โ๐= ๐๐ ร ๐๐๐๐[๐ ๐๐จ๐ฌ ๐๐๐๐ โ ๐ ๐ฌ๐ข๐ง ๐๐๐๐]
= ๐๐๐๐๐ โ๐๐ + ๐๐ ๐๐๐ /๐๐๐
= ๐๐๐๐๐ ๐๐๐ /๐๐๐
= ๐๐ ร ๐. ๐๐ฒ ๐๐๐ /๐๐๐
โ๐=โ๐
๐๐ = ๐. ๐๐ฒ๐ฏ๐
Phase deviation โ๐ is proportional to ๐ฝ๐
โ๐ = ๐๐ โ๐๐๐ + ๐๐๐
= ๐๐ ร ๐๐ = ๐๐๐๐ ๐๐๐
Option (d)
15. In a FM system, a carrier of 100 MHz is modulated by a sinusoidal signal of
5 KHz. The bandwidth by Carsonโs approximation is 1MHz. If y(t) =
(modulated waveform)3, then by using Carsonโs 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 ๐๐ = ๐๐ฒ๐ฏ๐
๐ฉ๐พ = ๐(โ๐ + ๐๐) = ๐๐ด๐ฏ๐
โ๐ + ๐๐ = ๐๐๐ ๐ฒ๐ฏ๐
โ๐= ๐๐๐ ๐ฒ๐ฏ๐
The nth order non-linearity makes the carrier frequency and frequency
deviation increased by n-fold, with baseband frequency fm unchanged.
(โ๐)๐๐๐ = ๐ ร ๐๐๐
= ๐๐๐๐ ๐ฒ๐ฏ๐
๐ต๐๐ ๐ฉ๐พ = ๐(๐๐๐๐ + ๐) ร ๐๐๐
= ๐. ๐๐ ๐ด๐ฏ๐
โ ๐ ๐ด๐ฏ๐
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 ฮฒ is multiplied by n in ntimes
frequency multiplier.
๐บ๐, ๐ทโฒ = ๐ ร ๐
= 27
Option (d)
17. A signal ๐ฅ(๐ก) = 2 cos(๐. 104๐ก) 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/ฯ
(d) 2/ฯ
[GATE 1989: 2 Marks]
Soln. Modulation index
๐ท =๐ฒ๐๐จ๐
๐๐
๐ฒ๐ = ๐๐๐ฒ๐ฏ๐/๐๐๐๐
Am is the amplitude of modulating signal
fm is the modulating frequency
๐ท =๐๐ ร ๐๐๐ ร ๐
๐ ร๐๐๐
๐๐
= ๐
Option (a)
18. A carrier ๐ด๐ถ cos ๐๐ ๐ก is frequency modulated by a signal ๐ธ๐ cos ๐๐ ๐ก.The
modulation index is mf. The expression for the resulting FM signal is
(a) ๐ด๐ cos[๐๐๐ก + ๐๐ sin ๐๐๐ก]
(b) ๐ด๐ cos[๐๐๐ก + ๐๐ cos ๐๐๐ก]
(c) ๐ด๐ cos[๐๐๐ก + 2๐ ๐๐ sin ๐๐๐ก]
(d) ๐ด๐ถ cos [๐๐๐ก +2๐๐๐๐ธ๐
๐๐cos ๐๐๐ก]
[GATE 1989: 2 Marks]
Soln. The frequency modulated signal
๐ฝ๐ญ๐ด(๐) = ๐จ๐ ๐๐จ๐ฌ[๐๐๐ + ๐ฒ๐ โซ ๐(๐)๐ ๐]
Kf is the frequency sensitivity of the modulator
โซ ๐(๐)๐ ๐ = โซ ๐ฌ๐ ๐๐จ๐ฌ ๐๐๐ ๐ ๐ =๐ฌ๐ ๐ฌ๐ข๐ง ๐๐๐
๐๐
๐ฝ๐ญ๐ด(๐) = ๐จ๐ ๐๐จ๐ฌ [๐๐๐ +๐ฒ๐๐ฌ๐
๐๐๐ฌ๐ข๐ง ๐๐๐]
= ๐จ๐ ๐๐จ๐ฌ[๐๐๐ + ๐๐ ๐ฌ๐ข๐ง ๐๐๐]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) ๐๐ cos ๐
2
(b) ๐๐
4
(c) ๐๐๐ ๐๐2๐
4
(d) ๐๐๐๐๐ 2๐
4
[GATE 2000: 2 Marks]
Soln. Output of the multiplier = ๐(๐) ๐๐จ๐ฌ ๐๐๐ ๐๐๐(๐๐๐ + ๐ฝ)
=๐(๐)
๐[๐๐จ๐ฌ(๐๐๐๐ + ๐ฝ) + ๐๐จ๐ฌ ๐ฝ]
Output of LPF ๐ฝ๐(๐) =๐(๐)
๐๐๐จ๐ฌ ๐ฝ
=๐
๐๐๐จ๐ฌ ๐ฝ ๐(๐)
Power of output signal
= ๐ฅ๐ข๐ฆ๐ปโโ
๐
๐ปโซ ๐ฝ๐
๐(๐)๐ ๐
๐ป
๐
= ๐ฅ๐ข๐ฆ๐ปโโ
๐
๐ปโซ
๐ช๐๐๐๐ฝ
๐๐๐(๐)๐ ๐
=๐๐๐๐๐ฝ
๐๐ฅ๐ข๐ฆ๐ปโโ
๐
๐ปโซ ๐๐(๐)๐ ๐
๐ป
๐
=๐๐๐๐๐ฝ
๐๐ท๐
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[2๐(1008 ร 103๐ก)] in the FM signal (in terms of the Bessel
coefficients) is
(a) 5๐ฝ4(3)
(b) 5
2๐ฝ8(3)
(c) 5
2๐ฝ8(4)
(d) 5๐ฝ4(6)
[GATE 2003: 2 Marks]
Soln.
๐ฝ๐ญ๐ด(๐) = โ ๐จ๐ฑ๐(๐๐) ๐๐จ๐ฌ(๐๐ + ๐๐๐)๐
โ
๐=โโ
Peak frequency deviation of FM signal is three times the bandwidth of
AM signal
๐น๐ = ๐ ร ๐๐๐ = ๐๐๐
Modulation index
๐๐ =๐น๐
๐๐=
๐๐๐
๐๐= ๐
๐ ๐๐จ๐ฌ[๐๐ (๐๐๐๐ ร ๐๐๐๐)] = ๐ ๐๐จ๐ฌ[๐๐ (๐๐๐๐ + ๐ ร ๐) ร ๐๐๐]
๐ = ๐
The required coefficient is ๐๐ฑ๐(๐)
Option (d)
21. Consider a system shown in the figure. Let ๐(๐) ๐๐๐ ๐(๐) denote the
Fourier transforms of ๐ฅ(๐ก)๐๐๐ ๐ฆ(๐ก) 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
๐๐ ยฑ ๐, ๐๐ ยฑ (โ๐)๐๐ ยฑ ๐ = ๐๐ ยฑ ๐
= ๐๐ ๐๐๐ ๐ ๐ฒ๐ฏ๐
๐๐ ยฑ (โ๐) = ๐๐ ยฑ (โ๐) = ๐ ๐ฒ๐ฏ๐ ๐๐๐ ๐๐ ๐ฒ๐ฏ๐
9 MHz will be filtered out by the HPF
After passing through 13 KHz balanced modulator, the signal will have
๐๐ ยฑ ๐๐ frequencies.
๐(๐) = ๐๐ ๐ฒ ๐๐๐ ๐ ๐ฒ
Option (b)
Common Data for Questions 22 & 23
Consider the following Amplitude Modulated (AM) signal,
Where ๐๐ < ๐ต ๐๐ด๐(๐ก) = 10(1 + 0.5 sin 2๐ ๐๐๐ก) cos 2๐๐๐๐ก
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
๐ท๐บ๐ฉ = ๐ท๐
๐๐๐
๐
๐ท๐ โ ๐๐๐๐๐๐๐ ๐๐๐๐๐
๐๐ โ ๐๐๐ ๐๐๐๐๐๐๐ ๐๐๐ ๐๐
๐ท๐ =๐จ๐๐
๐=
๐๐๐
๐
= ๐๐๐
๐๐ = ๐. ๐
So,
๐ท๐บ๐ฉ = ๐๐(๐. ๐)๐
๐
=๐๐ ร ๐. ๐๐
๐
= ๐. ๐๐ ๐๐๐๐๐
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
8๐0๐ต
(b) 25
4๐0๐ต
(c) 25
2๐0๐ต
(d) 25
๐0๐ต
[GATE 2006: Marks]
Soln. The AM signal gets added to a noise with spectral density ๐๐(๐)
The noise power
๐ท๐ป = โซ ๐๐(๐)๐ ๐
โ
โโ
= ๐ โซ ๐๐(๐)๐ ๐
โ
๐
Noise power = ๐ [๐
๐ร
๐ฉ
๐ร
๐ต๐
๐]
= ๐ต๐๐ฉ
Power in sidebands ๐ท๐บ๐ฉ =๐๐
๐ ๐๐๐๐๐
๐ท๐บ๐ฉ
๐๐๐๐๐ ๐๐๐๐๐=
๐๐
๐๐ต๐๐ฉ ๐๐๐๐๐๐ (๐)