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Ch 5 Angle Modulations and Demodulations ENGR 4323/5323 igital and Analog Communication Engineering and Physics University of Central Oklahoma Dr. Mohamed Bingabr
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Ch 5 Angle Modulations and Demodulations

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ENGR 4323/5323 Digital and Analog Communication. Ch 5 Angle Modulations and Demodulations. Engineering and Physics University of Central Oklahoma Dr. Mohamed Bingabr. Chapter Outline. Nonlinear Modulation Bandwidth of Angle Modulation Generating of FM Waves Demodulation of FM Signals - PowerPoint PPT Presentation
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Page 1: Ch  5 Angle Modulations and Demodulations

Ch 5Angle Modulations and Demodulations

ENGR 4323/5323Digital and Analog Communication

Engineering and PhysicsUniversity of Central Oklahoma

Dr. Mohamed Bingabr

Page 2: Ch  5 Angle Modulations and Demodulations

Chapter Outline

• Nonlinear Modulation

• Bandwidth of Angle Modulation

• Generating of FM Waves

• Demodulation of FM Signals

• Effects of Nonlinear Distortion and Interference

• Superheterodyne Analog AM/FM Receivers

• FM Broadcasting System

2

Page 3: Ch  5 Angle Modulations and Demodulations

Baseband Vs. Carrier CommunicationsAngle Modulation: The generalized angle θ(t) of a sinusoidal signal is varied in proportion the message signal m(t).

Two types of Angle Modulation• Frequency Modulation: The frequency of the carrier signal

is varied in proportion to the message signal.

• Phase Modulation: The phase of the carrier signal is varied in proportion to the message signal.

𝜑 (𝑡 )=𝐴𝑐𝑜𝑠 𝜃(𝑡 )𝜑 (𝑡 )=𝐴cos (𝜔𝐶𝑡+𝜃0 ) 𝑡1<𝑡<𝑡2

Instantaneous Frequency

𝜔 𝑖(𝑡 )=𝑑𝜃𝑑𝑡

3

Page 4: Ch  5 Angle Modulations and Demodulations

Frequency and Phase Modulation

Phase Modulation:

Frequency Modulation:

𝜃 (𝑡 )=𝜔𝐶𝑡+𝑘𝑝𝑚 (𝑡 )𝜑𝑃𝑀 (𝑡 )=𝐴 cos [𝜔𝐶𝑡+𝑘𝑝𝑚 (𝑡)]

𝜔 𝑖 (𝑡 )=𝜔𝑐+𝑘 𝑓𝑚(𝑡)

𝜃 (𝑡 )=∫− ∞

𝑡

𝜔𝑖 (𝑡 )𝑑𝑡=𝜔𝑐 𝑡+𝑘 𝑓 ∫− ∞

𝑡

𝑚 (𝛼 )𝑑𝛼

𝜑𝐹𝑀 (𝑡 )=𝐴cos [𝜔𝐶𝑡+𝑘 𝑓 ∫− ∞

𝑡

𝑚 (𝛼 )𝑑𝛼]Power of an Angle-Modulated wave is constant and equal A2/2. 4

𝜑 (𝑡 )=𝐴𝑐𝑜𝑠 𝜃(𝑡)

Page 5: Ch  5 Angle Modulations and Demodulations

Example of FM and PM Modulation

Sketch FM and PM waves for the modulating signal m(t). The constants kf and kp are 2π x105 and 10π, respectively, and the carrier frequency fc is 100 MHz.

5

Page 6: Ch  5 Angle Modulations and Demodulations

Example of FM and PM Modulation

Sketch FM and PM waves for the digital modulating signal m(t). The constants kf and kp are 2π x105 and π/2, respectively, and the carrier frequency fc is 100 MHz.

Frequency Shift Keying (FSK) Phase Shift Keying (PSK)

Note: for discontinuous signal kp should be small to restrict the phase change kpm(t) to the range (-π,π). 6

Page 7: Ch  5 Angle Modulations and Demodulations

PSK

7

Page 8: Ch  5 Angle Modulations and Demodulations

Bandwidth of Angle Modulated Waves

where

The bandwidth of a(t), a2(t), an(t) are B, 2B, and nB Hz, respectively. From the above equation it seems the bandwidth of angle modulation is infinite but for practical reason most of the power reside at B Hz since higher terms have small power because of n!.

𝜑𝐹𝑀 (𝑡 )=𝐴cos [𝜔𝐶𝑡+𝑘 𝑓 𝑎(𝑡)] 𝑎(𝑡)=∫−∞

𝑡

𝑚 (𝛼 )𝑑𝛼

𝜑𝐹𝑀 (𝑡 )=𝐴ℜ¿¿Expand

𝑒 𝑗𝑘 𝑓 𝑎 (𝑡 )using power series expansion

𝜑𝐹𝑀 (𝑡 )=𝐴ℜ[{1+ 𝑗 𝑘𝑓 𝑎 (𝑡 )−𝑘 𝑓

2

2!𝑎2 (𝑡 )+…+ 𝑗𝑛 𝑘 𝑓

𝑛

𝑛 !𝑎𝑛 (𝑡 )+…}𝑒 𝑗 𝜔𝐶𝑡 ]

𝜑𝐹𝑀 (𝑡 )=𝐴 [𝑐𝑜𝑠𝜔𝑐 𝑡−𝑘 𝑓 𝑎 (𝑡 )𝑠𝑖𝑛𝜔𝑐𝑡−𝑘 𝑓

2

2!𝑎2 (𝑡 )𝑐𝑜𝑠𝜔𝑐𝑡+

𝑘𝑓3

3 !𝑎3 (𝑡 )𝑠𝑖𝑛𝜔𝑐𝑡+…]

8

Page 9: Ch  5 Angle Modulations and Demodulations

Narrowband PM and FM

When kf a(t) << 1

𝜑𝐹𝑀 (𝑡 )=𝐴 [𝑐𝑜𝑠𝜔𝑐 𝑡−𝑘 𝑓 𝑎 (𝑡 )𝑠𝑖𝑛𝜔𝑐𝑡−𝑘 𝑓

2

2!𝑎2 (𝑡 )𝑐𝑜𝑠𝜔𝑐𝑡+

𝑘𝑓3

3 !𝑎3 (𝑡 )𝑠𝑖𝑛𝜔𝑐𝑡+…]

𝜑𝐹𝑀 (𝑡 ) ≈ 𝐴 [𝑐𝑜𝑠 𝜔𝑐 𝑡−𝑘 𝑓 𝑎 (𝑡 )𝑠𝑖𝑛𝜔𝑐 𝑡 ]The above signal is Narrowband FM and its bandwidth is 2B Hz.

𝜑𝑃𝑀 (𝑡 )≈ 𝐴 [𝑐𝑜𝑠𝜔𝑐𝑡−𝑘𝑝𝑚 (𝑡 )𝑠𝑖𝑛𝜔𝑐 𝑡 ]

9

Same steps can be carried out to find the Narrowband PM.

Note: the above equation is similar to amplitude modulation but the waveform are different.

Page 10: Ch  5 Angle Modulations and Demodulations

𝜑𝑃𝑀 (𝑡 )=𝐴 cos [𝜔𝐶𝑡+𝑘𝑝𝑚 (𝑡)]𝜑𝐹𝑀 (𝑡 )=𝐴cos [𝜔𝐶𝑡+𝑘 𝑓 ∫− ∞

𝑡

𝑚 (𝛼 )𝑑𝛼]

m(t)

fc = 300 FM modulation kf = 120 kf a(t) = 6 NBFM (kf a(t)max <<1)

fc = 300 PM modulation kp = 10 kp mp = 31.4 NBPM (kpm(t)max << 1)

Page 11: Ch  5 Angle Modulations and Demodulations

fc = 300 FM modulation kf = 10 kf a(t) = 0.5 NBFM (kf a(t)max <<1)

fc = 300 PM modulation kp = 0.1 kp mp = 0.314 NBPM (kpm(t)max << 1)

m(t)

𝜑𝑁𝐵𝐹𝑀 (𝑡 )≈ 𝐴 [𝑐𝑜𝑠𝜔𝑐𝑡−𝑘 𝑓 𝑎 (𝑡 ) 𝑠𝑖𝑛𝜔𝑐𝑡 ]𝜑𝑁𝐵𝑃𝑀 (𝑡 ) ≈ 𝐴 [𝑐𝑜𝑠𝜔𝑐 𝑡−𝑘𝑝𝑚 (𝑡 )𝑠𝑖𝑛𝜔𝑐𝑡 ]

𝑎 (𝑡 )=∫− ∞

𝑡

𝑚 (𝛼 )𝑑𝛼

Page 12: Ch  5 Angle Modulations and Demodulations

12

ts=1.e-4; kf = 10*pi; kp = 0.1*pi;fc = 300; t=-0.05:ts:0.05; Ta = 0.02;triangle = @(z)(1-abs(z)).*(z>=-1).*(z<1); m_sig = 1*(triangle((t+0.01)/Ta) - triangle((t-0.01)/Ta));m_intg = ts*cumsum(m_sig);s_fm = cos(2*pi*fc*t + kf*m_intg);s_pm = cos(2*pi*fc*t + kp*m_sig);subplot(311);plot(t, m_sig)subplot(312);plot(t, s_fm)subplot(313);plot(t, s_pm)s_nbfm = cos(2*pi*fc*t) - kf*m_intg.*sin(2*pi*fc*t);s_nbpm = cos(2*pi*fc*t) - kp*m_sig.*sin(2*pi*fc*t);figure;subplot(311);plot(t, m_sig)subplot(312);plot(t,s_nbfm)subplot(313);plot(t, s_nbpm)figure; plot(t, m_intg)

Code to demonstrate the effect of the condition for NBFM (kf a(t)max <<1) and NBPM (kpm(t)max << 1)

Page 13: Ch  5 Angle Modulations and Demodulations

Wideband FM (WBFM)

In many application FM signal is meaningful only if its frequency deviation is large enough, so kf a(t) << 1 is not satisfied, and narrowband analysis is not valid.

𝑟𝑒𝑐𝑡 (2𝐵𝑡 )𝑐𝑜𝑠 [{𝜔𝑐+𝑘𝑓 𝑚( 𝑡𝑘 ) }𝑡 ]Fourier Transform

Hz

13

+

𝐵𝐹𝑀=2(𝑘 𝑓𝑚𝑝

2𝜋 +2𝐵) Hz

Page 14: Ch  5 Angle Modulations and Demodulations

Wideband FM (WBFM)

Hz

¿2𝐵 (𝛽+1 ) 𝛽=∆ 𝑓𝐵 is the deviation ratioWhere

When Δf >> B the modulation is WBFM and the bandwidth is BFM = 2 Δf

When Δf << B the modulation is NBFM and the bandwidth is BFM = 2B 14

A better estimate: Carson’s rule

Hz

∆ 𝑓 =𝑘 𝑓𝑚𝑝

2𝜋Peak frequency deviation in hertz𝐵𝐹𝑀=2(𝑘 𝑓𝑚𝑝

2𝜋 +2𝐵)

Page 15: Ch  5 Angle Modulations and Demodulations

Wideband PM (WBPM)

The instantaneous frequency

𝜔 𝑖=𝜔𝑐+𝑘𝑝�̇�(𝑡)

∆ 𝑓 =𝑘𝑝�̇�𝑝

2𝜋 Hz

15

Page 16: Ch  5 Angle Modulations and Demodulations

Example

a) Estimate BFM and BPM for the modulating signal m(t) for kf =2π x105 and kp = 5π. Assume the essential bandwidth of the periodic m(t) as the frequency of its third harmonic.

b) Repeat the problem if the amplitude of m(t) is doubled.

c) Repeat the problem if the time expanded by a factor of 2: that is, if the period of m(t) is 0.4 msec.

16

Page 17: Ch  5 Angle Modulations and Demodulations

Example

An angle-modulated signal with carrier frequency ωc = 2 105 is described by the equation

𝜑𝐸𝑀 (𝑡 )=10𝑐𝑜𝑠 (𝜔𝑐𝑡+5 𝑠𝑖𝑛3000 𝑡+10 𝑠𝑖𝑛2 000𝜋𝑡 )

a) Find the power of the modulated signal.b) Find the frequency deviation Δf.c) Find the deviation ration .d) Find the phase deviation Δø.e) Estimate the bandwidth of .

17

Read the Historical Note: Edwin H. Armstrong (page 270)

Page 18: Ch  5 Angle Modulations and Demodulations

Generating FM Waves

Two Methods: 1) Indirect method using NBFM Generation 2) Direct method

𝜑𝐹𝑀 (𝑡 ) ≈ 𝐴 [𝑐𝑜𝑠 𝜔𝑐 𝑡−𝑘 𝑓 𝑎 (𝑡 )𝑠𝑖𝑛𝜔𝑐 𝑡 ]𝜑𝑃𝑀 (𝑡 )≈ 𝐴 [𝑐𝑜𝑠𝜔𝑐𝑡−𝑘𝑝𝑚 (𝑡 )𝑠𝑖𝑛𝜔𝑐 𝑡 ]

18With the NBFM generation, the amplitude of the NBFM modulator will have some amplitude variation due to approximation.

NBFM Generation

Page 19: Ch  5 Angle Modulations and Demodulations

Distortion with NBFM Generation

Example 5.6: Discuss the nature of distortion inherent in the Armstrong indirect FM generator.

Amplitude and frequency distortions.

19

𝜑𝐹𝑀 (𝑡 )=𝐴 [𝑐𝑜𝑠𝜔𝑐 𝑡−𝑘 𝑓 𝑎 (𝑡 )𝑠𝑖𝑛𝜔𝑐 𝑡 ]𝜑𝐹𝑀 (𝑡 )=𝐴𝐸 (𝑡 )𝑐𝑜𝑠 [𝜔𝑐 𝑡+𝜃(𝑡 ) ]

𝐸 (𝑡 )=√1+𝑘 𝑓2 𝑎2(𝑡) 𝜃 (𝑡 )=𝑡𝑎𝑛−1 [𝑘 𝑓 𝑎(𝑡)]

𝜔 𝑖 (𝑡 )=𝜔𝑐+𝑑𝜃𝑑𝑡 =𝜔𝑐+

𝑘 𝑓 �̇�(𝑡)1+𝑘 𝑓

2 𝑎2(𝑡)=𝜔𝑐+

𝑘 𝑓𝑚 (𝑡)1+𝑘 𝑓

2 𝑎2(𝑡)

𝜔 𝑖 (𝑡 )=𝜔𝑐+𝑘 𝑓𝑚 (𝑡 ) [1 −𝑘 𝑓2 𝑎2 (𝑡 )+𝑘 𝑓

4 𝑎4 (𝑡 )− … ]

Page 20: Ch  5 Angle Modulations and Demodulations

NBFM Generation

Bandpass Limiter

𝜃 (𝑡 )=𝜔𝑐𝑡+𝑘𝑓 ∫−∞

𝑡

𝑚 (𝛼 )𝑑𝛼

𝑣 𝑖 (𝑡 )=𝐴 (𝑡 )𝑐𝑜𝑠 𝜃(𝑡 )

𝑣𝑜(𝜃)={+1𝑐𝑜𝑠 𝜃>0− 1𝑐𝑜𝑠 𝜃<0

𝑣𝑜 (𝜃 )= 4𝜋 (cos𝜃− 1

3cos 3𝜃+1

5cos5 𝜃+…)

20

Page 21: Ch  5 Angle Modulations and Demodulations

Generating FM Waves

𝑣𝑜 (𝜃 )= 4𝜋 {cos [𝜔𝑐𝑡+𝑘𝑓 ∫

−∞

𝑡

𝑚 (𝛼 )𝑑𝛼 ]− 13

cos3 [𝜔𝑐𝑡+𝑘 𝑓 ∫−∞

𝑡

𝑚 (𝛼 )𝑑𝛼]+15

cos5 [𝜔𝑐𝑡+𝑘 𝑓 ∫−∞

𝑡

𝑚 (𝛼 )𝑑𝛼]+…}𝑒𝑜 (𝑡 )= 4

𝜋 𝑐𝑜𝑠[𝜔𝑐 𝑡+𝑘 𝑓 ∫− ∞

𝑡

𝑚 (𝛼 )𝑑𝛼 ] 21

The output of the bandpass filter

𝑣𝑜 (𝜃 )= 4𝜋 (cos𝜃− 1

3cos 3𝜃+1

5cos5 𝜃+…)

𝜃 (𝑡 )=𝜔𝑐𝑡+𝑘𝑓 ∫−∞

𝑡

𝑚 (𝛼 )𝑑𝛼

Page 22: Ch  5 Angle Modulations and Demodulations

Indirect Method of Armstrong

NBFM is generated first and then converted to WBFM by using additional frequency multipliers.

Example of frequency multiplier is a nonlinear device

22

𝑦 (𝑡 )=𝑎2𝑥2(𝑡 )

𝑦 (𝑡 )=𝑎2 cos2 [𝜔𝑐𝑡+𝑘 𝑓∫𝑚 (𝛼 )𝑑𝛼 ]𝑦 (𝑡 )=0.5 𝑎2+0.5𝑎2cos [2𝜔𝑐𝑡+2𝑘 𝑓∫𝑚 (𝛼 )𝑑𝛼 ]𝑦 (𝑡 )=𝑎0+𝑎1𝑥 (𝑡 )  +𝑎2𝑥2 (𝑡 )+…+𝑎𝑛 𝑥𝑛 (𝑡 )

Devices of higher multiplier

Page 23: Ch  5 Angle Modulations and Demodulations

Indirect Method of Armstrong

For NBFM << 1. For speech fmin = 50Hz, so if Δf = 25 then 25/50 = 0.5,

23

𝜑𝐹𝑀 (𝑡 ) ≈ 𝐴 [𝑐𝑜𝑠 𝜔𝑐 𝑡−𝑘 𝑓 𝑎 (𝑡 )𝑠𝑖𝑛𝜔𝑐 𝑡 ]𝜑𝐹𝑀 (𝑡 )=𝐴𝐸 (𝑡 )𝑐𝑜𝑠 [𝜔𝑐 𝑡+𝜃(𝑡 ) ]

Page 24: Ch  5 Angle Modulations and Demodulations

Example

Design an Armstrong indirect FM modulator to generate an FM signal with carrier frequency 97.3 MHz and Δf = 10.24 kHz. A NBFM generator of fc1 = 20 kHz and Δf = 5 Hz is available. Only frequency doublers can be used as multipliers. Additionally, a local oscillator (LO) with adjustable frequency between 400 and 500 kHz is readily available for frequency mixing.

24

Page 25: Ch  5 Angle Modulations and Demodulations

Direct Generation

1. The frequency of a voltage-controlled oscillator (VCO) is controlled by the voltage m(t).

𝜔 𝑖(𝑡)=𝜔𝑐+𝑘 𝑓𝑚(𝑡 )

25

2. Use an operational amplifier to build an oscillator with variable resonance frequency ωo. The resonance frequency can be varied by variable capacitor or inductor. The variable capacitor is controlled by m(t).

𝜔𝑜=1

√𝐿𝐶=

1

√𝐿 (𝐶0 −𝑘𝑚(𝑡) )=

1

√𝐿𝐶0(1 −𝑘𝑚 (𝑡 )𝐶0 )

𝜔𝑜=1

√𝐿𝐶0[1− 𝑘𝑚 (𝑡 )𝐶0 ]

1 /2 ≈ 1√𝐿𝐶0

[1+𝑘𝑚 (𝑡 )2𝐶0 ] 𝑘𝑚 (𝑡 )

𝐶0≪1

Page 26: Ch  5 Angle Modulations and Demodulations

Direct Generation

26

𝜔𝑜=𝜔𝑐 [1+𝑘𝑚 (𝑡 )2𝐶0 ] 𝜔𝑐=

1√𝐿𝐶0

𝜔𝑜=𝜔𝑐+𝑘 𝑓𝑚(𝑡 ) 𝑘 𝑓=𝑘𝜔𝑐

2𝐶0

𝐶=𝐶0 −𝑘𝑚(𝑡)The maximum capacitance deviation is

∆𝐶=𝑘𝑚𝑝=2𝑘 𝑓 𝐶0𝑚𝑝

𝜔𝑐

∆𝐶𝐶0

=2𝑘 𝑓𝑚𝑝

𝜔𝑐=2 ∆ 𝑓

𝑓 𝑐

In practice Δf << fc

Page 27: Ch  5 Angle Modulations and Demodulations

Demodulation of FM Signals

A frequency-selective network with a transfer function of the form |H(f)|=2af + b over the FM band would yield an output proportional to the instantaneous frequency.

�̇�𝐹𝑀 (𝑡 )= 𝑑𝑑𝑡 {𝐴cos [𝜔𝐶𝑡+𝑘 𝑓 ∫

− ∞

𝑡

𝑚 (𝛼 ) 𝑑𝛼]}

27

𝜑𝐹𝑀 (𝑡 )=𝐴cos [𝜔𝐶𝑡+𝑘 𝑓 ∫− ∞

𝑡

𝑚 (𝛼 )𝑑𝛼]

�̇�𝐹𝑀 (𝑡 )=𝐴 [𝜔𝑐+𝑘 𝑓𝑚 (𝑡)] sin [𝜔𝐶𝑡+𝑘 𝑓 ∫− ∞

𝑡

𝑚 (𝛼 )𝑑𝛼 ]

Page 28: Ch  5 Angle Modulations and Demodulations

Demodulation of FM Signals

�̇�𝐹𝑀 (𝑡 )=𝐴 [𝜔𝑐+𝑘 𝑓𝑚 (𝑡)] sin [𝜔𝐶𝑡+𝑘 𝑓 ∫− ∞

𝑡

𝑚 (𝛼 )𝑑𝛼 ]

28

Differentiator

Page 29: Ch  5 Angle Modulations and Demodulations

Practical Frequency Demodulators

𝐻 ( 𝑓 )= 𝑗 2𝜋 𝑓𝑅𝐶1+ 𝑗2𝜋 𝑓𝑅𝐶 ≈ 𝑗2𝜋 𝑓𝑅𝐶 if 2𝜋 𝑓𝑅𝐶≪1

𝜔𝑐𝑎𝑟𝑟𝑖𝑒𝑟<𝜔𝑐𝑢𝑡𝑜𝑓𝑓=1

𝑅𝐶

The slope is linear over small band, so distortion occurs if the signal band is larger than the linear band.Zero-crossing detectors: First step is to use amplitude limiter and then the zero-crossing detector.

29Instantaneous frequency = the rate of zero crossing

Page 30: Ch  5 Angle Modulations and Demodulations

Effect of Nonlinear Distortion and Interference

30

Immunity of Angle Modulation to Nonlinearities

𝑦 (𝑡 )=𝑎0+𝑎1𝑥 (𝑡 )+𝑎2𝑥2 (𝑡 )+…+𝑎𝑛𝑥𝑛(𝑡)

𝑥 (𝑡 )=𝐴 cos [𝜔𝐶𝑡+𝜓 (𝑡 )]

Vulnerability of Amplitude Modulation to Nonlinearities

𝑦 (𝑡 )=𝑎𝑚 (𝑡 ) cos (𝜔𝐶𝑡)+𝑏𝑚3(𝑡)𝑐𝑜𝑠3(𝜔𝐶𝑡)

𝑦 (𝑡 )=[𝑎𝑚 (𝑡 )+ 3𝑏4

𝑚3(𝑡)] cos (𝜔𝐶 𝑡 )+𝑏4 𝑚3 (𝑡 )𝑐𝑜𝑠(3𝜔𝐶𝑡)

Page 31: Ch  5 Angle Modulations and Demodulations

Interference Effect

31

Angle Modulation is less vulnerable than AM to small-signal interference from adjacent channels.

𝑟 (𝑡 )=𝐴 cos𝜔𝑐𝑡+𝐼𝑐𝑜𝑠¿𝑟 (𝑡 )=( 𝐴+ 𝐼 𝑐𝑜𝑠𝜔𝑡)cos𝜔𝑐 𝑡− 𝐼 𝑠𝑖𝑛𝜔𝑡 sin𝜔𝑐𝑡𝑟 (𝑡 )=( 𝐴+ 𝐼 𝑐𝑜𝑠𝜔𝑡 )cos𝜔𝑐 𝑡− 𝐼 𝑠𝑖𝑛𝜔𝑡 sin𝜔𝑐𝑡

𝜓𝑑 (𝑡 )=𝑡𝑎𝑛−1 𝐼 𝑠𝑖𝑛𝜔𝑡𝐴+𝐼 𝑐𝑜𝑠𝜔𝑡 for I << A

The output of ideal phase and frequency demodulators are For PM

For FM

Interference is inversely proportional to the carrier amplitude (capture effect).

Page 32: Ch  5 Angle Modulations and Demodulations

Preemphasis and Deemphasis in FM

For audio signal the PSD is concentrated at low frequency below 2.1 kHz, so interference at high frequency will greatly deteriorate the quality of audio signal.

PreemphasisFilter

DeemphasisFilterNoise

𝐼𝐴

𝐼𝜔𝐴With white noise, the

amplitude interference is constant for PM but increase with ω for FM.

Page 33: Ch  5 Angle Modulations and Demodulations

Preemphasis and Deemphasis (PDE) in FM

Preemphasis Filter

Deemphasis Filter

2.1 kHz

30 kHz

Page 34: Ch  5 Angle Modulations and Demodulations

Preemphasis and Deemphasis (PDE) in FM

𝐻𝑝 ( 𝑓 )=𝐾𝑗2𝜋 𝑓 +𝜔1

𝑗2𝜋 𝑓 +𝜔2

𝐻𝑑 ( 𝑓 )=𝜔1

𝑗 2𝜋 𝑓 +𝜔1

For 2f << ω1 𝐻𝑝 ( 𝑓 )≅ 1

For ω1 << 2f << ω2𝐻𝑝 ( 𝑓 )≅ 𝑗2𝜋 𝑓

𝜔1

Where K is the gain and = ω2 / ω1

PDE is used in many applications such as recording of audiotape and photograph recording, where PDE depends on the band.

Page 35: Ch  5 Angle Modulations and Demodulations

FM Broadcasting Standard

Federal Communications Commission (FCC) specifications for FM communication- Frequency range = 88 to 108 MHz- Channel separations = 200 kHz,- Peak frequency deviation = 75 kHz- Transmitted signal should be received by monophonic and

stereophonic receivers.

9088 108 MHz90.2

200 KHz

75 KHz150 KHz

Filter

89.8

Page 36: Ch  5 Angle Modulations and Demodulations

Superheterodyne Analog AM/FM Receivers

fIF = 455KHz (AM radio); 10.7 MHz (FM); 38 MHz (TV)

AM stations that are 2 fIF apart are called image stations and both would appear simultaneously at the IF output.RF filter eliminates undesired image station, while IF filter eliminates undesired neighboring stations.

90 desired signal90.2111.4

90+10.7=100.7

fc = 10.7

10.7 , 190.710.5 , 190.9

9090.2

10.7, 212.1

10.7

Page 37: Ch  5 Angle Modulations and Demodulations

FM Broadcasting System

𝑚 (𝑡 )=(𝐿+𝑅 )′+(𝐿− 𝑅 )′ cos𝜔𝑐𝑡+𝛼𝑐𝑜𝑠𝜔𝑐

2𝑡

Page 38: Ch  5 Angle Modulations and Demodulations

FM Broadcasting System

𝑚 (𝑡 )=(𝐿+𝑅 )′+(𝐿− 𝑅 )′ cos𝜔𝑐𝑡+𝛼𝑐𝑜𝑠𝜔𝑐

2𝑡

Page 39: Ch  5 Angle Modulations and Demodulations

39

Homework Problem 5.4-2

Page 40: Ch  5 Angle Modulations and Demodulations

40

Homework Problem 5.6-2