Lecture 15: Frequency and Phase Modulation (FM and PM) FM.pdf · Angle Modulation (FM and PM) •...

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Lecture 15: Frequency and Phase Modulation (FM and PM)

Dr. Mohammed HawaElectrical Engineering Department

University of Jordan

EE421: Communications I: Lecture 15. For more information read Chapter 5 in your textbook or visit http://wikipedia.org/.

Copyright © Dr. Mohammed Hawa Electrical Engineering Department, University of Jordan

Just like earlier…

• Time domain expression.

• Time domain sketch.

• Average power of modulated signal.

• Frequency domain representation.

• Bandwidth of modulated signal.

• Signal-to-Noise Ratio and Quality.

• Practical Applications.

• Modulators and Demodulators (hardware).

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Angle Modulation (FM and PM)

• θ(t) is generalized angle of the modulated signal.

• ωi(t) is instantaneous frequency of modulated signal.

• θi(t) is instantaneous phase of modulated signal.

�� = � cos�� + �0�

�� = � cos ��

�� ≜ ��

�� ≜ �� − ��

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Frequency Modulation (FM)

• The instantaneous frequency of the modulated signal changes in proportion to the message.

�� = �� + � ��

�� = �� + � ! ��−∞

�� = � cos #�� + � ! ��−∞

$

�� = � ! ��−∞

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Phase Modulation (PM)

• The instantaneous phase of the modulated signal changes in proportion to the message.

�� = �%��

�� = �� + �% ��

�� = � cos &�� + �% ��'

�� = �� + ��

� = �� + ��′��

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FM and PM Equivalence

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• PM– Constant

amplitude �– Constant carrier

frequency �)– Variable

instantaneousfrequency �* ∝ �′

– Variable instantaneous phase �* ∝ �

• FM– Constant

amplitude �– Constant carrier

frequency �)– Variable

instantaneousfrequency �* ∝ �

– Variable instantaneous phase �* ∝ , � �-

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Example 1• For the following message signal m(t) and a 100 MHz

carrier:a) Sketch the FM modulated signal. Use kf = 2π×105 rad/s/V.

b) Sketch the PM modulated signal. Use kp = 5π rad/V.

c) Find ∆f for both modulated signals.

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Solution: FM

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PM

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m’(t)

t

0 V

ϕPM(t)

t

0 V

0° 0°

80,000

−80,000

200 µs

m(t)

t0 V

4

− 4

200 µs

A

−A

fmax fmin fmax fmin


FM vs. PM

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Copyright © Dr. Mohammed Hawa Electrical Engineering Department, University of Jordan 11

Peak Frequency Deviation

• For FM:

∆ ≜ /01 − /*22 = �425 6 � /01 − � /*22∆ = �445 6 � 89:89 ;<=>

• For PM:

∆ ≜ /01 − /*22 = �825 6 �′ /01 − �′ /*22∆ = �845 6 �′ 89:89 ;<=>


Example 2

• For the following message signal m(t) and a 100 MHz carrier:a) Sketch the FM modulated signal. Use kf = 2π×105 rad/s/V.b) Sketch the PM modulated signal. Use kp = π/4 rad/V.c) Find ∆f for both modulated signals.

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Solution: FM or FSK

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PMor

BPSK

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m’(t)

t

0 V

ϕPM(t)

t

0 V

χ° χ°

200 µs

m(t)

t0 V

2

− 2

200 µs

A

−A

χ°

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Homework: P.5.1-2• For the following message signal m(t) and a

200 MHz carrier:a) Sketch the FM modulated signal. Use kf = 2000π rad/s/V.b) Sketch the PM modulated signal. Use kp = π/2 rad/V.c) Try other kf and kp values. What is the effect?d) Find ∆f for both modulated signals.

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Hint: For PM

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Rules of Thumb

• Smooth change in instantaneous frequency always means smooth change in instantaneous phase.

• Sudden change in instantaneous frequency (i.e., unit step change) does not mean a sudden change in phase, i.e., it means 0° sudden phase shift.

• Impulse change in instantaneous frequency (i.e., infinity frequency) might cause a sudden change in phase. To determine the sudden phase shift (or lack thereof) see kpm(t) for PM or kf ∫m(t)dt for FM.

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FM and PM Average Power

?�� = � cos #�� + � ! ��−∞

$

?�� = � cos &�� + �% ��'

��2 ��@@@@@@@@@ = �22

��2 ��@@@@@@@@@ = �22

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FM and PM Bandwidth

• Mathematically speaking:– BFM = ∞– BPM = ∞

• Practically speaking, use Carson’s Rule:– BFM ≈ 2∆f + 2B = 2B(β+1)– BPM ≈ 2∆f + 2B = 2B(β+1)

• FM Modulation Index: – A = Δ /D– Narrow-Band FM (NBFM) has A ≪ 1 or Δ ≪ D– Wide-Band FM (WBFM) has A ≫ 1 or Δ ≫ D– FM radio uses WBFM with A = 5

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FM Bandwidth: Semi-proof

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Copyright © Dr. Mohammed Hawa Electrical Engineering Department, University of Jordan 21


Bandwidth: Example 1

• Estimate the bandwidth BFM and BPM for the modulating signal m(t) shown below. Assume kf = π×104 rad/s/V and kp = π/4 rad/V.

• Answers: BFM = 60 kHz; BPM = 40 kHz.

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Bandwidth: Example 2

• Estimate the bandwidth BFM and BPM for the modulating signal m(t) shown below. Assume kf = π×105 rad/s/V and kp = 5π rad/V.

• Answers: 220 kHz; 20 kHz; 104 kHz; 4 kHz;

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FM Signal-to-Noise Ratio

IJK�� = #3A2��2 $ I��J0D

I�� = �2��@@@@@@@ = �22

��2 = �%2�2��@@@@@@@@

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FM (and PM) vs. AM

• FM (and PM) Advantages:

– FM is capable of exchanging SNR for bandwidth.

– The constant amplitude of FM makes it less susceptible to nonlinearities.

– Due to the constant amplitude, FM is less vulnerable than AM to adjacent-channel interference.

– The constant amplitude of FM gives it a kind of immunity against rapid fading (even with the larger bandwidth).

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FM (and PM) vs. AM

• FM (and PM) Disadvantages:

–WBFM (which provides better quality) requires large transmission bandwidth.

–FM modulators and demodulators are relatively more expensive that AM hardware.

–PM demodulation requires synchronous detection (relatively expensive).

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Applications: FM Radio

• FM + FDM– The baseband message is 15 kHz (voice + music).

– With β = 5, the bandwidth of each FM station is 200 kHz (both U.S. and Europe).

– The broadcast range is 88 – 108 MHz.

• FM radio sounds better than AM radio:– m(t) has a larger bandwidth.

– WBFM: exchanging SNR for bandwidth.

– Pre-emphasis/De-emphasis improves SNR.

– Stereo FM.

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FM Superhetrodyne Receiver

• IF frequency = 10.7 MHz

• L.O. frequency = 88 + 10.7 MH to 108 + 10.7 MHz

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FM Superhetrodyne Receiver

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Stereo FM vs. Mono FM

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Lecture 15: Frequency and Phase Modulation (FM and PM) FM.pdf · Angle Modulation (FM and PM) •...

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Transcript of Lecture 15: Frequency and Phase Modulation (FM and PM) FM.pdf · Angle Modulation (FM and PM) •...