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
• Effects of Nonlinear Distortion and Interference
• Superheterodyne Analog AM/FM Receivers
• FM Broadcasting System
2

Baseband Vs. Carrier Communications
Angle Modulation: The generalized angle θ(t) of a sinusoidal signal is varied in proportion the message signal m(t).
𝜑 𝑡 = 𝐴 𝑐𝑜𝑠𝜃(𝑡) 𝜑 𝑡 = 𝐴 cos 𝜔
𝐶 𝑡 + 𝜃
0 𝑡
1
< 𝑡 < 𝑡
2
Instantaneous Frequency 𝜔 𝑖
(𝑡) = 𝑑𝜃 𝑑𝑡
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.
3

Frequency and Phase Modulation 𝜑 𝑡 = 𝐴 𝑐𝑜𝑠𝜃(𝑡)
Phase Modulation: 𝜃 𝑡 = 𝜔
𝐶 𝑡 + 𝑘 𝑝 𝑚(𝑡) 𝜑
𝑃𝑀 𝑡 = 𝐴 cos 𝜔
𝐶 𝑡 + 𝑘 𝑝 𝑚(𝑡)
Frequency Modulation: 𝜔 𝑖 𝑡 = 𝜔 𝑐
+ 𝑘 𝑓 𝑚(𝑡) 𝑡 𝑡 𝜃 𝑡 = 𝜔 𝑖
−∞ 𝑡 𝑑𝑡 = 𝜔 𝑐 𝑡 + 𝑘 𝑓 𝑡
−∞ 𝑚 𝛼 𝑑𝛼 𝜑
𝐹𝑀 𝑡 = 𝐴 cos 𝜔
𝐶 𝑡 + 𝑘 𝑓 𝑚 𝛼 𝑑𝛼
−∞
Power of an Angle-Modulated wave is constant and equal A
2
/2.
4

Example of FM and PM Modulation
Sketch FM and PM waves for the modulating signal m ( t ). The constants k f and k carrier frequency f p c are 2 π x10 5 and 10 π , respectively, and the is 100 MHz.
5

Example of FM and PM Modulation
Sketch FM and PM waves for the digital modulating signal m ( t ). The constants k f and k p and the carrier frequency f c are 2 π x10 5 and π/2 , respectively, is 100 MHz.
Frequency Shift Keying
FSK
Phase Shift Keying
PSK
Note: for discontinuous signal k p should be small to restrict the phase change k p m ( t ) to the range ( -π,π ).
6

PSK and DPSK
7

Bandwidth of Angle Modulated Waves 𝑡 𝜑
𝐹𝑀 𝑡 = 𝐴 cos 𝜔
𝐶 𝑡 + 𝑘 𝑓 𝑎(𝑡) where 𝑎(𝑡) = 𝜑
𝐹𝑀 𝑡 = 𝐴 Re 𝑒 𝑗 [𝜔
𝐶 𝑡+𝑘 𝑓 𝑎(𝑡)]
= 𝐴 Re 𝑒 𝑗𝑘 𝑓 𝑎(𝑡) 𝑒
−∞ 𝑗 𝜔
𝐶 𝑡
Expand 𝑒 𝑗𝑘 𝑓 𝑎(𝑡) using power series expansion 𝑚 𝛼 𝑑𝛼 𝜑
𝐹𝑀 𝜑
𝐹𝑀 𝑡 = 𝐴 Re 1 + 𝑗𝑘 𝑓 𝑎 𝑡 − 𝑘
2 𝑓 𝑎 2
2!
𝑡 + ⋯ + 𝑗 𝑡 = 𝐴 𝑐𝑜𝑠𝜔 𝑐 𝑡 − 𝑘 𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡 − 𝑘 2 𝑓 𝑎 2
2!
𝑛 𝑘 𝑛 𝑓 𝑛!
𝑡 𝑐𝑜𝑠𝜔 𝑐 𝑎 𝑛 𝑡 + 𝑡 + ⋯ 𝑒 𝑘 3 𝑓 𝑎 3
3!
𝑗 𝜔
𝐶 𝑡 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡 + ⋯
The bandwidth of a ( t ), a 2 ( t ), a n ( t ) are B , 2 B , 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 !
.
8

Narrowband PM and FM 𝜑
𝐹𝑀 𝑡 = 𝐴 𝑐𝑜𝑠𝜔 𝑐 𝑡 − 𝑘 𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡 − 𝑘
2 𝑓 𝑎 2
2!
𝑡 𝑐𝑜𝑠𝜔 𝑐 𝑡 + 𝑘
3 𝑓 𝑎 3
3!
𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡 + ⋯
When k f a ( t ) << 1 𝜑
𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔 𝑐 𝑡 − 𝑘 𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡
The above signal is Narrowband FM and its bandwidth is 2 B Hz.
Same steps can be carried out to find the Narrowband PM . 𝜑
𝑃𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔 𝑐 𝑡 − 𝑘 𝑝 𝑚 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡
9

Wideband FM (WBFM)
In many application FM signal is meaningful only if its frequency deviation is large enough, so k f a ( narrowband analysis is not valid. t ) << 1 is not satisfied, and 𝑟𝑒𝑐𝑡 2𝐵𝑡 𝑐𝑜𝑠 𝜔 𝑐 𝑡 + 𝑘 𝑓 𝑚 𝑡 𝑘 𝑡
Fourier Transform
1
2 𝑠𝑖𝑛𝑐 𝜔+𝜔 𝑐
+𝑘 𝑓 𝑚(𝑡 𝑘
)
4𝐵
+
1
2 𝑠𝑖𝑛𝑐 𝜔−𝜔 𝑐
−𝑘 𝑓 𝑚(𝑡 𝑘
)
4𝐵
𝐵
𝐹𝑀
1
=
2𝜋
2𝑘 𝑓 𝑚 𝑝
+ 8𝜋𝐵
Hz
10

Wideband FM (WBFM)
𝐵
𝐹𝑀
1
=
2𝜋
2𝑘 𝑓 𝑚 𝑝
+ 8𝜋𝐵 = 2 𝑘 𝑓 𝑚 𝑝
2𝜋
+ 2𝐵 = 2 ∆𝑓 + 2𝐵 Hz
∆𝑓 = 𝑘 𝑓 𝑚 𝑝
Peak frequency deviation in hertz
2𝜋
A better estimate: Carson’s rule
𝐵
𝐹𝑀
= 2 ∆𝑓 + 𝐵 Hz
∆𝑓
= 2𝐵 𝛽 + 1 Where 𝛽 = is the deviation ratio
𝐵
When Δ f >> B the modulation is WBFM and the bandwidth is
B
FM
= 2 Δ f
When Δ f << B the modulation is NBFM and the bandwidth is
B
FM
= 2B
11

Wideband PM (WBPM)
The instantaneous frequency 𝜔 𝑖
= 𝜔 𝑐
+ 𝑘 𝑝 𝑚(𝑡)
∆𝑓 = 𝑘 𝑝 𝑚 𝑝
2𝜋
𝐵
𝑃𝑀
= 2 ∆𝑓 + 𝐵 Hz
12

Example a) Estimate B
FM x10 5 and k p and B
PM for the modulating signal m ( t ) for k
= 5 π.
Assume the essential bandwidth of the f periodic m ( t ) as the frequency of its third harmonic.
=2 π 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.
13

Example
An angle-modulated signal with carrier frequency ω c described by the equation
= 2
 
10 5 is 𝜑
𝐸𝑀 𝑡 = 10𝑐𝑜𝑠 𝜔 𝑐 𝑡 + 5 𝑠𝑖𝑛3000𝑡 + 10 𝑠𝑖𝑛2000𝜋𝑡 a) Find the power of the modulated signal.
b) Find the frequency deviation Δ f .
c) Find the deviation ration

.
d) Find the phase deviation Δø .
e) Estimate the bandwidth of 𝜑
𝐸𝑀 𝑡 .
14

Indirect NPFM Generation
NBFM generation 𝜑
𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔 𝑐 𝑡 − 𝑘 𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡 𝜑
𝑃𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔 𝑐 𝑡 − 𝑘 𝑝 𝑚 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡
With this method of generation, the amplitude of the NBFM modulator will have some amplitude variation due to
15 approximation.

Generating FM Waves
Bandpass Limiter 𝑣 𝑖 𝑡 = 𝐴 𝑡 𝑐𝑜𝑠𝜃(𝑡) 𝑡 𝜃 𝑡 = 𝜔 𝑐 𝑡 + 𝑘 𝑓 𝑚 𝛼 𝑑𝛼
−∞ 𝑣 𝑜
(𝜃) =
+1 𝑐𝑜𝑠𝜃 > 0
−1 𝑐𝑜𝑠𝜃 < 0 𝑣 𝑜
4 𝜃 = 𝜋 cos 𝜃 −
1
3 cos 3𝜃 +
1
5 cos 5𝜃 + ⋯ 𝑡 𝑣 𝑜 𝜃(𝑡) = 𝑣 𝑜 𝜔 𝑐 𝑡 + 𝑘 𝑓 𝑚 𝛼 𝑑𝛼
−∞
16

Generating FM Waves 𝑡 𝜃 𝑡 = 𝜔 𝑐 𝑡 + 𝑘 𝑓 𝑚 𝛼 𝑑𝛼
−∞ 𝑣 𝑜
4 𝜃 = 𝜋 cos 𝜃 −
1
3 cos 3𝜃 +
1
5 cos 5𝜃 + ⋯ 𝑣 𝑜 𝜃 =
4 𝜋
cos 𝜔 𝑐 𝑡 + 𝑘 𝑓 𝑡 𝑚 𝛼 𝑑𝛼 −
1
3 cos 3 𝜔 𝑐 𝑡 + 𝑘 𝑓
−∞ 𝑡
−∞ 𝑚 𝛼 𝑑𝛼 𝑒 𝑡
+
1
5 cos 5 𝜔 𝑐 𝑡 + 𝑘 𝑓
The output of the bandpass filter
−∞ 𝑚 𝛼 𝑑𝛼 + ⋯ 𝑜 𝑡 = 𝑡
4 𝜋 𝑐𝑜𝑠 𝜔 𝑐 𝑡 + 𝑘 𝑓
−∞ 𝑚 𝛼 𝑑𝛼
17

Indirect Method of Armstrong
NBFM is generated first and then converted to WBFM by using additional frequency multipliers.
For NBFM

<< 1. For speech
 
25/50 = 0.5, f min
= 50Hz, so if Δ f = 25 then 𝜑
𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔 𝑐 𝑡 − 𝑘 𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡
18

Example 5.6
Discuss the nature of distortion inherent in the Armstrong indirect FM generator.
Amplitude and frequency distortions.
𝜑
𝐹𝑀 𝑡 = 𝐴 𝑐𝑜𝑠𝜔 𝑐 𝑡 − 𝑘 𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔 𝑐 𝑡 𝜑
𝐹𝑀 𝑡 = 𝐴𝐸 𝑡 𝑐𝑜𝑠 𝜔 𝑐 𝑡 + 𝜃(𝑡)
𝐸 𝑡 = 1 + 𝑘 2 𝑓 𝑎 2 (𝑡) 𝜃 𝑡 = 𝑡𝑎𝑛 −1 𝑘 𝑓 𝑎(𝑡) 𝜔 𝑖 𝑡 = 𝜔 𝑐
+ 𝑑𝜃 𝑑𝑡
= 𝜔 𝑐
+ 𝑘 𝑓
𝑎(𝑡)
1 + 𝑘 2 𝑓 𝑎 2 (𝑡)
= 𝜔 𝑐
+ 𝑘 𝑓 𝑚 (𝑡)
1 + 𝑘 2 𝑓 𝑎 2 (𝑡) 𝜔 𝑖 𝑡 = 𝜔 𝑐
+ 𝑘 𝑓 𝑚 𝑡 1 − 𝑘
2 𝑓 𝑎 2 𝑡 + 𝑘
4 𝑓 𝑎 4 𝑡 − ⋯
19

Direct Generation
1. The frequency of a voltage-controlled oscillator (VCO) is controlled by the voltage m ( t ).
𝜔 𝑖
(𝑡) = 𝜔 𝑐
+ 𝑘 𝑓 𝑚(𝑡)
2. Use an operational amplifier to build an oscillator with variable resonance frequency ω o
. The resonance frequency can be varied by variable capacitor or inductor.
The variable capacitor is controlled by m ( t ).
𝜔 𝑜
=
1
𝐿𝐶
=
1
𝐿 𝐶
0
− 𝑘𝑚(𝑡)
=
𝐿𝐶
0
1
1 − 𝑘𝑚 𝑡
𝐶
0 𝜔 𝑜
=
1
𝐿𝐶
0
1 − 𝑘𝑚 𝑡
𝐶
0
1/2
≈
1
𝐿𝐶
0
1 + 𝑘𝑚 𝑡
2𝐶
0 𝑘𝑚 𝑡
𝐶
0
≪ 1
20

Direct Generation 𝜔 𝜔 𝑜 𝑜
= 𝜔 𝑐
= 𝜔 𝑐
1 + 𝑘𝑚 𝑡
2𝐶
0
+ 𝑘 𝑓 𝑚(𝑡) 𝜔 𝑐
=
1
𝐿𝐶
0 𝑘 𝑓
= 𝑘𝜔 𝑐
2𝐶
0
The maximum capacitance deviation is
𝐶 = 𝐶
0
− 𝑘𝑚(𝑡)
∆𝐶 = 𝑘𝑚
∆𝐶
𝐶
0 𝑝
=
2𝑘 𝑓 𝑚 𝑝 𝜔 𝑐
=
2𝑘 𝑓
𝐶
0 𝑚 𝑝 𝜔 𝑐
=
2∆𝑓 𝑓 𝑐
In practice Δf << f c
21

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 f c1
= 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.
22

Demodulation of FM Signals
A frequency-selective network with a transfer function of the form | H ( f )|=2 a
 f + b over the FM band would yield an output proportional to the instantaneous frequency.
𝑡 𝜑
𝐹𝑀 𝑡 = 𝐴 cos 𝜔
𝐶 𝑡 + 𝑘 𝑓 𝑚 𝛼 𝑑𝛼
−∞ 𝑡 𝜑
𝐹𝑀 𝑑 𝑡 = 𝑑𝑡
𝐴 cos 𝜔
𝐶 𝑡 + 𝑘 𝑓
−∞ 𝑚 𝛼 𝑑𝛼
23

Demodulation of FM Signals 𝑡 𝜑
𝐹𝑀 𝑑 𝑡 = 𝑑𝑡
𝐴 cos 𝜔
𝐶 𝑡 + 𝑘 𝑓 𝑚 𝛼 𝑑𝛼
−∞ 𝑡 𝜑
𝐹𝑀 𝑡 = 𝐴 𝜔 𝑐
+ 𝑘 𝑓 𝑚(𝑡) sin 𝜔
𝐶 𝑡 + 𝑘 𝑓 𝑚 𝛼 𝑑𝛼
−∞
24

Practical Frequency Modulators 𝑗2𝜋𝑓𝑅𝐶
𝐻 𝑓 =
1 + 𝑗2𝜋𝑓𝑅𝐶
≈ 𝑗2𝜋𝑓𝑅𝐶 𝜔 𝑐
< 𝜔
0
=
1
𝑅𝐶 if 2𝜋𝑓𝑅𝐶 ≪ 1
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.
25 Instantaneous frequency = the rate of zero crossing

Effect of Nonlinear Distortion and
Interference
Immunity of Angle Modulation to Nonlinearities 𝑦 𝑡 = 𝑎
0
+ 𝑎
1 𝑥 𝑡 + 𝑎
2 𝑥 2 𝑡 + ⋯ + 𝑎 𝑛 𝑥 𝑛 (𝑡) 𝑥 𝑡 = 𝐴 cos 𝜔
𝐶 𝑡 + 𝜓(𝑡) 𝑦 𝑡 = 𝑐
0
+ 𝑐
1 cos 𝜔
𝐶 𝑡 + 𝜓(𝑡) + 𝑐
2 cos 2𝜔
𝐶 𝑡 + 2𝜓(𝑡)
+ ⋯ + 𝑐 𝑛 cos 𝑛𝜔
𝐶 𝑡 + 𝑛𝜓(𝑡)
Vulnerability of Amplitude Modulation to Nonlinearities 𝑦 𝑡 = 𝑎 𝑚 𝑡 cos(𝜔
𝐶 𝑡) + 𝑏𝑚 3 (𝑡)𝑐𝑜𝑠 3 (𝜔
𝐶 𝑡) 𝑦 𝑡 = 𝑎 𝑚 𝑡 +
3𝑏
4 𝑚 3 (𝑡) cos 𝜔
𝐶 𝑡 + 𝑏
4 𝑚 3 𝑡 𝑐𝑜𝑠(3𝜔
𝐶 𝑡)
26

Interference Effect
Angle Modulation is less vulnerable than AM to small-signal interference from adjacent channels.
𝑟 𝑡 = 𝐴 cos 𝜔 𝑐 𝑡 + 𝐼𝑐𝑜𝑠( 𝜔
𝐶
+ 𝜔)𝑡 𝑟 𝑡 = (𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡) cos 𝜔 𝑐 𝑡 − 𝐼 𝑠𝑖𝑛 𝜔𝑡 sin 𝜔 𝑐 𝑡 𝑟 𝑡 = (𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡) cos 𝜔 𝑐 𝑡 − 𝐼 𝑠𝑖𝑛 𝜔𝑡 sin 𝜔 𝑐 𝑡 𝑟 𝑡 = 𝐸 𝑟
(𝑡) cos [𝜔 𝑐 𝑡 + 𝜓 𝑑
(𝑡)]
Instantaneous frequency is 𝜔 𝑐 𝑑
(𝑡) 𝜓 𝑑 𝑡 = 𝑡𝑎𝑛 −1
𝐼 𝑠𝑖𝑛 𝜔𝑡
𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡 for I << A 𝜓 𝑑 𝑡 ≈
𝐼
𝐴 𝑠𝑖𝑛 𝜔𝑡
The output of ideal phase and frequency demodulators are 𝑦 𝑑 𝑡 =
𝐼
𝐴 𝑠𝑖𝑛 𝜔𝑡 for PM 𝑦 𝑑 𝑡 =
𝐼𝜔 𝑐𝑜𝑠 𝜔𝑡 for FM
𝐴
27

Preemphasis and Deemphasis in FM
With white noise, the amplitude interference is constant for PM but increase with ω for 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.
Preemphasis
Filter Noise
Deemphasis
Filter

Preemphasis and Deemphasis (PDE) in FM
Preemphasis Filter
Deemphasis Filter

Preemphasis and Deemphasis (PDE) in FM
𝐻 𝑝 𝑓 = 𝐾 𝑗2𝜋𝑓 + 𝜔
1 𝑗2𝜋𝑓 + 𝜔
2
For 2
 f << ω
1
For ω
1
<< 2
 f << ω
2
Where K is the gain and =
𝐻 𝑝 𝑓 ≅ 1
𝐻 𝑝 𝑓 = 𝑗2𝜋𝑓 𝜔
1
ω
2
/ ω
1
𝐻 𝑑 𝑓 = 𝐾 𝜔
1 𝑗2𝜋𝑓 + 𝜔
1
PDE is used in many applications such as recording of audiotape and photograph recording, where PDE depends on the band.

Superheterodyne Analog AM/FM Receivers f
IF
= 455KHz (AM radio);
10.7 MHz (FM);
38 MHz (TV)
AM stations that are 2 f
IF apart are called image stations and both would appear simultaneously at the IF output.
RF section filter out undesired image station, while IF section filter out undesired stations.

FM Broadcasting System
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.

FM Broadcasting System 𝑚 𝑡 = 𝐿 + 𝑅 ′ + 𝐿 − 𝑅 ′ cos 𝜔 𝑐 𝑡 + 𝛼𝑐𝑜𝑠 𝜔 𝑐 𝑡
2

% fs=400; % sampling rate
% Window = 4; % length of window in second
% t=0:1/fs:Window;
% w =-fs/2:1/Window:fs/2;
% f0= 10; % fundamental frequency
% x=2*sin(2*pi*f0*t) + sin(2*pi*3*f0*t);
% plot(t, x)
[x fs]=wavread('Fill.wav');
Window=length(x)/fs; w =-fs/2:1/Window:fs/2;
FFTx= fftshift(fft(x)); figure; plot(abs(FFTx))
FFTx(1197:1202)=0;
FFTx(900:905)=0; xM=ifft(ifftshift(FFTx)); sound(xM,fs) 34