Transcript Continuous-Time Signal Analysis: The Fourier Transform

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
• 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 𝜔𝐶 𝑡 + 𝜃0
𝑡1 < 𝑡 < 𝑡2
Instantaneous Frequency
𝑑𝜃
𝜔𝑖 (𝑡) =
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 = 𝜔𝐶 𝑡 + 𝑘𝑝 𝑚(𝑡)
𝜑𝑃𝑀 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑝 𝑚(𝑡)
Frequency Modulation:
𝜔𝑖 𝑡 = 𝜔𝑐 + 𝑘𝑓 𝑚(𝑡)
𝑡
𝜃 𝑡 =
𝑡
𝜔𝑖 𝑡 𝑑𝑡 = 𝜔𝑐 𝑡 + 𝑘𝑓
−∞
𝑡
𝜑𝐹𝑀 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓
𝑚 𝛼 𝑑𝛼
−∞
𝑚 𝛼 𝑑𝛼
−∞
Power of an Angle-Modulated wave is constant and equal A2/2.
4
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
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
6
phase change kpm(t) to the range (-π,π).
PSK and DPSK
7
Bandwidth of Angle Modulated Waves
𝑡
𝜑𝐹𝑀 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓 𝑎(𝑡)
where 𝑎(𝑡) =
𝜑𝐹𝑀 𝑡 = 𝐴 Re 𝑒 𝑗 [𝜔𝐶 𝑡+𝑘𝑓 𝑎(𝑡)] = 𝐴 Re
𝑚 𝛼 𝑑𝛼
−∞
𝑒 𝑗𝑘𝑓𝑎(𝑡) 𝑒 𝑗 𝜔𝐶 𝑡
Expand 𝑒 𝑗𝑘𝑓 𝑎(𝑡) using power series expansion
𝜑𝐹𝑀 𝑡 = 𝐴 Re
𝜑𝐹𝑀
𝑛
𝑘𝑓2 2
𝑘
𝑓 𝑛
1 + 𝑗𝑘𝑓 𝑎 𝑡 − 𝑎 𝑡 + ⋯ + 𝑗 𝑛
𝑎 𝑡 + ⋯ 𝑒 𝑗 𝜔𝐶 𝑡
2!
𝑛!
𝑘𝑓2 2
𝑘𝑓3 3
𝑡 = 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 − 𝑎 𝑡 𝑐𝑜𝑠𝜔𝑐 𝑡 + 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 + ⋯
2!
3!
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!.
8
Narrowband PM and FM
𝜑𝐹𝑀
𝑘𝑓2 2
𝑘𝑓3 3
𝑡 = 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 − 𝑎 𝑡 𝑐𝑜𝑠𝜔𝑐 𝑡 + 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 + ⋯
2!
3!
When kf a(t) << 1
𝜑𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡
The above signal is Narrowband FM and its bandwidth is 2B Hz.
Same steps can be carried out to find the Narrowband PM.
𝜑𝑃𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑝 𝑚 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡
9
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 𝑡 + 𝑘𝑓 𝑚 𝑡𝑘 𝑡
Fourier Transform
𝜔+𝜔𝑐 +𝑘𝑓 𝑚(𝑡𝑘 )
1
𝑠𝑖𝑛𝑐
2
4𝐵
𝐵𝐹𝑀 =
1
2𝜋
+
𝜔−𝜔𝑐 −𝑘𝑓 𝑚(𝑡𝑘 )
1
𝑠𝑖𝑛𝑐
2
4𝐵
2𝑘𝑓 𝑚𝑝 + 8𝜋𝐵
Hz
10
Wideband FM (WBFM)
𝐵𝐹𝑀 =
1
2𝜋
2𝑘𝑓 𝑚𝑝 + 8𝜋𝐵
=2
𝑘𝑓 𝑚𝑝
2𝜋
+ 2𝐵 = 2 ∆𝑓 + 2𝐵 Hz
𝑘𝑓 𝑚𝑝
Peak frequency deviation in hertz
∆𝑓 =
2𝜋
A better estimate: Carson’s rule
𝐵𝐹𝑀 = 2 ∆𝑓 + 𝐵 Hz
∆𝑓
is the deviation ratio
= 2𝐵 𝛽 + 1
Where 𝛽 =
𝐵
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
11
Wideband PM (WBPM)
The instantaneous frequency
𝜔𝑖 = 𝜔𝑐 + 𝑘𝑝 𝑚(𝑡)
𝑚𝑝
∆𝑓 = 𝑘𝑝
2𝜋
𝐵𝑃𝑀 = 2 ∆𝑓 + 𝐵 Hz
12
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.
13
Example
An angle-modulated signal with carrier frequency ωc = 2 105 is
described by the equation
𝜑𝐸𝑀 𝑡 = 10𝑐𝑜𝑠 𝜔𝑐 𝑡 + 5 𝑠𝑖𝑛3000𝑡 + 10 𝑠𝑖𝑛2000𝜋𝑡
a)
b)
c)
d)
e)
Find the power of the modulated signal.
Find the frequency deviation Δf.
Find the deviation ration .
Find the phase deviation Δø.
Estimate the bandwidth of 𝜑𝐸𝑀 𝑡 .
14
Indirect NPFM Generation
NBFM generation
𝜑𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡
𝜑𝑃𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑝 𝑚 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡
With this method of generation, the amplitude of the NBFM
modulator will have some amplitude variation due to
approximation.
15
Generating FM Waves
Bandpass Limiter
𝑣𝑖 𝑡 = 𝐴 𝑡 𝑐𝑜𝑠𝜃(𝑡)
𝑡
𝜃 𝑡 = 𝜔𝑐 𝑡 + 𝑘𝑓
𝑚 𝛼 𝑑𝛼
−∞
+1
𝑣𝑜 (𝜃) =
−1
𝑐𝑜𝑠𝜃 > 0
𝑐𝑜𝑠𝜃 < 0
4
1
1
𝑣𝑜 𝜃 =
cos 𝜃 − cos 3𝜃 + cos 5𝜃 + ⋯
𝜋
3
5
𝑡
𝑣𝑜 𝜃(𝑡) = 𝑣𝑜 𝜔𝑐 𝑡 + 𝑘𝑓
𝑚 𝛼 𝑑𝛼
−∞
16
Generating FM Waves
𝑡
𝜃 𝑡 = 𝜔𝑐 𝑡 + 𝑘𝑓
4
1
1
𝑣𝑜 𝜃 =
cos 𝜃 − cos 3𝜃 + cos 5𝜃 + ⋯
𝜋
3
5
𝑚 𝛼 𝑑𝛼
−∞
4
𝑣𝑜 𝜃 =
cos 𝜔𝑐 𝑡 + 𝑘𝑓
𝜋
𝑡
−∞
1
𝑚 𝛼 𝑑𝛼 − cos 3 𝜔𝑐 𝑡 + 𝑘𝑓
3
𝑡
𝑚 𝛼 𝑑𝛼
−∞
𝑡
1
+ cos 5 𝜔𝑐 𝑡 + 𝑘𝑓
𝑚 𝛼 𝑑𝛼 + ⋯
5
−∞
The output of the bandpass filter
4
𝑒𝑜 𝑡 = 𝑐𝑜𝑠 𝜔𝑐 𝑡 + 𝑘𝑓
𝜋
𝑡
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 fmin = 50Hz, so if Δf = 25 then
  25/50 = 0.5,
𝜑𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡
18
Example 5.6
Discuss the nature of distortion inherent in the Armstrong
indirect FM generator.
Amplitude and frequency distortions.
𝜑𝐹𝑀 𝑡 = 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡
𝜑𝐹𝑀 𝑡 = 𝐴𝐸 𝑡 𝑐𝑜𝑠 𝜔𝑐 𝑡 + 𝜃(𝑡)
𝐸 𝑡 =
1 + 𝑘𝑓2 𝑎2 (𝑡)
𝜃 𝑡 = 𝑡𝑎𝑛−1 𝑘𝑓 𝑎(𝑡)
𝑘𝑓 𝑎(𝑡)
𝑘𝑓 𝑚 (𝑡)
𝑑𝜃
𝜔𝑖 𝑡 = 𝜔𝑐 +
= 𝜔𝑐 +
= 𝜔𝑐 +
2
2
𝑑𝑡
1 + 𝑘𝑓 𝑎 (𝑡)
1 + 𝑘𝑓2 𝑎2 (𝑡)
𝜔𝑖 𝑡 = 𝜔𝑐 + 𝑘𝑓 𝑚 𝑡 1 − 𝑘𝑓2 𝑎2 𝑡 + 𝑘𝑓4 𝑎4 𝑡 − ⋯
19
Direct Generation
1. The frequency of a voltage-controlled oscillator (VCO) is
controlled by the voltage m(t).
𝜔𝑖 (𝑡) = 𝜔𝑐 + 𝑘𝑓 𝑚(𝑡)
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 =
𝜔𝑜 =
1
𝐿𝐶
=
1
𝐿 𝐶0 − 𝑘𝑚(𝑡)
1
𝑘𝑚 𝑡
𝐿𝐶0 1 −
𝐶0
1/2
≈
1
=
𝐿𝐶0 1 −
1
𝐿𝐶0
1+
𝑘𝑚 𝑡
𝐶0
𝑘𝑚 𝑡
2𝐶0
𝑘𝑚 𝑡
≪1
𝐶0
20
Direct Generation
𝑘𝑚 𝑡
𝜔𝑜 = 𝜔𝑐 1 +
2𝐶0
𝜔𝑜 = 𝜔𝑐 + 𝑘𝑓 𝑚(𝑡)
𝜔𝑐 =
1
𝐿𝐶0
𝑘𝜔𝑐
𝑘𝑓 =
2𝐶0
The maximum capacitance deviation is
𝐶 = 𝐶0 − 𝑘𝑚(𝑡)
2𝑘𝑓 𝐶0 𝑚𝑝
∆𝐶 = 𝑘𝑚𝑝 =
𝜔𝑐
∆𝐶 2𝑘𝑓 𝑚𝑝 2∆𝑓
=
=
𝐶0
𝜔𝑐
𝑓𝑐
In practice Δf << fc
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 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.
22
Demodulation of FM Signals
A frequency-selective network with a transfer function of the
form |H(f)|=2af + b over the FM band would yield an output
proportional to the instantaneous frequency.
𝑡
𝜑𝐹𝑀 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓
𝑚 𝛼 𝑑𝛼
−∞
𝜑𝐹𝑀
𝑑
𝑡 =
𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓
𝑑𝑡
𝑡
𝑚 𝛼 𝑑𝛼
−∞
23
Demodulation of FM Signals
𝜑𝐹𝑀
𝑑
𝑡 =
𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓
𝑑𝑡
𝑡
𝑚 𝛼 𝑑𝛼
−∞
𝑡
𝜑𝐹𝑀 𝑡 = 𝐴 𝜔𝑐 + 𝑘𝑓 𝑚(𝑡) sin 𝜔𝐶 𝑡 + 𝑘𝑓
𝑚 𝛼 𝑑𝛼
−∞
24
Practical Frequency Demodulators
𝑗2𝜋𝑓𝑅𝐶
𝐻 𝑓 =
≈ 𝑗2𝜋𝑓𝑅𝐶
1 + 𝑗2𝜋𝑓𝑅𝐶
𝜔𝑐𝑎𝑟𝑟𝑖𝑒𝑟 < 𝜔𝑐𝑢𝑡𝑜𝑓𝑓
if
2𝜋𝑓𝑅𝐶 ≪ 1
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.
Instantaneous frequency = the rate of zero crossing
25
Effect of Nonlinear Distortion and
Interference
Immunity of Angle Modulation to Nonlinearities
𝑦 𝑡 = 𝑎0 + 𝑎1 𝑥 𝑡 + 𝑎2 𝑥 2 𝑡 + ⋯ + 𝑎𝑛 𝑥 𝑛 (𝑡)
𝑥 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝜓(𝑡)
𝑦 𝑡 = 𝑐0 + 𝑐1 cos 𝜔𝐶 𝑡 + 𝜓(𝑡) + 𝑐2 cos 2𝜔𝐶 𝑡 + 2𝜓(𝑡)
+ ⋯ + 𝑐𝑛 cos 𝑛𝜔𝐶 𝑡 + 𝑛𝜓(𝑡)
Vulnerability of Amplitude Modulation to Nonlinearities
𝑦 𝑡 = 𝑎 𝑚 𝑡 cos(𝜔𝐶 𝑡) + 𝑏𝑚3 (𝑡)𝑐𝑜𝑠 3 (𝜔𝐶 𝑡)
3𝑏 3
𝑏 3
𝑦 𝑡 = 𝑎𝑚 𝑡 +
𝑚 (𝑡) cos 𝜔𝐶 𝑡 + 𝑚 𝑡 𝑐𝑜𝑠(3𝜔𝐶 𝑡)
4
4
26
Interference Effect
Angle Modulation is less vulnerable than AM to small-signal
interference from adjacent channels.
𝑟 𝑡 = 𝐴 cos 𝜔𝑐 𝑡 + 𝐼𝑐𝑜𝑠( 𝜔𝐶 + 𝜔)𝑡
𝑟 𝑡 = (𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡) cos 𝜔𝑐 𝑡 − 𝐼 𝑠𝑖𝑛 𝜔𝑡 sin 𝜔𝑐 𝑡
𝑟 𝑡 = (𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡) cos 𝜔𝑐 𝑡 − 𝐼 𝑠𝑖𝑛 𝜔𝑡 sin 𝜔𝑐 𝑡
𝑟 𝑡 = 𝐸𝑟 (𝑡) cos [𝜔𝑐 𝑡 + 𝜓𝑑 (𝑡)]
𝜓𝑑 𝑡 =
𝑡𝑎𝑛−1
𝐼 𝑠𝑖𝑛 𝜔𝑡
𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡
for I << A
𝜓𝑑 𝑡 ≈
𝐼
𝑠𝑖𝑛
𝐴
𝜔𝑡
The output of ideal phase and frequency demodulators are
For PM 𝑦𝑑 𝑡 =
𝐼
𝑠𝑖𝑛
𝐴
For FM 𝑦𝑑 𝑡 =
𝐼𝜔
𝑐𝑜𝑠
𝐴
𝜔𝑡
𝜔𝑡
27
Preemphasis and Deemphasis in FM
𝐼
For PM 𝑦𝑑 𝑡 = 𝐴 𝑠𝑖𝑛 𝜔𝑡
For FM 𝑦𝑑 𝑡 =
𝐼𝜔
𝑐𝑜𝑠
𝐴
𝜔𝑡
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
Deemphasis
Filter
Noise
Preemphasis and Deemphasis (PDE) in FM
30 kHz
2.1 kHz
Preemphasis Filter
Deemphasis Filter
Preemphasis and Deemphasis (PDE) in FM
𝑗2𝜋𝑓 + 𝜔1
𝐻𝑝 𝑓 = 𝐾
𝑗2𝜋𝑓 + 𝜔2
Where K is the gain and = ω2/ ω1
For 2f << ω1
𝐻𝑝 𝑓 ≅ 1
For ω1 << 2f << ω2
𝑗2𝜋𝑓
𝐻𝑝 𝑓 ≅
𝜔1
𝜔1
𝐻𝑑 𝑓 =
𝑗2𝜋𝑓 + 𝜔1
PDE is used in many applications such as recording of audiotape
and photograph recording, where PDE depends on the band.
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.
150 KHz
88
90
200 KHz
75 KHz
90.2
108 MHz
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 section filter out undesired image station, while IF section
filter out undesired neighboring stations.
FM Broadcasting System
′
𝑚 𝑡 = 𝐿+𝑅 + 𝐿−𝑅
′ cos 𝜔 𝑡
𝑐
𝜔𝑐
+ 𝛼𝑐𝑜𝑠 𝑡
2
FM Broadcasting System
𝑚 𝑡 = 𝐿 + 𝑅 ′ + 𝐿 − 𝑅 ′ cos 𝜔𝑐 𝑡 + 𝛼𝑐𝑜𝑠
𝜔𝑐
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)
35
Homework Problem 5.4-2
36
Homework Problem 5.6-2
37