EE456 Digital Communications Professor Ha Nguyen September 2016 - - PowerPoint PPT Presentation
Chapter 3: Frequency Modulation (FM) EE456 Digital Communications Professor Ha Nguyen September 2016 EE456 Digital Communications 1 Chapter 3: Frequency Modulation (FM) ANALOG MODULATION Amplitude Modulation (AM) & Frequency
Chapter 3: Frequency Modulation (FM)
Professor Ha Nguyen September 2016
EE456 – Digital Communications 1
Chapter 3: Frequency Modulation (FM)
EE456 – Digital Communications 2
Chapter 3: Frequency Modulation (FM)
Modulation refers to a process that puts the message signal into a specific frequency band in order to match the transmission characteristics of the physical channel (e.g. telephone channel, wireless LAN channel, etc.) Modulation can be classified into baseband and passband. The term “baseband” refers to the frequency band of the original message signal, which is usually near the zero frequency. For example, the band of audio (voice) signals is between 0 to 3.5kHz, the video baseband occupies 0 to 4.3 MHz. In baseband modulation, the message signals are directly transmitted over the channels (e.g. twisted pairs of copper wires, coaxial cables). Passband modulation is also known as carrier modulation, in which the spectrum
sinusoidal carrier. Two important advantages of carrier modulation are: (i) To ease radio-frequency (RF) transmission, and (ii) to allow frequency division multiplexing. Carrier modulation can be analog or digital. Traditional communications such as AM/FM radios are based on analog modulation, while many current and new communications systems are all digital (cellular phone systems, HDTV, etc.)
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Chapter 3: Frequency Modulation (FM)
Let m(t) be the message signal, whose Fourier transform is denoted as M(f). To move the frequency band of m(t) to a new frequency band centered at fc, the message signal m(t) is simply multiplied by a sinusoid of frequency fc (i.e., the carrier): sDSB−SC(t) = m(t) cos(2πfct). The above signal is seen as a sinusoid of frequency fc whose amplitude is varied according to the message signal, hence the name amplitude modulation (AM). The spectrum of the AM signal s(t) is obtained as a convolution between the spectrum of m(t) and the spectrum of cos(2πfct). Since the spectrum of cos(2πfct) is 1
2δ(f + fc) + 1 2 δ(f − fc), one has
SDSB−SC(f) = M(f) ∗ 1 2 δ(f + fc) + 1 2δ(f − fc)
1 2 M(f − fc) + 1 2M(f + fc).
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Chapter 3: Frequency Modulation (FM)
( ) M f f B B − ( ) m t ( ) cos(2 )
c
c t f t π =
DSB-SC( )
( ) cos(2 )
c
s t m t f t π = × ( ) m t ( ) C f f
1 2 c
f
1 2 c
f − f
2 K c
f
2 K c
f −
DSB-SC( )
S f
c
f B +
c
f B − 2B t t
( )cos(2 )
c
m t f t π ( ) m t ( ) m t −
The process of amplitude modulation shifts the spectrum of the modulating signal to the left and to the right by fc. If the bandwidth of m(t) is B Hz, then the modulated AM signal has bandwidth
There are upper sideband (USB) and lower sideband (LSB) portions of the AM spectrum. The modulated signal does not contain a discrete component of the carrier frequency fc. For this reason, the modulated signal in this scheme is double-sideband, suppressed-carrier (DSB-SC) modulation.
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Chapter 3: Frequency Modulation (FM)
( ) cos(2 )
c
c t f t π = ( ) 2 m t 2 ( ) m t
DSB-SC
( ) ( ) r t s t = ( ) e t f
c
f
2 K c
f − ( ) E f 2
c
f
4 K
2
c
f −
4 K
f B B −
2 K
B B −
e(t) = m(t) cos2(2πfct) = 1 2m(t) + 1 2m(t) cos(2π(2fc)t) E(f) = 1 2 M(f) + 1 4[M(f + 2fc) + M(f − 2fc)] This method of AM demodulation is called synchronous detection, or coherent detection, where the receiver requires to have a carrier of exactly the same frequency (and phase) as the carrier used for modulation. In practice, such a requirement is typically fulfilled with a phase-locked loop (PLL) circuit.
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Chapter 3: Frequency Modulation (FM)
Consider a single tone baseband signal m(t) = cos(ωmt) = cos(2πfmt). Find the DSB-SC signal and sketch its spectrum. Identify the USB and LSB. Also verify that the coherent demodulation works and recovers m(t). (Partial) Solution:
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Chapter 3: Frequency Modulation (FM)
( ) cos(2 )
c
c t f t π =
DSB-SC( )
( ) cos(2 )
c
s t m t f t π = × ( ) m t cos(2 ( ) )
c
f f t π + ∆ 1 ( )cos(2 ) 2
d
m t t ft π θ − ∆ + ( ) ( ) cos(2 ( ))
c
r t m t t f t t π = − × − ( ) e t
To see the effect on the demodulated signal when the local oscillator at the receiver is not synchronized in frequency and phase with the incoming carrier, consider the case that the received signal at the receiver is a delayed version of the transmitted AM signal (due to propagation time): r(t) = m(t − t0) cos[2πfc(t − t0)] = m(t − t0) cos(2πfct − θd), where θd = 2πfct0 is the equivalent phase shift. Furthermore, assume that, due to the lack of a good PLL circuit, the oscillator in the receiver generates cos(2π(fc + ∆f)t), i.e., there is a frequency offset of ∆f compared to the carrier generated by the oscillator at the transmitter.
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Chapter 3: Frequency Modulation (FM)
Then the signal e(t) is e(t) = m(t − t0) cos(2πfct − θd) cos(2π(fc + ∆f)t) = 1 2m(t − t0) cos(2π∆ft + θd) + 1 2 m(t − t0) cos(2π(2fc + ∆f)t − θd) The first component of e(t) is the message signal modulated with a very small “carrier” frequency of ∆f, while the second component is the message signal modulated with a high carrier frequency of 2fc + ∆f. Thus, the output of the LPF will be the first component only, i.e., v(t) = 1
2m(t − t0) cos(2π∆ft + θd).
In the frequency domain, it can be shown that the spectrum of v(t) is V (f) = 1 4 ejθdM(f − ∆f)e−j2π(f−∆f)t0 + 1 4e−jθdM(f + ∆f)e−j2π(f+∆f)t0 In essence, the spectrum at the output of the LPF is the spectrum of the original message moved to ±∆f. In Lab #1 you will hear the effect of this frequency translation for different values
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Chapter 3: Frequency Modulation (FM)
( ) M f f B B − ( ) m t ( ) cos(2 )
c
c t f t π =
DSB( )
[ ( ) ] cos(2 )
c
s t t t A m f π = + × ( ) m t ( ) C f f
1 2 c
f
1 2 c
f − f
2 K c
f
2 K c
f −
DSB( )
S f
c
f B +
c
f B − 2B
∑
( ( ) ) A M f f δ + f B B − ( ) m t A +
2 A 2 A
t t t t
C
R ( ) m t t
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Chapter 3: Frequency Modulation (FM)
To enable a simpler receiver than the coherent receiver, the transmitter can send a carrier along with the modulated signal: sAM(t) = A cos(2πfct)+m(t) cos(2πfct) = [A+m(t)] cos(2πfct), where A > 0. The spectrum of sAM(t) is basically the same as that of sDSB−SC(t) = m(t) cos(2πfct), except for two additional impulses at ±fc: SAM(f) = 1 2M(f − fc) + 1 2M(f + fc) + A 2 δ(f − fc) + A 2 δ(f + fc). If the carrier component is large enough, the message signal m(t) can be recovered with a very simple envelope detector (see the next slides). The option of AM with carrier is very desirable in broadcasting systems as it
inexpensive receivers.
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Chapter 3: Frequency Modulation (FM)
The time constant of the RC circuit should be such that the capacitor discharges very little between positive peaks and in a similar rate of the AM envelope variation: 1/ωc ≪ RC < 1/(2πB),
2πB < 1 RC ≪ ωc, where B is the bandwidth of m(t). The envelope detector output is A + m(t) with a ripple frequency ωc. The DC term can be blocked by a simple highpass filter, while the ripple may be further reduced by another lowpass filter.
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Chapter 3: Frequency Modulation (FM)
By definition, the envelope of the AM signal is |A + m(t)|. If A is large enough to ensure that A + m(t) ≥ 0 for all t, then the envelope has the same shape as the message m(t). This means that we can detect the desired signal m(t) by detecting the envelope of the AM signal! If A + m(t) < 0 for some time t, then the envelope |A + m(t)| does not have the same shape as the message m(t) and envelope detection does not work correctly.
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Chapter 3: Frequency Modulation (FM)
Typically, a message signal m(t) has no DC (i.e., zero offset) since a DC does not carry any useful information. Let ±mp be the maximum and the minimum values of m(t). This means that m(t) ≥ −mp for all t. Then, the condition of envelope detection is A + m(t) ≥ 0, for all t ⇔ A + min
∀t {m(t)} ≥ 0
⇔ A + (−mp) ≥ 0 ⇔ A ≥ mp . The modulation index is defined as µ = mp A For envelope detection to work correctly, the condition is A ≥ mp and the modulation index satisfies 0 ≤ µ ≤ 1,
0 ≤ µ ≤ 100% When A < mp, µ > 100% (overmodulation) and envelope detection no longer works correctly.
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Chapter 3: Frequency Modulation (FM)
µ = mp A = (Amax − Amin)/2 (Amax + Amin)/2 = Amax − Amin Amax + Amin .
s(t) A + m(t) Amax Amin A t
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Chapter 3: Frequency Modulation (FM)
Consider the case of tone modulation in which the message signal is a pure sinusoid (e.g. a test tone) m(t) = mp cos(ωmt). Sketch the AM signals for modulation indices
Solution: sAM(t) = [A+m(t)] cos(ωct) = A
A cos(ωmt)
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Chapter 3: Frequency Modulation (FM)
Recall that the AM signal has two components: the carrier and sidebands sAM(t) = A cos(ωct)
+ m(t) cos(ωct)
The carrier does not carry any information, hence its power is wasteful from this point of view. The carrier power is Pc = A2/2. The sideband power is the power of m(t) cos(ωct). If Pm is the power of message m(t), then the sideband power is well approximated by Pm/2. The power efficiency of an AM signal is defined as: η = useful power total power = sideband power sideband power + carrier power ≈ Pm/2 Pm/2 + A2/2 = Pm A2 + Pm Example: For the special case of tone modulation, m(t) = mp cos(ωmt). Thus Pm = m2
p/2 = (µA)2/2. It then follows that η =
µ2 2 + µ2 100% Observations: Under the condition that 0 ≤ µ ≤ 1, η increases monotonically with µ and reaches a maximum value of ηmax = 33% when µ = 100%. Thus, for tone modulation, under the best condition of µ = 100%, only one-third of the transmitted power is used for carrying the message! Quiz: Determine η for tone modulation when (a) µ = 0.5, and (b) µ = 0.25.
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Chapter 3: Frequency Modulation (FM)
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Chapter 3: Frequency Modulation (FM)
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Chapter 3: Frequency Modulation (FM)
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Chapter 3: Frequency Modulation (FM)
The signals m1(t) and m2(t), both band-limited to 5 kHz, are to be transmitted simultaneously over a channel by a multiplexing scheme shown below. The signal at point b is the multiplexed signal, which then modulates a carrier of frequency 20
A
1( )
M f f 5 kHz 5 kHz − A
2( )
M f f 5 kHz 5 kHz −
1( )
m t
∑
2( )
m t 2cos(20,000 ) t π 2cos(40,000 ) t π
a b c
(a) Sketch the signal spectra at points a , b , and c . (b) What must be the bandwidth of an ideal channel to pass all the frequency content of the signal at point c undistorted? (c) Design a receiver to recover signals m1(t) and m2(t) from the modulated signal at point c .
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Chapter 3: Frequency Modulation (FM)
Solution to Part (a):
A 5 10 15 (kHz) f 5 − 10 − 15 − a Point A 5 10 15 (kHz) f 5 − 10 − 15 − b Point A 5 (kHz f 5 − 20 − c Point 20
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Chapter 3: Frequency Modulation (FM)
Solution to Part (b): The required channel bandwidth is 30 kHz. Solution to Part (c): LPF 0-15 kHz c LPF 0-5 kHz HPF LPF 0-5 kHz cos(20,000 ) t π
1( )
m t
2( )
m t cos(40,000 ) t π
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Chapter 3: Frequency Modulation (FM)
Transmitter: [ ] m n + A/D DSP D/A RF Filter NCO
[ ] 2 m n 1 2 1 2
ˆ (cycles/sample)
c
f ˆ cos( )
cn
ω
1 [ ] 2
ˆ cos( )
m n cn
ω
+ 1 [ ] 2 m n +
full-scale input + microphone
∑
ˆ ˆ 2 (radians/sample)
c c
f ω π = You will be able to set the carrier frequency and the DC level.
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Chapter 3: Frequency Modulation (FM)
Receiver: This is known as a “square-law” receiver. It does require (i) a local
carrier; (ii) Modulation index less than 100%. message A/D DSP D/A NCO RF Filter
ˆ ˆ cos[( ) ]
c
n ω ω θ + ∆ + ˆ ˆ sin[( ) ]
c
n ω ω θ + ∆ +
LPF
ˆ ˆ
c
f f + ∆
LPF
1 [ ] 2
ˆ cos( )
m n cn
ω
+
[ ]
c
x n [ ]
s
x n [ ]
c
y n [ ]
s
y n
2[ ] c
y n
2[ ] s
y n
1 [ ] 4 m n +
this does not need to synchronize to the incoming carrier
∑ You will be able to set the value of the carrier frequency offset.
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Chapter 3: Frequency Modulation (FM)
xc[n] = 1 + m[n] 2 cos(ˆ ωcn) cos[(ˆ ωc + ∆ˆ ω)n + θ] = 1 + m[n] 2 cos[(ˆ ωc + ˆ ωc + ∆ˆ ω)n + θ] 2
+ 1 + m[n] 2 cos(∆ˆ ωn + θ) 2 yc[n] = 1 4(1 + m[n]) cos(∆ωn + θ) xs[n] = 1 + m[n] 2 cos(ˆ ωcn) sin[(ˆ ωc + ∆ˆ ω)n + θ] = 1 + m[n] 2 sin[(ˆ ωc + ˆ ωc + ∆ˆ ω)n + θ] 2
+ 1 + m[n] 2 sin((−∆ˆ ω)n − θ) 2 ys[n] = − 1 4(1 + m[n]) sin(∆ˆ ωn + θ)
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Chapter 3: Frequency Modulation (FM)
=
c[n] + y2 s[n]
= 1 4(1 + m[n]) 2 cos2(∆ˆ ωn + θ) +
4(1 + m[n]) 2 sin2(∆ˆ ωn + θ) = 1 4(1 + m[n]) 2 = 1 4(1 + m[n]) Observations: The receiver only works correctly if (1 + m[n]) > 0 for all n. This is the same condition required by the envelope detector implemented in analog approach. The receiver is noncoherent since it does not require the NCO to lock in frequency and phase to the incoming carrier. The discrete-time output has a DC component. A DC blocking filter should be
4
. A gain of 4 would follow the blocking filter. Then the signal sent to the D/A is m[n], which is a full-scale signal.
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