1B Paper 6: Communications - University of Cambridge · 1B Paper 6: Communications Handout 2:...
Transcript of 1B Paper 6: Communications - University of Cambridge · 1B Paper 6: Communications Handout 2:...
1B Paper 6: CommunicationsHandout 2: Analogue Modulation
Ramji Venkataramanan
Signal Processing and Communications LabDepartment of [email protected]
Lent Term 2016
1 / 32
Modulation
Modulation is the process by which some characteristic of a carrierwave is varied in accordance with an information bearing signal
A commonly used carrier is a sinusoidal wave, e.g., cos(2πfct).fc is called the carrier frequency.
• We are allotted a certain bandwidth centred around fc for ourinformation signal
• E.g. BBC Cambridgeshire: fc = 96 MHz, informationbandwidth ≈ 200 KHz
• Q: Why is fc usually large?
A: Antenna size ∝ λc ⇒ larger frequency, smaller antennas!
2 / 32
Analogue vs. Digital ModulationAnalogue Modulation: A continuous information signal x(t)(e.g., speech, audio) is used to directly modulate the carrier wave.
We’ll study two kinds of analogue modulation:
1. Amplitude Modulation (AM) : Information x(t) modulates theamplitude of the carrier wave
2. Frequency Modulation (FM): Information x(t) modulates thefrequency of the carrier wave
We’ll learn about:
– Power & bandwidth of AM & FM signals
– Tx & Rx design
In the last 4 lectures, we will study digital modulation:
• x(t) is first digitised into bits
• Digital modulation then used to transport bits across thechannel
3 / 32
Amplitude Modulation (AM)
• Information signal x(t), carrier cos(2πfct)
• The transmitted AM signal is
sAM(t) = [a0 + x(t)] cos(2πfct)
• a0 is a positive constant chosen so that maxt |x(t)| < a0
• The modulation index of the AM signal is defined as
mA =maxt |x(t)|
a0
“The percentage that the carrier’s amplitude varies above andbelow its unmodulated level”
Why is the modulation index important ?ma < 1 is desirable because we can extract the information signalx(t) from the modulated signal by envelope detection.
4 / 32
When modulation index > 1:
• Phase reversals occur
• x(t) cannot be detected bytracing the +ve envelope
5 / 32
AM Receiver - Envelope Detector
sAM(t) Vout(t)RLC
• On the positive half-cycle of the input signal, capacitor Ccharges rapidly up to the peak value of input sAM(t)
• When input signal falls below this peak, diode becomesreverse-biased: capacitor discharges slowly through loadresistor RL
• In the next positive half-cycle, when input signal becomesgreater than voltage across the capacitor, diode conductsagain until next peak value
• Process repeats . . .
Very inexpensive receiver, but envelope detection needs mA < 1.
6 / 32
Circuit Diagram
AM wave input Envelope detector output
7 / 32
Spectrum of AM
Next, let’s look at the spectrum of sAM(t) = [a0 + x(t)] cos(2πfct)
SAM(f ) = F [sAM(t)]
= F[[a0 + x(t)]
(e j2πfc t + e−j2πfc t)
2
]
=a0
2[δ(f − fc) + δ(f + fc)]
︸ ︷︷ ︸carrier
+1
2[X (f − fc) + X (f + fc)]
︸ ︷︷ ︸information
(F [.] denotes the Fourier transform operation)
8 / 32
Example
SAM(f ) =a0
2[δ(f − fc) + δ(f + fc)] +
1
2[X (f − fc) + X (f + fc)]
−fc fc −W
SAM(f)
f
fc +W−fc +W−fc −W fc
0 W−W
X(f)
C
C/2 C/2
a0/2a0/2
9 / 32
Properties of AM
sAM(t) = [a0 + x(t)] cos(2πfct)
SAM(f ) =a0
2[δ(f − fc) + δ(f + fc)] +
1
2[X (f − fc) + X (f + fc)]
1. Bandwidth: From spectrum calculation, we see that if x(t) isa baseband signal with (one-sided) bandwidth W , the AMsignal sAM(t) is passband with bandwidth
BAM = 2W
2. Power: We now prove that the power of the AM signal is
PAM =a20
2+
PX
2
where PX is the power of x(t)
10 / 32
Power of AM signal
PAM = limT→∞
1
T
∫ T
0[a0 + x(t)]2 cos2(2πfct) dt
= limT→∞
1
T
∫ T
0[a0 + x(t)]2
1 + cos(4πfct)
2dt
=a20
2+
PX
2+ lim
T→∞1
T
∫ T
0
[a0 + x(t)]2
2cos(4πfct) dt
We now show that the last the last term is 0.• cos(4πfct) is a high-frequency sinusoid with period Tc = 1
2fc.
• g(t) = (a0+x(t))2
2 is a baseband signal which changes muchmore slowly than cos(4πfct). Hence, with T = nTc , we have
1
T
∫ T
0
g(t) cos(4πfct) dt ≈ 1
nTc
( ∫ Tc
0
g(0) cos(4πfct) dt +
+
∫ 2Tc
Tc
g(Tc) cos(4πfct) dt . . . +
∫ nTc
(n−1)Tc
g((n − 1)Tc) cos(4πfct) dt)
= 0
Hence PAM =a202 + PX
2 .11 / 32
Double Sideband Suppressed Carrier (DSB-SC)The power of AM signal is
PAM =a20
2︸︷︷︸carrier
+PX
2
• The presence of a0 makes envelope detection possible, but
requires extra power ofa202 corresponding to the carrier
• In DSB-SC, we eliminate the a0:
We transmit only the sidebands, and suppress the carrier
−fc fc −W
Sdsb−sc(f)
fc +W−fc +W−fc −W fc
f0 W−W
X(f)
Upper SidebandLower Sideband
12 / 32
DSB-SC ctd.
The transmitted DSB-SC wave is
sdsb-sc(t) = x(t) cos(2πfct)
−fc fc −W
Sdsb−sc(f)
fc +W−fc +W−fc −W fc
f0 W−W
X(f)
Upper SidebandLower Sideband
How to recover x(t) at the receiver?
Phase reversals ⇒ cannot use envelope detection
13 / 32
DSB-SC receiver
DSB-SC Receiver: Product Modulator + Low-pass filter
× Low-passfilter
cos(2πfct)
sdsb−sc(t)v(t) x̂(t)
1. Multiplying received signal by cos(2πfct) gives
v(t) = x(t) cos2(2πfct) =x(t)
2︸︷︷︸low freq.
+x(t) cos(4πfct)
2︸ ︷︷ ︸high freq.
2. Low-pass filter eliminates the high-frequency component
Ideal low-pass filter has H(f ) = constant for −W ≤ f ≤ W ,and zero otherwise
14 / 32
Properties of DSB-SC
sdsb-sc(t) = x(t) cos(2πfct)
Sdsb-sc(f ) =1
2(X (f + fc) + X (f − fc))
• Bandwidth of DSB-SC is Bdsb-sc = 2W , same as AM
• Power of DSB − SC is Pdsb-sc = PX2
(follows from AM power calculation)
• DSB-SC requires less power than AM as the carrier is nottransmitted
• But DSB-SC receiver is more complex than AM !
We assumed that receiver can generate locally generate afrequency fc sinusoid that is synchronised perfectly in phaseand frequency with transmitter’s carrier
• Effect of phase mismatch at Rx is explored in Examples paper
15 / 32
Single Sideband Suppressed Carrier (SSB-SC)DSB-SC transmits less power than AM. Can we also savebandwidth?
• x(t) is real ⇒ X (−f ) = X ∗(f )
⇒ Need to specify X (f ) only for f > 0• In other words, transmission of both sidebands is not strictly
necessary: we could obtain one sideband from the other!
−fc
Sssb−sc(f)
fc +W−fc −W fc
f0 W−W
X(f)
Upper Sideband
• Bandwidth is Bssb-sc = W , half of that of AM or DSB-SC!
• Power is is Pssb-sc = PX4 , half of DSB-SC
16 / 32
Summary: Amplitude Modulation
−fc fc −W
SAM(f)
f
fc +W−fc +W−fc −W fc
0 W−W
X(f)
−fc fc −W
Sdsb−sc(f)
fc +W−fc +W−fc −W fc
−fc
Sssb−sc(f)
fc +W−fc −W fc
Information signal
AM
DSB-SC
SSB-SC
17 / 32
You can now do Questions 1–5 onExamples Paper 8.
18 / 32
Frequency Modulation (FM)In FM, the information signal x(t) modulates the instantaneousfrequency of the carrier wave.
The instantaneous frequency f (t) is varied linearly with x(t):
f (t) = fc + kf x(t)
This translates to an instantaneous phase θ(t) given by
θ(t) = 2π
∫ t
0f (u)dt = 2πfct + 2πkf
∫ t
0x(u)du
The modulated FM signal
sFM(t) = Ac cos(θ(t)) = Ac cos(2πfct + 2πkf
∫ t
0x(u)du
)
• Ac is the carrier amplitude
• kf is called the frequency-sensitivity factor
19 / 32
ExampleWhat information signal does this FM wave correspond to?
0 1 2 3 4 5 6 7 8 9 10−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
t
s FM
(t)
(a) a constant, (b) a ramp, (c) a sinusoid, (d) no clue20 / 32
FM DemodulationAt the receiver, how do we recover x(t) from the FM wave?(ignoring effects of noise)
sFM(t) = Ac cos(2πfct + 2πkf
∫ t
0x(u)du
)
The derivative is
dsFM(t)
dt= −2πAc [fc + kf x(t)] sin
(2πfct + 2πkf
∫ t
0x(u)du
)
• The derivative is a passband signal with amplitude modulationby [fc + kf x(t)]
• If fc large enough, we can recover x(t) by envelope detection
of dsFM(t)dt !
• Hence FM demodulator is a differentiator + envelope detector
• Differentiator: ddt
F−→ j2πf (frequency response). SeeHaykin-Moher book for details on how to build a differentiator
21 / 32
Properties of FM
sFM(t) = Ac cos(2πfct + 2πkf
∫ t
0x(u)du
)
• Power of FM signal = A2c
2 , regardless of x(t)
• Non-linearity: FM(x1(t) + x2(t)) ̸= FM(x1(t)) + FM(x2(t))
• FM is more robust to additive noise than AM.
Intuitively, this is because the message is “hidden” in thefrequency of the signal rather than the amplitude.
• But this robustness comes at the cost of increasedtransmission bandwidth
• What is the bandwidth of the FM signal sFM(t)?
The spectral analysis is a bit complicated, but we will do it fora simple case . . . where x(t) is a sinusoid (a pure tone)
22 / 32
FM modulation of a toneConsider FM modulation of a tone x(t) = ax cos(2πfx t). We have
f (t) = fc + kf ax cos(2πfx t)
θ(t) = 2πfct +kf ax
fxsin(2πfx t)
• ∆f = kf ax is called the frequency deviation
∆f is the max. deviation of the carrier frequency f (t) from fc
• β = kf axfx
= ∆ffx
is called the modulation index
β is the max. deviation of the carrier phase θ(t) from 2πfct
Then the FM signal becomes
sFM(t) = Ac cos (2πfct + β sin(2πfx t))
23 / 32
The spectrum of the FM signalWe want to understand the frequency spectrum of
sFM(t) = Ac cos (2πfct + β sin(2πfx t))
We can write
sFM(t) = Re[Ace
j2πfc t + jβ sin(2πfx t)]
= Re[s̃(t)e j2πfc t
]
wheres̃(t) = Ace
jβ sin(2πfx t)
• s̃(t) is periodic with period 1/fx
• Can express using Fourier series (Fundamental frequency fx)
s̃(t) =∞∑
n=−∞cn e j2πfxnt
24 / 32
FM Spectrum ctd.
You will show in Examples Paper 8 that the Fourier seriescoefficients of s̃(t) are
cn = Ac Jn(β)
where Jn(β) = 12π
∫ π−π e j(β sin u−nu) du
Jn(.) is called the nth order Bessel function of the first kind
Thus
s̃(t) = Acejβ sin(2πfx t)
=∞∑
n=−∞cn e j2πfxnt = Ac
∑
n
Jn(β)e j2πfxnt .
25 / 32
Therefore
sFM(t) = Re[s̃(t)e j2πfc t
]= Re
[ ∞∑
n=−∞cne
j2πfxnt e j2πfc t
]
= Re
[ ∞∑
n=−∞Ac Jn(β) e j2π(fc+nfx )t
]
= Ac
∞∑
n=−∞Jn(β) cos(2π(fc + nfx)t)
Taking Fourier Transforms, the spectrum of the FM signal is
SFM(f ) =Ac
2
∞∑
n=−∞Jn(β) [ δ(f − fc − nfx) + δ(f + fc + nfx) ]
26 / 32
Plots of Jn(β) vs β
n
n
nn
n
27 / 32
ExampleWhat is the spectrum of the FM signal when x(t) is a pure toneand the modulation index β = 5 ?
Jn(β) vs n for β = 5
0 5 10 15−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
n
Jn(β
)
28 / 32
Example ctd.
The spectrum is
SFM(f ) =Ac
2
∞∑
n=−∞Jn(5) [ δ(f − fc − nfx) + δ(f + fc + nfx) ]
-fx
-fc fc
X(f)
|SFM(f)|
fx
29 / 32
Bandwidth of FM signalsTo summarise, when x(t) has only a single frequency fx , thespectrum of the FM wave is rather complicated:
• There is a carrier component at fc , and components locatedsymmetrically on either side of fc at fc ± fx , fc ± 2fx ,. . .
• The absolute bandwidth is infinite, but . . . the sidecomponents at fc ± nfx become negligible for large enough n
Carson’s rule for the effective bandwidth of FM signals:
1. The bandwidth of an FM signal generated by modulating asingle tone is
BFM ≈ 2∆f + 2fx = 2∆f(1 + 1
β
)
2. For an FM signal generated by modulating a general signalx(t) with bandwidth W , the bandwidth BFM ≈ 2∆f + 2W
(Recall: for any FM wave, ∆f is the frequency deviation around fc)
30 / 32
Example
BBC Radio Cambridgeshire: fc = 96 MHz and ∆f = 75 kHz.Assuming that the voice/music signals have W = 15 kHz, we have
β =∆f
W=
75
15= 5
and the bandwidth
BFM = 2(∆f + W ) = 2(75 + 15) = 180 kHz,
whileBAM = 2W = 30 kHz
FM signals have larger bandwidth than AM, but have betterrobustness against noise.
31 / 32
Summary: Analogue Modulation
Amplitude Modulation with information signal of bandwidth W
• AM modulated signal: Bandwidth 2W , high power, simple Rxusing envelope detection
• DSB-SC: Bandwidth 2W , lower power, more complex Rx
• SSB-SC: Bandwidth W , even lower power, Rx similar toDSB-SC
Frequency Modulation with information signal of bandwidth W :
• FM signal has constant carrier amplitude ⇒ constant power
• Bandwidth of FM signal depends on both β and W
Can be significantly greater than 2W
• Better robustness to noise than AM – the information is“hidden” in the phase
32 / 32