1B Paper 6: Communications - University of Cambridge · 1B Paper 6: Communications Handout 2: Analogue Modulation Ramji Venkataramanan Signal Processing and Communications Lab ...

1B Paper 6: CommunicationsHandout 2: Analogue Modulation

Ramji Venkataramanan

Signal Processing and Communications LabDepartment of [email protected]

Lent Term 2016

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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!

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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

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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.

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When modulation index > 1:

• Phase reversals occur

• x(t) cannot be detected bytracing the +ve envelope

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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.

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Circuit Diagram

AM wave input Envelope detector output

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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)

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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

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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)

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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

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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

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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

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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

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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

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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

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You can now do Questions 1–5 onExamples Paper 8.

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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

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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

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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)

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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))

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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

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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 .

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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) ]

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Plots of Jn(β) vs β

n

n

nn

n

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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

)

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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

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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)

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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.

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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

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