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EE4512 Analog and Digital Communications Chapter 6 Chapter 6 Chapter 6 Analog Modulation Analog Modulation and Demodulation and Demodulation

Ch6 Analog Modulation

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EE4512 Analog and Digital Communications Chapter 6

Chapter 6Chapter 6

Analog ModulationAnalog Modulationand Demodulationand Demodulation

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EE4512 Analog and Digital Communications Chapter 6

Chapter 6Chapter 6

Analog ModulationAnalog Modulationand Demodulationand Demodulation

•• Amplitude Modulation Amplitude Modulation•• Pages 306Pages 306--309309

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EE4512 Analog and Digital Communications Chapter 6

•• The analytical signal for double sideband, large carrier The analytical signal for double sideband, large carrier amplitude modulation (DSBamplitude modulation (DSB--LC AM) is:LC AM) is:

ssDSBDSB--LCLC AMAM(t(t) =) = AACC (c +(c + s(ts(t)))) coscos (2(2ππ f f CC t)t)

wherewhere c c is theis the DC biasDC bias or or offset offset andand A AC C  is the carrier is the carrier 

amplitude. The continuous analog signalamplitude. The continuous analog signal s(t)s(t) is a basebandis a baseband

signal with the information content (voice or music) to besignal with the information content (voice or music) to betransmitted.transmitted.

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EE4512 Analog and Digital Communications Chapter 6

•• The baseband power spectral density (PSD) spectrum of The baseband power spectral density (PSD) spectrum of the information signalthe information signal s(ts(t) or ) or S(f S(f ) for voice has significant) for voice has significant

components below 500 Hz and a bandwidth of < 8 kHz:components below 500 Hz and a bandwidth of < 8 kHz:

S(f S(f ) =) = F F (s(t(s(t))))TheThe singlesingle--sided sided spectrum of the modulated signal is:spectrum of the modulated signal is:

F F (A(ACC (c +(c + s(ts(t)))) coscos (2(2ππ f f CC tt)) =)) = S(f S(f  – – f f CC))

Power Spectral Density of s(t)Power Spectral Density of s(t)500 Hz500 Hz 8 kHz8 kHz

dBdB

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EE4512 Analog and Digital Communications Chapter 6

•• TheThe singlesingle--sided sided (positive frequency axis) spectrum of the(positive frequency axis) spectrum of themodulated signal modulated signal replicates the baseband spectrum as areplicates the baseband spectrum as a

doubledouble--sided sided spectrum about the carrier frequency.spectrum about the carrier frequency.

Carrier 25 kHzCarrier 25 kHzDoubleDouble--sided spectrumsided spectrum

Baseband spectrumBaseband spectrum

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EE4512 Analog and Digital Communications Chapter 6

•• TheThe doubledouble--sided modulated spectrum about the carrier sided modulated spectrum about the carrier frequency has anfrequency has an lower lower ((LSBLSB) and) and upper upper ((USBUSB) sideband.) sideband.

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EE4512 Analog and Digital Communications Chapter 6

•• The modulated DSBThe modulated DSB--LC AM signal shows anLC AM signal shows an outer envelopeouter envelopethat follows the polar baseband signal s(t)that follows the polar baseband signal s(t)..

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EE4512 Analog and Digital Communications Chapter 6

•• The analytical signal for double sideband, suppressedThe analytical signal for double sideband, suppressedcarrier amplitude modulation (DSBcarrier amplitude modulation (DSB--SC AM) is:SC AM) is:

ssDSBDSB--SCSC AMAM(t(t) =) = AACC s(t)s(t) coscos (2(2ππ f f CC t)t)

wherewhere A AC C  is the carrier amplitude. Theis the carrier amplitude. The singlesingle--sidedsided

spectrum of the modulated signal replicates the basebandspectrum of the modulated signal replicates the baseband

spectrum as a doublespectrum as a double--sided spectrum about the carrier sided spectrum about the carrier frequencyfrequency butbut without without a carrier component.a carrier component.

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EE4512 Analog and Digital Communications Chapter 6

•• The analytical signal for double sideband, suppressedThe analytical signal for double sideband, suppressedcarrier amplitude modulation (DSBcarrier amplitude modulation (DSB--SC AM) is:SC AM) is:

ssDSBDSB--SCSC AMAM(t(t) =) = AACC s(t)s(t) coscos (2(2ππ f f CC t)t)

wherewhere A AC C  is the carrier amplitude. The modulated signalis the carrier amplitude. The modulated signal

ssDSBDSB--SCSC AMAM(t(t) looks similar to s(t) but has a temporal but not) looks similar to s(t) but has a temporal but not

spectral carrier component.spectral carrier component.

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EE4512 Analog and Digital Communications Chapter 6

•• The DSBThe DSB--LC AM and the DSBLC AM and the DSB--SC AM modulated signalsSC AM modulated signalshave the samehave the same sidebandssidebands..

Carrier 25 kHzDSB-LC AM

DSB-SC AM No carrier  

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EE4512 Analog and Digital Communications Chapter 6

•• The modulated DSBThe modulated DSB--LC AM and the DSCLC AM and the DSC--SC AM signalsSC AM signalsare different.are different.

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EE4512 Analog and Digital Communications Chapter 6

•• The modulated DSBThe modulated DSB--SC AM signal has anSC AM signal has an envelopeenvelope thatthatfollows the polar baseband signal s(t) but not an outer follows the polar baseband signal s(t) but not an outer 

envelopeenvelope..

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EE4512 Analog and Digital Communications Chapter 6

Chapter 6Chapter 6

Analog ModulationAnalog Modulationand Demodulationand Demodulation

•• Coherent DemodulationCoherent Demodulationof AM Signalsof AM Signals

•• Pages 309Pages 309--315315

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EE4512 Analog and Digital Communications Chapter 6

•• The DSBThe DSB--SC AMSC AM coherent receiver coherent receiver has ahas a bandpass filter bandpass filter centered at f centered at f CC and with a bandwidth of and with a bandwidth of twicetwice the bandwidththe bandwidth

of s(t) because of the LSB and USBof s(t) because of the LSB and USB. The output of the. The output of the

multiplier ismultiplier is lowpass filtered lowpass filtered with a bandwidth equal towith a bandwidth equal to

the bandwidth of s(t).the bandwidth of s(t).

r(tr(t) =) = γγ ssDSBDSB--SCSC(t(t) +) + n(tn(t))

•• The DSBThe DSB--SC AM received signal is r(t) =SC AM received signal is r(t) = γγ ssDSBDSB--SCSC

(t(t) +) + n(tn(t).).

The bandpass filter passes the modulated signal but filtersThe bandpass filter passes the modulated signal but filters

the noise:the noise:

z(tz(t) =) = γγ ssDSBDSB--SCSC(t(t) + n) + noo(t) S&M Eq. 6.3(t) S&M Eq. 6.3

nnoo(t) has a Gaussian distribution. The bandpass filter has a(t) has a Gaussian distribution. The bandpass filter has a

center frequency of f center frequency of f CC = 25 kHz and a= 25 kHz and a --3 dB bandwidth of 83 dB bandwidth of 8kHz (25kHz (25 ±± 4 kHz).4 kHz).

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EE4512 Analog and Digital Communications Chapter 5

•• The filter noise nThe filter noise noo(t) has a(t) has a flat power spectral density flat power spectral density within the bandwidth of the bandpass filter:within the bandwidth of the bandpass filter:

PSD

no(t)

f C = 25 kHz

21 kHz 29 kHz

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EE4512 Analog and Digital Communications Chapter 6

•• The filter noise nThe filter noise noo(t) can be described as a(t) can be described as a quadraturequadraturerepresentation:representation:

nnoo(t) =(t) = WW(t)(t) coscos (2(2ππ f f CCtt) +) + ZZ(t) sin ((t) sin (22ππ f f CCtt) S&M Eq. 5.62R) S&M Eq. 5.62R

In the coherent receiver the noise is processed:In the coherent receiver the noise is processed:

nnoo(t)(t) coscos (2(2ππ f f CCtt)) == WW(t) cos(t) cos22 (2(2ππ f f CCtt) + S&M Eq. 6.5) + S&M Eq. 6.5ZZ(t)(t) coscos (2(2ππ f f CCtt)) sin (2sin (2ππ f f CCtt))

PSD

f C = 25 kHz

21 kHz 29 kHz

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EE4512 Analog and Digital Communications Chapter 6

•• Applying theApplying the trignometric trignometric identity identity the filter noise nthe filter noise noo(t) is:(t) is:

nnoo(t)(t) coscos (2(2ππ f f CCtt)) == ½½ WW(t)(t) + ½½ WW(t)(t) coscos (4(4ππ f f CCtt) +) +

½½ ZZ(t) sin (4(t) sin (4ππ f f CCtt) S&M Eq. 6.5) S&M Eq. 6.5

After the lowpass filter in the receiver the demodulatedAfter the lowpass filter in the receiver the demodulated

signal is:signal is:

ssdemoddemod(t) =(t) = ½½ γγ AACC s(t) +s(t) + ½½ WW(t)(t) S&M Eq. 6.7S&M Eq. 6.7

PSD

f C = 25 kHz

21 kHz 29 kHz

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EE4512 Analog and Digital Communications Chapter 6

•• The transmitted DSBThe transmitted DSB--SC AM signal is:SC AM signal is:

ssDSBDSB--SCSC AMAM(t(t) =) = AACC s(t)s(t) coscos (2(2ππ f f CC t)t)

The average normalized biThe average normalized bi--sided power of sided power of ssDSBDSB--SCSC(t(t) is) is

found in the spectral domain withfound in the spectral domain with S(f S(f ) =) = F F  (s(t)):(s(t)):

[ ]−∫ 2

2trans C C1P = A S(f f ) + S(f + f ) df  

2S&M Eq. 6.8S&M Eq. 6.8

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EE4512 Analog and Digital Communications Chapter 6

•• The dualThe dual--sided spectral do notsided spectral do not overlapoverlap (at zero frequency)(at zero frequency)and theand the cross termscross terms are zero so that:are zero so that:

where Pwhere Pss is the average normalized power of s(t).is the average normalized power of s(t).

[ ]−∫ 2

2

trans DSB-SC C DSB-SC C

2

trans s

1P = A S (f f ) + S (f + f ) df  

2A

P = P2

S&M Eq. 6.9S&M Eq. 6.9

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EE4512 Analog and Digital Communications Chapter 6

•• TheThe average normalized power of s(t) is found in theaverage normalized power of s(t) is found in thespectral domain:spectral domain:

In aIn a noiseless channel noiseless channel the power in the demodulatedthe power in the demodulated

DSBDSB--SC AM signal is:SC AM signal is:

∫ ∫ 2 2

s CP = S(f) df = S(f + f ) df   S&M Eq. 6.10S&M Eq. 6.10

= =

2

2 2demod, noiseless s trans1 γP γ A P P

4 2S&M Eq. 6.11S&M Eq. 6.11

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EE4512 Analog and Digital Communications Chapter 6

•• TheThe average normalized power of the processed noise is:average normalized power of the processed noise is:

The signalThe signal--toto--noise power ratio then is:noise power ratio then is:

=processed noise o

1P N (2 B)

4

= =

2

2transtrans

coherent DSB-SCo

o

γP γ P2SNR

1 N BN (2 B)4

S&M Eq. 6.12S&M Eq. 6.12

2 B2 B

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EE4512 Analog and Digital Communications Chapter 6

•• The DSBThe DSB--SC AM coherent receiver SC AM coherent receiver requires a phaserequires a phaseand frequency synchronous reference signal. If theand frequency synchronous reference signal. If the

reference signal has areference signal has a phase error  phase error φφ then:then:

S&M Eq. 6.17S&M Eq. 6.17ϕ 

=coherent DSB-SC phase error 

2 2

trans

o

SNRγ cos P

N B

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EE4512 Analog and Digital Communications Chapter 6

•• The DSBThe DSB--SC AM coherent receiver SC AM coherent receiver requires a phaserequires a phaseand frequency synchronous reference signal. If theand frequency synchronous reference signal. If the

reference signal has areference signal has a frequency error frequency error  ∆ ∆f f  then:then:

SSdemoddemod frequencyfrequency error error (t(t) =) = ½½ γγ AACC s(t) cos (2s(t) cos (2ππ  f t)f t)++ ½½ XX(t(t) cos (2) cos (2ππ  f t)f t)

++ ½½ Y Y(t(t) sin (2) sin (2ππ  f t) S&M Eq. 6.18f t) S&M Eq. 6.18

•• Although the noise component remains the same, theAlthough the noise component remains the same, the

amplitudeamplitude of the demodulated signal varies withof the demodulated signal varies with   f f ::

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EE4512 Analog and Digital Communications Chapter 6

Chapter 6Chapter 6

Analog ModulationAnalog Modulationand Demodulationand Demodulation

•• NonNon--coherent Demodulationcoherent Demodulationof AM Signalsof AM Signals

•• Pages 315Pages 315--326326

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EE4512 Analog and Digital Communications Chapter 6

•• TheThe nonnon--coherent AM coherent AM (DSB(DSB--LC) receiver uses anLC) receiver uses an envelopeenvelopedetector detector implemented as aimplemented as a semiconductor diodesemiconductor diode and aand a low low --

 pass filter  pass filter ::

The DSBThe DSB--LC AM analytical signal is:LC AM analytical signal is:

ssDSBDSB--LCLC AMAM(t(t) = A) = ACC (c +(c + s(ts(t)) cos (2)) cos (2ππ f f CC t)t)

where c is the DC bias (offset).where c is the DC bias (offset).

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EE4512 Analog and Digital Communications Chapter 5

•• The envelope detector is aThe envelope detector is a half half --wave rectifier wave rectifier andandprovides aprovides a DC biasDC bias ((c c ) to the processed DSB) to the processed DSB--LC AMLC AM

signal :signal :

c = DC bias

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EE4512 Analog and Digital Communications Chapter 5

•• The output of the half The output of the half --wave diode rectifier is lowwave diode rectifier is low--passpassfiltered to remove the carrier frequency and outputs thefiltered to remove the carrier frequency and outputs the

envelope which is the information:envelope which is the information:

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EE4512 Analog and Digital Communications Chapter 6

•• The DSBThe DSB--LC AM signal can be decomposed as:LC AM signal can be decomposed as:

ssDSBDSB--LCLC AMAM(t(t) =) = s(ts(t)) coscos (2(2ππ f f CC t) + At) + ACC cc coscos (2(2ππ f f CC t)t)

S&M Eq. 6.20RS&M Eq. 6.20R

The average normalized power of the information term:The average normalized power of the information term:

=

2

Cinfo term S

AP P2 S&M Eq. 6.23S&M Eq. 6.23

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EE4512 Analog and Digital Communications Chapter 6

•• The average normalized transmitted power is:The average normalized transmitted power is:

SinceSince s(ts(t) + c must be >= 0 to avoid distortion in the) + c must be >= 0 to avoid distortion in theDSBDSB--LC AM signal: cLC AM signal: c ≥≥ | min [| min [s(ts(t)] | or c)] | or c22 ≥≥ ss22(t) for all t.(t) for all t.

[ ]=

=

∫ T

2

carrier term C C

0

2 2

Ccarrier term

1P A c cos(2πf t) dt

T

A cP

2S&M Eq. 6.24S&M Eq. 6.24

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EE4512 Analog and Digital Communications Chapter 6

••

Therefore cTherefore c22

≥≥

PP

ss and for DSBand for DSB

--LC AMLC AM

::

The power efficiencyThe power efficiency ηη of a DSBof a DSB--LC AM signal is:LC AM signal is:

≥carrier term info termP P S&M Eq. 6.28S&M Eq. 6.28

η ≤info term info term

carrier term info term trans DSB-LC AM term

P P= = 0.5

P + P P

S&M Eq. 6.29S&M Eq. 6.29

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EE4512 Analog and Digital Communications Chapter 6

••

The DSBThe DSB

--LC AM signalLC AM signal

wasteswastes

at least half theat least half the

transmitted power because the power in the carrier termtransmitted power because the power in the carrier term

has no information:has no information:

TheThe modulation index modulation index m is defined as:m is defined as:≥carrier term info termP P η ≤ 0.5

[ ] [ ][ ] [ ]

−max s(t) + c min s(t) + cm = max s(t) + c + min s(t) + c S&M Eq. 6.30S&M Eq. 6.30

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EE4512 Analog and Digital Communications Chapter 6

••

TheThe

modulation index modulation index 

m defines the power efficiency butm defines the power efficiency but

mm must be less than 1. If m > 1 then min [must be less than 1. If m > 1 then min [s(ts(t) + c] < 0 and) + c] < 0 and

distortion occurs.distortion occurs.

[ ] [ ][ ] [ ]

−max s(t) + c min s(t) + cm =max s(t) + c + min s(t) + c

S&M Eq. 6.30S&M Eq. 6.30

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EE4512 Analog and Digital Communications Chapter 6

••

TheThe

average normalized power of the demodulationaverage normalized power of the demodulation

noiseless DSBnoiseless DSB--LC AM signal is:LC AM signal is:

Then the signalThen the signal--toto--noise power ratio for the DSBnoise power ratio for the DSB--LC AMLC AMsignal is:signal is:

= 2

demod, noiseless info termP 2 γ P

= = η

2 2

info term trans DSB-LCnoncoherent DSB-LC

o o

PSNRN (2 B) N B

S&M Eq. 6.40S&M Eq. 6.40

2 B2 B

S&M Eq. 6.39S&M Eq. 6.39

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EE4512 Analog and Digital Communications Chapter 6

••

TheThe

nonnon

--coherent AM coherent AM 

(DSB(DSB

--LC) receiver is theLC) receiver is the

crystal crystal 

radioradio which needs no batteries! Power for the highwhich needs no batteries! Power for the high--

impedance ceramic earphone is obtained directly from theimpedance ceramic earphone is obtained directly from the

transmitted signaltransmitted signal. For simplicity, the RF BPF is omitted. For simplicity, the RF BPF is omitted

and the audio frequency filter is a simpleand the audio frequency filter is a simple RC RC network.network.

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EE4512 Analog and Digital Communications Chapter 6

Chapter 6Chapter 6

Analog ModulationAnalog Modulationand Demodulationand Demodulation

••

Frequency Modulation and Frequency Modulation and 

Phase ModulationPhase Modulation

•• Pages 334Pages 334--343343

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EE4512 Analog and Digital Communications Chapter 6

••

The analytical signal for anThe analytical signal for an

analog phase modulated analog phase modulated 

(PM)(PM)

signal is:signal is:

ssPMPM(t(t) =) = AACC coscos [2[2ππ f f CC t +t + αα s(ts(t)] S&M Eq. 6.53)] S&M Eq. 6.53

wherewhere α α  is theis the phase modulation constant  phase modulation constant radrad/V and/V and A AC C  isisthe carrier amplitude. The continuous analog signalthe carrier amplitude. The continuous analog signal s(t)s(t) is ais a

baseband signal with the information content (voice or baseband signal with the information content (voice or 

music) to be transmitted.music) to be transmitted.

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EE4512 Analog and Digital Communications Chapter 6

••

The analytical signal for anThe analytical signal for an

analog frequency modulated analog frequency modulated 

(FM) signal is:(FM) signal is:

ssFMFM(t(t) =) = AACC coscos{ 2{ 2ππ [f [f CC + k+ k s(ts(t)] t +)] t + φφ] S&M Eq. 6.53] S&M Eq. 6.53

wherewhere k k is theis the frequency modulation constant frequency modulation constant Hz / V,Hz / V, A AC C 

is the carrier amplitude andis the carrier amplitude and φφ is the initial phase angle atis the initial phase angle at

t = 0.t = 0. The continuous analog signalThe continuous analog signal s(t s(t  ) ) is a basebandis a basebandsignal with the information content.signal with the information content.

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EE4512 Analog and Digital Communications Chapter 6

•• TheThe instantaneous phaseinstantaneous phase of the PM signal is:of the PM signal is:

ΨΨPMPM(t(t) = 2) = 2ππ f f CC t +t + αα s(ts(t) S&M Eq. 6.56) S&M Eq. 6.56

TheThe instantaneous phaseinstantaneous phase of the FM signal is:of the FM signal is:

ΨΨFMFM(t(t) =) = 22ππ [f [f CC + k+ k s(ts(t)] t +)] t + φφ] S&M Eq. 6.57] S&M Eq. 6.57

The instantaneous phase is also call theThe instantaneous phase is also call the angleangle of the signalof the signal..

TheThe instantaneous frequency instantaneous frequency is the time rate of change of is the time rate of change of 

the angle:the angle:

f(tf(t) = (1/2) = (1/2ππ)) ddΨΨ(t) /(t) / dtdt S&M Eq. 6.58S&M Eq. 6.58

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EE4512 Analog and Digital Communications Chapter 6

•• To avoid ambiguity and distortion in PM signals due toTo avoid ambiguity and distortion in PM signals due to

phase wrapping:phase wrapping:

--ππ << αα s(ts(t)) ≤≤ ππ radians for all t S&M Eq. 6.61radians for all t S&M Eq. 6.61

Since FM and PM are both change the angle of the carrier Since FM and PM are both change the angle of the carrier 

signal as a function of the analog information signalsignal as a function of the analog information signal s(ts(t), FM), FM

and PM are calledand PM are called angle modulationangle modulation..

For example, is this signal FM, PM or neither:For example, is this signal FM, PM or neither:

ttx(tx(t) = A) = ACC coscos { 2{ 2ππ f f CCtt ++ ∫∫ k s(k s(λλ) d) dλλ ++ φφ}} S&M Eq. 6.60S&M Eq. 6.60

--∞∞

C C

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EE4512 Analog and Digital Communications Chapter 6

•• The instantaneous phase of the signal is:The instantaneous phase of the signal is:

tt

ΨΨxx(t(t) = 2) = 2ππ f f CCtt ++ ∫∫ k s(k s(λλ) d) dλλ ++ φφ S&M Eq. p. 336S&M Eq. p. 336

--∞∞which iswhich is not not a linear function of a linear function of s(ts(t) so the signal is) so the signal is not PM not PM ..The instantaneous frequency of the signal is:The instantaneous frequency of the signal is:

f f xx(t(t) = (1/2) = (1/2ππ)) ddΨΨxx(t(t) /) / dtdt == f f CC + k+ k s(ts(t) / 2) / 2ππ

and the frequency differenceand the frequency difference f f xx – – f f CC is a linear function of is a linear function of s(ts(t))

so the signal is FM.so the signal is FM.

TheThe maximum phase deviationmaximum phase deviation of a PM signal isof a PM signal is

max |max | ααs(ts(t) |. The) |. The maximum frequency deviationmaximum frequency deviation of a FMof a FM

signal issignal is   f = max | kf = max | k s(ts(t) |.) |.

EE4512 A l d Di it l C i ti Ch t 6

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EE4512 Analog and Digital Communications Chapter 6

•• The spectrum of a PM or FM signal can be developed asThe spectrum of a PM or FM signal can be developed as

follows: S&Mfollows: S&M EqsEqs. 6.64 through 6.71. 6.64 through 6.71

C C m

C m

C m

C m C m

C

v(t) = A sin(2π f t +β sin 2π f t)

v(t) = Re { exp(j 2π f t + jβ sin 2π f t) }exp(j 2π f t + jβ sin 2π f t) =

cos (2π f t +β sin 2π f t) + j sin (2π f t +β sin 2π f t)

v(t) = Im { A exp(2π

now 

( )

C m

m n m

n = -

m n m

n = -

f t + jβ sin 2π f t) }

exp(jβ sin 2π f t) = c exp(j 2π n f t)

exp(jβ sin 2π f t) = J β exp(j 2π n f t)

now 

after further development  Bessel function of the first kindBessel function of the first kind

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EE4512 Analog and Digital Communications Chapter 6

•• Bessel functions of the first kindBessel functions of the first kind J J nn( (  β β ) ) are tabulated for FMare tabulated for FM

with single tone f with single tone f mm angle modulation (S&M Table 6.1):angle modulation (S&M Table 6.1):

n

β

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EE4512 Analog and Digital Communications Chapter 6

•• For single tone f For single tone f mm

angle modulation the spectrum is periodicangle modulation the spectrum is periodic

and infinite in extentand infinite in extent::

n

β

( )∞

∑C n m C

n = -

v(t) = A J β sin[2π (n f + f ) t] S&M Eq. 6.72S&M Eq. 6.72

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EE4512 Analog and Digital Communications Chapter 6

•• TheThe complexity complexity of the Bessel function solution for theof the Bessel function solution for the

spectrum of a single tone angle modulation can bespectrum of a single tone angle modulation can be

simplified by thesimplified by the CarsonCarson’ ’ s Rule approximations Rule approximation for thefor the

bandwidthbandwidth BB. Since. Since ββ == ∆∆f / f f / f mm::

B = 2 (B = 2 (ββ + 1) f + 1) f mm = 2 (= 2 (∆∆f + f f + f mm) Hz) Hz

n

β

S&M Eq. 6.74S&M Eq. 6.74

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EE4512 Analog and Digital Communications Chapter 6

•• CarsonCarson’ ’ s Rules Rule for the approximate bandwidth of an anglefor the approximate bandwidth of an angle

modulated signal was developed by John R. Carson inmodulated signal was developed by John R. Carson in

1922 while he worked at AT&T. Prior to this in 1915 he1922 while he worked at AT&T. Prior to this in 1915 he

presaged the concept of presaged the concept of bandwidthbandwidth

efficiency efficiency in AM by proposingin AM by proposingthe suppression of a sidebandthe suppression of a sideband

(see S&M p. 326(see S&M p. 326--333)333)::

B = 2 (B = 2 (ββ + 1) f + 1) f mm = 2 (= 2 (∆∆f + f f + f mm) Hz) Hz

18861886--19401940

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EE4512 Analog and Digital Communications Chapter 6

•• TheThe normalized power normalized power within the Carsonwithin the Carson’’s Rule bandwidths Rule bandwidth

for a single tone angle modulated signals is:for a single tone angle modulated signals is:

Note that JNote that J--nn((ββ) =) = ±± JJnn((ββ) so that) so that JJ--nn22((ββ) = J) = Jnn

22((ββ) and for the) and for the

normalized power calculation the sign of J(normalized power calculation the sign of J(ββ)) is not used.is not used.

( )= ∑2 β+1

2Cin-band, sinusoid n

n = -(β+1)

AP J β  

2

S&M Eq. 6.75S&M Eq. 6.75

Spectrum of 

single tone FM

modulation

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EE4512 Analog and Digital Communications Chapter 6

•• The analog FM power spectral density PSD of the voiceThe analog FM power spectral density PSD of the voice

signal has a bandwidth predicted only by Carsonsignal has a bandwidth predicted only by Carson’’s Rules Rule

since it is not a single tone.since it is not a single tone.

PSDPSD

VoiceVoice

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EE4512 Analog and Digital Communications Chapter 6

•• Here f Here f maxmax

= 4 kHz, k = 25 Hz/V and= 4 kHz, k = 25 Hz/V and ∆∆f f maxmax

= 40(25) = 1 kHz.= 40(25) = 1 kHz.

The CarsonThe Carson’’s Rule approximate maximum bandwidths Rule approximate maximum bandwidth

B = 2 (B = 2 (∆∆f + f f + f mm) = 10 kHz or ) = 10 kHz or ±± 5 kHz (but seems wrong!)5 kHz (but seems wrong!)

PSDPSD

VoiceVoice

40

Bandwidth

f C

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EE4512 Analog and Digital Communications Chapter 6

•• A 200 HzA 200 Hz single tonesingle tone FM signal has a PSD with periodicFM signal has a PSD with periodic

terms at f terms at f CC ±± n f n f mm = 25= 25 ±± 0.2 n kHz.0.2 n kHz.

PSDPSD

f C

200 Hz200 Hz

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EE4512 Analog and Digital Communications Chapter 6

•• Here f Here f mm

= 200 Hz, k = 25 Hz/V and= 200 Hz, k = 25 Hz/V and ∆∆f f maxmax

= 40(25) = 1= 40(25) = 1

kHz. The CarsonkHz. The Carson’’s Rule approximate maximum bandwidths Rule approximate maximum bandwidth

B = 2 (B = 2 (∆∆f + f f + f mm) = 2.4 kHz or ) = 2.4 kHz or ±± 1.2 kHz:1.2 kHz:

PSDPSD

f C

200 Hz200 Hz

BandwidthBandwidth

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EE4512 Analog and Digital Communications Chapter 6

•• SinceSince ββ == ∆∆f / f f / f mm

= 1 kHz / 0.2 kHz = 5 and the Bessel= 1 kHz / 0.2 kHz = 5 and the Bessel

function predicts a bandwidth of 2 n f function predicts a bandwidth of 2 n f mm = 2(12)(200) == 2(12)(200) =

4.8 kHz (since n = 12 for 4.8 kHz (since n = 12 for ββ = 5 from Table 6.1):= 5 from Table 6.1):

PSDPSD

f C

200 Hz200 Hz

BandwidthBandwidth

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EE4512 Analog and Digital Communications Chapter 6

Chapter 6Chapter 6

Analog ModulationAnalog Modulationand Demodulationand Demodulation

•• Noise in FM and PM SystemsNoise in FM and PM Systems

•• Pages 347Pages 347--355355

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EE4512 Analog and Digital Communications Chapter 6

•• A general angle modulated transmitted signal, whereA general angle modulated transmitted signal, where ΨΨ(t)(t)

is the instantaneous phase,is the instantaneous phase, is:is:

ssangleangle--modulatedmodulated(t(t) = A) = ACC coscos [[ΨΨ(t(t)])] S&M Eq. 6.86S&M Eq. 6.86

The received signals is:The received signals is:

r r angleangle--modulatedmodulated(t(t) =) = γγ AACC coscos [[ΨΨ(t)] +(t)] + n(tn(t) S&M Eq. 6.87) S&M Eq. 6.87

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EE4512 Analog and Digital Communications Chapter 6

•• The analytical signal for PM is:The analytical signal for PM is:

ssPMPM(t(t) = A) = ACC coscos [[ΨΨ(t)](t)] = AACC coscos [2[2ππ f f CC tt ++ αα s(ts(t)])]

S&M Eq. 6.53S&M Eq. 6.53

After development the SNR for demodulated PM is:After development the SNR for demodulated PM is:

SNRSNRPMPM = (= (αγαγ AACC))22 PPSS / (2 N/ (2 Noo f f maxmax)) S&M Eq. 6.98S&M Eq. 6.98

wherewhere −−ππ << αα s(ts(t)) ≤≤ ππ for all t.for all t.

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EE4512 Analog and Digital Communications Chapter 6

•• The analytical signal for FM is:The analytical signal for FM is:

ssFMFM(t(t) = A) = ACC coscos [[ΨΨ(t)](t)] = AACC coscos [2[2ππ f f CC tt ++ ∫∫ k s(k s(λλ) d) dλλ]]

S&M Eq. 6.53S&M Eq. 6.53

After development the SNR for demodulated FM is:After development the SNR for demodulated FM is:

SNRSNRFMFM = 1.5 (k= 1.5 (k γγ AACC /(2/(2ππ)) ))22 PPSS / (N/ (Noo f f maxmax33)) S&M Eq. 6.98S&M Eq. 6.98

where kwhere k s(ts(t)) ≤≤ f f CC for all t.for all t.

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EE4512 Analog and Digital Communications Chapter 6

End of Chapter 6End of Chapter 6

Analog ModulationAnalog Modulationand Demodulationand Demodulation