Test 2 Review - University of Torontodkundur/course_info/316/KundurTest2Review... · Test 2 Review Professor Deepa Kundur University of Toronto Professor Deepa Kundur (University

Test 2 Review

Professor Deepa Kundur

University of Toronto

Professor Deepa Kundur (University of Toronto) Test 2 Review 1 / 57

Test 2 Review

Reference:

Sections:4.1, 4.2, 4.3, 4.4, 4.6, 4.7, 4.8 of5.1, 5.2, 5.3, 5.4, 5.56.1, 6.2, 6.3, 6.4, 6.5, 6.6

S. Haykin and M. Moher, Introduction to Analog & Digital Communications, 2nded., John Wiley & Sons, Inc., 2007. ISBN-13 978-0-471-43222-7.


Chapter 4: Angle Modulation


Angle Modulation

I Consider a sinusoidal carrier:

c(t) = Ac cos(2πfct + φc︸︷︷︸angle

) = Ac cos(θi(t))

θi(t) = 2πfct + φc = 2πfct for φc = 0

fi(t) =1

2π

dθi(t)

dt= fc

I Angle modulation: the message signal m(t) is piggy-backed onθi(t) in some way.


Angle ModulationI Phase Modulation (PM):

θi (t) = 2πfct + kpm(t)

fi (t) =1

2π

dθi (t)

dt= fc +

kp2π

dm(t)

dtsPM(t) = Ac cos[2πfct + kpm(t)]

I Frequency Modulation (FM):

θi (t) = 2πfct + 2πkf

∫ t

0

m(τ)dτ

fi (t) =1

2π

dθi (t)

dt= fc + kfm(t)

sFM(t) = Ac cos

[2πfct + 2πkf

∫ t

0

m(τ)dτ

]


PM vs. FM

sPM(t) = Ac cos[2πfct + kpm(t)]

sFM(t) = Ac cos

[2πfct + 2πkf

∫ t

0

m(τ)dτ

]sPM(t) = Ac cos

[2πfct + kp

dg(t)

dt

]sFM(t) = Ac cos [2πfct + 2πkf g(t)]

FMwave

Modulatingwave Integrator

PhaseModulator

PMwave

Modulatingwave

Di�eren-tiator

FrequencyModulator


carrier

message

amplitudemodulation

phasemodulation

frequencymodulation


carrier

message

amplitudemodulation

phasemodulation

frequencymodulation


carrier

message

amplitudemodulation

phasemodulation

frequencymodulation


Angle Modulation

Integrator PhaseModulator

m(t) s (t)FM

Di�erentiator FrequencyModulator

m(t) s (t)PM


Properties of Angle Modulation

1. Constancy of transmitted power

2. Nonlinearity of angle modulation

3. Irregularity of zero-crossings

4. Difficulty in visualizing message

5. Bandwidth versus noise trade-off


Constancy of Transmitted Power: PM


Constancy of Transmitted Power: FM


Constancy of Transmitted Power: AM


Nonlinearity of Angle Modulation

Consider PM (proof also holds for FM).

I Suppose

s1(t) = Ac cos [2πfct + kpm1(t)]

s2(t) = Ac cos [2πfct + kpm2(t)]

I Let m3(t) = m1(t) + m2(t).

s3(t) = Ac cos [2πfct + kp(m1(t) + m2(t))]

6= s1(t) + s2(t)

∵ cos(2πfct + A + B) 6= cos(2πfct + A) + cos(2πfct + B)

Therefore, angle modulation is nonlinear.


Irregularity of Zero-Crossings

I Zero-crossing: instants of time at which waveform changesamplitude from positive to negative or vice versa.


Zero-Crossings: PM


Zero-Crossings: FM


Zero-Crossings: AM


Difficulty of Visualizing Message

I Visualization of a message refers to the ability to glean insightsabout the shape of m(t) from the modulated signal s(t).


Visualization: PM


Visualization: FM


Visualization: AM


Bandwidth vs. Noise Trade-Off

I Noise affects the message signal piggy-backed as amplitudemodulation more than it does when piggy-backed as anglemodulation.

I The more bandwidth that the angle modulated signal takes,typically the more robust it is to noise.


carrier

message

amplitudemodulation

phasemodulation

frequencymodulation


AM vs. FM

I AM is an older technology first successfully carried out in themid 1870s than FM was developed in the 1930s (by EdwinArmstrong).

I FM has better performance than AM because it is lesssusceptible to noise.

I FM takes up more transmission bandwidth than AM; Recall,

BT ,FM = 2∆f + 2fm vs. BT ,AM = 2W or W

I AM is lower complexity than FM.


Narrow Band Frequency Modulation

I Suppose m(t) = Amcos(2πfmt).

fi (t) = fc + kf Amcos(2πfmt) = fc + ∆f cos(2πfmt)

∆f = kf Am ≡ frequency deviation

θi (t) = 2π

∫ t

0

fi (τ)dτ

= 2πfct +∆f

fmsin(2πfmt) = 2πfct + βsin(2πfmt)

β =∆f

fmsFM(t) = Ac cos [2πfct + βsin(2πfmt)]

For narrow band FM, β � 1.


Narrowband FM

Modulation:

sFM(t) ≈ Ac cos(2πfct)︸︷︷︸carrier

−β Ac sin(2πfct)︸︷︷︸−90oshift of carrier

sin(2πfmt)︸︷︷︸2πfmAm

∫ t0 m(τ)dτ︸︷︷︸

DSB-SC signal

Modulatingwave

IntegratorNarrow-band

FM waveProduct

Modulator

-90 degreePhase Shifter carrier

+-


Transmission Bandwidth of FM Waves

A significant component of the FM signal is within the followingbandwidth:

BT ≈ 2∆f + 2fm = 2∆f

(1 +

1

β

)I called Carson’s Rule

I ∆f is the deviation of the instantaneous frequency

I fm can be considered to be the maximum frequency of themessage signal

I For β � 1, BT ≈ 2∆f = 2kfAm

I For β � 1, BT ≈ 2∆f 1β

= 2∆f∆f /fm

= 2fm


Generation of FM Waves

Narrow bandModulator

FrequencyMultiplier

m(t) s(t) s’(t)Integrator

CrystalControlledOscillator

frequency isvery stable

Narrowband FM modulator

widebandFM wave


Demodulation of FM Waves

Ideal EnvelopeDetector

ddt

I Frequency Discriminator: uses positive and negative slopecircuits in place of a differentiator, which is hard to implementacross a wide bandwidth

I Phase Lock Loop: tracks the angle of the in-coming FM wavewhich allows tracking of the embedded message


Chapter 5: Pulse Modulation


Pulse Modulation

I the variation of a regularly spaced constant amplitude pulsestream to superimpose information contained in a message signal

tT

TsNote: T < Ts

A

I Three types:

1. pulse amplitude modulation (PAM)2. pulse duration modulation (PDM)3. pulse position modulation (PPM)


Pulse Amplitude Modulation (PAM)

tT

Ts

t

m(t)m(0)m(-T )s

m(T )s

m(2T ) s

t

m(t)

s(t)

T

Ts

m(0)m(-T )sm(T )s

m(2T ) s


Pulse Duration Modulation (PDM)

tT

Ts

t

m(t)m(0)m(-T )s

m(T )s

m(2T ) s

t

m(t) s(t)m(0)m(-T )s

m(T )s

m(2T ) s

PDM


Pulse Position Modulation (PPM)s

t

m(t)m(0)m(-T )s

m(T )s

m(2T ) s

t

m(t) s(t)m(0)m(-T )s

m(T )s

m(2T ) s

PDM

m(t) s(t)

PPM


Summary of Pulse ModulationLet g(t) be the pulse shape.

I PAM:

sPAM(t) =∞∑

n=−∞kam(nTs)g(t − nTS)

where ka is an amplitude sensitivity factor; ka > 0.

I PDM:

sPDM(t) =∞∑

n=−∞g

(t − nTs

kdm(nTs) + Md

)where kd is a duration sensitivity factor; kd |m(t)|max < Md .

I PPM:

sPPM(t) =∞∑

n=−∞g(t − nTs − kpm(nTs))

where kp is a position sensitivity factor; kp |m(t)|max < (Ts/2).


Pulse-Code Modulation

I Most basic form of digital pulse modulation

PCM Data Sequence

ChannelOutput

Transmitter ReceiverTranmissionPathSO

URC

E

DES

TIN

ATIO

N


PCM Transmitter

PCM Data Sequence

Anti-aliasedCts-timeSignal

Discrete-timeSignal

Digitalsignal

Continuous-timeMessageSignal

Source Sampler Quantizer EncoderLow-passFilter { {

Anti-aliasingFilter

Sampling aboveNyquist with

Narrow RectangularPAM Pulses

Maps Numbersto Bit Sequences

{ {Using a

Non-uniformQuantizer


PCM Transmitter: Sampler

PCM Data Sequence


Discrete-timeSignal

Digitalsignal



Anti-aliasingFilter




{ {

Using aNon-uniform

Quantizer

ts(t)

T

Ts

m(0)m(-T )sm(T )s

m(2T ) sm(t)

t

m(0)m(-T )sm(T )s

m(2T ) s


PCM Transmitter: Non-Uniform Quantizer

PCM Data Sequence


Discrete-timeSignal

Digitalsignal



Anti-aliasingFilter




{ {

Using aNon-uniform

Quantizer

AmplitudeCompressor

v

m2 3-2-3 -

23

-2

-3

-

UniformQuantizer

-1 10-2-3 2 3n

1

v[n] m[n]

0 0.25 0.5 0.75 1.0

0.25

0.5

0.75

1.0

Normalized input |m|

Nor

mal

ized

out

put |

v| large mu


PCM Transmitter: Encoder

PCM Data Sequence


Discrete-timeSignal

Digitalsignal



Anti-aliasingFilter




{ {

Using aNon-uniform

Quantizer

Quantization-Level Index Binary Codeword (R = 3)

0 0001 0012 0103 0114 1005 1016 1107 111


Q: A signal g(t) with bandwidth 20 Hz is sampled at the Nyquist rate isuniformly quantized into L = 16, 777, 216 levels and then binary coded. (a) Howmany bits are required to encode one sample? (b) What is the number ofbits/second required to encode the audio signal?

A: (a) Number of bits per sample = log2(16777216) = 24 bits.A: (b) Number of bits per second = Number of bits per sample × Number ofsamples/second = 24× 2× 20 = 960 bits per second


Q: The signal g(t) has a peak voltage of 8 V (i.e., ranges between ±8 V). Whatis the power of the uniform quantization noise eq(t)?Note: The power of quantization noise from a uniform quantizer is:

EQN =∆2

12Watts

where ∆ ≡ separation of quantization levels.

-1 10-2-3 2 3n

1

x [n]q x[n]

A: To compute ∆:

∆ =range of signal

L=

8− (−8)

16777216=

1

1048576=

1

220

EQN =∆2

12Watts =

(2−20)2

12=

2−40

12Watts


Q: The original un-quantized signal has average power of 10 W. What is theresulting ratio of the signal to quantization noise (SQNR) of the uniformquantizer output in decibels?

A: It is given that ES = 10. Therefore,

SQNR = 10 log10

(ES

EQN

)= 10 log10

(10

2−40

12

)≈ 141.2 dB


PCM: Transmission Path

PCM Data Sequence

ChannelOutput

Transmitter ReceiverTranmissionPathSO

URC

E

DES

TIN

ATIO

N

ChannelOutput

TranmissionLine

RegenerativeRepeater

TranmissionLine

RegenerativeRepeater

TranmissionLine

...PCM Data Shaped for Transmission

Decision-making

DeviceAmpli�er-Equalizer

TimingCircuit

DistortedPCMWave

RegeneratedPCMWave


PCM: Regenerative RepeaterDecision-making

DeviceAmpli�er-Equalizer

TimingCircuit

DistortedPCMWave

RegeneratedPCMWave

t

t

t

t

t

THRESHOLD

BIT ERROR

“0” “0” “0”“1” “1” “1”

Original PCM Wave


PCM: Receiver

Two Stages:

1. Decoding and Expanding:

1.1 regenerate the pulse one last time1.2 group into code words1.3 interpret as quantization level1.4 pass through expander (opposite of compressor)

2. Reconstruction:

2.1 pass expander output through low-pass reconstruction filter(cutoff is equal to message bandwidth) to estimate originalmessage m(t)


Chapter 6: Baseband Data Transmission


Baseband Transmission of Digital Data

ThresholdBinary input sequence

LineEncoder

0011100010Source

Threshold

Decision-Making Device

Sampleat time

DestinationTransmit-Filter G(f )

ChannelH(f )

Receive-�lter Q(f )

Outputbinary data

Threshold


Sampleat time

Pulse Spectrum P(f )

bk = {0, 1} and ak =

{+1 if bk is symbol 1−1 if bk is symbol 0

s(t) =∞∑

k=−∞

akg(t − kTb)

x(t) = s(t) ? h(t)

y(t) = x(t) ? q(t) = s(t) ? h(t) ? q(t)

=∞∑

k=−∞

akg(t − kTb) ? h(t) ? q(t) =∞∑

k=−∞

akp(t − kTb)




LineEncoder

0011100010Source

Threshold


Sampleat time

DestinationTransmit-Filter G(f )

ChannelH(f )


Outputbinary data

Threshold


Sampleat time


∴ y(t) =∞∑

k=−∞

akp(t − kTb)

where p(t) = g(t) ? h(t) ? q(t)

P(f ) = G (f ) · H(f ) · Q(f ).



Threshold


Sampleat time

Transmit-Filter G(f )

ChannelH(f )



LineEncoder

0011100010Source Destination

Outputbinary data

Threshold


Sampleat time


P(f ) = G(f )H(f )Q(f )



Threshold


Sampleat time


P(f ) = G(f )H(f )Q(f )

yi = y(iTb) and pi = p(iTb)

yi =√Eai︸︷︷︸

signal to detect

+∞∑

k=−∞,k 6=i

akpi−k︸︷︷︸intersymbol interference

for i ∈ Z

To avoid intersymbol interference (ISI), we need pi = 0 for i 6= 0.


The Nyquist Channel

I Minimum bandwidth channel

I Optimum pulse shape:

popt(t) =√E sinc(2B0t)

Popt(f ) =

{ √E

2B0−B0 < f < B0

0 otherwise, B0 =

1

2Tb

Note: No ISI.

pi = p(iTb) =√E sinc(2B0iTb)

√E sinc

(2 · 1

2TbiTb

)=√E sinc(i) = 0 for i 6= 0.

Disadvantages: (1) physically unrealizable (sharp transition in freq domain); (2)

slow rate of decay leaving no margin of error for sampling times.


Raised-Cosine Pulse SpectrumI has a more graceful transition in the frequency domain

I more practical pulse shape:

p(t) =√E sinc(2B0t)

(cos(2παB0t)

1− 16α2B02t2

)

P(f ) =

√E

2B00 ≤ |f | < f1

√E

4B0

{1 + cos

[π(|f |−f1)(B0−f1)

]}f1 < f < 2B0 − f1

0 2B0 − f1 ≤ |f |

α = 1− f1B0

BT = B0(1 + α) where B0 =1

2Tband fv = αB0

Note: No ISI. ∵ pi = 0 for i 6= 0.Professor Deepa Kundur (University of Toronto) Test 2 Review 55 / 57

Raised-Cosine Pulse Spectrum

f (kHz)

A= sqrt(E)/2B

A/2

0

0B0-B 02B0-2B0B /2 0B /2 0B /2 0B /2

Raised-Cosine,α=0.5

Raised-Cosine,α=1

NyquistPulse, α=0

Nyquist PulseBandwidth

Trade-off: larger bandwidth than Nyquist pulse.Professor Deepa Kundur (University of Toronto) Test 2 Review 56 / 57

The Eye Pattern

Slope dictates sensitivityto timing error

Best samplingtime

Distortion at sampling time

NO

ISE

MA

RGIN

Time interval over which waveis best sampled.

ZERO-CROSSINGDISTORTION

Note: an “open” eye denotes a larger noise margin, lowerzero-crossing distortion and greater robustness to timing error. �


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Test 2 Review - University of Torontodkundur/course_info/316/KundurTest2Review... · Test 2 Review Professor Deepa Kundur University of Toronto Professor Deepa Kundur (University