final exam review - university of toronto · final exam review professor deepa ... professor deepa...

17
Final Exam Review Professor Deepa Kundur University of Toronto Professor Deepa Kundur (University of Toronto) Final Exam Review 1 / 67 Final Exam Review Reference: Sections: 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 4.1, 4.2, 4.3, 4.4, 4.6, 4.7, 4.8 5.1, 5.2, 5.3, 5.4, 5.5 6.1, 6.2, 6.3, 6.4, 6.5, 6.6 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8 10.1, 10.2 of S. Haykin and M. Moher, Introduction to Analog & Digital Communications, 2nd ed., John Wiley & Sons, Inc., 2007. ISBN-13 978-0-471-43222-7. Professor Deepa Kundur (University of Toronto) Final Exam Review 2 / 67 Chapter 3: Amplitude Modulation Professor Deepa Kundur (University of Toronto) Final Exam Review 3 / 67 Amplitude Modulation I In modulation need two things: 1. a modulated signal: carrier signal: c (t ) 2. a modulating signal: message signal: m(t ) I carrier: I c (t )= A c cos(2πf c t ); phase φ c = 0 is assumed. I message: I m(t ) (information-bearing signal) I assume bandwidth/max freq of m(t ) is W Professor Deepa Kundur (University of Toronto) Final Exam Review 4 / 67

Upload: phamphuc

Post on 25-May-2018

239 views

Category:

Documents


0 download

TRANSCRIPT

Final Exam Review

Professor Deepa Kundur

University of Toronto

Professor Deepa Kundur (University of Toronto) Final Exam Review 1 / 67

Final Exam Review

Reference:

Sections:2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.73.1, 3.2, 3.3, 3.4, 3.5, 3.64.1, 4.2, 4.3, 4.4, 4.6, 4.7, 4.85.1, 5.2, 5.3, 5.4, 5.56.1, 6.2, 6.3, 6.4, 6.5, 6.67.1, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.810.1, 10.2 of

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 2 / 67

Chapter 3: Amplitude Modulation

Professor Deepa Kundur (University of Toronto) Final Exam Review 3 / 67

Amplitude Modulation

I In modulation need two things:

1. a modulated signal: carrier signal: c(t)2. a modulating signal: message signal: m(t)

I carrier:I c(t) = Ac cos(2πfct); phase φc = 0 is assumed.

I message:I m(t) (information-bearing signal)I assume bandwidth/max freq of m(t) is W

Professor Deepa Kundur (University of Toronto) Final Exam Review 4 / 67

Amplitude Modulation

Three types studied:

1. Amplitude Modulation (AM)(yes, it has the same name as the class of modulation techniques)

2. Double Sideband-Suppressed Carrier (DSB-SC)

3. Single Sideband (SSB)

Professor Deepa Kundur (University of Toronto) Final Exam Review 5 / 67

Amplitude Modulation TechniquesAM:

sAM(t) = Ac [1 + kam(t)] cos(2πfct)

SAM(f ) =Ac

2[δ(f − fc) + δ(f + fc)] +

kaAc

2[M(f − fc) + M(f + fc)]

f

S (f )

f

S (f )

f

S (f )

f

S (f )

2W 2W

2W 2W

W W

W W

AM

USSB

LSSB

DSB

I highest power

I BT = 2W

I lowest complexity

Professor Deepa Kundur (University of Toronto) Final Exam Review 6 / 67

Amplitude Modulation TechniquesAM: For:

1. 1 + kam(t) > 0 (envelope is always positive); and

2. fc �W (message moves slowly compared to carrier)

m(t) can be recovered with an envelope detector.

AMwave

Output+

-+

-

Professor Deepa Kundur (University of Toronto) Final Exam Review 7 / 67

Amplitude Modulation TechniquesDSB-SC:

sDSB(t) = Ac cos(2πfct)m(t)

SDSB(f ) =Ac

2[M(f − fc) + M(f + fc)] f

S (f )

f

S (f )

f

S (f )

f

S (f )

2W 2W

2W 2W

W W

W W

AM

USSB

LSSB

DSB

I lower power

I BT = 2W

I higher complexity

Professor Deepa Kundur (University of Toronto) Final Exam Review 8 / 67

Amplitude Modulation Techniques

DSB-SC:

I An envelope detector will not be able to recover m(t); it willinstead recover |m(t)|.

I Coherent demodulation is required.

ProductModulator

Low-pass�lter

Local Oscillaor

DemodulatedSignal

v (t)0s(t)

Professor Deepa Kundur (University of Toronto) Final Exam Review 9 / 67

Amplitude Modulation TechniquesSSB:

sUSSB(t) =Ac

2m(t) cos(2πfct)− Ac

2m̂(t) sin(2πfct)

SUSSB(f ) =

{Ac

2 [M(f − fc) + M(f + fc)] |f | ≥ fc0 |f | < fc

sLSSB(t) =Ac

2m(t) cos(2πfct) +

Ac

2m̂(t) sin(2πfct)

SLSSB(f ) =

{0 |f | > fcAc

2 [M(f − fc) + M(f + fc)] |f | ≤ fc

H(f ) = -j sgn(f )M(f ) M(f )

f

H(f )j

-j

h(t) = 1/( t)m(t) m(t)

Professor Deepa Kundur (University of Toronto) Final Exam Review 10 / 67

Amplitude Modulation TechniquesSSB:

upper SSB

lower SSB

f

S (f )

f

S (f )

f

S (f )

f

S (f )

2W 2W

2W 2W

W W

W W

AM

USSB

LSSB

DSB

I lowest power

I BT = W

I highest complexity

Professor Deepa Kundur (University of Toronto) Final Exam Review 11 / 67

Amplitude Modulation Techniques

SSB:

I Coherent demodulation works here as well.

ProductModulator

Low-pass�lter

Local Oscillaor

DemodulatedSignal

v (t)0s(t)

Professor Deepa Kundur (University of Toronto) Final Exam Review 12 / 67

Costas Receiver

-90 degreePhase Shifter

Voltage-controlledOscillator

ProductModulator

Low-passFilter

PhaseDiscriminator

ProductModulator

Low-pass�lter

DSB-SC wave

DemodulatedSignal

Coherent Demodulation

Circuit for Phase Locking

v (t)0local oscillator output

Professor Deepa Kundur (University of Toronto) Final Exam Review 13 / 67

Costas Receiver

-90 degreePhase Shifter

Voltage-controlledOscillator

ProductModulator

Low-passFilter

PhaseDiscriminator

ProductModulator

Low-pass�lter

DSB-SC wave

DemodulatedSignal

Q-Channel (quadrature-phase coherent detector)

I-Channel (in-phase coherent detector)

v (t)I

v (t)Q

Professor Deepa Kundur (University of Toronto) Final Exam Review 14 / 67

Quadrature Amplitude Modulation

-90 degreePhase Shifter

ProductModulator

ProductModulator

MultiplexedSignal

Messagesignal

Messagesignal

s(t) = Acm1(t) cos(2πfct) + Acm2(t) sin(2πfct)

Professor Deepa Kundur (University of Toronto) Final Exam Review 15 / 67

Quadrature Amplitude Modulation

s(t) = Acm1(t) cos(2πfct) + Acm2(t) sin(2πfct)

ProductModulator

Low-passFilter

ProductModulator

Low-pass�lter

MultiplexedSignal

-90 degreePhase Shifter

Professor Deepa Kundur (University of Toronto) Final Exam Review 16 / 67

Quadrature Amplitude Modulation

s(t) = Acm1(t) cos(2πfct) + Acm2(t) sin(2πfct)

upper SSB

lower SSB

f

S (f )DSB

f

S (f )

f

S (f )

f

S (f )

2W 2W

2W 2W

W W

W W

USSB

LSSB

QAM

Professor Deepa Kundur (University of Toronto) Final Exam Review 17 / 67

Chapter 4: Angle Modulation

Professor Deepa Kundur (University of Toronto) Final Exam Review 18 / 67

Angle ModulationI Phase Modulation (PM):

θi (t) = 2πfct + kpm(t)

fi (t) =1

2pi

dθi (t)

dt= 2πfckp

dm(t)

dt

sPM(t) = Ac cos[2πfct + kpm(t)]

I Frequency Modulation (FM):

θi (t) = 2πfct + 2πkf

∫ t

0

m(τ)dτ

fi (t) =1

2pi

dθi (t)

dt= fc + kfm(t)

sFM(t) = Ac cos

[2πfct + 2πkf

∫ t

0

m(τ)dτ

]

Professor Deepa Kundur (University of Toronto) Final Exam Review 19 / 67

carrier

message

amplitudemodulation

phasemodulation

frequencymodulation

Professor Deepa Kundur (University of Toronto) Final Exam Review 20 / 67

Angle Modulation

PM and FM:

Integrator PhaseModulator

m(t) s (t)FM

Di�erentiator FrequencyModulator

m(t) s (t)PM

Professor Deepa Kundur (University of Toronto) Final Exam Review 21 / 67

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 22 / 67

Narrowband FM

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.

Professor Deepa Kundur (University of Toronto) Final Exam Review 23 / 67

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

+-

Professor Deepa Kundur (University of Toronto) Final Exam Review 24 / 67

Carson’s Rule

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

BT ≈ 2∆f + 2fm = 2∆f

(1 +

1

β

)

I For β � 1, BT ≈ 2∆f = 2kfAm

I For β � 1, BT ≈ 2∆f 1β

= 2∆f∆f /fm

= 2fm

Professor Deepa Kundur (University of Toronto) Final Exam Review 25 / 67

Generation of FM Waves: Armstrong Modulator

Narrow bandModulator

FrequencyMultiplier

m(t) s(t) s’(t)Integrator

CrystalControlledOscillator

frequency isvery stable

Narrowband FM modulator

widebandFM wave

Professor Deepa Kundur (University of Toronto) Final Exam Review 26 / 67

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 27 / 67

Chapter 5: Pulse Modulation

Professor Deepa Kundur (University of Toronto) Final Exam Review 28 / 67

Pulse Modulation

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

tTTs

A

I Three types:

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 29 / 67

Pulse Amplitude Modulation (PAM)

tTTs

At

TTs

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

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 30 / 67

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 31 / 67

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

m(t) s(t)

PPM

Professor Deepa Kundur (University of Toronto) Final Exam Review 32 / 67

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).

Professor Deepa Kundur (University of Toronto) Final Exam Review 33 / 67

Pulse-Code Modulation

I Most basic form of digital pulse modulation

PCM Data Sequence

ChannelOutput

Transmitter ReceiverTranmissionPathSO

URC

E

DES

TIN

ATIO

N

Professor Deepa Kundur (University of Toronto) Final Exam Review 34 / 67

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 35 / 67

PCM Transmitter: Sampler

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 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

Professor Deepa Kundur (University of Toronto) Final Exam Review 36 / 67

PCM Transmitter: Non-Uniform Quantizer

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 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

Professor Deepa Kundur (University of Toronto) Final Exam Review 37 / 67

PCM Transmitter: Encoder

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 aNon-uniform

Quantizer

Quantization-Level Index Binary Codeword (R = 3)

0 0001 0012 0103 0114 1005 1016 1107 111

Professor Deepa Kundur (University of Toronto) Final Exam Review 38 / 67

Encoder: Example

Professor Deepa Kundur (University of Toronto) Final Exam Review 39 / 67

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 40 / 67

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 41 / 67

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)

Professor Deepa Kundur (University of Toronto) Final Exam Review 42 / 67

Chapter 6: Baseband Data Transmission

Professor Deepa Kundur (University of Toronto) Final Exam Review 43 / 67

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

Decision-Making Device

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)

Professor Deepa Kundur (University of Toronto) Final Exam Review 44 / 67

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

Decision-Making Device

Sampleat time

Pulse Spectrum P(f )

∴ y(t) =∞∑

k=−∞

akp(t − kTb)

where p(t) = g(t) ∗ h(t) ∗ q(t)

P(f ) = G (f ) · H(f ) · Q(f ).

Professor Deepa Kundur (University of Toronto) Final Exam Review 45 / 67

Baseband Transmission of Digital Data

Threshold

Decision-Making Device

Sampleat time

Transmit-Filter G(f )

ChannelH(f )

Receive-�lter Q(f )

ThresholdBinary input sequence

LineEncoder

0011100010Source Destination

Outputbinary data

Threshold

Decision-Making Device

Sampleat time

Pulse Spectrum P(f )

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 46 / 67

Baseband Transmission of Digital Data

Threshold

Decision-Making Device

Sampleat time

Pulse Spectrum P(f )

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.

Professor Deepa Kundur (University of Toronto) Final Exam Review 47 / 67

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.

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

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 48 / 67

Raised-Cosine Pulse Spectrum

I 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.Professor Deepa Kundur (University of Toronto) Final Exam Review 49 / 67

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

NyquistPulse

Professor Deepa Kundur (University of Toronto) Final Exam Review 50 / 67

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

Professor Deepa Kundur (University of Toronto) Final Exam Review 51 / 67

Chapter 7: Digital Band-Pass ModulationTechniques

Professor Deepa Kundur (University of Toronto) Final Exam Review 52 / 67

Binary Modulation Schemes

c(t) = Ac cos(2πfct + φc)

I Binary amplitude-shift keying (BASK): carrier amplitude is keyed between

two possible values (typically√Eb and 0 to represent 1 and 0, respectively);

carrier phase and frequency are held constant.

I Binary phase-shift keying (BPSK):carrier phase is keyed between twopossible values (typically 0 and π to represent 1 and 0, respectively); carrieramplitude and frequency are held constant.

I Binary frequency-shift keying (BFSK): carrier frequency is keyed betweentwo possible values (typically f1 and f2 to represent 1 and 0, respectively);carrier amplitude and phase are held constant.

Professor Deepa Kundur (University of Toronto) Final Exam Review 53 / 67

Preliminaries

c(t) = Ac cos(2πfct + φc)

I Tb represents the bit duration

I Eb represents the energy of the transmitted signal per bit

I In digital communications the carrier amplitude is normalized to have unitenergy in one bit duration; thus we set

Ac =

√2

Tb

I The carrier frequency fc = kTb

for k ∈ Z to ensure an integer number ofcarrier cycles in a bit duration.

Professor Deepa Kundur (University of Toronto) Final Exam Review 54 / 67

Carrier for Digital Communications

Therefore

c(t) =

√2

Tbcos(2πfct + φc).

For k = 4

0t

Professor Deepa Kundur (University of Toronto) Final Exam Review 55 / 67

Binary Amplitude-Shift Keying (BASK)

Let φc = 0 and the carrier frequency is fc .

b(t) =

{ √Eb for binary symbol 1

0 for binary symbol 0

c(t) =

√2

Tbcos(2πfct + φc) =

√2

Tbcos(2πfct)

s(t) = b(t) · c(t)

=

{ √2Eb

Tbcos(2πfct) for symbol 1

0 for symbol 0

Professor Deepa Kundur (University of Toronto) Final Exam Review 56 / 67

BASK Transmitter and Receiver

Transmitter

BASKTransmittedSignal

LevelEncoding

ProductModulator

carrier

EnvelopeDetector

Low-passFilterBASK

wave

Output+

-

Threshold

Decision-Making Device

Binary inputsequence

t0011100010

Sampleat time

Receiver

EnvelopeDetector

Low-passFilterBASK

wave

Output+

-

Threshold

Decision-Making Device

BASKTransmittedSignal

LevelEncoding

ProductModulator

carrier

Binary inputsequence

t0011100010

Sampleat time

Professor Deepa Kundur (University of Toronto) Final Exam Review 57 / 67

Binary Phase-Shift Keying (BPSK)

si(t) =

2Eb

Tbcos(2πfct) for symbol 1 (i = 1)√

2Eb

Tbcos(2πfct + π) for symbol 0 (i = 2)

=

2Eb

Tbcos(2πfct) for symbol 1 (i = 1)

−√

2Eb

Tbcos(2πfct) for symbol 0 (i = 2)

Professor Deepa Kundur (University of Toronto) Final Exam Review 58 / 67

BPSK Transmitter and Receiver

Transmitter

Low-passFilter

Threshold

Decision-Making Device

Sampleat time

BPSKTransmittedSignal

Binary inputsequence

LevelEncoding

ProductModulator

carrier

ProductModulator

local carrierfrom PLL

(Non-return-to zero)

t0011100010

Receiver

Low-passFilter

Threshold

Decision-Making Device

Sampleat time

BPSKTransmittedSignal

Binary inputsequence

LevelEncoding

ProductModulator

carrier

ProductModulator

local carrierfrom PLL

(Non-return-to zero)

t0011100010

Professor Deepa Kundur (University of Toronto) Final Exam Review 59 / 67

Binary Frequency-Shift Keying (BFSK)

Let φc = 0, |f1 − f2| = 1Tb

and fi = kiTb

(integer number of cycles in a

bit duration).

si(t) =

2Eb

Tbcos(2πf1t) for symbol 1 (i = 1)√

2Eb

Tbcos(2πf2t) for symbol 0 (i = 2)

0t

0 1 0 1 1 0 0 0

4 cycles 4 cycles 4 cycles 4 cycles 4 cycles5 cycles 5 cycles 5 cycles

Professor Deepa Kundur (University of Toronto) Final Exam Review 60 / 67

BFSK Transmitter and Receiver

Transmitter

BFSKTransmittedSignal

Binary inputsequence 0011100010

controlsswitchposition

localoscillator

localoscillator

Receiver

EnvelopeDetector

EnvelopeDetector

Band-pass�lter, f1

Comparator

Band-pass�lter, f2

v1

v2

{ 1 if v1 > v20 if v2 > v1

Professor Deepa Kundur (University of Toronto) Final Exam Review 61 / 67

Summary

BASK s1(t) =√

2Eb

Tbcos(2πfct) for symbol 1

s2(t) = 0 for symbol 0

BPSK s1(t) =√

2Eb

Tbcos(2πfct + 0) for symbol 1

s2(t) =√

2Eb

Tbcos(2πfct + π) for symbol 0

BFSK s1(t) =√

2Eb

Tbcos(2πf1t) for symbol 1

s2(t) =√

2Eb

Tbcos(2πf2t) for symbol 0

Professor Deepa Kundur (University of Toronto) Final Exam Review 62 / 67

Summary: Phasor Diagrams

BASK

BPSK

BFSK

symbol1

symbol1

symbol1

symbol0

symbol0

symbol0

Professor Deepa Kundur (University of Toronto) Final Exam Review 63 / 67

Summary: Phasor Diagrams

BASK

BPSK (antipodal)

BFSK (orthogonal)

symbol1

symbol1

symbol1

symbol0

symbol0

symbol0

Professor Deepa Kundur (University of Toronto) Final Exam Review 64 / 67

M-ary Digital Modulation SchemesFor M = 2m and T = mTb,

I M-ary Phase-Shift Keying

si (t) =

√2E

Tcos

(2πfc +

Mi

)i = 0, 1, . . . ,M − 1, 0 ≤ t ≤ T .

I M-ary Quadrature Amplitude Modulation

si (t) =

√2E0

Tai cos(2πfct)−

√2E0

Tbi sin(2πfct)

i = 0, 1, . . . ,M − 1, 0 ≤ t ≤ T .

I M-ary Frequency-Shift Keying

si (t) =

√2E

Tcos( πT

(n + i)t)

i = 0, 1, . . . ,M − 1, 0 ≤ t ≤ T .

Professor Deepa Kundur (University of Toronto) Final Exam Review 65 / 67

Signal Constellations

M-PSK M-QAM M-FSK

1. For a given basis set: φ1, φ2, . . . , φn. Compute correlation of si (t) witheach of the bases members.

2. Graph the result in signal space.

Professor Deepa Kundur (University of Toronto) Final Exam Review 66 / 67

Important Identities

cos(A + B) = cos(A) cos(B)− sin(A) sin(B)

cos(A) cos(B) =1

2cos(A + B) +

1

2cos(A− B)

cos(A) sin(B) =1

2sin(A + B)− 1

2sin(A− B)

cos(A) = sin(A +

π

2

)cos(A + π) = − cos(A)

cos(A) = cos(−A) sin(A) = − sin(−A)

cos2(A) =1

2+

1

2cos(2A)

cos2(A) + sin2(A) = 1

cos(A) ≈ 1 for |A| � 1

sin(A) ≈ A for |A| � 1

�Professor Deepa Kundur (University of Toronto) Final Exam Review 67 / 67