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TELE4653 Digital Modulation & Coding Digital Modulation Wei Zhang [email protected] School of Electrical Engineering and Telecommunications The University of New South Wales

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Page 1: Tele4653 l2

TELE4653 Digital Modulation &Coding

Digital Modulation

Wei Zhang

[email protected]

School of Electrical Engineering and Telecommunications

The University of New South Wales

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Outline

PAM (ASK)

PSK

QAM

FSK

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.1/25

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Modulation

Source information to be transmitted is usually in the form of

a binary data stream.

The transmission medium, i.e., communication channel

suffers from noise, attenuation, distortion, fading, and

interference.

Digital Modulation - To generate a signal that represents the

binary data stream and matches the characteristics of the

channel

Modulation with Memoryless or with Memory

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.2/25

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Definitions

Signaling Interval: Ts. Signaling (Symbol) Rate: Rs = 1Ts

.

Bit Interval: Tb = Ts

kfor a signal carrying k bits of

information.

Bit Rate: R = kRs = Rs logM2 .

Average signal energy: Eavg =∑M

m=1 pmEm with pm being

the probability of the mth signal.

Average energy per bit: Ebavg =Eavg

k=

Eavg

logM2

.

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.3/25

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PAM

The signal waveform may be represented as

sm(t) = Amp(t), 1 ≤ m ≤ M (1)

where p(t) is a pulse of duration T and Am denotes the

amplitude with the mth value, given by

Am = 2m − 1 − M, 1 ≤ m ≤ M (2)

i.e., the amplitudes are ±1,±3,±5, · · · ,±(M − 1).

Digital amplitude modulation is usually called amplitude-shift

keying (ASK).

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.4/25

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PAM

The energy of signal sm(t) is given by

Em =

∫ ∞

−∞A2

mp2(t)dt = A2mEp (3)

The average signal energy is

Eavg =

M∑

m=1

pmEm =Ep

M

M∑

m=1

A2m

=(M2 − 1)Ep

3. (4)

and the average energy per bit is

Ebavg =(M2 − 1)Ep

3 logM2

. (5)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.6/25

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Bandpass PAM

The bandpass PAM signals are carrier-modulated bandpass

signals with lowpass equivalents of the form sml(t) = Amg(t),

where Am and g(t) are real. The signal waveform is

sm(t) = <[

sml(t)ej2πfct

]

= Amg(t) cos(2πfct) (6)

The energy of signal sm(t) is given by

Em =A2

m

2Eg. (7)

Moreover,

Eavg =(M2−1)Eg

6 , Ebavg =(M2−1)Eg

6 logM2

. (8)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.7/25

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Bandpass PAM - Expansion

For basedband PAM, the expansion of sm(t) = Amp(t) is

sm(t) = Am

Epφ(t) (9)

whereφ(t) =

p(t)√

Ep

. (10)

For bandpass PAM, the expansion of sm(t) = <[

sml(t)ej2πfct

]

is [Tutorial 1]

sm(t) = Am

Eg

2φ(t) (11)

where

φ(t) =

2

Eg

g(t) cos(2πfct). (12)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.9/25

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Bandpass PAM - dmin

The bandpass PAM can be represented as the one-dimensional

vector: sm = Am

Eg

2 , where Am = ±1,±3, · · · ,±(M − 1).

The Euclidean distance between any pair of signal points is

dmn =√

‖sm − sn‖2 = |Am − An|√

Eg

2(13)

For adjacent signal points |Am − An| = 2, it has

dmin =√

2Eg =

12 log2 M

M2 − 1Ebavg (14)

where in the last equality Eq. (8) is used.

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.10/25

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Phase Modulation

The bandpass PM signal waveform may be represented as

sm(t) = <[

g(t)ej2π(m−1)

M ej2πfct]

, 1 ≤ m ≤ M

= g(t) cos

[

2π(m − 1)

M+ 2πfct

]

(15)

Let θm = 2π(m−1)M

, m = 1, 2, · · · ,M . Then,

sm(t) = g(t) cos θm cos(2πfct) − g(t) sin θm sin(2πfct). (16)

Digital phase modulation is usually called phase-shift keying

(PSK).

The PSK signals have equal energy, Eavg = Em = 12Eg.

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.11/25

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Phase Modulation - Expansion

The expansion of PM signal s(t) = <[

g(t)ej2π(m−1)

M ej2πfct]

is

sm(t) =

Eg

2cos(θm)φ(t) +

Eg

2sin(θm)φ̃(t) (17)

where [see Tutorial 1]

φ(t) =

2

Eg

g(t) cos(2πfct) (18)

φ̃(t) = −√

2

Eg

g(t) sin(2πfct) (19)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.13/25

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Phase Modulation - dmin

The bandpass PM can be represented as the two-dimensional

vector: sm =

(

Eg

2 cos θm,√

Eg

2 sin θm

)

, m = 1, 2, · · · ,M . Note

θm = 2π(m−1)M

for m = 1, 2, · · · ,M .

The Euclidean distance between any pair of signal points is

dmn =√

‖sm − sn‖2 =

Eg

2

| cos θm − cos θn|2 + | sin θm − sin θn|2

=√

Eg [1 − cos(θm − θn)]. (20)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.14/25

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Phase Modulation - dmin

For adjacent signal points |m − n| = 1, it has

dmin =

Eg

[

1 − cos(2π

M)

]

(21)

=

2Eg sin2( π

M

)

(22)

= 2

Ebavg log2 M × sin2( π

M

)

(23)

For large values of M , we have sin(

πM

)

≈ πM

, and then

dmin ≈ 2

Ebavg log2 M × π2

M2(24)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.15/25

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QAM

The quadrature amplitude modulation (QAM) signal waveform

may be expressed as

sm(t) = <[

(Ami + jAmq)g(t)ej2πfct]

, m = 1, 2, · · · ,M

= Amig(t) cos(2πfct) − Amqg(t) sin(2πfct), (25)

where Ami and Amq are the information-bearing signal

amplitudes of the quadrature carriers and g(t) is the signal pulse.

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.16/25

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QAM

Alternatively, the QAM signal may be expressed as

sm(t) = <[

rmg(t)ejθmej2πfct]

= rmg(t) cos(θm + 2πfct), (26)

where rm =√

A2mi + A2

mq and θm = tan−1(Amq/Ami).

It is apparent that QAM signal can be viewed as combined

amplitude rm and phase θm modulation.

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.17/25

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QAM - Expansion

Similar to PSK case, φ(t) in (18) and φ̃(t) in (19) can be used as

orthonormal basis for expansion of QAM signals [Tutorial 1]

sm(t) = Ami

Eg

2φ(t) + Amq

Eg

2φ̃(t) (27)

which results in vector representation of the form

sm = (sm1, sm2) =

(

Ami

Eg

2, Amq

Eg

2

)

(28)

and

Em = ‖sm‖2 =Eg

2

(

A2mi + A2

mq

)

. (29)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.19/25

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QAM - dmin

The Euclidean distance between any pair of signal vectors in

QAM is

dmn =√

‖sm − sn‖2

=

Eg

2[(Ami − Ani)2 + (Amq − Anq)2]. (30)

In the case when the signal amplitude take values of

±1,±3, · · · ,±(√

M − 1) on both Ami and Amq, the signal space

diagram is rectangular, as shown in Fig. on next page. In this

case,

dmin =√

2Eg. (31)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.20/25

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Square QAM

In the case of square QAM (i,e., M = 4, 16, 64, 256, · · · ) with

amplitudes of ±1,±3, · · · ,±(√

M − 1)

Eavg =1

M

Eg

2

√M∑

m=1

√M∑

n=1

(A2m + A2

n)

=Eg

2M× 2M(M − 1)

3=

M − 1

3Eg (32)

and

Ebavg =M − 1

3 log2 MEg (33)

dmin =

6 log2 M

M − 1Ebavg (34)

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.22/25

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PAM, PSK, QAM

For bandpass PAM, PSK, and QAM, the signaling schemes are

of the general form

sm(t) = <[

Amg(t)ej2πfct]

, m = 1, 2, · · · ,M (35)

For PAM, Am is real, equal to ±1,±3, · · · ,±(M − 1)

For PSK, Am is complex and equal to ej 2πM

(m−1)

For QAM, Am = Ami + jAmq.

Therefore, PAM and PSK can be considered as special cases of

QAM.

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.23/25

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FSK

The FSK signal waveform is

sm(t) = <[

sml(t)ej2πfct

]

, 1 ≤ m ≤ M, 0 ≤ t ≤ T

=

2ET

cos (2πfct + 2πm∆ft) (36)

where sml(t) =√

2ET

ej2πm∆ft, 1 ≤ m ≤ M and 0 ≤ t ≤ T .

FSK signaling is a nonlinear modulation scheme, whereas ASK,

PSK, and QAM are linear modulation schemes.

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.24/25

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FSK

FSK is an orthogonal signaling if < [〈sml(t), snl(t)〉] = 0, m 6= n.

〈sml(t), snl(t)〉 =2ET

∫ T

0ej2π(m−n)∆ftdt (37)

and

< [〈sml(t), snl(t)〉] = 2Esinc(2T (m − n)∆f) (38)

FSK is an orthogonal signaling when ∆f = k/2T . The minimum

frequency separation ∆f that guarantees orthogonality is

∆f = 12T

.

TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.25/25