tele4653 l2
TRANSCRIPT
TELE4653 Digital Modulation &Coding
Digital Modulation
Wei Zhang
School of Electrical Engineering and Telecommunications
The University of New South Wales
Outline
PAM (ASK)
PSK
QAM
FSK
TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.1/25
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
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
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
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
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)
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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
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
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
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
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
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
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
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
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
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
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
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
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
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