1. TELE4653 Digital Modulation &
Coding
Digital Modulation
Wei Zhang
w.zhang@unsw.edu.au
School of Electrical Engineering and Telecommunications
The University of New South Wales

2. Outline
PAM (ASK)
PSK
QAM
FSK
TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.1/2

3. 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
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4. Definitions
1
Signaling Interval: Ts . Signaling (Symbol) Rate: Rs = Ts .
Ts
Bit Interval: Tb = k for a signal carrying k bits of
information.
Bit Rate: R = kRs = Rs logM .
2
M
Average signal energy: Eavg = m=1 pm Em with pm being
the probability of the mth signal.
Eavg Eavg
Average energy per bit: Ebavg = k = logM
.
2
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5. PAM
The signal waveform may be represented as
sm (t) = Am p(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.4/2

7. PAM
The energy of signal sm (t) is given by
∞
Em = A2 p2 (t)dt = A2 Ep
m m (3)
−∞
The average signal energy is
M M
Ep
Eavg = p m Em = A2
m
M
m=1 m=1
(M 2 − 1)Ep
= . (4)
3
and the average energy per bit is
(M 2 − 1)Ep
Ebavg = M
. (5)
3 log2
TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.6/2

8. Bandpass PAM
The bandpass PAM signals are carrier-modulated bandpass
signals with lowpass equivalents of the form s ml (t) = Am g(t),
where Am and g(t) are real. The signal waveform is
sm (t) = sml (t)ej2πfc t = Am g(t) cos(2πfc t) (6)
The energy of signal sm (t) is given by
A2
Em = m Eg . (7)
2
Moreover,
(M 2 −1)Eg (M 2 −1)Eg
Eavg = 6 , Ebavg = 6 logM
. (8)
2
TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.7/2

10. Bandpass PAM - Expansion
For basedband PAM, the expansion of sm (t) = Am p(t) is
sm (t) = Am Ep φ(t) (9)
where p(t)
φ(t) = . (10)
Ep
For bandpass PAM, the expansion of sm (t) = sml (t)ej2πfc t
is [Tutorial 1]
Eg
sm (t) = Am φ(t) (11)
2
where
2
φ(t) = g(t) cos(2πfc t). (12)
Eg
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11. Bandpass PAM - dmin
The bandpass PAM can be represented as the one-dimensional
E
vector: sm = Am 2g , where Am = ±1, ±3, · · · , ±(M − 1).
The Euclidean distance between any pair of signal points is
2
Eg
dmn = s m − sn = |Am − An | (13)
2
For adjacent signal points |Am − An | = 2, it has
12 log2 M
dmin = 2Eg = E
2 − 1 bavg
(14)
M
where in the last equality Eq. (8) is used.
TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.10/2

12. Phase Modulation
The bandpass PM signal waveform may be represented as
2π(m−1)
j
sm (t) = g(t)e M ej2πfc t , 1 ≤ m ≤ M
2π(m − 1)
= g(t) cos + 2πfc t (15)
M
2π(m−1)
Let θm = M , m = 1, 2, · · · , M . Then,
sm (t) = g(t) cos θm cos(2πfc t) − g(t) sin θm sin(2πfc t). (16)
Digital phase modulation is usually called phase-shift keying
(PSK).
The PSK signals have equal energy, Eavg = Em = 1 Eg .
2
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14. Phase Modulation - Expansion
2π(m−1)
j
The expansion of PM signal s(t) = g(t)e M ej2πfc t is
Eg Eg ˜
sm (t) = cos(θm )φ(t) + sin(θm )φ(t) (17)
2 2
where [see Tutorial 1]
2
φ(t) = g(t) cos(2πfc t) (18)
Eg
˜ 2
φ(t) = − g(t) sin(2πfc t) (19)
Eg
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15. Phase Modulation - dmin
The bandpass PM can be represented as the two-dimensional
Eg Eg
vector: sm = 2 cos θm , 2 sin θm , m = 1, 2, · · · , M . Note
θm = 2π(m−1) for m = 1, 2, · · · , M .
M
The Euclidean distance between any pair of signal points is
2
Eg
dmn = sm − sn = | cos θm − cos θn |2 + | sin θm − sin θn |2
2
= Eg [1 − cos(θm − θn )]. (20)
TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.14/2

16. Phase Modulation - dmin
For adjacent signal points |m − n| = 1, it has
2π
dmin = Eg 1 − cos( ) (21)
M
2 π
= 2Eg sin (22)
M
2 π
= 2 Ebavg log2 M × sin (23)
M
π π
For large values of M , we have sin M ≈ M, and then
π2
dmin ≈ 2 Ebavg log2 M × 2 (24)
M
TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.15/2

17. QAM
The quadrature amplitude modulation (QAM) signal waveform
may be expressed as
sm (t) = (Ami + jAmq )g(t)ej2πfc t , m = 1, 2, · · · , M
= Ami g(t) cos(2πfc t) − Amq g(t) sin(2πfc t), (25)
where Ami and Amq are the information-bearing signal
amplitudes of the quadrature carriers and g(t) is the signal pulse.
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18. QAM
Alternatively, the QAM signal may be expressed as
sm (t) = rm g(t)ejθm ej2πfc t
= rm g(t) cos(θm + 2πfc t), (26)
where rm = A2 + A2 and θm = tan−1 (Amq /Ami ).
mi mq
It is apparent that QAM signal can be viewed as combined
amplitude rm and phase θm modulation.
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20. QAM - Expansion
˜
Similar to PSK case, φ(t) in (18) and φ(t) in (19) can be used as
orthonormal basis for expansion of QAM signals [Tutorial 1]
Eg Eg ˜
sm (t) = Ami φ(t) + Amq φ(t) (27)
2 2
which results in vector representation of the form
Eg Eg
sm = (sm1 , sm2 ) = Ami , Amq (28)
2 2
and
2 Eg
Em = s m = A2 + A 2 .
mi mq (29)
2
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21. QAM - dmin
The Euclidean distance between any pair of signal vectors in
QAM is
dmn = s m − sn 2
Eg
= [(Ami − Ani )2 + (Amq − Anq )2 ]. (30)
2
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)
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23. Square QAM
In the case of square QAM (i,e., M = 4, 16, 64, 256, · · · ) with
√
amplitudes of ±1, ±3, · · · , ±( M − 1)
√ √
M M
1 Eg
Eavg = (A2 + A2 )
m n
M 2
m=1 n=1
Eg 2M (M − 1) M −1
= × = Eg (32)
2M 3 3
and
M −1
Ebavg = Eg (33)
3 log2 M
6 log2 M
dmin = Ebavg (34)
M −1
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24. PAM, PSK, QAM
For bandpass PAM, PSK, and QAM, the signaling schemes are
of the general form
sm (t) = Am g(t)ej2πfc t , m = 1, 2, · · · , M (35)
For PAM, Am is real, equal to ±1, ±3, · · · , ±(M − 1)
j 2π (m−1)
For PSK, Am is complex and equal to e M
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.23/2

25. FSK
The FSK signal waveform is
sm (t) = sml (t)ej2πfc t , 1 ≤ m ≤ M, 0 ≤ t ≤ T
2E
= cos (2πfc t + 2πm∆f t) (36)
T
2E j2πm∆f t
where sml (t) = T e , 1 ≤ m ≤ M and 0 ≤ t ≤ T .
FSK signaling is a nonlinear modulation scheme, whereas ASK,
PSK, and QAM are linear modulation schemes.
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26. FSK
FSK is an orthogonal signaling if [ sml (t), snl (t) ] = 0, m = n.
T
2E
sml (t), snl (t) = ej2π(m−n)∆f t dt (37)
T 0
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
1
∆f = 2T .
TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.25/2