1

Digital Communications
Digital Modulation

• Digital Baseband Modulation

• Digital Bandpass Modulation
 2

Digital Communications
Analog Signal
Bit sequence
t 0001101110……

A-D Digital Digital

Source Baseband Bandpass
Conversion
SOURCE Modulation Modulation

t t

Baseband Bandpass
Channel Channel

D-A Digital Digital

User Baseband Bandpass
Conversion Demodulation Demodulation
 3

Digital Modulation

t
Digital
Baseband
Baseband
Channel
Modulation

Bit sequence
Modulated signal
0001101110……
Digital Bandpass
Bandpass Channel
Modulation
t

• How to choose proper digital waveforms to “carry” the digits?

 4

Digital Modulation

Bit sequence Digital Modulated Baseband/Bandpass

Baseband/Bandpass
0001101110…… Signal Channel
Modulation

• Bit Rate: number of bits transmitted in unit time

• Required channel bandwidth: determined by the bandwidth of the
modulated signal.

• Bandwidth Efficiency:

Information Bit Rate Rb


Required Channel Bandwidth Bh
 5

Digital Baseband Modulation

• Pulse Amplitude Modulation (PAM)

• Pulse Shaping
 6

Digital Baseband Modulation

• Choose baseband signals to carry the digits.

– Each baseband signal can carry multiple bits.

• Each baseband signal carries 1 bit.

Binary • Bit Rate: Rb  1/ 
• Totally 2 baseband signals are required.

• Each baseband signal carries a symbol (with log2M bits).

M-ary • Symbol Rate: R  1/ 
s Bit Rate: Rb  log 2 M / 
• Totally M baseband signals are required.
 7

Digital Baseband Modulation

• Focus on “amplitude modulation”

– The baseband signals have the same shape, but different amplitudes.

– Time-domain representation of the modulated signal:


s (t )  Z
n 
n  v(t  n )

where Zn is a discrete random variable with Pr{Z n  ai }  1/ M , i  1,..., M ,

v(t) is a unit baseband signal.

– Power spectrum of the modulated signal:

1  2  Z2 
 m 
Gs ( f )  V ( f )    Z     f  
2

   m    
 8

Pulse Amplitude Modulation

(PAM)

• Binary PAM
• Binary On-Off Keying (OOK)
• 4-ary PAM
 9

Binary PAM

1: a positive rectangular pulse 0: a negative rectangular pulse

with amplitude A and width  with amplitude -A and width 


1 1 0 1 0 0 1 …
s(t) s (t )  Z
n 
n  v(t  n )

 Pr{Z n  1}  1/ 2
 A, 0  t  
 v(t )  
0, otherwise
 10

Power Spectrum of Binary PAM

1  2  Z2 
 m 
Gs ( f )  V ( f )    Z  f 
2

   m     
With Binary PAM: V ( f )  A sinc( f  ) GBPAM ( f )  A2 sinc 2 ( f  )
 Z  0,  Z2  1

GBPAM ( f )
A2

-  0   f

See Textbook or Reference [Proakis & Salehi] for more details.

 11

Effective Bandwidth of Binary PAM

GBPAM(f)  A2 sinc2 ( f  )
A2

-  0   f

90% bandwidth: 1/ 
90% power
95% bandwidth: 2 / 
95% power

• Suppose 90% of signal power must pass through the channel (90% in-band power):
Required Channel Bandwidth: Bh _ 90%  1/ 
Bh _ 90%  Rb
Bit rate: Rb  1/ 

• Suppose 95% of signal power must pass through the channel (95% in-band power):

Required Channel Bandwidth: Bh _ 95%  2 /   2 Rb

 12

Bandwidth Efficiency of Binary PAM

Information Bit Rate Rb

• Bandwidth Efficiency :  
Required Channel Bandwidth Bh

• Bandwidth Efficiency of Binary PAM:

Rb  1/ 
 BPAM  1 with 90% in-band power
Bh _ 90%  1/ 

Bh _ 95%  2 / 
 BPAM  0.5 with 95% in-band power

What if the two pulses have unsymmetrical amplitudes?

 13

Binary On-Off Keying (OOK)

1: a positive rectangular pulse 0: nothing (can be regarded as a

with amplitude A and width  pulse with amplitude 0)
A


1 1 0 1 0 0 1 …
s(t) s(t )  Z
n 
n  v(t  n )

 Pr{Z n  1}  Pr{Z n  0}  1/ 2

 A, 0  t  
 v(t )  
0, otherwise
 14

Power Spectrum of Binary OOK

1  2  Z2 
 m 
Gs ( f )  V ( f )    Z  f 
2

   m     
1
GBOOK ( f )   A sinc( f  ) 
2
With Binary OOK: V ( f )  A sinc( f  ) 
 Z  1/ 2,  Z2  1/ 4 1 1 
 m 
 
 4 4

m 
f 
   

GBOOK ( f )

1 2
A
4
… …
  0   f
 15

Bandwidth Efficiency of Binary OOK

Information Bit Rate Rb

• Bandwidth Efficiency : 
Required Channel Bandwidth Bh

• Bandwidth Efficiency of Binary OOK:

Rb  1/ 
 BOOK  1 with 90% in-band power
Bh _ 90%  1/ 

Bh _ 95%  2 / 
 BOOK  0.5 with 95% in-band power

Can we improve the bandwidth efficiency without sacrificing the in-band power?
 16

4-ary PAM

• 4-ary PAM: Each waveform carries 2-bit information.

11: 10: 01: 00:

1 1 0 1 0 0 1 0 0 0 …
s(t )  Z
n 
n  v(t  n )
s(t)
 Pr{Z n  1}  Pr{Z n  1/ 3}
 Pr{Z n  1}  Pr{Z n  1/ 3}
 1/ 4
 A, 0  t  
 v (t )  
0, otherwise
 17

Power Spectrum of 4-ary PAM

1  2  Z2 
 m 
Gs ( f )  V ( f )    Z  f 
2

   m     
With 4-ary PAM: V ( f )  A sinc( f  ) G4 PAM ( f )  95 A2 sinc 2 ( f  )
 Z  0,  Z2  5 / 9

G4PAM(f)
5
9 A2

-  0   f

• Required channel bandwidth with 90% in-band power: Bh _ 90%  1/ 

• Required channel bandwidth with 95% in-band power: Bh _ 95%  2 / 
 18

Bandwidth Efficiency of 4-ary PAM

• Symbol rate: RS  1/ 

• Bit rate: Rb  2  RS  2 / 

• Require channel bandwidth:

1
with 90% in-band power: Bh _ 90%  1/   RS  Rb
2
with 95% in-band power: Bh _ 95%  2 /   2 RS  Rb

 4 PAM  2 with 90% in-band power

 4 PAM  1 with 95% in-band power

4-ary PAM achieves higher bandwidth efficiency than binary PAM!

 19

Bandwidth Efficiency of M-ary PAM

• Suppose there are totally M distinct amplitude (power) levels.
• How many bits are carried by each symbol?
M  2k k  log 2 M
• What is the relationship between symbol rate RS and bit rate Rb?
RS  Rb / k or Rb  kRS
• What is the required channel bandwidth with 90% in-band power?
Bh _ 90%  RS  Rb / k
Tradeoff between bandwidth efficiency
• Bandwidth Efficiency of M-ary PAM and fidelity performance

 MPAM  k  log 2 M with 90% in-band power

• A larger M also leads to a smaller minimal amplitude difference – higher error

probability (to be discussed).
 20

Pulse Shaping

• Inter-Symbol Interference (ISI)

• Sinc-Shaped Pulse and Raised-Cosine Pulse
 21

Transmission over Bandlimited Channel

• Frequency domain
Baseband Channel
PAM signal H(f) GY ( f )  GPAM ( f ) | H ( f ) |2
GPAM(f)
-Bh 0 Bh f The signal distortion
incurred by channel is
0 f
always non-zero!!

• Time domain


PAM signal Baseband Channel

y (t )  s(t )  h(t )  Z
n 
n  x(t  n )

h(t) x(t )  v(t )  h(t )
s (t )  Z
n 
n  v(t  n )

Sample y(t) at m, m=1,2,…, we have Inter-symbol


Interference
y (m )  Z
n 
n  x(m  n )  Z m  x(0)   Z n  x(m  n )
nm (ISI)!
 22

ISI and Eye Diagram

• An eye diagram is
constructed by plotting
overlapping k-symbol
segments of a baseband
signal.
• An eye diagram can be
displayed on an oscillo-
scope by triggering the
time sweep of the
oscilloscope.
See Reference [Ziemer &
Tranter] (Sec. 4.6) for
more details about eye
diagram.
• ISI is caused by insufficient channel bandwidth.
• Any better choice than rectangular pulse? Sinc-Shaped pulse
 23

Sinc-Shaped Pulse

v(t) V(f)  A sinc( f  )

A A

-/2 0 /2 t -2/ -1/  1/ 2/ f


Rectangular Pulse

v(t)=Asinc(t/) V(f)
A A

     t -1/(2) 0 1/(2) f

Sinc-Shaped Pulse
 24

Binary Sinc-Shaped-Pulse Modulated Signal

1: a positive sinc-shaped pulse 0: a negative sinc-shaped pulse

with amplitude A and first with amplitude -A and first
crossing-zero point  crossing-zero point 
A
  0   t

  0   t -A

s(t) 1 1 1 0 0 1 1
… …
t
…  …
  Pr{Z n  1}  1/ 2
s (t )  Z
n 
n  v(t  n )
 v(t )  Asinc(t /  )
 25

Power Spectrum of Sinc-Shaped-Pulse Modulated Signal

1  2  Z2 
 m 
Gs ( f )  V ( f )    Z  f 
2

   m     
With Binary Sinc-Shaped-  Z  0,   1 2 GBSSP ( f )  A2 , | f | 1
2
Z
Pulse Modulated Signal: V ( f )  A , | f | 21

GBSSP(f)
A2

-1/(2) 0 1/(2) f

Bit Rate: Rb  1/   BSSP  2

Required channel bandwidth: Bh  1/(2 )  Rb / 2 (with 100% in-band power)
 26

Sinc-Shaped-Pulse Modulated Signal over Bandlimited Channel

• Frequency domain
Baseband Channel
Binary Sinc-Shaped-Pulse signal H(f) GY ( f )  GBSSP ( f ) | H ( f ) |2
GBSSP(f)
GY(f)
A2
A2
-1/(2) 0 1/(2) f
-Bh 0 Bh f
-1/(2) 0 1/(2) f

• Time domain
Binary Sinc-Shaped-Pulse signal Baseband Channel y (t )  s (t )  h(t )

s (t )  Z n  v(t  n ) h(t)
n  Zero ISI at t=m!
y(t) 1 1 1 0 0 1 1
… …
t
…  …
Are there any other (better) choices to achieve zero ISI?
 27

Nyquist Pulse-Shaping Criterion for Zero ISI

Nyquist pulse-shaping criterion for zero ISI
A necessary and sufficient condition for x(t) to satisfy
1, n  0
x(n )  
0, n  0 
is that its Fourier transform X(f) satisfies 
m 
X ( f  m )   .

Suppose that there is a baseband channel with the channel bandwidth W. To

pass a modulated signal with symbol rate 1/ through the channel:
• If 1/-W>W, there is no way to satisfy the Nyquist pulse-shaping criterion
for zero ISI. 

 X ( f  m )  
m 

…… ……
f
-2/ -1/ 0 W 1/-W 1/ 2/
 28

Nyquist Pulse-Shaping Criterion for Zero ISI

According to Nyquist pulse-shaping criterion for zero ISI:

 If the symbol rate 1/>2W, there is no way that we can design a system
with zero ISI.
 , | f | W
 If the symbol rate 1/=2W, we must have X ( f )  
0, otherwise
• The maximum symbol rate for zero ISI is 2W.

• In the binary case, the highest bandwidth efficiency for zero-ISI is

2, which is achieved by the binary sinc-shaped-pulse modulated signal.

 If the symbol rate 1/<2W, we have numerous choices. One of them is

called Raised-Cosine Pulse.

See Reference [Proakis & Salehi] (Sec. 8.3.1) for more details.
 29

Raised-Cosine Pulse: Tradeoff between Bandwidth

Efficiency and Robustness

Sinc-Shaped Pulse: v(t)=Asinc(t/) V(f)

A A

     t 
1
0
1
f
2 2
• Strong ISI at t  n .
• Perfect synchronization is required at the receiver side. 0 
1
1 2

 cos(2 t )  2
Raised-Cosine Pulse: v (t )  Asinc(t /  )  1  (4 t ) 2  V(f)
  A
A

     t 
1 0 1 f
More robust 2 2
• Larger 
Larger bandwidth
 30

Summary I: Digital Baseband Modulation

Complexity Bandwidth Efficiency

Binary PAM Low 1 (90% in-band power)

PAM
4-ary PAM Low 2 (90% in-band power)

Binary Sinc- High

Shaped-Pulse (Susceptible to 2 (100% in-band power)
Modulation timing jitter)

Rb
Binary Raised- 1 2
2 Rb  
1
Cosine-Pulse Moderate
Modulation (100% in-band power)
 31

Digital Bandpass Modulation

• Binary ASK
• Binary FSK
• Binary PSK
• Quaternary PSK
 32

Digital Bandpass Modulation

• How to transmit a baseband signal over a bandpass channel?

1 0 1 0

Amplitude Binary Amplitude Shift Keying

modulation (BASK)
Carrier Signal
t
cos(2fct)
Frequency Binary Frequency Shift Keying
modulation (BFSK)
t
t

Phase
modulation Binary Phase Shift Keying
(BPSK)

t
 33

Binary Amplitude Shift Keying (ASK)

• Generate a binary ASK signal:

– Send the carrier signal if the information bit is “1”;
– Send 0 volts if the information bit is “0”.

1 0 1 0

sBASK (t )  sBOOK (t ) cos(2 f c t ) t

Binary On-Off Keying sBOOK (t )

t
x
cos(2fct) t
 34

Power Spectrum of BASK

• Power spectrum of Binary OOK:

1 1 1 
 m 
GBOOK ( f )   A sinc( f  )      f 
2

  4 4 m     

• Power spectrum of Binary ASK:

GBASK ( f )  14 [GBOOK ( f  f c )  GBOOK ( f  f c )]

GBASK ( f )

… … … …
-fc -fc -fc 0 fc fc fc f
2/
2 Rb
 35

Bandwidth Efficiency of BASK

GBASK ( f )

-fcRb -fc -fcRb 0 fcRb fc fcRb f

2Rb

The bandwidth of BASK signal is twice of that of its baseband signal

(binary On-Off Keying)!

• The required channel bandwidth for 90% in-band power:

Bh _ 90%  2 Rb
• Bandwidth Efficiency of BASK:  BASK  0.5 with 90% in-band power
 BASK  0.25 with 95% in-band power
 36

Binary Frequency Shift Keying (BFSK)

• Generate a binary FSK signal: Frequency offset

– Send the signal A cos(2 ( f c  f )t ) if the information bit is “1”;

– Send the signal A cos(2 ( f c  f )t ) if the information bit is “0”.

sBFSK (t )  sb1, BFSK (t ) cos(2 ( f c  f )t )  sb 2, BAFSK (t ) cos(2 ( f c  f )t )

A bi  1 0 bi  1
sb1, BFSK (t )   sb 2, BFSK (t )  
bi  0 A bi  0
0
Binary On-Off Keying Binary On-Off Keying

0 1 0 1

t
 37

Bandwidth Efficiency of BFSK

GBFSK ( f )  14 [Gb1, BFSK ( f  ( f c  f ))  Gb1, BFSK ( f  ( f c  f ))]

 14 [Gb 2, BFSK ( f  ( f c  f ))  Gb 2, BFSK ( f  ( f c  f ))]

GBFSK ( f )

( f c  f ) ( f c  f ) 0 f c  f f c  f f
2f
2f  2 Rb
• The required channel bandwidth for 90% in-band power:
Bh _ 90%  2f  2 Rb
• Bandwidth efficiency of BFSK:   0.5 
1
 0.5   BASK
1  f / Rb
BFSK
(with 90% in-band power)
The bandwidth efficiency of BFSK signal is lower than that of BASK signal!
 38

Binary Phase Shift Keying (BPSK)

• Generate a binary PSK signal:

– Send the signal A cos(2 f c t ) if the information bit is “1”;
– Send the signal A cos(2 f c t   ) if the information bit is “0”.
  A cos(2 f c t )

1 0 1 0
sBPSK (t )  sBPAM (t ) cos(2 f c t )
t

Binary PAM sBPAM (t )

t
x
cos(2fct) t
 39

Bandwidth Efficiency of BPSK

GBPSK ( f )  14 [GBPAM ( f  f c )  GBPAM ( f  f c )]

GBPSK ( f )

… … … …
-fc 0 fc f
2 Rb
• The required channel bandwidth for 90% in-band power:
Bh _ 90%  2 Rb

• Bandwidth Efficiency of BPSK:  BPSK  0.5 with 90% in-band power

 BPSK  0.25 with 95% in-band power

The bandwidth efficiency of BPSK signal is the same as that of BASK signal!
 40

M-ary PSK

• M-ary PSK: transmitting pulses with M possible different carrier

phases, and allowing each pulse to represent log2M bits.

 Binary PSK: “1” s1 (t )  A cos(2 f c t )

“0” s2 (t )  A cos(2 f ct   )

 Quaternary PSK: “11” s1 (t )  A cos(2 f ct  ( / 4))

(QPSK) “10” s2 (t )  A cos(2 f c t   / 4)
“00” s3 (t )  A cos(2 f c t  3 / 4)

“01” s4 (t )  A cos(2 f c t  5 / 4)
 41

QPSK
A A
“1 1” s1 (t )  A cos(2 f c t   / 4)   cos(2 f c t )  sin(2 f c t )
2 2
A A
“1 0” 2s ( t )  A cos(2 f c t   / 4)   cos(2 f c t )  sin(2 f c t )
2 2
A A
s
“0 0” 3 (t )  A cos(2 f c t  3 / 4)   cos(2 f c t )  sin(2 f c t )
2 2
A A
s
“0 1” 4 ( t )  A cos(2 f c t  5 / 4)   cos(2 f c t )  sin(2 f c t )
2 2
A QPSK signal can be decomposed into the sum of two PSK signals:
an in-phase component and a quadrature component.
A A
sQPSK (t )  d I cos(2 f c t )  dQ sin(2 f c t )
2 2

1 if b2i 1  1 1 if b2i  1
dI   dQ  
1 if b2i 1  0 1 if b2i  0
 42

QPSK Modulator

A A
sQPSK (t )  d I cos(2 f c t )  dQ sin(2 f c t )
2 2

1 if b2i 1  1 1 if b2i  1
dI   dQ  
1 if b2i 1  0 1 if b2i  0

A
cos(2 f c t )
• Modulator Mapping 2
b2i-1 dI
1 if b2i 1  1
Bit series
dI  
 1 if b2i 1  0
x sQPSK
{bi}
Serial to Parallel
b2i
+
Mapping dQ
1 if b2i  1 x
dQ  
1 if b2i  0 A
sin(2 f c t )
2
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Bandwidth Efficiency of QPSK

GQPSK ( f )

… … … …
-fc -fc -fc 0 fc fc fc f
2/

• Symbol rate: RS ,QPSK  1/  • Bit rate: Rb ,QPSK  2 RS ,QPSK  2 / 

• Required Channel Bandwidth: Bh _ 90%  2 RS ,QPSK  Rb ,QPSK

Bh _ 95%  4 RS ,QPSK  2 Rb ,QPSK

• Bandwidth Efficiency:
 QPSK  1 with 90% in-band power
 QPSK  0.5 with 95% in-band power

QPSK achieves higher bandwidth efficiency than BPSK!

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Summary II: Digital Bandpass Modulation

Bandwidth Efficiency
(90% in-band power)

Binary ASK 0.5

1
Binary FSK 0.5 
1  f / Rb

Binary PSK 0.5

QPSK 1