Transcript Pass-band Data Transmission

Pass-band Data Transmission
Dr. Teerasit Kasetkasem
Block Diagram

Functional model of pass-band data transmission
system.
Signaling

Illustrative waveforms for the three basic forms of
signaling binary information. (a) Amplitude-shift keying.
(b) Phase-shift keying. (c) Frequency-shift keying with
continuous phase.
What do we want to study?

We are going to study and compare
different modulation techniques in terms of



Probability of errors
Power Spectrum
Bandwidth efficiency
Rb

B
Bits/s/Hz
Coherent PSK

Binary Phase Shift Keying (BPSK)


Consider the system with 2 basis functions
and
2
1 t  
cos2f c t
Tb
2
2 t  
sin 2f c t
Tb
BPSK

If we want to fix that for both symbols (0 and 1) the
transmitted energies are equal, we have
2
s0
s1
1
s0
We place s0 to minimize
probability of error
BPSK

We found that phase of s1 and s0 are
180 degree difference. We can rotate s1
and s0
2
s1
1
s0
Rotate
BPSK
2
s0
s1
1

We observe that 2 has nothing to do with
signals. Hence, only one basis function is
sufficient to represent the signals
BPSK

Finally, we have
s1 t   Eb 1 (t ) 
2 Eb
cos 2f c t
Tb
2 Eb
s0 t    Eb 1 (t )  
cos 2f c t
Tb
BPSK

Signal-space diagram for coherent binary PSK
system. The waveforms depicting the transmitted
signals s1(t) and s2(t), displayed in the inserts, assume
nc  2.
BPSK

Probability of error calculation. In the case of
equally likely (Pr(m0)=Pr(m1)), we have
 d ik
1
Pe  erfc
2 N
2
0

 Eb 
1

 erfc
 N 
2
0 





BPSK

Block diagrams for (a) binary PSK transmitter and (b)
coherent binary PSK receiver.
Quadriphase-Shift Keying (QPSK)
2E


si t  
cos2f c t  2i  1 ; 0  t  T
T
4


T is symbol duration
 E is signal energy per symbol
 There are 4 symbols for i = 1, 2, 3, and 4
QPSK
 2
 2


si t   E cos2i  1 
cos2f c t   E sin 2i  1 
sin 2f c t 
4 T
4 T






 E cos2i  1 1 t   E sin 2i  1  2 t ; 0  t  T
4
4



Which we can write in vector format as
 



 E cos 2i  1 4 
si  

 E sin 2i  1 
4

QPSK
i
Input Dibit Phase of
QPSK
signaling
1
10
 /4
2
00
3 / 4
3
01
5 / 4
4
11
7 / 4
Coordinate of
Message point
si1
si2
E/2
 E/2
 E/2
E/2
 E/2
 E/2
E/2
E/2
QPSK
2
(01)
s3
s2
(00)
s4 (11)
s1
(10)
1
QPSK signals
QPSK

Block
diagrams of
(a) QPSK
transmitter
and (b)
coherent
QPSK
receiver.
QPSK: Error Probability QPSK

Consider signal
constellation
given in the
figure
2
Z3
(10)
s3
Z4
s4 (11)
E/2
 E/2
s2
Z2
(00)
Z1
 E/2
E/2
s1
(10)
1
QPSK
We can treat QPSK as the combination
of 2 independent BPSK over the interval
T=2Tb
 since the first bit is transmitted by 1 and
the second bit is transmitted by 2.
 Probability of error for each channel is
given by



P 
 E
d
1
1
erfc 12   erfc
2 N  2
2
0 
 2N0





QPSK

If symbol is to be received correctly both bits
must be received correctly.
 Hence, the average probability of correct
decision is given by Pc  1  P 2
 Which gives the probability of errors equal to
 E  1
 E 
2
  erfc 

Pe  1  PC  erfc



2
N
4
2
N
0
0




 E
 erfc
 2N 0




QPSK

Since one symbol of QPSK consists of two
bits, we have E = 2Eb.
 Eb 

Peper symbol  erfc
 N 
0 


The above probability is the error probability
per symbol. The avg. probability of error per bit
 Eb 
1
1

Peper bit   Peper symbol  erfc


2
2
 N0 

Which is exactly the same as BPSK .
BPSK vs QPSK
Power spectrum density of BPSK vs. QPSK
2
BPSK
QPSK
1.8
1.6
Normalized PSD, Sf/2E
b
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Normalized frequency,fTb
1.6
1.8
2
QPSK

Conclusion
QPSK is capable of transmitting data twice
as faster as BPSK with the same energy per
bit.
 We will also learn in the future that QPSK
has half of the bandwidth of BPSK.

OFFSET QPSK
90 degree shift in phase
2
(01)
s3
s2
(00)
s4 (11)
s1
(10)
180 degree shift in phase
1
OFFSET QPSK
OFFSET QPSK
Whenever both bits are changed
simultaneously, 180 degree phase-shift
occurs.
 At 180 phase-shift, the amplitude of the
transmitted signal changes very rapidly
costing amplitude fluctuation.
 This signal may be distorted when is
passed through the filter or nonlinear
amplifier.

OFFSET QPSK
2
1
0
-1
-2
0
1
2
3
4
5
6
7
8
Original Signal
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
0
1
2
3
4
5
Filtered signal
6
7
8
OFFSET QPSK
To solve the amplitude fluctuation
problem, we propose the offset QPSK.
 Offset QPSK delay the data in
quadrature component by T/2 seconds
(half of symbol).
 Now, no way that both bits can change at
the same time.

OFFSET QPSK

In the offset QPSK, the phase of the
signal can change by 90 or 0 degree
only while in the QPSK the phase of the
signal can change by 180 90 or 0
degree.
OFFSET QPSK
Inphase
QPSK
1
0.5
0
0
1
0
1
-0.5
-1
0
1
2
3
4
5
6
7
8
1
Q phase
QPSK
0
0.5
0
-0.5
1
0
0
-1
0
1
2
3
4
5
4
5
6
7
8
2
1
QPSK
10
0
-1
-2
0
01
1
2
3
10
00
6
7
8
1
Inphase
Offset QPSK
0.5
0
0
1
0
1
-0.5
-1
0
1
2
3
4
5
6
7
8
7
8
1
Q phase
Offset QPSK
0.5
1
0
0
0
-0.5
-1
0
1
2
3
4
5
6
2
1
Offset QPSK
-1
10
10
01
0
00
-2
0
1
2
3
4
5
6
7
8
Offset QPSK
Possible paths for switching between the
message points in (a) QPSK and (b) offset QPSK.
OFFSET QPSK
Bandwidths of the offset QPSK and the
regular QPSK is the same.
 From signal constellation we have that

 E 

Pe  erfc

2
N
0



Which is exactly the same as the regular
QPSK.
/4-shifted QPSK

Try to reduce amplitude fluctuation by
switching between 2 signal constellation
/4-shifted QPSK

As the result,
the phase of the
signal can be
changed in
order of /4 or
3/4
/4-shifted QPSK

Since the phase of the next will be varied in
order of /4 and 3/4, we can designed the
differential /4-shifted QPSK as given below
Gray-Encoded Input Data
Phase Change in radians
00
+/4
+3/4
-3/4
-/4
01
11
10
/4-shifted QPSK:00101001
Step
1
Initial Input Dibit Phase Transmitted
phase
change
phase
/4
00
/4
/2
2
/2
10
-/4
/4
3
/4
10
-/4
0
4
0
01
3/4
3/4
/4-shifted QPSK
01
10
QPSK
01
10
2
0
-2
0
0.5
1
1.5
2
2.5
OFFSET
QPSK
3
3.5
4
4.5
0
0.5
1
1.5
2
2.5
D OFFSET
QPSK
3
3.5
4
4.5
0
0.5
1
1.5
3
3.5
4
4.5
2
0
-2
1
0
-1
2
2.5
/4-shifted QPSK

Since we only measure the phase
different between adjacent symbols, no
phase information is necessary. Hence,
non-coherent receiver can be used.
Block diagram of the /4-shifted DQPSK
detector.
/4-shifted QPSK

Illustrating the
possibility of phase
angles wrapping
around the positive
real axis.
M-array PSK

At a moment, there are M possible
symbol values being sent for M different
phase values, i  2i  1 / M
si t  
2E
2




cos 2f c t 
i  1 ,
T
M


i  1,2,, M
M-array PSK


Signal-space diagram
for octaphase-shift
keying (i.e., M  8). The
decision boundaries are
shown as dashed lines.
Signal-space diagram
illustrating the
application of the union
bound for octaphaseshift keying.
M-array PSK

Probability of errors
 d12  d18  2 E sin / M 
 E

Pe  erfc
sin  / M ;
 N0

M 4
M-ary PSK
0
10
-10
Probability of Symbol errors
10
-20
10
-30
10
-40
10
QPSK
8-ary PSK
16-ary PSK
-50
10
0
5
10
15
Eb/N0 dB
20
25
30
M-array PSK

Power Spectra (M-array)
S PSK ( f )  2 E sinc2 Tf 
 2 Eb log2 M sinc2 Tb f log2 M 

M=2, we have
S BPSK ( f )  2 Eb sinc 2 Tb f 
M-array PSK

Power spectra of M-ary PSK
signals for M  2, 4, 8.
Tbf
M-array PSK

Bandwidth efficiency:
 We only consider the bandwidth of the main lobe
(or null-to-null bandwidth)
2Rb
2
2
B 

T Tb log2 M log2 M

Bandwidth efficiency of M-ary PSK is given by
Rb
Rb


log2 M  0.5 log2 M
B 2Rb
M-ary QAM
QAM = Quadrature Amplitude
Modulation
 Both Amplitude and phase of carrier
change according to the transmitted
symbol, mi.

2 E0
2 E0
si t  
ai cos2f c t  
bi sin 2f c t ; 0  t  T
T
T
where ai and bi are integers.
M-ary QAM

Again, we have
2
1 t  
cos2f c t ;0  t  T
T
2
2 t  
sin 2f c t 0  t  T
Tb
as the basis functions
M-ary QAM

QAM square Constellation
Having even number of bits per symbol,
denoted by 2n.
 M=L x L possible values
 Denoting L  M

16-QAM
 (3,3) (1,3) (1,3) (3,3) 
 (3,1)

(

1
,
1
)
(
1
,
1
)
(
3
,
1
)

ai , bi   
(3,1) (1,1) (1,1) (3,1) 


(3,3) (1,3) (1,3) (3,3)
L-ary, 4-PAM
16-QAM
16-QAM

Calculation of Probability of errors
Since both basis functions are orthogonal,
we can treat the 16-QAM as combination of
two 4-ary PAM systems.
 For each system, the probability of error is
given by

 d
 1
Pe  1  erfc
2 N
 L
0

 


  1  1 erfc E0 
 N 
 
M
0 


16-QAM

A symbol will be received correctly if data
transmitted on both 4-ary PAM systems are
received correctly. Hence, we have
Pc symbol   1  Pe 2

Probability of symbol error is given by
Pe sym bol  1  Pc sym bol  1  1  Pe 2
 1  1  2Pe  Pe 2  2Pe
16-QAM

Hence, we have
 E0 
1 


Pe sym bol  21 
erfc

N
M 

0



But because average energy is given by
 2 E0 L / 2
2  2M  1E0
Eav  2
 2i  1  
3
 L i 1


We have

3E av
1 


Pe sym bol  21 
erfc
M 

 2M  1N 0




Coherent FSK
FSK = frequency shift keying
 Coherent = receiver have information on
where the zero phase of carrier.
 We can treat it as non-linear modulation
since information is put into the
frequency.

Binary FSK

Transmitted signals are
 2 Eb
cos2f i t , 0  t  Tb

si t    Tb

elsewhere
0,

where
nc  i
fi 
; i  1,2
Tb
Binary FSK
S1(t) represented symbol “1”.
 S2(t) represented symbol “0”.

This FSK is also known as Sunde’s FSK.
 It is continuous phase frequency-shift
keying (CPFSK).

Binary FSK

There are two basis functions written as
 2
cos2f i t , 0  t  Tb

i t    Tb

elsewhere
0,

As a result, the signal vectors are
 Eb 
 0 
s1  
 and s 2  

E
0
 b


BFSK
From the figure, we have d12  2Eb
 In case of Pr(0)=Pr(1), the probability of
error is given by

 Eb 
1

Pe  erfc
 2N 
2
0 


We observe that at a given value of Pe,
the BFSK system requires twice as
much power as the BPSK system.
TRANSMITTER
RECEIVER
Power Spectral density of BFSK

Consider the Sunde’s FSK where f1 and f2 are different
by 1/Tb. We can write
si t  



2 Eb
t 
cos 2f c t  
Tb
Tb 

 t
2 Eb
cos 
Tb
 Tb

 t 
2 Eb
 cos2f c t  
sin    sin 2f c t 
Tb

 Tb 
We observe that in-phase component does not
depend on mi since
 t 
 t 
2 Eb
2 Eb
cos   
cos 
Tb
Tb
 Tb 
 Tb 
Power Spectral density of BFSK

We have
Half of the symbol power
2

 t 

Eb  
1 
1 
 2Eb







S BI  f   F 
cos  
 f 
  f 

2Tb  
2Tb 
2Tb 

 Tb 

 Tb


For the quadrature component
 t 
2 Eb
g t  
sin  
Tb
 Tb 
S BQ 
8 EbTb cos 2 Tb f 

2
 2 4Tb2 f 2  1
Power Spectral density of BFSK

Finally, we obtain
S B ( f )  S BI ( f )  S BQ ( f )
Phase Tree of BFSK

FSK signal is given by
st  

At t = 0, we have
s0 


2 Eb
t 
cos 2f c t  
Tb
Tb 


2 Eb
2 Eb
0
 
cos 2f c 0 
cos0
Tb
T
T
b 
b

The phase of Signal is zero.
Phase Tree of BFSK

At t = Tb, we have

2 Eb
Tb 
2 Eb

 
sTb  
cos 2f cTb 
cos  
Tb
T
T
b 
b


We observe that phase changes by  after
one symbol (Tb seconds). - for symbol “1”
and + for symbol “0”

We can draw the phase trellis as
Minimum-Shift keying (MSK)

MSK tries to shift the phase after one
symbol to just half of Sunde’s FSK
system. The transmitted signal is given
by
 2 Eb
cos2f1t   0 for "1"

2 Eb
 Tb
st  
cos2f c t   t   
Tb
 2 Eb
 T cos2f 2t   0 for "0"
b

MSK


Where
 t    0 
h
Tb
t
Observe that
h
f1  f c 
2Tb
and
h
f2  fc 
2Tb
1
f c   f1  f 2 
2
MSK





h = Tb(f1-f2) is called “deviation ratio.”
For Sunde’s FSK, h = 1.
For MSK, h = 0.5.
h cannot be any smaller because the
orthogonality between cos(2f1t) and
cos(2f2t) is still held for h < 0.5.
Orthogonality guarantees that both signal will
not interfere each other in detection process.
MSK

Phase trellis diagram for MSK signal 1101000
MSK

Signal s(t) of MSK can be decomposed
into
st  

2 Eb
cos2f c t   t 
Tb
2 Eb
2 Eb
cos t cos2f c t  
sin t sin 2f c t 
Tb
Tb
 s I t  cos2f c t   sQ t sin 2f c t 

where
 t    0 

2Tb
t ;0  t  Tb
MSK
Symbol
(0)
(Tb)
0
/2

-/2
0
-/2
1
0

/2
MSK

For the interval –Tb < t  0, we have
 t    0 

2Tb
t ;Tb  t  0
Let’s note here that the  for the interval
-Tb<t 0 and 0< tTb may not be the same.
 We know that


 t 
 t 
t 
  sin 0sin

cos 0 
  cos 0cos
2Tb 

 2Tb 
 2Tb 
MSK

Since (0) can be either 0 or  depending on
the past history. We have

 t 
 t 
t 
   cos

cos 0 
  cos 0cos
2Tb 

 2Tb 
 2Tb 
“+” for (0) = 0 and “-” for (0) = 
 Hence, we have

 t 
2 Eb
 ;Tb  t  Tb
s I (t )  
cos
Tb
 2Tb 
MSK

Similarly we can write
 t    Tb  



2Tb
t  Tb 
for 0< tTb and Tb < t2Tb. Note the “+” and “-”
may be different between these intervals.
Furthermore, we have that (Tb) can be /2
depending on the past history.
MSK

Hence, we have

  t  Tb  
  t  Tb  
 t  Tb 



sin  Tb  
 cos Tb sin 
  sin Tb cos

2Tb 

 2Tb 
 2Tb 
 t  
 t  
 sin Tb cos
   cos Tb sin 
 
 2Tb 2 
 2Tb 2 

we have that (Tb) can be /2 depending
on the past history.

 t  
 t 
 t  Tb 




sin  Tb  


cos



sin



2Tb 

 2Tb 2 
 2Tb 
MSK

Hence, we have
 t 
2 Eb
 ;0  t  2Tb
sQ (t )  
sin
Tb
 2Tb 
“+” for (Tb) = +/2 and “-” for (Tb) = -/2
 The basis functions change to

 t 
2
 cos2f c t  ;0  t  Tb
1 t  
cos
Tb
 2Tb 
2 t  
 t 
2
 sin 2f c t  ;0  t  Tb
sin
Tb
 2Tb 
MSK

We write MSK signal as
st  

2 Eb
2 Eb
cos t cos2f c t  
sin t sin 2f c t 
Tb
Tb
 2t 
 2t 
2 Eb
2 Eb
 cos2f c t  
 sin 2f c t 
cos 0cos
sin Tb sin 
Tb
Tb
 Tb 
 Tb 
 E b cos 01 (t )  E b sin Tb  2 (t )
 s11 (t )  s 2 2 (t )

Where
s1  Eb cos 0
and
s2   Eb sin Tb 
MSK
Symbol
(0)
s1
(Tb)
0
Eb
/2
 Eb

 Eb
-/2
Eb
0
Eb
-/2
 Eb
 Eb
/2
s2
1
0

Eb
 Eb 
1

Pe  erfc
 N 
2
0 

Phase:
0
/2 
/2

/2
0
-/2
MSK

We observe that MSK is in fact the
 t 
QPSK having cos   the pulse shape
 2Tb 

Block diagrams for transmitter and
receiver are given in the next two slides.
Tb
x1   x(t )1 (t )dt
Tb
2Tb
x 2   x(t ) 2 (t )dt
0
4
MSK
BPSK
QPSK
3.5
Normalized PSD, S(f)/E
b
3
2.5
2
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Normalized Frequency, fTb
1.6
1.8
2
MSK

Probability of error of MSK system is equal to
BPSK and QPSK
 This due to the fact that MSK observes the
signal for two symbol intervals whereas FSK
only observes for single signal interval.
 Bandwidth of MSK system is 50% larger than
QPSK.
2


32E b cos2Tb f 


S MSK ( f ) 
 2 16Tb2 f 2  1
Noncoherent Orthogonal Modulation

Noncoherent implies that phase information is
not available to the receiver.
 As a result, zero phase of the receiver can
mean any phase of the transmitter.
 Any modulation techniques that transmits
information through the phase cannot be used
in noncoherent receivers.
Noncoherent Orthogonal Modulation
sin(2ft)
sin(2ft)
cos(2ft)
cos(2ft)
Receiver
Transmitter
Noncoherent Orthogonal Modulation

It is impossible to draw the signal constellation
since we do not know where the axes are.
 However, we can still determine the distance
of the each signal constellation from the origin.
 As a result, the modulation techniques that put
information in the amplitude can be detected.
 FSK uses the amplitude of signals in two
different frequencies. Hence non-coherent
receivers can be employed.
Noncoherent Orthogonal Modulation

Consider the BFSK system where two
frequencies f1 and f2 are used to represented
two “1” and “0”.
 The transmitted signal is given by
2E
s(t ) 
cos2f i t    ; i  1,2,0  t  Tb
T

Problem is that  is unknown to the receiver.
For the coherent receiver,  is precisely known
by receiver.
Noncoherent Orthogonal Modulation

Furthermore, we have
s(t ) 
2E
cos2f i t   
T
2E
2E
cos  cos2f i t  
sin  sin 2f i t 
T
T
 si11 (t )  si 22 (t )


To get rid of the phase information (), we use
the amplitude
st   si21  si22  E cos2    E sin 2    E
Noncoherent Orthogonal Modulation

Where
T
T
0
T
0
T
0
0
si1   s (t )1 (t )dt  x1   x(t )1 (t )dt
si 2   s (t ) 2 (t )dt  x2   x(t ) 2 (t )dt

The amplitude of the received signal
2
2
 T

T
 
li    x(t ) cos2f i t dt     x(t ) sin 2f i t dt  


 

0
0



 

1/ 2
Quadrature Receiver using
correlators
Quadrature Receiver using Matched
Filter
Noncoherent Orthogonal Modulation

Decision rule: Let mˆ  mi if li > lk for all k. For
examples, decide mˆ  m1 if l1 > l2

This decision rule suggests that if the envelop
(amplitude) of the received signal described in
term of cos(2f1t) is greater than the envelop of
the received signal described in term of
cos(2f2t), we say s1(t) was sent.
Noncoherent Matched Filter
Noncoherent Orthogonal Modulation

Consider the output of matched filter of
cos(2fit).
T
y (t )  xt   cos2f i T  t    x( ) cos2f i T  t   d
0
T
y (t )  cos2f i (T  t ) x( ) cos2f i d
0
T
- sin2f i (T  t ) x( ) sin 2f i d
0
Noncoherent Orthogonal Modulation

Envelope at t=T is
2 1 / 2
 T

T





li   x( ) cos2f i d   x( ) cos2f i d  


 


0
 
 0
2

Which is exactly the same as in
correlator receiver
Generalized binary receiver for noncoherent orthogonal
modulation.
Quadrature receiver equivalent to either one of the two
matched filters in part
Noncoherent Orthogonal Modulation

Probability of Errors

1
E 


Pe  exp  
2
 2N0 
Noncoherent: BFSK

For BFSK, we have
 2 Eb
cos2f i t ; 0  t  Tb

si t    Tb

; elsewhere
0
Noncoherent: BFSK
Noncoherent: BFSK

Probability of Errors
 Eb 
1

Pe  exp  
2
 2N0 
DPSK

Differential PSK
Instead of finding the phase of the signal on
the interval 0<tTb. This receiver determines
the phase difference between adjacent time
intervals.
 If “1” is sent, the phase will remain the same
 If “0” is sent, the phase will change 180 degree.

DPSK


Or we have
and



s1 (t )  



Eb
cos2f c t ;
2Tb
0  t  2Tb
Eb
cos2f c t ;
2Tb
Tb  t  2Tb



s 2 (t )  



Eb
cos2f c t ;
2Tb
0  t  2Tb
Eb
cos2f c t   ;
2Tb
Tb  t  2Tb
DPSK

In this case, we have T=2Tb and E=2Eb

Hence, the probability of error is given by
 Eb 
1

Pe  exp 
2
 N0 
DPSK: Transmitter
d k  bk d k 1  bk d k 1
DPSK
{bk}
1 0 0 1 0 0 0 1 1
{dk-1}
1 1 0 1 1 0 1 0 0
Differential
encoded {dk}
1 1 0 1 1 0 1 0 0 0
Transmitted
Phase
0 0  0 0  0   
DPSK: Receiver
DPSK: Receiver

From the block diagram, we have that the
decision rule as
say1

l x   x I 0 x I 1  xQ 0 xQ1
0

say 0
If the phase of signal is unchanged (send “1”)
the sign (“+” or “-”) of both xi and xQ should not
change. Hence, the l(x) should be positive.
 If the phase of signal is unchanged (send “0”)
the sign (“+” or “-1”) of both xi and xQ should
change. Hence, the l(x) should be negative.

Signal-space diagram of received DPSK
signal.