Coherent phase shift keying
In coherent phase shift keying different phase modulation schemes will be covered i.e. binary PSK, quadrature phase shift keying and M-ary PSK
Binary PSK will be studied in the next slides
1

Binary Phase shift keying
In a coherent PSK system the pair of signals 𝑠
1 𝑡 and 𝑠
2 𝑡 are used to represent binary logics 1 and 0 respectively 𝑠
1 𝑡 =
2𝐸 𝑏
𝑇 𝑏 cos 2𝜋𝑓 𝑐 𝑡 𝑠
2 𝑡 =
2𝐸 𝑏
𝑇 𝑏 cos 2𝜋𝑓 𝑐 𝑡 + 𝜋 = −
2𝐸 𝑏
𝑇 𝑏 cos 2𝜋𝑓 𝑐 𝑡
2

Binary Phase shift keying
Where 0 ≤ 𝑡 ≤ 𝑇 𝑏
, and 𝐸 𝑏 signal energy per bit is the transmitted
The carrier frequency is selected such that 𝑛 𝑓 𝑐
=
𝑇 𝑏 so that each bit contains an integral number of cycles
From the pair of symbols 𝑠
1
(𝑡) and 𝑠
2
(𝑡) we can see only one basis function (carrier) is needed to represent both 𝑠
1
(𝑡) and 𝑠
2
(𝑡)
3

Binary Phase shift keying
The basis function is given by 𝜙
1
(𝑡) =
2
𝑇 𝑏 cos(2𝜋𝑓 𝑐 𝑡) 0 ≤ 𝑡 ≤ 𝑇 𝑏
Now we can rewrite 𝑠
1 𝑡 = 𝐸 𝑏 𝜙
1
(𝑡) and 𝑠
2 interval 0 ≤ 𝑡 ≤ 𝑇 𝑏 𝑡 = − 𝐸 𝑏 𝜙
1
(𝑡) on the
4

Signal constellation for binary
Phase shift keying
In order to draw the constellation diagram we need to find the projection of each transmitted symbol on the basis function
The projection of the logic (1); 𝑆
1 by 𝑆
11
=
0
𝑇 𝑏 𝑆
1 𝑡 𝜙
1 𝑡 𝑑𝑡 = + 𝐸
(𝑡) 𝑏
; is given
The projection of the second symbol 𝑆
2 the basis function is given by
0
𝑇 𝑏 𝑆
2 𝑡 𝜙
1 𝑡 𝑑𝑡 = − 𝐸 𝑏
𝑆
21
=
(𝑡) on
5

Signal constellation for binary
Phase shift keying
If we plot the transmitted symbols for
BPSK we may got the following constellation diagram
6

Error probability of BPSK
In order to compute the error probability of
BPSK we partition the constellation diagram of the BPSK (see slide 6) into two regions
If the received symbol falls in region Z
1
, the receiver decides in favor of symbol
1) was received
S
1
( logic
If the received symbol falls in region Z
2
, the receiver decides in favor of symbol
0) was received
S
2
(logic
7

Error probability of BPSK-
Receiver model
The receiver in the pass band can be modeled as shown
The received signal vector 𝑥(𝑡) = 𝑠 (𝑡) + 𝑛(𝑡)
8

Error probability of BPSK
The observable element 𝑥
1
(symbol zero was sent and the detected sample was read in zone 1) is given by
𝑇 𝑏 𝑥
1
= 𝑥
1
0
𝑇 𝑏 𝑡 𝜙
1 𝑡 𝑑𝑡 𝑥
1
=
𝑇 𝑏
0 𝑠
2 𝑡 + 𝑛 𝑡 𝜙
1 𝑡 𝑑𝑡 𝑥
1
= 𝑠
2
0 𝑡 𝜙
1 𝑡 𝑑𝑡 = 𝑆
21
= − 𝐸 𝑏
9

Error probability of BPSK
To calculate the probability of error that symbol 0 was sent and the receiver detect 1 mistakenly in the presence of AWGN with 𝜎 2 𝑥
=
𝑁
0
, we need to find the conditional
2 probability density of the random variable 𝑥
1
, given that symbol 0, 𝑠
2 𝑡 was transmitted as shown below
10

Error probability of BPSK
The conditional probability of the receiver deciding in favor of symbol 1, given that symbol zero was transmitted is given by
11

Error probability of BPSK
By letting the above integral for 𝑝
10 as can be rewritten
12

Error probability of error
In similar manner we can find probability of error that symbol 1 was sent and the receiver detect 0 mistakenly
The average probability as we did in the baseband can be computed as
This average probability is equivalent to the bit error rate
13

Generation of BPSK signals
To generate a binary PSK signal we need to present the binary sequence in polar form
The amplitude of logic 1 is + 𝐸 𝑏 whereas the amplitude of logic 0 is − 𝐸 𝑏
This signal transmission encoding is performed by using polar NRZ encoder
14

Generation of BPSK signals
The resulting binary wave and the carrier
(basis function) are applied to product modulator as shown below
15

Detection of BPSK signals
To detect the original binary sequence we apply the received noisy PSK signal 𝑥(𝑡) = 𝑠 (𝑡) + 𝑛(𝑡) to a correlator followed by a decision device as shown below
The correlator works a matched filter
16

Power spectra of binary PSK signals
The power spectral density of the binary
PSK signal can be found as described for the bipolar NRZ signaling (see problem
3.11 (a) Haykin)
This assumption is valid because the
BPSK is generated by using bipolar NRZ signaling
17

Power spectra of binary PSK signals
The power spectral density can be found as
18

Quadrature phase shift keying
QPSK
In quadrature phase shift keying 4 symbols are sent as indicated by the equation 𝑠 𝑖 𝑡 =
2𝐸
𝑇 𝑐𝑜𝑠 2𝜋𝑓 𝑐 𝑡 + 2𝑖 − 1 𝜋
4
0 ≤ 𝑡 ≤ 𝑇
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
Where 𝑖 = 1, 2, 3, 4 ; 𝐸 is the transmitted signal energy per symbol, and 𝑇 is the symbol duration
The carrier frequency is 𝑛 𝑐
𝑇 𝑛 𝑐 for some fixed integer
19

Signal space diagram of QPSK
If we expand the QPSK equation using the trigonometric identities we got the following equation s i


E
E cos



2 i

1


4


2
T cos



2 i

1


4



1
  cos


2
 f
E c t


E sin



2 i

1


4

 sin



2 i

1


4



2
 
; 0
 t
T
2 sin

2
 f c t


T
Which we can write in vector format as s i







E
E cos sin


2 i
2 i


1

1


4

4





20

Signal space diagram of QPSK
There are four message points defined by
According to this equation , a QPSK has a two-dimensional signal constellation (i.e.
𝑁 = 2 or two basis functions)
21

Detailed message points for
QPSK i
3
4
1
2
Input
Dibit
10
00
01
11
Phase of QPSK signalin g

/ 4
3

/ 4
5

/ 4
Coordinate of
Message point s i1
E / 2

E / 2

E / 2 s i2

E / 2

E / 2
E / 2
7

/ 4 E / 2
E / 2
22

Signal space diagram of QPSK
(01) s
3

2 s
4 (11)

1
(00) s
2 s
1
(10)
23

Signal space diagram of QPSK with decision zones
The constellation diagram may appear as shown below

2
Z
3
(10) s
3
Z
4 s
4
(11)
E / 2

1

E / 2
Z
2
(00) s
2

Z
1
E / 2
E / 2 s
1
(10)
24

Example
Sketch the QPSK waveform resulting from the input binary sequence 01101000 solution
25

Error probability of QPSK
In coherent QPSK, the received signal 𝑥 𝑡 is defined by 𝑥 𝑡 = 𝑠 𝑖 𝑡 + 𝑤(𝑡)
0 ≤ 𝑡 ≤ 𝑇 𝑖 = 1, 2, 3, 4
Where 𝑤(𝑡) is the sample function of
AWGN with zero mean and power spectral density of σ 2 𝑥
=
𝑁
0
2
26

Error probability of QPSK
The observation vector 𝑥 has two elements, 𝑥
1 and
𝑇 𝑥
2 defined by 𝑥
1
= 𝑥 𝑡 𝜙
1 𝑡 𝑑𝑡 𝑥
1
0
= 𝐸𝑐𝑜𝑠 (2𝑖 − 1) 𝜋
4
+ 𝑤
1 𝑥
1
= ±
𝐸
2
+ 𝑤
1
27

Error probability of QPSK
𝑇 𝑥
2
= 𝑥 𝑡 𝜙
2 𝑡 𝑑𝑡 𝑥
2
0
= 𝐸𝑠𝑖𝑛 (2𝑖 − 1) 𝜋
4
+ 𝑤
2 𝑥
2
= ∓
𝐸
2
+ 𝑤
2
28

Error probability decision rule
If the received signal point associated with the observation vector 𝑥 falls inside region
𝑍
1
, the receiver decide that 𝑠
1
(𝑡) was transmitted
Similarly the receiver decides that 𝑠
2
(𝑡) was transmitted if 𝑥 falls in region 𝑍
2
The same rule is applied for 𝑠
3
(𝑡) and 𝑠
4
(𝑡)
29

Error probability of QPSK
We can treat QPSK as the combination of
2 independent BPSK over the interval 𝑇 =
2𝑇 𝑏 since the first bit is transmitted by ϕ
1 and the second bit is transmitted by ϕ
2
Probability of error for each channel is given by
P
 
1
2 erfc
2 d
12
N
0

1
2 erf c
 E
2 N
0
30

Error probability of QPSK
If symbol is to be received correctly both bits must be received correctly
Hence, the average probability of correct decision is given by
P c


1

P
 
2
Which gives the probability of errors equal to
P e

1

P
C
 erfc
E
2 N
0

1
4 erfc
2
E
2 N
0
 erfc
E
2 N
0
31

Error probability of QPSK
Since one symbol of QPSK consists of two bits, we have E = 2 E b
P e
 per symbol

 erfc
E b
N
0
The above probability is the error probability per symbol
With gray encoding the average probability of error per bit
P e
 per bit


1
2
P e
 per symbol


1
2 erfc
E b
N
0
Which is exactly the same as BPSK
32

Error probability of QPSK summery
We can state that a coherent QPSK system achieves the same average probability of bit error as a coherent PSK system for the same bit rate and the same
𝐸 𝑏 but uses only half the channel
𝑁
0 bandwidth
33

Generation and detection of
QPSK
Block diagrams of
( a ) QPSK transmitter and ( b ) coherent
QPSK receiver.
34

Power spectra of QPSK
Power spectrum density of BPSK vs. QPSK
1
0.8
0.6
0.4
0.2
0
0
2
1.8
1.6
1.4
1.2
BPSK
QPSK
0.2
0.4
0.6
0.8
1 1.2
Normalized frequency,fT b
1.4
1.6
1.8
2
35

(01)
OFFSET QPSK s
3

2
90 degree shift in phase s
4 (11)
(00) s
2 s
1
(10)
180 degree shift in phase

1
36

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.
37

OFFSET QPSK
2
1
0
-1
-2
0
2
1. 5
1
0. 5
0
-0. 5
-1
-1. 5
-2
0
1
1
2 3 4
Original Signal
5
2
6 7
3 4 5
Filtered signal
6 7
8
8
38

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.
39

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.
40

OFFSET QPSK
Inphase
1
QPSK 0.5
0
-0.5
-1
0
1
0.5
Q phase
0
-0.5
QPSK
-1
0
2
1
QPSK 0
-1
-2
0
1
0.5
Inphase
0
-0.5
Offset QPSK
-1
0
1
0.5
Q phase
0
-0.5
Offset QPSK
-1
0
2
1
0
Offset QPSK
-1
-2
0
1
0
1
1
01
1
1
1
1
0
2
2
2
1
2
2
01
2
3
1
3
0
10
3
4
4
4
3
3
3
1
4
0
4
10
4
1
5
5
5
5
1
5
0
5
10
6
6
6
6
0
6
10
6
0
0
7
7
00
7
7
0
7
7
00
8
8
8
8
8
8
41

Offset QPSK
Possible paths for switching between the message points in ( a ) QPSK and ( b ) offset QPSK.
42

OFFSET QPSK
Bandwidths of the offset QPSK and the regular QPSK is the same.
From signal constellation we have that
P e
 erfc
E
2 N
0
Which is exactly the same as the regular
QPSK.
43


/4-shifted QPSK
Try to reduce amplitude fluctuation by switching between 2 signal constellation
44


/4-shifted QPSK
As the result, the phase of the signal can be changed in order of

/4 or

3

/4
45


/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
01
11
10
+

/4
+3

/4
-3

/4
-

/4
46


/4-shifted QPSK:00101001
Step Initial
1 phase

/4
Input Dibit Phase
00 change

/4
Transmitted phase

/2
2

/2 10 -

/4

/4
3
4

/4
0
10
01
-

/4
3

/4 3
0

/4
47

0
-2
0
1
0
-1
0
2
0
-2
0
2
0.5
0.5
0.5
01

/4-shifted QPSK
01
10
QPSK
10
1 1.5
1 1.5
2
2.5
2.5
2.5
3 3.5
3 3.5
3 3.5
1 1.5
4 4.5
4 4.5
4 4.5
48


/4-shifted QPSK
Since we only measure the phase different between adjacent symbols, no phase information is necessary. Hence, noncoherent receiver can be used.
49


50


/4-shifted QPSK
Illustrating the possibility of phase angles wrapping around the positive real axis.
51

M-array PSK
At a moment, there are M possible symbol values being sent for M different phase values,  i

2
  
/ M s i

2 E
T
 cos

2
 f c t

2

M
  

, i

1 , 2 ,  , M
52

M-array PSK
Signal-space diagram for octa phase-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 octa phase shift keying.
53

M-array PSK
Probability of errors
 d
12
 d
18

2 E sin
 
/ M

P e
 erfc
E
N
0 sin
 
/ M

; M

4
54

M-ary PSK
10
0
10
-10
10
-20
10
-30
10
-40
10
-50
0
QPSK
8-ary PSK
16-ary PSK
5 10 15
E b
/N
0
dB
20 25 30
55

M-array PSK
Power Spectra (M-array)
S
PSK
( f )

2 E sinc
2

2 E b log
2
M sinc
2

T b f log
2
M

M=2, we have
S
BPSK
( f )

2 E b sinc
2

T b f

56

M-array PSK
Power spectra of M -ary PSK signals for M

2, 4, 8.
T b f
57

M-array PSK
Bandwidth efficiency:

We only consider the bandwidth of the main lobe (or null-to-null bandwidth)
B

2
T

2
T b log
2
M

2 R b log
2
M

Bandwidth efficiency of M-ary PSK is given by
 
R b
B

R b
2 R b log
2
M

0 .
5 log
2
M
58

M-ary QAM
QAM = Quadrature Amplitude Modulation
Both Amplitude and phase of carrier change according to the transmitted symbol, m i s i

2 E
0 a i
T cos

2
 f c t


2 E
0 b i
T sin

2
 f c t ;

0
 t

T where a i and b i are integers, E0 is the energy of the signal with the lowest amplitude
59

M-ary QAM
Again, we have

1

2 cos 2
 f c t
T
; 0
 t

T

2

2 sin 2
 f c t
T
; 0
 t

T as the basis functions
There are two QAM constellations, square constellation and rectangular
60

M-ary QAM
QAM square Constellation

Having even number of bits per symbol, denoted by 2 n


M= L x L possible values
Denoting L

M
61

16-QAM
 a i
, b i






 (
(
(
(


3
3
,
,
3
1 )
)

3 ,

1 )

3 ,

3 )
(

1 , 3 )
(

1 , 1 )
(

1 ,

1 )
(

1 ,

3 )
( 1 , 3 )
( 1 , 1 )
( 1 ,

1 )
( 1 ,

3 )
(
(
(
3
3
3
,
,
,

3
1 )
1
)
( 3 ,

3 )
)





62

16-QAM
L-ary, 4-PAM
63

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
P e
 
1
1
L erfc
2 d
N
0
 


1

1
M


 erfc
E
0
N
0
64

16-QAM

A symbol will be received correctly if data transmitted on both 4-ary PAM systems are received correctly. Hence, we have
P c
 symbol
 
1

P e
 
2

Probability of symbol error is given by
P e
 symbol



1

P c
1

1

 symbol


1


1

P e
 
2
2 P e
 
 
2 
2 P e

65

16-QAM

Hence, we have
P e
 symbol


2



1

1
M


 erfc
E
0
N
0


But because average energy is given by
E av

2



2 E
0
L i
L /

2

1

2 i

1

2




2

M

1

E
0
3
We have
P e
 symbol


2



1

1
M


 erfc
2

M
3 E
 av
1

N
0

66

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.
67

Binary FSK
Transmitted signals are s i




0 ,
2
T
E b b cos

2
 f i t

, 0
 t

T b elsewhere where f i
 n c
T b
 i
; i

1 , 2
68

Binary FSK
S
1
(t) represented symbol “1”.
S
2
(t) represented symbol “0”.
This FSK is also known as Sunde’s FSK.
It is continuous phase frequency-shift keying (CPFSK).
69

Binary FSK
There are two basis functions written as
 i




0 ,
2
T b cos

2
 f i t

, 0
 t

T b elsewhere
As a result, the signal vectors are s
1




E b
0


 and s
2



0
E b


70

71

BFSK
From the figure, we have d
12

2 E b
In case of Pr(0)=Pr(1), the probability of error is given by
P e

1
2 erfc
 E b
2 N
0
We observe that at a given value of P e
, the
BFSK system requires twice as much power as the BPSK system.
72

TRANSMITTER
73

RECEIVER
74

Power Spectral density of BFSK
Consider the Sunde’s FSK where f
1 by 1/T b
. We can write and f
2 are different s i


2 E b
T b
2 E b
T b cos 


2
 f c t


T b t cos 




T b t





 cos

2
 f c t


2 E b
T b sin





T b t


 sin

2
 f c t

We observe that in-phase component does not depend on m i since
2 E b
T b cos




T b t



2 E b
T b cos



T b t


75

Power Spectral density of BFSK
S
BI

F
Half of the symbol power
We have
2 E b
T b cos



T b t




2

E b
2 T b

 

 f

1
2 T b


 

 f

1
2 T b




For the quadrature component g

2 E b
T b sin




T b t


S
BQ

8 E b
 2
T b  cos
4 T b
2 f
2
2
 
T b f


1

2
76

Power Spectral density of
BFSK
Finally, we obtain S
B
( f )

S
BI
( f )

S
BQ
( f )
77

Phase Tree of BFSK
FSK signal is given by s
 

2 E b
T b cos

 2
 f c t


T b t


At t = 0, we have s

2 E b
T b
 cos

2
 f c
0


T b
0



2 E b
T b cos
 
The phase of Signal is zero.
78

Phase Tree of BFSK
At t = T b
, we have s

2 E b
T b cos

 2
 f c
T b


T b
T b



2 E b
T b cos
 
We observe that phase changes by
 after one symbol (T seconds). -
 for symbol “1” and +  b for symbol “0”
We can draw the phase trellis as
79

80

Minimum-Shift keying (MSK) s
MSK tries to shift the phase after one symbol to just half of Sunde’s FSK system.
The transmitted signal is given by

2 E b
T b cos

2
 f c t
 







2 E b
T b
2 E b
T b cos

2
 f
1 t
  cos

2
 f
2 t
 

 for "1" for "0"
81

MSK
Where
    

 h
T b t
Observe that f
1
 f c
 h
2 T b and f
2
 f c
 h
2 T b f c

1
2
 f
1
 f
2

82

MSK h = T b
( f
1
f
2
) 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
 f
1 t ) and cos(2
 f
2 t ) is still held for h < 0.5.
Orthogonality guarantees that both signal will not interfere each other in detection process.
83

MSK
Phase trellis diagram for MSK signal 1101000
84

MSK
Signal s (t) of MSK can be decomposed into s

2 E b
T b cos

2
 f c t
 


 s
I
2 E b cos
  cos

2
 f c t


2 E
T b
2
 f 2
 f
T b
   c t

 s
Q
   c t
 b sin
  sin

2
 f c t
 where
 

2 T b t ; 0
 t

T b
85

Symbol
1
0
MSK

(0)
0

0


(T b
)

/2
-

/2
-

/2

/2
86

MSK
For the interval – T b
< t

0, we have
 
 t ;

T b
 t

0
2 T b
Let’s note here that the  for the interval -
T b
< t

0 and 0< t

T b may not be the same.
We know that cos


 
 t
2 T b


 cos
  cos


 t
2 T b

  sin
  sin


 t
2 T b


87

MSK
Since

(0) can be either 0 or
 depending on the past history. We have cos


 
 t
2 T b


 cos
  cos


 t
2 T b


  cos


 t
2 T b


“+” for  (0) = 0 and “-” for 
(0) =

Hence, we have s
I
( t )
 
2 E b
T b cos


 t
2 T b

 ;

T b
 t

T b
88

MSK
Similarly we can write
    


 t

T b

2 T b for 0< t

T b and T b
< t

2 T b
. Note the “+” and “-” may be different between these intervals.
Furthermore, we have that

( T b
) can be

/2 depending on the past history.
89

MSK
Hence, we have sin



    t

2 T b
T b




 sin


 cos


   t
2

T b
T b




 sin
   cos



 t
2 T b


2 


 cos


 sin


   t
2

T b
T b




 cos
   sin


  t
2 T b


2 

 we have that

( T b
) can be

/2 depending on the past history.
sin


    t

2 T b
T b
 

  cos


 t
2 T b


2


  sin


 t
2 T b


90

MSK
Hence, we have s
Q
( t )
 
2 E b
T b sin


 t
2 T b

 ; 0
 t

2 T b
“+” for 
( T b
) = +
 /2 and “-” for 
( T b
) = -

/2
The basis functions change to

1

2
T b cos


 t
2 T b

 cos

2
 f c t

; 0
 t

T b

2

2
T b sin


 t
2 T b

 sin

2
 f c t

; 0
 t

T b
91

MSK
We write MSK signal as s

2 E b cos
  cos

2
 f c t


2 E b sin
  sin

2
 f c t

T b
T b



2
E
E b
T b b cos
  cos
 

1
 cos

( t )

2
 t
T b
E

 b cos

2
 f c t
 sin





2 s
1

1
( t )
 s
2

2
( t )
2 E b
( t )
T b sin


 sin


2
 t
T b

 sin

2
 f c t
 s
2
 
E b sin



92

MSK
Symbol
1
0

(0)
0

0
 s
1
E b

E b
E b

E b

(T b
)

/2
-

/2
-

/2

/2 s
2

E b
E b

E b
E b
93

P e

1
2 erfc
 E b
N
0
94

Phase: 0

/2
 
/2
 
/2 0 -

/2
95

MSK
We observe that MSK is in fact the QPSK t having the pulse shape
2 T b

Block diagrams for transmitter and receiver are given in the next two slides.
96

97

x
1


T b

T b x ( t )

1
( t ) dt x
2

2 T b
 x ( t )

2
0
( t ) dt
98

4
3.5
3
MSK
BPSK
QPSK
2.5
2
1.5
1
0.5
0
0 0.2
0.4
0.6
0.8
1 1.2
1.4
1.6
1.8
Normalized Frequency, fT b
2
99

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.
S
MSK
( f )

32 E b
 2

 cos

 16 T b
2
2

T b f
2  f
 

1 
2
100

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.
101

Noncoherent Orthogonal Modulation sin(2
 ft) sin(2
 ft) cos(2
 ft) cos(2
 ft)
Receiver
Transmitter
102

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.
103

Noncoherent Orthogonal Modulation
Consider the BFSK system where two frequencies f
1 and f two “1” and “0”.
2 are used to represented
The transmitted signal is given by s ( t )

2 E cos

2
 f i t
  
T
; i

1 , 2 , 0
 t

T b
Problem is that
 is unknown to the receiver. For the coherent receiver,
 is precisely known by receiver.
104

Noncoherent Orthogonal Modulation
Furthermore, we have s ( t )

2 E cos

2
 f i t
  
T


2 E cos
  
2
 f i t

T s i 1

1
( t )
 s i 2

2
( t )

2 E sin
  
2
 f i t

T
To get rid of the phase information (

), we use the amplitude s
 
 s i
2
1
 s i
2
2

E cos
2 
E sin
2
 

E
105

Noncoherent Orthogonal Modulation
Where s i 1 s i 2


T
 s ( t )

1
( t ) dt

T
0 s ( t )

2
( t ) dt

 x
1 x
2


T

0
T
 x ( t )

1
( t ) dt x ( t )

2
( t ) dt
0 0
The amplitude of the received signal l i









T

0 x ( t ) cos

2
 f i t
 dt
2

T

0 x ( t ) sin

2
 f i t
 dt
2



 1 / 2
106

Quadrature Receiver using correlators
107

Quadrature Receiver using Matched
Filter
108

Noncoherent Orthogonal Modulation
Decision rule: Let if examples, decide if l l i
1
> l k
> l
2 for all k . For
This decision rule suggests that if the envelop
(amplitude) of the received signal described in term of cos(2
 f
1 t ) is greater than the envelop of the received signal described in term of cos(2
 f
2 t ), we say s
1
( t ) was sent.
109

Noncoherent Matched Filter
110

Noncoherent Orthogonal Modulation
Consider the output of matched filter of cos(2
 f i t ).
y ( t )
 x
 
 cos

2
 f i

T
 t
 

T

0 x (

) cos

2
 f i

T
 t
    d
 y ( t )
 cos

2
 f i
( T
 t )
 T

0 x (

) cos

2
 f i
  d

s in

2
 f i
( T
 t )

T

0 x (

) sin

2
 f i
  d

111

Noncoherent Orthogonal Modulation
Envelope at t=T is l i









T

0 x (

) cos

2
 f i
  d

2

T

0 x (

) cos

2
 f i
  d

2




1 / 2
Which is exactly the same as in correlator receiver
112

Generalized binary receiver for noncoherent orthogonal modulation.
113

Quadrature receiver equivalent to either one of the two matched filters in part
114

Noncoherent Orthogonal Modulation
Probability of Errors
P e

1
2 exp



E
2 N
0


115

Noncoherent: BFSK
For BFSK, we have s i




0
2
T
E b b cos

2
 f i t

; 0
 t

T
; elsewhere b
116

Noncoherent: BFSK
117

Noncoherent: BFSK
Probability of Errors
P e

1
2 exp



E b
2 N
0


118

DPSK
Differential PSK



Instead of finding the phase of the signal on the interval 0< t

T b.
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.
119

DPSK
Or we have s
1
( t )





 and
E b
2 T b cos

2
 f c t

;
E b
2 T b cos

2
 f c t

;
0
 t

2 T b
T b
 t

2 T b s
2
( t )






E b
2 T b cos

2
 f c t ;

E b
2 T b cos

2
 f c t
 
;

0
 t

2 T b
T b
 t

2 T b
120

DPSK
In this case, we have T =2 T b and E =2 E b
Hence, the probability of error is given by
P e

1
2 exp



E b
N
0


121

DPSK: Transmitter d k
 b k d k

1
 b k d k

1
122

DPSK
{ b k
}
{ d k -1
}
Differential encoded { d k
}
Transmitted
Phase
1
1
0
1
0
0
1
1
0
1
0
0
0
1
1
0
1
0
1 1 0 1 1 0 1 0 0 0
0 0

0 0

0
  
123

DPSK: Receiver
124

DPSK: Receiver
From the block diagram, we have that the decision rule as
 x
I 0 x
I 1
 x
Q 0 x
Q 1 say 1


0 say 0
If the phase of signal is unchanged (send “1”) the sign (“+” or “-”) of both x i and x
Q 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 x i and x
Q should change. Hence, the l ( x ) should be negative.
125

126