x Q

Report
Non-coherent Orthogonal Modulation
Non-coherent 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; such as
BPSK, M-ary PSK; must be demodulated
only coherently
1
Non-coherent Orthogonal Modulation
sin(2ft)
sin(2ft)
cos(2ft)
cos(2ft)
Receiver
Transmitter
2
Non-coherent 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 to demodulate FSK
signals
3
Non-coherent 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
si (t ) 
2E
cos2f i t    ; i  1,2 0  t  Tb
T
If the carrier phase  is unknown to the receiver,
we can still be able to demodulate the
transmitted signal using non coherent
demodulation
4
Non-coherent Orthogonal Modulation
The previous BFSK signal, can be rearranged as
shown below
2E
si (t ) 
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 )

The amplitude (envelope) of the transmitted
signal is given by
st   si21  si22  E cos 2    E sin 2    E
5
Non-coherent Orthogonal Modulation
Recall that
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 (envelope) of the received signal
is given by
2

 




li    x(t ) cos2f i t dt     x(t ) sin 2f i t dt 
0
 0

T
T
2
6
Non-coherent BFSK
There are three common implementations
for the equation li
1.
2.
3.
Quadrature receiver using correlator
Quadrature receiver using matched filter
Matched filter using envelope detector
7
Quadrature receiver using correlators
8
Quadrature receiver using Matched
Filter
9
Non-coherent matched filter
using envelope detector
Since the equation
2
T
 T

li    x(t ) cos2f i t dt     x(t ) sin 2f i t dt 
0
 0

2
Represent the envelope of the received
signal we can demodulate the received signal
using an envelope detector as shown in the
next slide
The envelope detector finds the square root
of 21 + 22
10
Non-coherent Matched Filter
envelope detector
11
Non-coherent matched filter
envelope detector
An alternate way to show the equivalence
between the non-coherent matched filter
using envelope detector and the noncoherent detector using quadrature
detector is illustrated in mathematically in
the next few slides
12
Noncoherent Orthogonal Modulation
Consider the output of y(t) a filter matched to
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
13
Noncoherent Orthogonal Modulation
The detected envelope at  =  is

 

li    x( ) cos2f i d     x( ) sin 2f i d 
 0
 0


T
2
T
2 1/ 2




Which is exactly the same as in correlator
receiver
14
Generalized binary receiver for noncoherent orthogonal
modulation.
15
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 envelope
(amplitude) of the received signal described in
terms of cos(2f1t) is greater than the envelope of
the received signal described in terms of
cos(2f2t), we say s1(t) was sent
16
Quadrature receiver equivalent to either one of the two
matched filters in part
17
Noncoherent Orthogonal Modulation
The average probability of error in noncoherent detector can be given by

1
E 

Pe  exp  
2
 2N0 
0
2
Where E is the symbol energy and is
the power spectral density of the noise
18
Noncoherent: BFSK
For BFSK, the transmitted symbols are defined
by
 2 Eb
cos2f i t ; 0  t  Tb

si t    Tb

; elsewhere
0
The receiver can be implemented as shown in
the block diagram detailed in the next slide
19
Non-coherent: BFSK
demodulator
20
Non-coherent BFSK, probability
of error
Probability of Errors
 Eb 
1

Pe  exp  
2
 2N0 
21
Differential PSK
Differential PSK is the non coherent
version of phase shift keying (PSK)
It eliminates the need for a coherent
reference signal by combining two basic
operations
1.
2.
Differential encoding of the input binary
stream (as explained at the end of Ch. 3)
Applying phase shift keying modulation to
the encoded binary data
22
Differential PSK
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 remains
unchanged
If “0” is sent, the phase changed by 180◦
23
Differential PSK
Suppose that the transmitted DPSK signal
equals to

cos(2 
2
for 0 ≤  ≤  , where 
is the bit duration and  is the signal energy per
bit
Let 1  denote the transmitted DPSK signal for
0 ≤  ≤ 2 for the case when symbol 1 is being
transmitted for the second part of the this time
interval  ≤  ≤ 2
The transmission of symbol 1 leaves the carrier
unchanged over the interval 0 ≤  ≤ 2
24
Differential PSK
The transmission of symbol 1 leaves the carrier
unchanged over the interval 0 ≤  ≤ 2 ,
therefore we can define 1  as
1  =

cos 2  ,
2
0 ≤  ≤ 

cos 2  ,
2
 ≤  ≤ 2
25
Differential PSK
Similary we can define 2  for the transmission
of symbol 0 for DPSK encoding over the time
interval 0 ≤  ≤ 2 as
2  =

cos 2  ,
2
0 ≤  ≤ 

cos 2  +  ,
2
 ≤  ≤ 2
26
Average probability of error of
DPSK signals
The average probability of error of DPSK signal
is given by  =
1

2

−
0
A comparison between the average probability of
error of DPSK and non-coherent BFSK shows that
the DPSK has a better average probability of error
compared with non-coherent BFSK
27
Generation of DPSK
Let the binary sequence  represents
the binary data to be transmitted using
DPSK
Let the sequence  represents the
differentially encoded sequence
The encoding rule can be as follows (there
is a reference bit as explained in Ch.3)
28
Generation of DPSK


If the incoming binary symbol  is 1, leave
the symbol  unchanged with respect to the
previous bit
If the incoming binary symbol  is 0, change
the symbol  with respect to the previous bit
The differentially encoded sequence 
is used to phase-shift a carrier with phase
angles 0 and  representing symbols 1
and 0, respectively
29
Example of differential encoding
process
Consider the binary sequence shown
below find the differentially encoded
sequence and the transmitted phase of the
carrier
30
Generation of DPSK
Differential PSK modulator can be realized
by the block diagram shown below, where
 =  −1 ⨁ −1 (Xnor operation)
31
DPSK: Receiver
32
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.
33
Signal-space diagram of received DPSK
signal.
34
Comparison of digital modulation
schemes using a single carrier
35
36

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