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 𝑠1 𝑡 and 𝑠2 𝑡 are used to
represent binary logics 1 and 0
respectively
𝑠1 𝑡 =
𝑠2 𝑡 =
2𝐸𝑏
cos 2𝜋𝑓𝑐 𝑡
𝑇𝑏
2𝐸𝑏
2𝐸𝑏
cos 2𝜋𝑓𝑐 𝑡 + 𝜋 = −
cos 2𝜋𝑓𝑐 𝑡
𝑇𝑏
𝑇𝑏
2
Binary Phase shift keying
Where 0 ≤ 𝑡 ≤ 𝑇𝑏 , and 𝐸𝑏 is the transmitted
signal energy per bit
The carrier frequency is selected such that
𝑛
𝑓𝑐 = so that each bit contains an integral
𝑇𝑏
number of cycles
From the pair of symbols 𝑠1 (𝑡) and 𝑠2 (𝑡) we
can see only one basis function (carrier) is
needed to represent both 𝑠1 (𝑡) and 𝑠2 (𝑡)
3
Binary Phase shift keying
The basis function is given by
𝜙1 (𝑡) =
2
cos(2𝜋𝑓𝑐 𝑡)
𝑇𝑏
0 ≤ 𝑡 ≤ 𝑇𝑏
Now we can rewrite
𝑠1 𝑡 = 𝐸𝑏 𝜙1 (𝑡) and 𝑠2 𝑡 = − 𝐸𝑏 𝜙1 (𝑡) on
the interval 0 ≤ 𝑡 ≤ 𝑇𝑏
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); 𝑆1 (𝑡); is given
𝑇𝑏
by 𝑆11 = 0 𝑆1 𝑡 𝜙1 𝑡 𝑑𝑡 = + 𝐸𝑏
The projection of the second symbol 𝑆2 (𝑡) on
the basis function is given by 𝑆21
𝑇𝑏
= 0 𝑆2 𝑡 𝜙1 𝑡 𝑑𝑡 = − 𝐸𝑏
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 28) into two regions
If the received symbol falls in region Z1, the
receiver decide in favor of symbol S1 ( logic
1) was received
If the received symbol falls in region Z2, the
receiver decide in favor of symbol S2 (logic 0)
was received
7
Error probability of BPSKReceiver model
The receiver in the pass band can be
modeled as shown
The received signal vector 𝑥𝑖 (𝑡) = 𝑠𝑖 (𝑡)
+ 𝑛𝑖 (𝑡)
8
Error probability of BPSK
The observable element 𝑥1 (symbol zero
was sent and the detected sample was
read in zone 1) is given by
𝑇𝑏
𝑥1 =
𝑇𝑏
𝑥1 =
𝑇𝑏
𝑥1 =
𝑥1 𝑡 𝜙1 𝑡 𝑑𝑡
0
𝑠2 𝑡 + 𝑛 𝑡 𝜙1 𝑡 𝑑𝑡
0
𝑠2 𝑡 𝜙1 𝑡 𝑑𝑡 = 𝑆21 = − 𝐸𝑏
0
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 𝜎2𝑥
𝑁0
= , we need to find the conditional
2
probability density of the random variable 𝑥1,
given that symbol 0, 𝑠2 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 𝑝10 can be rewritten
as
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 + 𝐸𝑏
whereas the amplitude of logic 0 is − 𝐸𝑏
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
multiplier 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
𝑠𝑖 𝑡 =
2𝐸
𝜋
𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + 2𝑖 − 1
𝑇
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
𝑛𝑐
𝑛𝑐
𝑇
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
 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
2
i

1


4
si  

 E sin 2i  1 
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
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
22
Signal space diagram of QPSK
2
(01)
s3
s2
(00)
s4 (11)
s1
1
(10)
23
Signal space diagram of QPSK
with decision zones
The constellation diagram may appear as
shown below
2
Z3
(10)
s3
Z4
s4 (11)
E/2
 E/2
s2
Z2
(00)
Z1
E/2
s1
1
 E / 2 (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
0≤𝑡≤𝑇
𝑥 𝑡 = 𝑠𝑖 𝑡 + 𝑤(𝑡)
𝑖 = 1, 2, 3, 4
Where 𝑤(𝑡) is the sample function of
AWGN with zero mean and power spectral
𝑁0
density of
2
26
Error probability of QPSK
The observation vector 𝑥 has two
elements, 𝑥1 and 𝑥2 defined by
𝑇
𝑥1 =
𝑥 𝑡 𝜙1 𝑡 𝑑𝑡
0
𝜋
𝑥1 = 𝐸𝑐𝑜𝑠 (2𝑖 − 1) + 𝑤1
4
𝐸
𝑥1 = ±
+ 𝑤1
2
27
Error probability of QPSK
𝑇
𝑥2 =
𝑥 𝑡 𝜙2 𝑡 𝑑𝑡
0
𝜋
𝑥2 = 𝐸𝑠𝑖𝑛 (2𝑖 − 1) + 𝑤2
4
𝐸
𝑥2 = ∓
+ 𝑤2
2
28
Error probability decision rule
If the received signal point associated with
the observation vector 𝑥 falls inside region
𝑍1, the receiver decide that 𝑠1(𝑡) was
transmitted
Similarly the receiver decides that 𝑠2(𝑡)
was transmitted if 𝑥 falls in region 𝑍2
The same rule is followed for 𝑠3(𝑡) and
𝑠4(𝑡)
29
Error probability of QPSK
We can treat QPSK as the combination of
2 independent BPSK over the interval 𝑇 =
2𝑇𝑏
since the first bit is transmitted by ϕ1 and
the second bit is transmitted by ϕ2
Probability of error for each channel is
given by P  1 erfc d12   1 erfc E 
2
2 N 
0 

2
 2N 
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 Pc  1  P 2
Which gives the probability of errors equal
to Pe  1  PC  erfc E   1 erfc2  E 
 2N 
0 

4
 2N 
0 

 E 

 erfc

2
N
0


31
Error probability of QPSK
Since one symbol of QPSK consists of two
bits, we have E = 2Eb Peper symbol  erfc Eb 
 N 
0 

The above probability is the error
probability per symbol
With gray encoding the average 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
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
𝐸𝑏
but uses only half the channel
𝑁0
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
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
35