MODULASI PSK (PHASE
SHIFT KEYING)
Binary Shift Keying (BPSK)
Perubahan Parameter Fasa dari sinyal pembawa sesuai dengan sinyal informasi.
Menggunakan alternatif-alternatif fasa gelombang sinus utk mengkodekan bit-bit dimana Fasa dipisahkan 180 derajat Sederhana utk diimplementasikan, tidak efisien dalam
penggunaan bandwidth.
Blok Sistem BPSK
"
0
"
)
(
;
)
2
cos(
"
1
"
)
(
;
)
0
2
cos(
)
(
t
b
t
f
A
t
b
t
f
A
t
s
c c
Data
Carrier
Carrier+
BPSK waveform
1 1 0 1 0 1
Konstelasi sinyal BPSK
Fasa dipisahkan 180 derajat.
'0'
binary
)
2
cos(
)
(
'1'
binary
)
2
cos(
)
(
2 1
t
f
A
t
s
t
f
A
t
s
c c
c c
Q
0
f
S(f)
f
cf
c+ R
f
c- R
f
c+ 2R
f
c- 2R
BW
minBW
min=2B
N= 2.Rb/2
BN=Bandwidth Nyquist
Quadrature Phase Shift Keying
•
Teknik modulasi multilevel : 2 bit per symbol
•
Lebih efisien spektrum, lebih kompleks
receiver.
•
Dua kali lebih efisien bandwidth daripada
BPSK
Q
00 State
11 State 10 State 01 State
4 bentuk gelombang berbeda:
-1.5 -1 -0.5 0 0.5 1 1.5
0 0.2 0.4 0.6 0.8 1 -1.5 -1 -0.5 0 0.5 1 1.5
0 0.2 0.4 0.6 0.8 1
-1.5 -1 -0.5 0 0.5 1 1.5
0 0.2 0.4 0.6 0.8 1 -1.5 -1 -0.5 0 0.5 1 1.5
0 0.2 0.4 0.6 0.8 1
00 01
11 10
cos+sin
-cos+sin
cos-sin -cos-sin
Bandpass modulation: Signal Space & Vector
Bandpass modulation:
The process of converting data
signal to a sinusoidal waveform where its amplitude,
phase or frequency, or a combination of them, is
varied in accordance with the transmitting data.
Bandpass signal (
General Condition
):
where is the baseband pulse shape with energy .
We assume here (otherwise will be stated):
is a rectangular pulse shape with unit energy.
Gray coding is used for mapping bits to symbols.
denotes average symbol energy given by
t
i
t
t
t
T
T
E
t
h
t
s
i(
)
(
)
2
icos
c
(
1
)
i(
)
0
)
(
t
h
E
hs
E
s
iME
iM
E
1
1
Phase Shift Keying (PSK)
I.
PSK signal waveform (transmitted signal):
Phase (examples):
Symbol energy and symbol interval
0 , phase: ,
2
( ) cos[ ( )] i( ) 2 0 , 1, , ,
i i
E
s t t t t i
T
M t T i M
7 / 2
1/ 2
3
5
3
, ,
, 0
2
7
,
,
,
BPSK(
2) :
, 0
QPSK(
4
4
4
4
) :
o
2
r
4
i
i
i
M
M
2 0
is symbol energy. is symbol inter
.
Can you show that
va
(
?
l
)
T i
E
s t dt
T
II. Signal space representation
Note: decomposition of PSK signal waveform:
1) Orthonormal basis:
2) Signal vector:
3) Constellations:
2 2
cos , si
( ) : n , 1, ,
i i
i i
E E
s t i
M M M
s
1 0 2 0
2
2
( )
t
cos(
t
),
( )
t
sin(
t
)
T
T
0 0
2
2
( )
cos[ ( )]cos(
)
sin[ ( )]sin(
)
i i i
E
E
s t
t
t
t
t
T
T
Signal space representation (cont.)
4)
A proof of signal space representation
Bases are orthonormal
Signal space vector for each waveform si(t)
15 2 0 2 2 2 2
0 0 0
1 0 0
1 2 0 0
0
0 0
0
0
( ) 2 sin ( ) 2 [1 cos(
( ) 2 2
cos( )sin( ) sin(2 ) /
1.
.
( ) ( ) 2
cos ( ) [1 cos
2
2 )]/ 2 1
(2 )] 2 0. / 2 T T T T T T
t dt T t dt T t
t dt T dt T
t t t
dt
t t dt T
t t
d
d
d
t
t T t
1 1 0 0 0
2 2 0 0 0
0
0
( ) ( ) 2 T
T c cos[ ( )]
( ) ( ) 2 T cos[ ( )]sin( ) T sin[ ( )] sin[2 (
cos[ ( )]cos( )
os[ ( )] cos[2 (
] n ) ) i ] s [ T i i T
i i i
T
i i i
T
i
a s t t dt
a s t t dt E dt
E dt
E t t t dt
E t t t d
E t
t
t E
t t
t t t
Signal space representation (cont.)
5) What happens if baseband pulse-shaping h(t) is considered?
Signal waveform:
Use basis (note that h(t) can be assumed normalized):
Signal space vector is still:
Conclusion: there is no difference in signal space whether pulse-shaping is considered. We can study only PSK instead of the more general PM.
1 0 2 0
2
2
( )
t
h t
( ) cos(
t
),
( )
t
h t
( )sin(
t
)
T
T
2
2
cos
,
si
( ) :
n
,
1,
,
i i
i
i
E
E
s t
i
M
M
M
s
0
2
( )
( ) cos[
( )]
i i
E
s t
h t
t
t
T
PSK modulator
Special case: BPSK modulator
General case: M-ary PSK modulator
i
a
E
( )
h t
0
2
T
cos(
t
)
( )
i
s t
binary
bits
1
i
a
2
i
a
( )
h t
( )
h t
0
2
T
cos(
t
)
0
2
T
sin(
t
)
( )
i
s t
Note:
Inputs are signal-space vector.
Carriers are in basis form.
1 0 2 0
1 2
( ) 2 cos( ) 2 sin( )
( , ) cos(2 ), sin(2 )
i i i
i i i
s t a T t a T t
a a E i M E i M
s
Bandwidth of PSK signal waveform
Just like DSB modulation:
Exercise :
Consider QPSK transmission with date rate
2000 bps. The magnitude of the signal si
(t) is √2E/T
=1 volt.
a) What is the minimum PSK signal bandwidth? b) Find the signal space points
c) Draw the constellation
d) Find signal waveform for transmitting {1001}.
18
PSK
2
baseband
W
W
2 PSK baseband,min 3
1 0
a) (log ) 2000 / 2 1000. 2 2 2 1000Hz. b) ( cos 2 4, sin 2 4), where / 2 0.5 10 , 1, , 4 d) Define mapping as: {00:0, 01: , 10: 2, 11: 3 2}.
Then {10} ( ) cos(
s b s
i
R R M W W R
E i E i E T i
s t t
s
2 0
2). {01} s t( ) cos( t )
Phase ( ) in ( ) is different from phase of
i t s ti
Demodulation and detection
Demodulation: The receiver signal is converted to baseband, filtered and sampled.
Detection: Sampled values are used for detection using a decision rule such as ML detection rule.
N
z
z
1z
T0
)
(
1t
T0
)
(
t
N
)
(
t
r
1
z
N
z
z
Decisioncircuits
(ML detector)
m
ˆ
Demodulations type:
Some notations
Carrier:
Modulation types with respect to carrier parameters
Modulatio n
Varying parameter
Demodulation
PSK
Coherent ornoncoherent
QAM
Coherent ornoncoherent
FSK
Coherent orNoncoherent
( )
t
0
both ( ) and ( )
A t
t
0 0 0
( )
( ) cos[
( )],
2
Two dimensional modulation,
demodulation and detection (M-PSK)
M-ary Phase Shift Keying (M-PSK)
21
M
i
t
T
E
t
s
i(
)
2
scos
c2
Two dimensional
mod…
(MPSK)
)
(
1t
2s
1s
bE
“0” “1” bE
)
(
2t
3s
7s
“110”)
(
1t
4s
s
2s
E
“000”)
(
2t
6s
s
81
s
5s
“001” “011” “010” “101” “111” “100”)
(
1t
2s
s
1Demodulation BPSK
BPSK with coherent detection:
T
0
)
(
1
t
)(t
r Decision
Circuits
Compare z with threshold.
m
ˆ
z
)
(
1
t
2
s
1
s
b
E
“0” “1”
b
E
b
E
2
2 1
s
Error probability …
BPSK with coherent detection (with perfect carrier
synchronization):
2
0
N
E
Q
P
B b2
/
2
/
0 2 1
N
Q
P
Bs
s
)
(
1
t
b
E
b
E
0
1
s
2
s
)
|
(
m
1p
zz
)
|
Demodulation M-PSK
Coherent detection of Q-PSK
25
Decision Region QPSK
s2
s4
decide s1
s1 decide s2
decide s3
decide s4
s3
1(
t
)
Power Spectra of M-Ary PSK
f
E
M
c
T
f
M
S
Tf
c
E
f
S
b b
B B
2 2
2
log
sin
log
2
sin
2
QPSK vs. BPSK
Let’s compare the two based on BER and bandwidth
BER Bandwidth
BPSK QPSK BPSK QPSK
R
b
R
b
/2
EQUAL
2
0
N
E
Q
P
B b2
0
N
E
Q
Error probability …
Coherent
detection
of M-PSK
28 8-PSK 3s
7s
“110” ) ( 1 t 4s
s
2s E “000” ) ( 2 t 6
s
s
81
s
5s
“001” “011” “010” “101” “111” “100” Decision variable r ˆ zCompute Choose smallest
X
Y
arctan
ˆ
|
ˆ
|
i
T
0
)
(
1t
T
0
)
(
2t
)
(
t
r
X
Tr
t
t
dt
r
0
1 1
(
).
(
)
m
ˆ
Y
Tr
t
t
dt
r
(
).
(
)
Error probability …
Coherent detection of MPSK …
The detector compares the phase of observation vector
to M-1 thresholds.
Due to the circular symmetry of the signal space, we
have:
where
It can be shown that (
for M > 4
)
29
p
d
P
P
M
M
P
M
P
M M c M m m c CE
(
)
1
(
)
1
(
)
1
1
)
(
1
)
(
/ / ˆ 1 1
s
s
M N E Q M P s E
sin 2 2 ) ( 0
M
N
E
M
Q
M
P
E(
)
2
2
log
2 bsin
or
2
|
|
;
sin
exp
)
cos(
2
)
(
2 0 0 ˆ
N
E
N
E
Probability of symbol error for
M-PSK
E
P
Note!
• M = 2k
• “The same average symbol