MODULASI PSK (PHASE SHIFT KEYING)

  MODULASI PSK (PHASE SHIFT KEYING) Sistem Komunikasi Prodi D3 TT

  Yuyun Siti Rohmah, ST.,MT

BINARY SHIFT KEYING

  2

  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.

   Sangat kokoh, sering digunakan secara extensif pada komunikasi satelit.

  3

  Blok Sistem BPSK A cos( 2 f t  ) ; b ( t )  " 1 "

    c s ( t ) 

   A cos( 2 f t  ) ; b ( t )  " "

    c

  

  4

  Data Carrier

  Carrier+  BPSK waveform 1 1 0 1 0 1

  Contoh Sinyal BPSK

  5

  

Konstelasi sinyal BPSK

 Fasa dipisahkan 180 derajat.

  6

  '0' binary )

) 2 cos( (

  '1' binary )

) 2 cos( (

  2

  1   

     A t f t s A t f t s c c c c

  Q State

  1 State

  7

  f S(f) f c f c + R f c - R f c + 2R f c - 2R BW min BW min =2B N = 2.Rb/2 B N

  =Bandwidth Nyquist

  Spektrum Sinyal BPSK

  MODULASI QPSK (QUADRATURE SHIFT KEYING)

  8

  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

  01 State

  11 State

  10 State Phase of Carrier: /4, 3/4, 5/4, 7/4

  9

• cos+sin cos-sin
• cos-sin

• 0.5
• 0.5
>1.5 <>1.5
• 1

  1.5 0.2 0.4 0.6 0.8

  QPSK

  cos+sin

  10

  11

  01

  00

  1

  1.5 0.2 0.4 0.6 0.8

  1

  0.5

  1

  1

  0.5

  1

  1.5 0.2 0.4 0.6 0.8

  1

  0.5

  1

  1.5 0.2 0.4 0.6 0.8

  1

  0.5

  4 bentuk gelombang berbeda:

• 0.5
>1.5 <>1.5
• 1
• 0.5

10 Bentuk Sinyal

  Blok Sistem QPSK

  11

  

Bandpass modulation: Signal Space &amp; 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 ):

2 E

  i s ( t )  h ( t ) cos t  ( i  1 )  t  ( t )  t  T

     

   i c i

  T h ( t ) E

  where is the baseband pulse shape with energy . h

   We assume here (otherwise will be stated): h ( t )  is a rectangular pulse shape with unit energy.

  M

  1  Gray coding is used for mapping bits to symbols.

  E  E E s i s

   i

  1 

  M

   denotes average symbol energy given by

  12

  

Phase Shift Keying (PSK)

I. PSK signal waveform (transmitted signal):

  , phase: ,

  Can you show that

   

  

  

     

          

        

2

is symbol energy. is symbol inter .

  va

  4 i i i

  ( ?

  l

  )

  T i E s t dt

  T E

  

  

  M M   

  

2

r

  2 ( ) cos[ ( )] , 1 (

    7 / 2 1/ 2

  ,

  2 ) , , i i i

  E s t i

t t t

  T t

  M T i M

     

       

  3

   Phase (examples):  Symbol energy and symbol interval

  5

  3 , , , 0

  2 7 , , ,

  BPSK( 2) : , 0 QPSK(

  4

  4

  4

  4 ) : o

II. Signal space representation

  Note: decomposition of PSK signal waveform:

  2 ( ) cos( ), ( ) sin( ) t t t t T T

  (examples)

   

     

  E E

s t t t t t

T T

  2 ( ) cos[ ( )]cos( ) sin[ ( )]sin( ) i i i

  2

       

  2

  1) Orthonormal basis: 2) Signal vector :

  2

  1

   s

    

     

  M M M   

  2 ( ) : cos , si , 1, , n i i i i E E s t i

  2

3) Constellations:

  Signal space representation (cont.)

4) A proof of signal space representation

  Signal space vector for each waveform s

  2 T cos[ ( )]sin( ) T sin[ ( )] sin[2 ( cos[ ( )]cos( ) os[ ( )] cos[2 (

  

  

  

       

     

  1

  1

  2

  2 ( ) ( )

  2 T T c cos[ ( )]

  ( ) ( )

  ] n )

      

  ) i ] s [

  

T

i i T i i i T i i i 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

             

   

   

     

      

  

    

   

       

     ( )] t

  

  

  i

  2

  (t)

  15

  2

  2

  2

  2

  2

  1

  1

  2 ( ) 2 sin ( ) 2 [1 cos(

  ( )

  2

  cos( )sin( ) sin(2 ) / 1.

   Bases are orthonormal 

  . ( ) ( )

  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

  



  

    

     

       

  Signal space representation (cont.)

  h(t) is considered?  Signal waveform:

5) What happens if baseband pulse-shaping

  1

  M M M  

     

  E s t h t t t T

  2 ( ) ( ) cos[ ( )] i i

  s

  

     

     

  i i i i E E s t i

  2

  , 1, , n

   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.

  2

  

      

  T T

  2 ( ) ( ) cos( ), ( ) ( )sin( ) t h t t t h t t

  2

  2 ( ) : cos , si PSK modulator s t

  ( ) a   E i i h t

  ( )

  2 T cos(  t ) Special case: BPSK modulator

   a

  binary h t

  ( ) i

  1

  s t ( )

  bits

  i

  Symbol Note:

  T t

  2 cos(  ) map

  Inputs are signal- a space vector. i

  2 General case: M-ary PSK modulator

   h t

  ( ) Carriers are in basis

  T t

  2 sin(  ) form.

  s t ( )  a

  2 T cos(  t )  a

  2 T sin(  t ) i i i

  1

  2 s

   a a  E  i M  E  i M ( , ) cos(2 ), sin(2 ) i i i

  1

  

2



   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 baseband

  2 W W 

  2 PSK baseband,min

  3

  1 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 2). {01} ( ) cos( ) s t t        Phase ( ) in ( ) is different from phase of (phase in signal space) i i i

   t s t s 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.

  1  t

  N z z Decision circuits (ML detector) m

  1 z

  ) (t r

  N 

  ) (t

   T

  T ) (

  19

   z 

  1

  

  

N

z z

     

       

  ˆ Demodulation Detection Demodulations type: Some notations 

   Carrier: s t A t  t  t   f

  ( )  ( ) cos[  ( )], 

  2  Modulation types with respect to carrier parameters

  Modulatio Varying Demodulation n parameter Coherent or t

   ( ) PSK noncoherent

  A t  t both ( ) and ( ) Coherent or

  QAM noncoherent Coherent or

   FSK

  Noncoherent Two dimensional modulation, demodulation and detection (M-PSK)

   M-ary Phase Shift Keying (M-PSK)

  2 sin 2 cos sin

    

    

    

     

    

    

  M i t a t a t s  s 

  E a M i E a t T t t T t

  E E M i

  i i s s i s i c c i i i

  , , ( 1 ) ) ( ) (

  2 ) (

  2 ) ( cos

  1

  21

  1

  2

  2

  1

  2

  1

  2

  2

  2 ) (    

    2 cos

  E t s c

s

i

  M i t T

     

     

   

       Two dimensional mod… (MPSK)

  22

  2 s

  1 s

  

5

s

  “001” “011” “010” “101”

“111”

“100”

  ) (

  1 t

  

  1 s _s

  6 s

  E “00” “11”

  ) (

  2 t

  

  3 s

  4 s “10” “01” QPSK (M=4) BPSK (M=2)

  8 s

  

  ) (

  

  1  t

  2 s

  1 s b

  E “0” “1” b

  E  ) (

  2 t

  3 s

  2 t

  7 s “110”

  ) (

  1  t

  4 s

  2 s s

  E “000”

  ) (

  8PSK (M=8) Demodulation BPSK

   BPSK with coherent detection:

  s

  2

  2

  E

  E  b

  “0” “1” b

  s _b E

  1

  2

  23

  1  t

  ) (

  ˆ z

  Decision Circuits Compare z with threshold. m

   ) (t r

  1 t

  ) (

   T

  

1  s s Error probability …

   BPSK with coherent detection (with perfect carrier synchronization):

  1 t

  ) | (

  1 m p z z

  ) | (

  2 s

  1 s

  E 

  E b

   b

  ) (

  24

    N Q P B s s

     

  

1

   

  2

  2 / 2 /

  N E Q P b B

     

  2     

  2 m p z z

  Demodulation M-PSK

  decide s

  2

  2

  2

    

  2 t

  ) (

  1  t

  3 ) (

  s

  4

  3

   Coherent detection of Q-PSK

  decide s

  2

  decide s

  1

  s

  1

  decide s

  4

  s

  2

  1

     

1 N

     

  1

  I BPSK C

  Q p p b e b b c e b

  Q p p N E

  E Q p N E Q N E

  25 Decision Region QPSK s

     

  1

  2

  2

  1

  2

      

  2

  2

     

      

  

     

     

     

     

        

     

     

  

Power Spectra of M-Ary PSK

     

   

M f T c M E f S

  Tf c E f S B b b B

  2

  2

  2 log sin log

  2 sin 2 

  

  26

  QPSK vs. BPSK

   Let’s compare the two based on BER and bandwidth

  BER Bandwidth

BPSK QPSK BPSK QPSK

  b R b /2

  EQUAL

  2     

     

  N E Q P b B

  2     

     

  N E Q P b B

27 R

  Error probability …  Coherent detection of M-PSK

  1 s

  3 s

  i

  ) ( ₁  t

  4 s

  2 s _s

  E “000”

  ) ( ₂ t 

  is a noisy estimate of the transmitted 

  ˆ

8-PSK

  5 s “001” “011” “010” “101” “111” “100”

    

  7 s “110”

  6 s

  8 s

  2 ) ( ). ( 

  2

  Y dt t t r r

     T

  ˆ 

  1 ) ( ). ( m

  1

  T X r dt t t r

  

  Decision variable  r   

   ) (t r

  2 t

  ) (

   T

  1  t

  28

   T

  | ˆ |    i

   ˆ

  X Y arctan

  Compute Choose smallest

  ˆ z

  ) (

  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: M

   / M

  1 P M   P M   P s   P s   p  d  ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( )

  E C c m c 1 ˆ

   

   / M  

  M m

  

  1 where

    E E 

  2 s s

  2 

    p    

  ( ) cos( ) exp sin ; | | ˆ

    

  N N 

  2    It can be shown that (for M &gt; 4)

  or  

2 E

   _s   

  

  2 log M E      b

  2

   

  P M Q

  

  ( )  2 sin _E   

  P ( M ) 

  2 Q sin

      E

  N M

  29   

    

  N M

    

   Probability of symbol error for M- PSK

• M = 2
• “The same average symbol energy for different sizes of signal space”

  30 E

  P / dB

  N E b

  Note!

  k

THANK YOU

  31