Transcript No Slide Title

Communication Systems : Prof. Ravi Warrier
FOURIER SERIES
Let g(t) be periodic; period = To . Fundamental frequency = fo = 1/ To Hz or o = 2/ To rad/sec.
Harmonics =n fo
, n =2,3 4, . . .
Trigonometric forms

g( t )  a o   a n cos(n ot )  bn sin(n ot ) 
n 1
ao 
Forn  0
1
To
t 1 To
an 

g( t )dt
 dc component
,
t1
2
To
t 1 To

g( t ) cos(n ot )dt , bn 
t1
2
To
t 1 To

g( t ) sin(n ot )dt
t1
Compactform:

g( t )  Co   Cn cos(n ot   n )
n 1
a o  Co
Forn  0
 bn
)
an
 n  tan 1 (
Cn  a n2  bn2
EXAMPLE :
g(t)
2
-1
-0.6
-0.2
0

g( t )  a o   a n cos(n ot )  bn sin(n ot ) ;
0.2
0.6
1
t
To  0 .8 sec,  o  2.5  rad / sec.
n 1
ao 
1
To
t 1 To

t1
Forn  0
bn 
2
To
t 1 To


g( t )dt
an 
2
To
t 1 To

1
0 .8
0 .2

2 dt ,
0 .2
g( t ) cos(n ot )dt 
t1
g( t ) sin(n ot )dt 
t1
2
0 .8
0 .2

2
0 .8
0 .2

2 cos(2.5 nt )dt 
0 .2
2 sin(2.5 nt )dt 
0 .2
g( t ) 
R
+
g(t)
+
C go(t)
a) For R = 1 M and C=1 µF , what is go(t) ?
b) For R = 1 M and C=0.1 µF , what is go(t) ?
-
1
Communication Systems : Prof. Ravi Warrier
EXERCISE :
g(t)
2
-5
-3
-1
0
1
3
5
t (sec)
-2
g(t)
go(t)
Filter
H()
What is go(t) if the frequency response of the filter is as shown ?
PHASE
0
1.2
-50
1
-100
0.8
-150
| H(jw)
|H(jw)|
MAGNITUDE
1.4
0.6
-200
0.4
-250
0.2
-300
0
0
10
Freq. in rad/s
-350
20
0
10
Freq. in rad/s
20
EXERCISE : a)Find the Fourier series in trigonometric compact form.
g(t)
2
-4
-2
-1
0
1
2
4
t (sec)
-2
2
Communication Systems : Prof. Ravi Warrier
Complex
Exponentia
l form:
g( t ) 


Dn e jn ot
Dn 
n  
t 1 To
1
To
EXAMPLE :
 g( t ) e
 jn ot
dt
t1
g(t)
2
0.5 
1.5 
0
1.5 
0.5 
t
Sketch the Fourier spectra.
Dn 
1
To
t 1 To
 g( t ) e
 jn ot
dt 
t1
1
2
0 .5  o
2 e
0 .5 
 jnt
dt To 
1
2
Suppose that g(t) is passed through a filter of frequency response as shown. What is the output signal ?
(Both positive and negative frequencies are shown here)
|H(j)|
1
g(t)
H(j)
go(t)
-3.2
0
H(j)
3.2 

3
Communication Systems : Prof. Ravi Warrier
ENERGY AND POWER
Eg 
Energy of g(t):

g
2
g(t) is an energy signal if Eg < .
( t ) dt

Power of g(t) :
1
P  lim
T  T
T
2
T g

2
( t ) dt
where T is an arbitrary time interval .
2
g(t) is a power signal if 0< P < .
g(t)
3
EXAMPLES : Let g(t) be as shown . The energy of g(t) is : Eg=54 J.
-1
4
5
t

-3
Let g(t) be a unit step function: g(t) = u(t). Is this a power or an energy signal ?
g(t)
Eg 
1

g
2

t


(t)dt   dt  
Not an energy signal.
0
1
P  lim
T
T 
T
2
Tg

T
2
(t)dt
12
1
 lim
 dt  2 w
T
T  0
2
A power signal.
Average Power of sinewaves
Let g(t)  Acos(o t)
T
T
T
o
o
2
2
1
1
1 2 A2
A2
2
2
2
P  lim
[1 cos(2o t)]dt 
 g (t)dt 
 A cos (o t)dt 

To T
To T 2
2
T T  T
o
o


2
2
2
POWER OF ANY PERIODIC FUNCTION IN TERMS OF FOURIER COEFFICIENTS :

 Cn2
Let g(t)  Co 
Cncos(no t  n ) Then P  C2
.
o 
n1 2
n1



2
2
Let g(t) 
Dn e jn o t Then P   Dn  Do2  2  Dn .
n-
n-
n1
C 

 why? C  D ,| D | n 
o
o n

2 

g(t)


2
EXAMPLE :
-1
-0.6
4

Wefoundthat g( t )  1   
sin(0 .5 n )  cos(2.5 nt )
n 1  n

2

1
4

ThenP  1   
sin(0 .5 n ) 
2 n 1  n

 1  0 .81206  0  0 .0901  0  ...
 2W

-0.2
0 0.2
0.6
-0.6

t
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Communication Systems : Prof. Ravi Warrier
Signal Comparison : CORRELATION
Let g1(t) and g2(t) be two signals. Their correlation is defined as

 g1g2 ( ) 
 g (t)g (t  )dt.
1
2

If g1(t) = g2(t) = g(t), this becomes autocorrelation function, given by

 g () 
 g(t)g(t  )dt.

We see that g(0)=Eg we ge the signal energy. That is, the signal energy =autocorrelation at  = 0.
FOURIER TRANSFORMS
F{g(t)} = G()
Definition :

G(  )   g(t)e- jt dt
F-1{G()} =g(t)
g(t) 

1 
G(  )e jt d
2  
TRANSFORM EXAMPLES :
G()
g(t)
g( t )  ( t )
t
0
G()  1,
  t  
2A().
g(t)
g( t )  A ,   t  
t
g(t)
t
g( t )  A rect 

A
 2
0
t

2
G()  2A().
g(t)  sincWt
g(t)
 2
0
g(t)  A ( t )

2
G()

0
A
  
G()  A sin c

 2 

A

0
A
0
1
G() 
2

G()

2


  
rect

W
 2W 

  
G()  A  sinc 
2
 4 
2
G()
t
sinc( x) 
sin(x )
x
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Communication Systems : Prof. Ravi Warrier
g(t)
g(t )  e at u(t ),a  0
G() 
1
a  j
g(t)  e  a|t| , a  0
G() 
2a
a  2
t
0
g(t)
0
t
2
G()
g(t)  e jot
G()  2    )
0 o

o 0  o

G()
t
g(t)  cos(o t )
g(t)  sin(o t )
g(t)  sgn(t )
1
G()      )      )
G( )  j     )      )
G()  j2
t
0
-1
g(t)  u( t )
G()  ()  j1
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Communication Systems : Prof. Ravi Warrier
PROPERTIES F{g(t)} = G()
g( t )  G()
FOR PROOF READ TEXT.
1. Symmetry :
G( t )  2g( )
2. Scaling :
g(at ) 
1
a
G( 
)
a
3. Time-shifting :
e  jt o represents a linear phase ; in time
g(t  t o )  e  jt G() (The term
domain it is delay).
4. Frequency shifting :
g(t) e jot  G(  o ) ; g(t ) e jot  G(  o )
o
What is F g( t ) cos(o t ) ?
e jo t  e  jo t 
F
g( t )

2


1
 2 G(  o )  G(  o )
Here g(t) is modulating the sinusoid amplitude - AMPLITUDE MODULATION.
g(t) is the modulating signal, cos(ot) is called the carrier.
Let g( t )  rectt .
EXAMPLE :
We will find the Fourier transform of g(t)cos(10t).

g( t )  rect t
1
TIME
cos(10t)
domain
g(t)cos(10t)


FREQUENCY
domain
1
2
0
1
2
t
=
t

1
2
0
1
2
t
G()
F g( t ) cos(10t )
1
2
2

1
2
Note : Multiplication in timedomain doesn’t transform to
multiplication in frequency
domain.
-10
0

10
Math:
 
2
F rect( t )  sinc 
 F rect( t ) cos(10t ) 
1 sinc   10   sinc   10 




2
2
2





EXERCISE : 1. What is F g( t ) cos(10t )?
2. Find the spectrum of a) g(t)  (2t ) cos(20t ).
b) g(t)  (2t ) sin(20t ).
Sketch the time functions and the spectra.
7
Communication Systems : Prof. Ravi Warrier
  jG()
 dng( t ) 
F n   ( j)n G().
 dt 
dg( t )
5. Differentiation
F dt
EXAMPLE : We find the Fourier transform of the triangular function shown using this property.
g(t)  A ( t )
g(t)
A
 2
0
d2g( t )
dt 2
2A

2A
 dg( t )
dt
t

2
0
 2
0


2
t


2

2

 d2g( t ) 
F  2   ( j)2 G()  2G()
 dt 
Taking the Fourier transf ormof theimpulse f unctions:
j 
 j  2 
 2G()  2A  e 2  2  e
 2A 2 cos(2)  2 


2 
8 A sin2 (  )
A sin ( 4 ) A

4
 G() 


sinc 2 ( 4)
2 (  )2
2
2



t
2A


4A

1 2 sin (
4A


2 
) 1
4
4
6. Integration :
 t


 G( )
F  g( )d   j  G(0)()


 


7. Convolution :
CONVOLUTION IN TIME DOMAIN ( )
g1( t )  g2 ( t ) 

 g1()g2 (t  )d





Fg1( t )  g2 ( t )   F  g1( )g2 ( t   )d   G1()G2 ()



MULTIPLICATION IN FREQUENCY DOMAIN
CONVOLUTION IN FREQUENCY DOMAIN
Fg1( t )g2 ( t )  

1
2
 G1( x)G2 (  x)dx  21 G1()  G2 ()

MULTIPLICATION IN TIME DOMAIN
8
Communication Systems : Prof. Ravi Warrier
Let g(t )  rect( t );
EXAMPLE:
We hav e G()   sinc( 2).
Find g(t )  g(t )
F {g( t )  g( t )}  G2 ()   2 sinc 2 ( 2)
From Fourier transf ormtable F {( 2t )}   sinc 2 ( 2)
 F-1 {g( t )  g( t )}   ( 2t ).
g(t)
1

 2

0
Note : g(t) has a pulse width of /2 sec
but g( t )  g( t ) has pulse width of  sec.
Convolution increases the width of the
function.
 g( t )  g( t )
 t
2

 t
0
EXERCISE : Let g(t) = sinc(50t). What is the spectral width (bandwidth) of g(t) ?
What is the bandwidth of g2(t) ?
EXAMPLE : What is the Fourier transform of a periodic function ?
Periodic functions can be expressed in time domain as sum of a dc term and sinusoids of fundamental
frequency and harmonics. Fourier transform of a sinusoid is a pair of impulse functions. Therefore, the Fourier
transform of a periodic function is a sum of impulse functions centered at zero frequency, fundamental
frequency and harmonics.

 D n e jnot
g(t) 
n 

 

D n e jno t   2  D n (  n o )

 n 

n 
F {g( t )}  F 
EXAMPLE : We find the Fourier transform of the periodic function shown.
The Fourier series of g(t) is given (in page 51, text) by
g(t)
g( t ) 
1
 2
2

2
0
2
t
1
2
 2 cos(t )  62 cos(3t )  102 cos(5t )
 142 cos(7t )  
We have F {cos(o t )}      )       )
The Fourier transform of g(t) is
G()  ()  2(  1)  (  1)  31 (  3)  (  3)  51 (  5)  (  5 
 71 (  7)  (  7)  
|G()|
-5
-4
-3
-2
-1
0 1
G()
2
3
4
5

0
2
3
4
5


-5
-4
-3
-2
-1

1
9
Communication Systems : Prof. Ravi Warrier
EXAMPLE : Consider the periodic function g(t) consisting of impulse functions at equal spaces of To sec.
We find the Fourier transform of g(t). We can express g(t) as
g(t)

 ( t  nTo )
g(t) 
-4To -3To -2To -To
0
To
t
2To 3To 4To
 T1
o


o
o
0

2o 3o 4o
Fourier series.
n 
F {g( t )}  F  T1
G()
-4o -3o -2o - o
n 

e  jno t
o

 e jnot
n 

 2 
  To  (  n o )   o  (  n o )

n 
n 

SIGNAL ENERGY AND ENEGY SPECTRAL DENSITY

We define signal energy as
Eg 

.
g2 (t)dt


If g(t) is complex we can express energy as
Eg 


2
g(t) dt 

 g(t)g (t)dt
*

Parseval’s theorem : Signal energy is

Eg 




2
g(t) dt  21

G()G* ()d  21

 G()
2
d

2
Energy Spectral Density (ESD):  g ()  G() is called the energy spectral density of g(t). The signal
energy is the integral of the energy spectral density ( multiplied by 2).

E g  21
  g ()d

ESD provides a way of computing energy from the Fourier transform of g(t).
EXAMPLE : Consider g(t)=e-0.5t u(t). Find the energy and the ESD of g(t).


Eg 

2
g(t) dt 

G() 
1
j0.5
-0.5t 2
0
0
 g ()  G()
2

Eg 

1
2
1
2 x 0.5
 1J.
-t

1
2
2
dt  1J.
1
2 0.25

  g ()d  


 e  dt   e
G( )
0.25
1
d
2 0.25


1
tan1( 0.5 )
2x 0.5

0


10
Communication Systems : Prof. Ravi Warrier
ENERGY OF MODULATED SIGNAL
Let g(t) be a baseband energy signal bandlimited to B Hz.
Let
( t )  g( t ) cos(o t )
() 
1
2
an amplitude modulated signal, where o  2B.
G(  o )  G(  o ).
ESD of ( t ) is   () 
G()
1  G(   )  G(   ) 2 .
o
o
4 

Since o  2B,
2
2
  ()  41  G(  o )  G(  o ) 


 41  g (  o )   g (  o )

2B
0 2B
Suppose that G() is as shown. Then

()  21 G(  o )  G(  o ).
o

()

o
4B
Then we f ind that E   21 E g .
The signal energy is reduced by 1/2 when it is multiplied by a sinewave of unit amplitude.
ESD OF A SYSTEM INPUT AND OUTPUT
G()
Y()
H()
Y()  H()G() 
Y()
2
2
2
 H() G()
 Y ()  H()  G ()  E y 
1
2
2

  Y ()d 

1
2

 H()
2
 G ()d

EXAMPLE : Find the input and output energies. R=200 and C=0.01 F. g(t)=sinc(t).
R
+
+
g(t)
C go(t)
-
G()   rect( 2 )
Eg 
H() 
Y()
2
2
 H() G()
2

1
4 2 1
1
jRC 1
1
1
2

x rect( )

2
2
  2 d   J
1
Y()  H()G()
1
j 2  1
 2
for  1 rad / s

  042 1 Otherwise


2
 Y ()  H()  G ()
Ey 

1
2
  Y ()d 



8
2 tan
1
1
1
2

1
 22
4 2 1
1

d 
1
8
1
  1 
1
2
1 2
2
d
(2) 1  0.87 J
EXERCISE : Redo the Example problem for R=200 and C=0.001 F.
11
Communication Systems : Prof. Ravi Warrier
Autocorrelation Function and ESD : For g(t) a real function
 g ( ) 


 g(t)g(t  )dt   g(t)g(t  )dt

Thus
(We can show this by letting   t    READ TEXT)

 g ( )   g (  )
an ev en f unction.
How is this related to conv olution ?
but they represent dif f erentoperations. Conv olution is a time operation.
Time autocorrelation is a f unctionof the time lag (  in this case).
Conv olution as a f unctionof  : g( )  g( ) 

 g(x)g(  x)dx

g( )  g(-) 




 g(x)g((  x))dx 
 g(x)g(x  )dx 
 g ( )
  g ( )  F{g()}F{g(-)}.
F{g()}  G()
F{g(-)} 

 g()e  j 

d 

 g()e j 
d  G(-)  G * ()

2
  g ()  G()G * ()  G() .
Energy Spectral Density is the Fourier Transform of the autocorrelation function.
SIGNAL POWER AND POWER SPECTRAL DENSITY(PSD)
Energy and energy spectral density are useful for energy signals. For power signals we define power and
power spectral density as follows :
Pg  lim T1
T 
T
2
Tg 2 (t)dt
2
 g( t ) for - T  t  T
Def iningg T ( t )  
2
2
 0 otherwise
Pg  lim T1
T 
G T ( )
T
T 
The power spectraldensity S g ()  lim



2
2
1E
1 1
g
(t)
dt

lim

lim
G
(

)
d


T
g
T
T


T  T
T  T  2 


 

2
.
THEN
Pg  21

 S g ()d

Time autocorrelation is a f unctionof a real power signal is def inedas
Rg ( ) 
lim 1
T  T
T
2
 g(t)g(t  )dt
 T2
Rg ( )  Rg ( )
F {Rg ( )}  S g ()
Rg (0)  Pg
Or
Rg ( )  F - 1 {S g ()}  21

 S g ()e j 
d

12
Communication Systems : Prof. Ravi Warrier
EXAMPLE : Let g(t)=A cos(ot) , a power signal.
Pg 
To
2
1
To

 g 2 (t)dt 
To  period
A2
2
To
2
Time autocorrelation is a f unctionis
To
2
T
2
Rg ( )  lim
1
T  T
 g(t)g(t  )dt 
 T2

 g(t)g(t  )dt 
1
To



2
2
A2
2

A2
2
cos( o )dt 
Pg 
2
cos( o (2t  )  )]dt
To
2
cos( o )
 S g ()d 

 A2


1
2
To
2
1
To
To
2
The power spectraldensity S g ()  F{ Rg ( )}
THEN
2
[cos( o )  cos( o (2t  )  )]dt
2

To
To
To
1
To

2
 A2

 A 2 cos( o t  ) cos( o (t  )  )dt
To
To
2
1
To
To
2
1
To

1
2


A 2
2

A 2
2
( -  o )  ( -  o )
( -  o )  ( -  o )d 
A2
2
Also Pg  Rg (0)

S g ( )
o
A 2
2
o
0

Input signal power , output signal power Let g(t) be a power signal applied to a system.
Y()  H()G() 
G()
H()
Y()
Y()
2
2
 H() G()
2
S Y ()  H() S G ()  Py 
1
2

 S Y ()d

EXAMPLE : Consider a noise signal n(t) with PSD Sn ()  K, - b    b
Sn ()
What is the output noise power ?
K
n(t)
H()=j 
H()  j,
y(t)
- b
H()
2
1
2
b

is the input to a differentiator.
H()
- b

2
S n ()  K H()
 S Y ()d 

0
2
b  
S y ()
 2
Sy()  H()
Py 
0
2
K3
b
3
2
- b    b
- b
0
b  
13
Communication Systems : Prof. Ravi Warrier
EXERCISE : 1 Consider a noise signal n(t) with PSD Sn ()  0.1, - 10    10 applied to a RC filter with
RC=1 sec. Determine the input noise power and output noise power. (Input power = 2W, Output power=0.25 W)
R
2. Suppose that the input to RC filter is
g(t)=2cos(0.5t), what is the input and out put signal
powers ? (Ans: power in=2W, power out=1.79W)
+
+
n(t)
C no(t)
-
3. Next consider the input filter to be g(t)+n(t). Find the (signal power)/(noise power) at the input and output.
This is called the signal-to-noise ratio.
Distortionless Transmission : The ideal goal of a communication system is to make sure the received and
transmitted signals are the same. That is, the received signal is not distorted. This means the communication
system transfer function should have a constant magnitude and linear phase characteristics, in the frequency
H()
region of interest.
G()
Y()
H()
H()  K e
-jt d
K
0
H()
0
REVIEW :


t d
1) Fourier transform inverse Fourier transform definitions
2) Properties : Important ones :- Symmetry, time-delay /phase shift, Modulation
3) Results : Fourier transform of periodic functions, Energy, ESD, autocorrelation, power,
PSD, autocorrelation, input energy-output energy, input power-output power.
What is the autocorrelation function of a sinewave ? What is the PSD of a sinewave ? What is
the average power of a sinewave ? Does phase shift affect the power and autocorrelation ?
What is the autocorrelation function of g(t) =cos(20t) ? What is the autocorrelation function of
g(t) =sin(20t) ?
READ TEXT BOOK A LOT.
EXERCISE : Find the Fourier transform and sketch the spectra of : i)g(t) = sinc(20t)cos(100t)
ii) g(t) = sinc(20t)cos2(100t).
14
g(t)=1 for binary 1, g(t) = -1 for binary 0

g( t )

Tb
2
Communication Systems : Prof. Ravi Warrier
Tb
2
Tb

2
T
b
For    

t

g( t   )
t
Tb

2
t
In the shaded region the product g(t)g(t - )  1
T
For 0    b
Tb
2
Area of the shaded region 

2
Tb
T
   0 the area under g(t)g(t - ) is b  
2
2
T
T
So we can write   b the area under g(t)g(t - ) is b  
2
2
Similarly we can show that, f or
15
Communication Systems : Prof. Ravi Warrier
Rg ( )  lim
N Tb
T
2
1
T  T
 g(t)g(t  )dt  lim
1
N  NTb
 T2
2

 g(t)g(t  )dt
N Tb
2
For N pulses the area is N( T2   )
b
N Tb
For  
Tb
2
2
Rg ( )  lim
1
N  NTb
 g(t)g(t  )dt  NT1
b

 N(
Tb
2

  )  ( 21 

Tb
)
N Tb
2
16