Transcript Lecture17

FOURIER ANALYSIS TECHNIQUES
FOURIER SERIES
Fourier series permit the extension of steady state analysis to general periodic
signal.
FOURIER SERIES
The Fourier series permits the representation of an arbitrary periodic signal as
a sum of sinusoids or complex exponentials
Periodic signal
The signal f (t ) is periodic iff there exists T  0 such that
f (t )  f (t  T ), t
The smallest T that satisfies the previous condition is called the (fundamental)
period of the signal
FOURIER SERIES RESULTS
If f (t ) is periodic, with period T0 , then f (t ) can be expressed in one of the following
equivalent forms
Phasor for n-th harmonic
Cosine expansion


n 1
n 1

f (t )  a0   Dn cos(n o t   n )  a0   Re Dn ne jn o t
f (t ) 

 cne jn0t

Complex exponential expansion

n 1
n 1
f (t )  a0   an cos n o t   bn sin n o t
c0  a0
2
T0
Dn n  2cn  an  jbn
e j  cos  j sin 
n  

0 
Trigonometric series
e j  e  j  2 cos
e j  e  j  2 j sin 
Relationship between exponential and trigonometric expansions
For n  0
cn e jn0t  cn e  jn0t  (cn  cn ) cos n0t  j (cn  cn ) sin n0t
an
bn
 2cn  an  jbn
2cn  an  jbn
If f (t ) is real - valued then cn  cn 
GENERAL STRATEGY:
. Approximate a periodic signal using a Fourier series
. Analyze the network for each harmonic using phasors or complex exponentials
. Use the superposition principle to determine the response to the periodic signal
*
Approximation with 4 terms
Original Periodic Signal
EXAMPLE OF
QUALITY OF
APPROXIMATION
f N (t ) 
N
jn  0 t
c
e
 n
n N
Approximation with 2 terms
2
2
n 1
n 1
f 2 (t )  a0   an cos n o t   bn sin n o t
Approximation with 100 terms
100
100
n 1
n 1
a0   an cos n o t   bn sin n o t
EXPONENTIAL FOURIER SERIES
Any “physically realizable” periodic signal, with period To, can be represented over
the interval t1  t  t1  T0 by the expression
f (t ) 
n 
 cne jn0t ; 0 
n 
2
T0
The sum of exponential functions is always a continuous function. Hence, the right
hand side is a continuous function.
Technically, one requires the signal, f(t), to be at least piecewise continuous. In that
case, the equality does not hold at the points where the signal is discontinuous
Computation of the exponential Fourier series coefficients
t1 T0
f (t ) 

 f ( t )e
 jk 0 t
t1
1
ck 
T0
 cne jn t
0
n  
t1
t1 T0
n
dt 
t1 T0

t1
 e  jk0t
t1 T0
 0 for n  k
j ( n  k ) 0 t
e
dt



T0 for n  k
t1
 n jn 0t   jk 0t
dt
  cn e
e
 n

t1 T0

f (t )e  jk 0t dt
t1
t1 is arbitray and can be chosen to make computations simpler
EXAMPLE
Determine the exponential Fourier series
n 
f (t ) 
0

cn   2V
n
sin
 n
2
T0  T   0 
t1  T / 2
cn 
1
T
T
2
 jn  t
 v (t )e 0 dt 

T
2

1
T
T
4


T
2
(V )e  jn0t dt 
T
4
n 
n even
n
odd
1
cn 
T0
4
2
T0
t1 T0

f (t )e  jn 0t dt
t1
A strategy :
1. Determine T0 and  0
2
T
2. Select a convenient t1
3. Do the integratio n
T
2
1
1
 jn  0 t
Ve
dt

Ve  jn0t dt


T T
TT

 cne jn0t ; 0 
4
V
 jn 0 t T / 4
 jn 0 t T / 4
 jn 0 t T / 2  This is for n  0!

cn 
e
e
e
T / 2
T / 4
T /4
 c0  0 in this case
T ( jn 0 ) 
e j  e  j
 2 sin 
j
T0  2
 T
 T
T 
 T
T 
T 
V   jn0   4   jn0   2   jn0  4   jn0   4   jn0  2   jn0  4  
e

cn 
e
e
e
e
e
jTn0 


n
n
 V 
j
V  j2

 n 
 jn 
jn 
2
cn 
 2e
e
e  
4
sin

2
sin(
n

)
 2e




j 2n 
 2 

 2n 