Transcript EE102 – SYSTEMS & SIGNALS

Review of methods for LTI systems
• Time domain methods (differential equations,
convolutions). Apply to all cases, but may be
cumbersome to compute.
• Laplace transforms. For systems starting at time t=0;
useful to study transients, instability.
• Fourier series. For studying stable systems in steady
state (after transients die down). Restricted to
periodic inputs. Key idea: superposition of sinusoids.
• Question: can we extend this steady-state analysis via
sinusoids to non-periodic inputs?
• Answer: Fourier transforms.
Fourier series as the period increases.
Consider the series f (t ) 

F e

n 
in0t
n
, for period T 
The frequencies involved in the expansion are n0 n  .
20
0
0

0
2
0

20
Now consider a function with twice the period, T  2T . Then
0 
0
2
 
. The frequencies involved in the expansion are n0
20
0
0
0 20 30
As the period gets longer, we get a more dense set of
frequencies. In the limit: include all frequencies.


n 
Fourier Transform
We wish to extend the Fourier series concept to a non-periodic f(t).
Intuitively: f(t) takes infinitely long to repeat itself, so we think of
f
it as a function of infinite period.
This suggests replacing
1
Fn 
T
T
2


t
f (t ) ein0t dt by an integral where T  .
T
2
Definition: The Fourier transform of time-domain function
f (t ), t   ,   is the function of frequency    ,  

defined by the integral F (i )  F  f (t )    f (t )ei t dt ,

assuming it converges.
Example I: f (t )  e
F (i ) 

0



 t i t
e e
t
F (i ) 
.
dt 

0




et eit dt   et eit dt
e(1i )t dt   e(1i )t dt  e
0
f (t )eit dt
0
(1i )t
1  i
0
e

 (1i )t
(1  i )


0
(1  i )  (1  i )
2




1  i 1  i
(1  i )(1  i )
1 2
1
1
t

t it
Example II: f (t )  e u(t ). F (i )   e e
0
dt 
1
1  i
Note: in case II, the Fourier transform F (i ) coincides with
the Laplace transform F ( s) evaluated at s  i .
Representation of the Fourier transform
F (i ) is a complex function of the real variable  . The "i "
in the argument of F (i ) plays no role, and could be omitted.
We only put it in for compatibility with Laplace transforms,
as in example II above (more on this later).
iF ( )
Magnitude and phase representation, F (i )  F (i ) e
Example:
F (i ) 
F (i ) 
F ( i )
1
1  i
1
2 1
 F ( )   Arctan( ).
 F ( )
Bode Plots
Represent the frequencies    in a logarithmic scale
(i.e. proportional to log( ).).
Also plot the magnitude in "decibels (dB)": 20log F (i )
20log F (i )
 F ( )
Widely used, but we will not emphasize them in this course.
The inverse Fourier transform
In Fourier series, we can reconstruct the function from

in0t
the Fourier coefficients via f (t ) 
Fn e
.

n
Similarly, for Fourier transforms we can reconstruct
the function using the inverse formula
1
f (t ) 
2


F (i )eit d
A complete derivation is mathematically involved, but we
can sketch a proof based on Fourier series case, letting the
period go to infinity.
F (i ) 


f (t )e
T
T
Let FT (i ) 
2

T
it
2
1
dt  f (t ) 
2
f
T


F (i )eit d
t
2
f (t )eit dt . Then F (i )  lim FT (i )
T 
 T T
Considered on the interval t    ,  , f (t ) (if "well-behaved")
 2 2

in0t
has a Fourier series expansion f (t ) 
Fn e
, where
2

n 
T
1
1
0
in0t
Fn 
f ( t )e
dt  FT (in0 ) 
FT (in0 )

T T
T
2
2
2

0
in0t
 T T
So we have f (t )  
FT (in0 ) e
, for t    , 
 2 2
n 2
 f (t ) 
1

i t
F (i )e

2 

0
in0t
d  
F
(
in

)

F
(
in

)
e
 T 0
0 
n  2

0
1
in0t
 
F (in0 ) e

2
n  2
As T   :

(I)

F (i )ei t d
( II)
 Term (I) goes to zero since lim FT (in0 )  F (in0 )
T 
 Term (II) goes to zero as shown in the next slide.

1
it d
So the left-hand side must be zero: f (t ) 
F
(
i

)
e

2 
Also, as T   all t 's will be included in the interval.
1
Term (II):
2


n 
in0t
F (in0 ) e
1
0 
2


F (i )eit d
F (i )eit
0
n0

The sum on the left is a staircase approximation to the
integral on the right. As T  ,  0  0 and the
approximation becomes exact, provided the integral exists.
So the difference tends to zero.
Recap
Fourier Series
Fourier Transforms
(T-periodic functions)
1
Fn 
T
f (t ) 
T

f (t ) e
in0t
dt
F (i ) 
0

in0t
Fn e

n
1
f (t ) 
2




f (t )eit dt
F (i )eit d
Note that the Fourier transform and its inverse are formally
almost identical: except for a sign change and the factor
2 , t and  could be interchanged. This is called duality.
Example: F (i )  u(  B)  u(  B).
F (i )
1
B
1
f (t ) 
2


1
i

t
F (i )e d 
i t
1 e

2 it

B
2
 B

  B
e

2 it
1
iBt
B
i t d
e

B
e
iBt

sin( Bt )

t
sin(t )
Notation: we define the function sinc(t ) 
t
B
Then we have f (t )  sinc(Bt )

sin(t )
Plot of sinc(t ) 
t
sinc(t )
1
t
1

t
sin(t )
Note that sinc(0)  lim
1
t 0
t
Example: f (t )   (t ).
F (i ) 

it

(
t
)
e
dt  1 for all  .



1
i t d
The inverse formula would say that  (t ) 
e

2 
Now the function ei t has magnitude 1 for all  .
Not an absolutely convergent integral!
We can interpret it as
B
1
i t d 
 (t )  lim
e
B  2 
B
B
 lim sinc(Bt )
B  
B

sinc(Bt )
Table of Basic Fourier Transforms
f (t )
F (i )
 (t )
1
1
B
sinc(Bt )

u(t  a)  u(t  a)
 (t   )
i0t
e
2  ( )
u(  B )  u(  B ).
Homework
ei
2 (  0 )
They can be easily obtained applying the Fourier definition,
or the inverse formula. Notice the time-frequency duality.
F (i )
f (t )
i0t
e
cos(0t ) 
i0t
e
e
2
i0t
2 (  0 )
  (  0 )   (  0 ) 
( )
( )
0
sin(0t ) 
i0t
e
e
2i
i0t
0

i   (  0 )   (   0 ) 
( i )
0
0

( i )