Transcript EE102 – SYSTEMS & SIGNALS

Cascade of linear time-varying systems
x
S1
y
S2
z
S1, S2 have impulse response functions h1(t, ), h2 (t, ).

z (t ) 
 h (t, ) y ( )d
2



  h2 (t ,  )   h1 ( , ) x( )d  d

 



Change order
of integration


   h2 (t, )h1 ( , )d  x( )d
h1,2 (t, )
Cascade of linear time-varying systems
x
y
S1
S2
z
S1, S2 have impulse response functions h1(t, ), h2 (t, ).
h1,2 (t , ) 

 h (t, )h ( , )d
2
1
Impulse response of the cascade.
Note: in general, h1,2  h2,1.
LTV systems do not commute.
Example: LTV systems do not commute
S1
x
y
y (t )  x  t  cos  t  ,
Applying x(t )   (t )
Therefore h1,2 (t,0)  u(t )
z (t ) 
v (t ) 
t
 x( )d ,
Applying x(t )   (t )
z
t
 y( )d
y (t )   (t ) cos(t )
  (t )
S2
x
S2
v
S1
z (t )  u (t )
w
w(t )  v  t  cos  t  .
v (t )  u (t )
w(t )  u(t )cos(t )
Therefore h2,1(t,0)  u(t )cos(t )  h1,2 (t,0)
The step response of an LTI system
y
x
y(t )  T  x(t )
The step response is defined as: g (t )  T u(t )
We assume the system is time invariant, with
impulse response h(t ). Then:
g (t )  h  u  u  h 
g (t ) 
t

t
- u(t   )h( )d  - h( )d
- h( )d ,
dg
h
dt
LTI systems and differentiation
dg
d
 du 
 h means that
T [u(t )]  T  
dt
dt
 dt 
d
 dx 
More generally,
T [ x(t )]  T   for any x(t ).
dt
 dt 
Proof: let y (t )  T  x(t ). Since T is LTI we have
y ( t   )  y ( t ) T  x (t   )   T  x (t ) 
 x (t   )  x (t ) 

T
.






dy
 dx 
Taking limit as   0, we have
 T  .
dt
 dt 
d
 dx 
T [ x(t )]  T   , for T linear time invariant.
dt
 dt 
d
dt
T
=
T
d
dt
Another way to see it: LTI systems commute,
d
and the "differentiator"
is also LTI:
dt
d
dx
 x (t   )   (t   )
dt
dt
Q: What is the impulse response of the differentiator?
d
A:
. But what is this?
dt
The derivative of delta.
d
 (t   )   ( t )
A natural definition would be
 lim
.
dt  0

This is the limit of a pair
of opposing impulses, of
increasing magnitude and
becoming close together in
time.
1
 
 

Strange object, rarely encountered in
physical models, or in the rest of this course.
t
 1
 
 
One example: an electric dipole: a pair of positive and
negative electric charges, becoming close together in space.
Laplace Transforms
• Time-domain tools for studying systems:
differential equations and convolutions.
• We want a more convenient analytical tool.
• Idea: transform time-domain functions to
functions in another domain.
L
f (t ) 
 F ( s)
• This mapping should be such that the system
operations become simpler.
Laplace Transform – Definition
Given a time-domain function f (t ), its Laplace
transform is the function of the complex variable s

defined by F ( s )   e  st f (t ) dt.
0
Remarks:
• The integral will be defined only for s in a region
of the complex plane (more on this later).
• The Laplace transform maps one function f(t) to the
other F(s). We denote this by
F ( s)  L  f (t )


0
0
Example: f (t )  et ; F ( s )   e st et dt   e( s1) t dt
Assume first that s  R..
e( s1)t
Then the integral converges
only for s  1.
s<-1
s=-1
s>-1
Now for complex s    i  :
t
e( s1) t = e( 1)t ei t  e( 1) t

e
( s1) t
dt
converges for   Re  s   1.
0

e
0
( s1) t
dt is absolutely convergent for Re  s   1.


0
0
Example: f (t )  et ; F ( s )   e st et dt   e( s1) t dt
Re  s   1 is called the
domain of convergence
or region of convergence
(DOC or ROC) of the
t 

Laplace transform L  e  .
1

Re  s  1
Inside this DOC, we compute

F ( s)   e
0
 ( s 1) t
dt 
e
 ( s 1) t
( s  1)


0 Re s 1

1 
1
 0  ( s  1)   ( s  1)


Laplace Transform – Remarks
 We integrate in t  [0, ). This is a "one-sided" Laplace
transform. Values of f (t ) for negative time are irrelevant.
 Some books also define the "bilateral" Laplace transform,
integrating on ( , ). We will not use it in this course.
 Still, to emphasize onesided-ness we often write L  uf (t )u(t ) .
This reaffirms that only [0, ) counts. For instance,
1
t
L  ue u(t )  
.
s 1
 The point t  0 is included in the integration; more precisely,
F ( s) 


0
0
 st
 st
e
f
(
t
)
dt
.
Example:
L

(
t
)

e



  (t)dt  1.
Table of Basic Laplace Transforms
f (t )
F ( s)
 (t )
1
at
e u(t )
u (t )
t n u(t )
cos( t ) u(t )
sin( t ) u(t )
1
sa
1
s
n!
s n1
s
s2   2

s2   2
DOC
All s  C
Re  s   Re[a ]
Re  s   0
Re  s   0
Re  s   0
Re  s   0
Example: f (t )  t u(t )
t

F ( s )   t e  st dt

0 u
parts
dv
te

Re[ s ]0

Re[ s ]0
 st


s
0
1
s
1
s

0
e st
s
dt
 e st
1
 st
 e dt  s   s
0

1 
1

 0 
 2
s  s



0



