Download + X(t)

Chapter 3
CT LTI Systems
Updated: 9/16/13
A Continuous-Time System
• How do we know the output?
X(t)
y(t)
System
LTI Systems
• Time Invariant
– X(t) y(t) & x(t-to)  y(t-to)
• Linearity
– a1x1(t)+ a2x2(t) a1y1(t)+ a2y2(t)
– a1y1(t)+ a2y2(t)= T[a1x1(t)+a2x2(t)]
• Meet the description of many physical systems
• They can be modeled systematically
– Non-LTI systems typically have no general
mathematical procedure to obtain solution
What is the input-output relationship for LTI-CT Systems?
Convolution Integral
• An approach (available tool or operation) to describe the
input-output relationship for LTI Systems
X(t)=d(t)
• In a LTI system
y(t)=h(t)
LTI System
 d(t) h(t)
– Remember h(t) is T[d(t)]
– Unit impulse function  the impulse response
• It is possible to use h(t) to solve for any input-output
relationship
X(t)
y(t)
LTI System: h(t)
• One way to do it is by using the Convolution Integral
Convolution Integral
• Remember
X(t)=Ad(t-kto)
y(t)=Ah(t-kto)
LTI System
• So what is the general solution for
X(t)
y(t)
LTI System
?
Convolution Integral
• Any input can be
expressed using the
unit impulse function
Proof:
d (t )  d (t )
Sifting
Property

  (t )d (t  t )dt   (t )
o
o


x(t ) 
 x( )d (t   )d

to and
integrate by d
 x(t )d (t  to )  x(to )d (t  to )




 x(t )d (t   )d   x( )d (t   )d

 x(t )  d (t   )d  x(t )(1)  x(t )

X(t)
y(t)
LTI System
Convolution Integral
• Given
X(t)
y(t)
LTI System
• We obtain Convolution
Integral
• That is: A system can be
characterized using its
impulse response:
y(t)=x(t)*h(t)
y (t )  T {x(t )}
h(t )  T {d (t )}
h(t   )  T {d (t   )}


y (t )  T   x( )d (t   )d 



Linearity : y (t ) 
 x( )T d (t   )d

X(t)
y(t)
LTI System: h(t)

 y (t ) 
 x( )h(t   )d

Do not confuse convolution with multiplication!
y(t)=x(t)*h(t)
By definition
Convolution Integral
X(t)
y(t)
LTI System: h(t)
Convolution Integral - Properties
x(t ) * h(t )  h(t ) * x(t )
[ x(t ) * h1 (t )] * h2 (t )  x(t ) *[h1 (t ) * h2 (t )]
• Commutative
• Associative
• Distributive
x(t ) *[h1 (t )  h2 (t )]  [ x(t ) * h1 (t )]  [ x(t ) * h2 (t )]
• Thus, using commutative property:
x(t ) 




 x( )h(t   )d   h( )x(t   )d
Next: We draw the block diagram representation!
Convolution Integral - Properties
•
•
•
Commutative
Associative
Distributive
x(t ) * h(t )  h(t ) * x(t )
[ x(t ) * h1 (t )] * h2 (t )  x(t ) *[h1 (t ) * h2 (t )]
x(t ) *[h1 (t )  h2 (t )]  [ x(t ) * h1 (t )]  [ x(t ) * h2 (t )]
Simple Example
• What if a step unit function is
the input of a LTI system?
u(t)
y(t)=S{u(t)}=s(t)
LTI System
• S(t) is called the Step
Response
y (t )  S{u (t )}
y (t )  s (t )  h(t )  x(t )  h(t )  u (t )




y (t )  s (t )   u ( )h(t   )d   h( )u (t   )d
t
  h( )d
Step response can be obtained by
integrating the impulse response!

Note : h(t )  y (t ) / dt  s(t ) / dt
Impulse response can be obtained by
differentiating the step response
Example 1
• Consider a CT-LTI system. Assume the impulse response
of the system is h(t)=e^(-at) for all a>0 and t>0 and input
x(t)=u(t). Find the output.
y (t )  h(t )  x(t )  h(t )  u (t )
u(t)
h(t)=e^-at

y (t )   h( )u (t   )d



 (e
 a
y(t)
Because t>0

 u ( ) u (t   )d

1
(

e
d


(e

a
t
 a
0

 at
 1)
Draw x(), h(), h(t-),etc.  next slide
1
(1  e  at )u (t )
a
The fact that a>0 is not an issue!
Example 1 – Cont.
y(t); for a=3
y(t)
*
t
t
y (t )  h(t )  x(t )  h(t )  u (t )

U(-(-t))
t<0
U(-(-t))
t>0
y (t )   h( )u (t   )d



 (e
 a

 u ( ) u (t   )d

Remember we are plotting it over 
and t is the variable
1
(

e
d


(e

a
t
 a
0

1
(1  e  at )u (t )
a
 at
 1)
Example 1 – Cont.
Graphical Representation (similar)
a=1
In this case!
http://www.wolframalpha.com/input/?i=convolution+of+two+functions&lk=4&num=4&lk=4&num=4
Example 1 – Cont.
Graphical Representation (similar)
Note in our case
We have u(t) rather
than rectangular
function!
http://www.jhu.edu/signals/convolve/
Example 2
• Consider a CT-LTI system. Assume the impulse response
of the system is h(t)=e^-at for all a>0 and t>0 and input
x(t)= e^at u(-t). Find the output.
x(t)
y(t)
h(t)=e^-at
y (t )  h(t )  x(t )  h(t )  u (t )

y (t ) 
 x( )h(t   )d



 (e
a
(

 u ( ) e  a (t  )  u (t   ) d
Note that for
t>0; x(t) =0; so
the integration
can only be
valid up to t=0

(   (e d  (e  21a e
t  0  e
 at
t
2 a
 at

2 at

1 at
e
2a
(   (e d  (e  21a [1  0]  21a e
t  0  e
 at
0
2 a
 at
 at

 y (t ) 
1 a t
e a  0
2a
Draw x(), h(), h(t-),etc.  next slide
Example – Cont.
x(t)= e^at u(-t)
*
y (t )  h(t )  x(t )  h(t )  u (t )

y (t )   h( )x(t   )d

?
y (t )  h(t )  x(t )  h(t )  u (t )
1 a t
 y (t ) 
e a  0
2a
h(t)=e^-at u(t)
Another Example
notes
Properties of CT LTI Systems
•
When is a CT LTI system memory-less (static)
h(t )  Kd (t )  y (t )  Kx(t )
•
When does a CT LTI system have an inverse system (invertible)?
h(t ) * hi (t )  d (t )
•
When is a CT LTI system considered to be causal? Assuming the
input is causal:
t
t
0
0
y(t )   x( )h(t   )d   h( )x(t   )d
•
When is a CT LTI system considered to be Stable?

notes
y (t ) 


h(t ) dt  
Example
• Is this an stable system?
3t
h(t )  e u (t )

y (t ) 

h(t ) dt  





0
 3t
 3t
e
u
(
t
)
dt

e

 dt  1 / 3
• What about this?
notes
h(t )  e3t u (t )
Differential-Equations Models
• This is a linear first order differential equation with
constant coefficients (assuming a and b are constants)
• The general nth order linear DE with constant equations
is
Is the First-Order DE Linear?
• Consider
• Does a1x1(t)+ a2x2(t) a1y1(t)+ a2y2(t)?
notes
• Is it time-invariant? Does input delay results in an output delay by
the same amount?
notes
• Is this a linear system?
+
X(t)
Sum
e(t)
a
Integrator
Y(t)
Example
• Is this a time invariant linear system?
R
V(t)
Ldi(t)/dt + Ri(t)= v(t)
a= -R/L
b=1/L
y(t)=i(t)
x(t) = V(t)
L
Solution of DE
• A classical model for the solution of DE is
called method of undermined coefficients
– yc(t) is called the complementary or natural
solution
– yp(t) is called the particular or forced solution
Solution of DE
Thus, for
x(t) =constant  yp(t)=P
x(t) =Ce^-7t  yp(t)= Pe^-7t
x(t) =2cos(3t)  yp(t)=P1cos(3t)+P2sin(3t)
Example
• Solve
– Assume x(t) = 2 and y(0) = 4
yc(t) = Ce^-2t; yp(t) = P; P = 1
y(t) = Ce^-2t + 1
y(0) = 4  C = 3  y(t) = 3e^-2t + 1
– What happens if
notes
Schaum’s Examples
• Chapter 2:
– 2, 4-6, 8, 10, 11-14, 18, 19, 48, 52, 53,