This preview shows pages 1–2. Sign up to view the full content.
, April 7, 2010
Convolution, or, the superposition of impulse response
1. General equivalent initial conditions for delta response
2. Two implications of LTI
3. Superposition of unit impulse responses
4. Convolution product
1. Equivalent for general deltaresponse ( t > 0 ) :
p(D) x = a_n x^(n) + .
... + a_0 x
( a_n
not zero)
p(D) w = delta(t)
with rest initial conditions specifies the
*unit impulse response*.
For
t > 0 , this equation is equivalent (for
t > 0 ) to
a_n x^(n) + .
.. + a_0 x = 0
with initial conditions
x(0) = .
.. = x^(n2)(0) = 0 ,
x^(n1)(0) = 1/a_n .
Ex:
n = 2 :
mx" + bx' + kx = delta(t) , rest initial conditions,
is equivalent to
mx" + bx' + kx = 0 ,
x(0) = 0 ,
x'(0) = 1/m .
[2] We work with a system modeled by an LTI operator
p(D) = a_n D^n + .
...
The time invariance has two important consequences:
(a) If
p(D) x
=
q(t) ,
then
p(D) x'
=
q'(t) .
This is because
D p(D) = p(D) D (because the coefficients are constant)
so
p(D) x' = p(D) Dx = D p(D) x = D q = q' .
In particular,
since
u'(t) = delta(t),
v'(t) = w(t) :
the derivative of the unit step response is the unit impulse response.
e.g.
D v = D ( u(t) (1/k) ( 1  e^{kt}) ) = e^{kt} = w .
(Notice that
v
is continuous: there are no delta terms in its
generalized derivative.)
(b) It doesn't really matter when you start the clock.
So, for example, we found that
the unit impulse response for
D+kI
is
w(t) = u(t) e^{kt}
and the unit step response is
v(t) = u(t) (1/k) ( 1  e^{kt} ) .
By time invariance the solution to
x' + kx = delta(ta)
with
x(t) = 0
for
t < a
is
x
=
u(ta) e^{k(ta)}) .
[3] We learn about a system by studying how it responds to various input
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09
 vogan

Click to edit the document details