Math 344, April 13, 2012
The Impulse Response and Convolution
1
The Impulse Response Function
The impulse response function for the linear system m y ''C b y 'C k y = f t is the solution when the driver
f t is the Dirac delta at t = 0 and the initial conditions are y 0 = 0, y ' 0 = 0. In that case, the transform of
the ode is
m s2 Y C b s Y C k Y = 1
1
so the impulse response function is h t = LK1
. If we define Z s = m s2 C b s C k , (the
2
m s Cb sCk
1
characteristic polynomial of the system) and let H s =
, then h t = LK1 H s .
Z s
Example 1. The impulse response function for the overdamped system y ''C 3 y 'C 2 y = f t is
1
h1 d unapply invlaplace 2
, s, t , t = t/KeK2 t C eKt
s C3 sC2
The impulse response function for the undamped system y ''C 2 y = f t is
1
1
h2 d unapply invlaplace 2
, s, t , t = t/
2 sin
2 t
2
s C2
Their graphs are displayed below.
display Array plot h1 t , t = 0 ..5, caption = "Overdamped" ,
plot h2 t , t = 0 ..5, caption = "Undamped"
0.2
0
0
1
2
3
t
Overdamped
4
5
1
2
t
Undamped
3
4
5
The Solution to a Linear System using the Impulse Response and the Convolution Integral
Consider now the system m y ''C b y 'C k y = f t with generic initial conditions
y 0 = y0 and y ' 0 = v0.
y0 is the initial position of the mass and v0 is its initial velocity. The transform of the ode is
m s2 Y K s y0 K vo C b s Y K y0 C k Y = F s
or
m s2 C b s C k Y K m s y0 K m v0 Kb y0 = F s
so
Consequently,
Y = m y0 s C m v0 C b y0 $H s C F s $H s .
y t = m y0 LK1 s H s
C m v0 C b y0 LK1 H s
C LK1 F s H s
.
Using the facts that LK1 H s = h t and LK1 s H s = h ' t , along with the Convolution Theorem to
handle LK1 F s H s , we obtain the following solution formula directly in terms of the forcing function
f t and the impulse response function h t .
t
f t h t K t dt
y t = m y0 h ' t C m v0 C b y0 h t C
0
Math 344, April 13, 2012
The Impulse Response and Convolution
2
Example 2. Consider the system y ''C 3 y 'C 2 y = sin t . In Example 1 its impulse response function was
found to be h t =KeK2 t C eKt. The driver is f t = sin t so the solution satisfying the initial conditions
y 0 = y ' 0 = 0, the so-called zero-state response, ZSR, is simply the convolution product of the driver
sin t with the impulse response function KeK2 t C eKt. Maple calculates it as follows.
t
tK t
sin t $ KeK2
C eK tK t
dt
0
1
10
K2 C 5 et K 3 e2 t cos t C e2 t sin t
eK2 t
(1)
0.2
It's graph: plot (1), t = 0 ..20 =
0.0
5
K0.2
10
t
15
20
Example 3. For the undamped system y ''C 2 y = sin t the impulse response function is
1
h t d
2 sin
2 t : . Therefore, the zero-state response is the convolution integral displayed (and
2
evaluated) below.
t
sin t $
0
1
2
2 sin
2
tKt
dt
1
2
K
2 sin
2 t C sin t
(2)
Its graph: plot (2), t = 0 ..40 =
1
0
10
20
t
K1
30
40
Example 4. If the undamped system in Example 3 is set into motion from y 0 = 2 with initial velocity
y ' 0 = 3, then the response is
2$h ' t C 3$h t C (2)
2 cos
2 t C
2 sin
2 t C sin t
(3)
Now the graph is plot (3), t = 0 ..40 =
2
0
K2
10
20
t
30
40