Digital Signal Processing Reference
In-Depth Information
Examples are now given that illustrate the convolution integral for certain
systems.
Impulse response of an integrator
EXAMPLE 3.2
Consider the system of Figure 3.3. The system is an integrator, in which the output is the in-
tegral of the input:
t
y(t) =
L
x(t) dt.
(3.20)
-
q
This equation is the mathematical model of the system. We use the integral symbol in a block
to denote the integrator. The system is practical and can be realized as an electronic circuit
with an operational amplifier, a resistor, and a capacitor, as described in Section 1.2. Inte-
grating amplifiers of this type are used extensively in analog signal processing and in closed-
loop control systems.
We see that the impulse response of this system is the integral of the unit impulse func-
tion, which is the unit step function:
t
0,
t 6 0
b
h(t) =
L
d(t) dt = u(t) =
t 7 0
.
1,
-
q
We will now use the convolution integral to find the system response for the unit ramp input,
From (3.14),
x(t) = tu(t).
q
y(t) = x(t)*h(t) = tu(t)*u(t) =
L
tu(t)u(t - t) dt.
-
q
In this integral,
t
is considered to be constant. The unit step
u(t)
is zero for
t 6 0;
hence, the
lower limit on the integral can be increased to zero with
u(t)
removed from the integrand:
q
y(t) =
L
tu(t - t) dt.
0
In addition, the unit step
u(t - t)
is defined as
0,
t 7 t
u(t - t) =
b
t 6 t
.
1,
The upper limit on the integral can then be reduced to
t
, and thus,
t
t
2
2
t
t
2
2
u(t).
y(t) =
L
t dt =
=
0
0
Integrator
x
(
t
)
y
(
t
)
t
y
(
t
)
x
( )
d
Figure 3.3
System for Example 3.2.
