Physically realizable, linear time-invariant systems can be
described by a set of linear differential equations (LDEs):
dndtnyt+
a
n
−
1
dn−1dtn−1yt+…+
a
1
ddtyt+
a
0
yt=
b
m
dmdtmft+…+
b
1
ddtft+
b
0
ft
n
t
y
t
a
n
−
1
n
1
t
y
t
…
a
1
t
y
t
a
0
y
t
b
m
m
t
f
t
…
b
1
t
f
t
b
0
f
t
Equivalently,
∑i=0n
a
i
didtiyt=∑i=0m
b
i
didtift
i
n
0
a
i
i
t
y
t
i
m
0
b
i
i
t
f
t
(1)
with
a
n
=1
a
n
1
.
It is easy to show that these equations define a system that is
linear and time invariant. A natural question to ask, then, is
how to find the system's output response
yt
y
t
to an input
ft
f
t
. Recall that such a solution can be written as
yt=
y
i
t+
y
s
t
y
t
y
i
t
y
s
t
We refer to
y
i
t
y
i
t
as the zero-input response -- the
homogeneous solution due only to the initial conditions of the
system. We refer to
y
s
t
y
s
t
as the zero-state response -- the
particular solution in response to the input
ft
f
t
. We now discuss how to solve for each of these
components of the system's response.
The zero-input response,
y
i
t
y
i
t
, is the system response due to initial conditions
only.
Close the switch in the circuit pictured
in Figure 2 at time t=0 and then leave everything else alone. The voltage response is shown in Figure 3.
Imagine a mass attached to a spring, as shown in Figure 4. When
you pull the mass up and let it go, you have an example
of a zero-input response. A plot of this response is shown in Figure 5.
There is no input, so we solve for
y
0
t
y
0
t
such that
∀
a
n
,
a
n
=1:∑i=0n
a
i
didti
y
0
t=0
a
n
a
n
1
i
n
0
a
i
i
t
y
0
t
0
(2)
If
DD is the derivative
operator, we can write the previous equation as:
Dn+
a
n
−
1
Dn−1+…+
a
0
y
0
t=0
D
n
a
n
−
1
D
n
1
…
a
0
y
0
t
0
(3)
Since we need the weighted sum of a bunch of
y
0
t
y
0
t
's derivatives to be
00
for all tt, then
y
0
tddt
y
0
td2dt2
y
0
t…
y
0
t
t
y
0
t
2
t
y
0
t
…
must all be of the same form.
Only the exponential,
ⅇst
s
t
where
s∈ℂ
s
, has this property (see your Differential
Equation's textbook for details). So we must assume that,
∀c,c≠0:
y
0
t=cⅇst
c
c
0
y
0
t
c
s
t
(4)
for some
cc and
ss.
Since
ddt
y
0
t=csⅇst
t
y
0
t
c
s
s
t
,
d2dt2
y
0
t=cs2ⅇst
2
t
y
0
t
c
s
2
s
t
, … we have
Dn+
a
n
−
1
Dn−1+…+
a
0
y
0
t=0
D
n
a
n
−
1
D
n
1
…
a
0
y
0
t
0
csn+
a
n
−
1
sn−1+…+
a
1
s+
a
0
ⅇst=0
c
s
n
a
n
−
1
s
n
1
…
a
1
s
a
0
s
t
0
(5)
Equation 5 holds for all
tt
only when
sn+
a
n
−
1
sn−1+…+
a
1
s+
a
0
=0
s
n
a
n
−
1
s
n
1
…
a
1
s
a
0
0
(6)
Where this equation is referred to as the
characteristic
equation of the system. The possible values of
ss are the roots of this polynomial
s
1
s
2
…
s
n
s
1
s
2
…
s
n
s−
s
1
s−
s
2
s−
s
3
…s−
s
n
=0
s
s
1
s
s
2
s
s
3
…
s
s
n
0
i.e. possible solutions are
c
1
ⅇ
s
1
t
c
1
s
1
t
,
c
2
ⅇ
s
2
t
c
2
s
2
t
,
……,
c
n
ⅇ
s
n
t
c
n
s
n
t
. Since the system is
linear, the
general solution if of the form:
y
0
t=
c
1
ⅇ
s
1
t+
c
2
ⅇ
s
2
t+…+
c
n
ⅇ
s
n
t
y
0
t
c
1
s
1
t
c
2
s
2
t
…
c
n
s
n
t
(7)
Then, solve for the
c
1
…
c
n
c
1
…
c
n
using the initial conditions.
See Lathi p.108 for a good example!
We generally assume that the IC's of a system are zero,
which implies
y
i
t=0
y
i
t
0
. However, the method of
solving for
y
i
t
y
i
t
will prove useful later on.
Solving a linear differential equation
∑i=0n
a
i
didtiyt=∑i=0m
b
i
didtift
i
n
0
a
i
i
t
y
t
i
m
0
b
i
i
t
f
t
(8)
given a specific input
ft
f
t
is a difficult task in general. More importantly,
the method depends entirely on the nature of
ft
f
t
; if we change the input signal, we must completely
re-solve the system of equations to find the system
response.
Convolution helps
to bypass these difficulties. In section 2, we explain how
convolution helps to determine the system's output, given
only the input
ft
f
t
and the system's impulse response,
ht
h
t
.
Before deriving the convolution procedure, we show that a
system's impulse response is easily derived from its linear,
differential equation (LDE). We will show the derivation
for the LDE below, where
m<n
m
n
:
dndtnyt+
a
n
−
1
dn−1dtn−1yt+…+
a
1
ddtyt+
a
0
yt=
b
m
dmdtmft+…+
b
1
ddtft+
b
0
ft
n
t
y
t
a
n
−
1
n
1
t
y
t
…
a
1
t
y
t
a
0
y
t
b
m
m
t
f
t
…
b
1
t
f
t
b
0
f
t
(9)
We can rewrite
Equation 9 as
Q
D
yt=
P
D
ft
Q
D
y
t
P
D
f
t
(10)
where
Q
D
·
Q
D
·
is an operator that maps
yt
y
t
to the left hand side of
Equation 9
Q
D
yt=dndtnyt+
a
n
−
1
dn−1dtn−1yt+…+
a
1
ddtyt+
a
0
yt
Q
D
y
t
n
t
y
t
a
n
−
1
n
1
t
y
t
…
a
1
t
y
t
a
0
y
t
(11)
and
P
D
·
P
D
·
maps
ft
f
t
to the right hand side of
Equation 9. Lathi shows (in Appendix 2.1) that the
impulse response of the system described by
Equation 9 is given by:
ht=
b
n
δt+
P
D
y
n
tμt
h
t
b
n
δ
t
P
D
y
n
t
μ
t
(12)
where for
m<n
m
n
we have
b
n
=0
b
n
0
. Also,
y
n
y
n
equals the zero input response with initial
conditions.
y
n
−
1
0=1
y
n
−
2
0=1…y0=0
y
n
−
1
0
1
y
n
−
2
0
1
…
y
0
0
"My introduction to signal processing course at Rice University."