Linear constant-coefficient differential equations - Continuous and Discrete Time Signals and Systems - page 809

where A 0 , A 1 , and A 2 are constants determined from the given initial conditions.
Substituting the initial conditions into Eq. (C.16) yields

A 0 + A 1 = 4 ,

− A 0 + A 1 − 2 A 2 =− 5 , (C.17)

A 0 − 2 A 1 + 4 A 2 = 9 ,

= 1, A 1 = 2, and A 2 = 3. The zero-input response for

which has solution A 0

Eq. (C.14) is therefore given by

− t − 2 t − 2 t .

y zi ( t ) = e + 2e + 3 t e (C.18)

C.1.2 Complex roots

Solving a characteristic equation may give rise to complex roots of the form
s = a + j b . Typically, a homogeneous differential equation, Eq. (C.3), with real
coefficients, has complex roots in conjugate pairs. In other words, if s = a + j b
is a root of the characteristic equation obtained from Eq. (C.3) then s = a − j b
must also be a root of the characteristic equation. For such complex roots, the
zero-input response can be modified to the following form:

A 0 e at cos( bt ) + A 1 e at sin( bt ) . y zi ( t ) = (C.19)

Example C.3
Compute the zero-input response of a system represented by the following
differential equation:

d 4 y
d t 4

+ 2 d 2 y
d t 2

+ 1 = x ( t ) , (C.20)

with initial conditions y (0) = 2 , y (0) = 2, y (0) = 0, y (0) =− 4.

Solution
Substituting y zi ( t ) = A e st in the homogeneous representation for Eq. (C.20)
results in the following characteristic equation:

s 4 + 2 s 2 + 1 = 0 . (C.21)

The roots of the characteristic equation are given by s = j, j, − j, and − j. Note
that the roots are not only complex but also repeated. The zero-input solution
is given by

y zi ( t ) = A 0 cos( t ) + A 1 t cos( t ) + A 2 sin( t ) + A 3 t sin( t ) , (C.22)