Time-domain analysis of LTIC systems - Continuous and Discrete Time Signals and Systems - page 108

where C is a constant. The particular component of the zero-state response of
Eq. (3.11) for input x ( t ) = sin(2 t ) is of the following form:

y (p)
zs

( t ) = K 1 cos(2 t ) + K 2 sin(2 t ) .

Substituting the particular component in Eq. (3.11) gives K 1 = 0 . 4 and K 2 =

0 . 2. The overall zero-state response of the system is as follows:

− 4 t

zero state response y zs ( t ) = C e + 0 . 2 sin(2 t ) + 0 . 4 cos(2 t ) ,

with zero initial condition, i.e. y zs ( t ) = 0. Substituting the initial condition in
the zero-state response yields C =− 0 . 4. The total response of the system is
the sum of the zero-input and zero-state responses and is given by

− 4 t + 0 . 2 sin(2 t ) + 0 . 4 cos(2 t ) .

y ( t ) = 1 . 6e (3.12)

Theorem 3.1 states the total response of a LTIC system modeled with a first-
order, constant-coefficient, linear differential equation.

Theorem 3.1 The output of a first-order differential equation,
d y
d t

+ f ( t ) y ( t ) = r ( t ) , (3.13)

resulting from input r(t ) is given by

− p

e p r d t + c

y ( t ) = e , (3.14)

where function p is given by

p ( t ) = f ( t )d t (3.15)

and c is a constant.

Using Theorem 3.1 to solve Eq. (3.11), we obtain p ( t ) = ∫ 4d t = 4 t . Substi-

tuting p ( t ) = 4 t into Eq. (3.14), we obtain

− 4 t

e 4 t 2 cos(2 t )d t + c y ( t ) = e ,

where the integral simplifies to (see Section A.5 of Appendix A)

= 2

2 2 + 4 2 [4e 4 t cos(2 t ) + 2e 4 t sin(2 t )] .
Based on Theorem 3.1, the output is therefore given by

e 4 t cos(2 t )d t 2

− 4 t + 0 . 2 sin(2 t ) + 0 . 4 cos(2 t ) .

y ( t ) = c e

The value of constant c in the above equation can be computed using the initial
condition. Substituting y (0 ) = 2 V gives c = 1 . 6. The result is, therefore, the

same as the solution in Eq. (3.12) obtained by following the formal procedure
outlined in Appendix C.

−