Linear constant-coefficient differential equations - Continuous and Discrete Time Signals and Systems - page 811

Table C.1. Zero-state response corresponding to common input signals

Particular component of the

Input zero-state response

Impulse function, K δ ( t ) C δ ( t )

Unit step function, Ku ( t ) Cu ( t )

− at

Exponential, K e − at C e

Sinusoidal, A cos( ω 0 t + φ ) C 0 cos( ω 0 t ) + C 1 sin( ω 0 t )

response is given by

y (h)

− t − 4 t ,

zs ( t ) =

B 0 e + B 1 e (C.26)

where B 0 and B 1 are constants.

Step 2 Determine the particular component y (p)

zs ( t ) The particular com-
ponent is obtained by consulting Table C.1. For the input signal x ( t ) =
cos tu ( t ), the particular component of the zero-state response is of the form
y (p)

zs ( t ) = C 0 cos t + C 1 sin t for t > 0. Substituting the particular component in
Eq. (C.25) yields

( − 5 C 0 + 3 C 1 ) sin t + (3 C 0 + 5 C 1 ) cos t = 3 cos t . (C.27)

Equating the cosine and sine terms on the left- and right-hand sides of the
equation, we obtain the following simultaneous equations:

− 5 C 0 + 3 C 1 = 0 ,

3 C 0 + 5 C 1 = 3 , (C.28)

= 15 / 34. The particular component y (p)

with solution C 0 = 9 / 34 and C 1

zs ( t )of

the zero-state response is given by

34 cos t + 15

9

y (p)

zs ( t ) =

34 sin t

for t > 0 . (C.29)

zs ( t ) + y (p)

y (h)

Step 3 Determine the zero-state response from y zs ( t ) =

zs ( t ) .
The zero-state response is the sum of the homogeneous and particular com-
ponents, and is given by

− 4 t ) + 9

34 cos t + 15

− t + B 1 e

y zs ( t ) = ( B 0 e

34 sin t ,

(C.30)

where B 0 and B 1 are obtained by inserting zero initial conditions, y (0) = 0 and
y (0) = 0. This leads to the following simultaneous equations:

=− 9

B 0 + B 1

34 ,

(C.31)

15
34 ,

B 0 + 4 B 1 =