Signals and Systems - Electrical Engineering

9.5 One-Sided Z-Transform Inverse 559

so that

y[n]=

(

−0.5−

1

6

(− 1 )n+

8

3

2 n

)

u[n]

which as expected will grow asnincreases — there is no steady-state response.

In a problem like this the chance of making computational errors is large, so it is important to

figure out a way to partially check your answer. In this case we can check the value ofy[0] using the

difference equation, which isy[0]= −y[−1]+ 4 y(− 2 )+ 4 y(− 3 )+ 3 = − 1 + 3 =2, and compare

it with the one obtained using our solution, which givesy[0]=− 3 / 6 − 1 / 6 + 16 / 6 =2. They

coincide. Another way to partially check your answer is to use the initial and the final values

theorems. n

Solution of Differential Equations
The solution of differential equations requires converting them into difference equations, which can
then be solved in a closed form by means of the Z-transform.

nExample 9.20

Consider an RLC circuit represented by the second-order differential equation

d^2 vc(t)

dt^2

+

dvc(t)

dt

+vc(t)=vs(t)

where the voltage across the capacitorvc(t)is the output and the sourcevs(t)=u(t)is the input.

Let the initial conditions be zero. Find the voltage across the capacitorvc(t).

Solution

The Laplace transform of the output is found from the differential equation as

Vc(s)=

Vs(s)

1 +s+s^2

=

1

s(s^2 +s+ 1 )

=

1

s((s+0.5)^2 + 3 / 4 )

where the final equation is obtained after replacingVs(s)= 1 /s. The solution of the differential

equation is of the general form

vc(t)=[A+Be−0.5tcos(

√

3 / 2 t+θ)]u(t)

for constantsA,B, andθ. To convert the differential equation into a difference equation we

approximate the first derivative as

dvc(t)

dt

≈

vc(t)−vc(t−Ts)

Ts