## Chapter 2

## Q. 2.6.6

## Q. 2.6.6

Impulse Response of a Second-Order Model

Obtain the unit-impulse response of the following model. The initial conditions are x(0−)=0, ẋ(0−) = 0. What are the values of x(0+) and ẋ(0+)?

\frac{X(s)}{F(s)}=\frac{1}{2s^2+14s+20}

## Step-by-Step

## Verified Solution

Because f(t) = 𝛿(t), F(s) = 1, and

X(s)=\frac{1}{2s^2+14s+20}F(s)=\frac{1}{2s^2+14s+20}=\frac{1}{6}\frac{1}{s+2}-\frac{1}{6}\frac{1}{s+5}

The response is x(t)=(e^{-2t}-e^{-5t})/6. This gives

x(0+)= \underset{t \rightarrow 0+}{\text{lim}} x(t)= \underset{t \rightarrow 0+}{\text{lim}} (\frac{e^{-2t}-e^{-5t}}{6})=0

and

\overset{.}{x}(0+)=\underset{t \rightarrow 0+}{\text{lim}} \overset{.}{x}(t) = \underset{t \rightarrow 0+}{\text{lim}} (\frac{-2e^{-2t}+5e^{-5t}}{6})=\frac{1}{2}

So the impulse input has not changed x between t = 0− and t = 0+ but has changed ẋ from 0 to 1/2. These results could have been obtained from the initial-value theorem:

x(0+) = \underset{t \rightarrow \infty}{\text{lim}} sX(s)= \underset{t \rightarrow \infty}{\text{lim}} s \frac{1}{2s^2+14s+20}=0

and, noting that x(0−) = 0,

\overset{.}{x}(0+)= \underset{t \rightarrow \infty}{\text{lim}} s [sX(s)-x(0-)]= \underset{t \rightarrow \infty}{\text{lim}} s \frac{s}{2s^2+14s+20}= \frac{1}{2}