## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 2.6.5

Impulse Response of a Simple Second-Order Model

Obtain the unit-impulse response of the following model in two ways: (a) by separation of variables and (b) with the Laplace transform. The initial conditions are x(0−) = 5 and $\overset{.}{x}$(0−) = 10. What are the values of x(0+) and $\overset{.}{x}$(0+)?

$\overset{..}{x}= \delta (t) \quad (1)$

## Verified Solution

a. Let v(t) = ẋ(t). Then equation (1) becomes v̇= δ(t), which can be integrated to obtain v(t) = v(0−) + 1 = 10 + 1 = 11. Thus, ẋ(0+) = 11 and is not equal to ẋ(0−). Now integrate ẋ = v = 11 to obtain x(t) = x(0−) + 11t = 5 + 11t. Thus, x(0+) = 5, which is the same as x(0−).

So for this model the unit-impulse input changes ẋ from t = 0− to t = 0+ but does not change x.

b. The transformed equation is

$s^2X(s)-sx(0-)-\overset{.}{x}(0-)=1$

or

$X(s)=\frac{sx(0-)+\overset{.}{x}(0-)+1}{s^2}=\frac{5s+11}{s^2}=\frac{5}{s}+\frac{11}{s^2}$

which gives the solution x(t) = 5 + 11t and ẋ(t) = 11. Note that the initial values used with the derivative property are the values at t = 0−

The initial-value theorem gives

$x(0+)=\underset{s \rightarrow \infty}{\text{lim}} sX(s)= \underset{s \rightarrow \infty}{\text{lim}} s \frac{5s+11}{s^2}=5$

and because $\mathcal{L}(\overset{.}{x})=sX(s)-x(0-),$

$\overset{.}{x}(0+)= \underset{s \rightarrow \infty}{\text{lim}} s[sX(s)=x(0-)]= \underset{s \rightarrow \infty}{\text{lim}} s (\frac{5s+11}{s}-5)=11$

as we found in part (a).