(a) let a mechanical or electrical system be described by the differential equation$\mathit{A}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{B}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{C}\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{,}\mathit{y}\mathit{o}\mathbf{=}{\mathit{y}}_{\mathbf{0}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$. As in Problem 10.16b, write the solution as a convolution (Assume$\mathit{a}\mathbf{\ne }\mathit{b}$). Let$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$be one of the functions in Figure 11.4 and Problem 5. Find Yand then let$\mathit{n}\mathbf{\to }\mathbf{\infty }$.

(b) Solve (a) with $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{\delta }\mathbf{(}\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{0}}\mathbf{)}$; your result should be the same as in (a).

(c) The solutionY as found in (a) and (b) is called the response of the system to a impulse. Show that the response of a system to a unit impulse at t₀=0is the inverse Laplace transform of the transfer function.

The differential equation $A{y}^{\text{'}\text{'}}+B{y}^{\text{'}}+Cy=f\left(t\right),y0=y{0}^{\text{'}}=0$ and the boundary condition $n\to \infty$.

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t).

The solution using convolution can be given as follows:

$y={\int }_0^{l}g(t-\tau )f\left(\tau \right)d\tau$, where$g\left(t\right)=\frac{1}{A}\left(e^{-at}-e^{-bt}\right)/(b-a)$.

Then,

$$\begin{gathered} y={\int }_0^{t}g(t-\tau )f\left(\tau \right)d\tau \\ y={\int }_{t_0}^{t_0+1/n}\frac{1}{A}\left(\frac{e^{-a}(t-\tau )-e^{-t}\left(\right(t-\tau \left)\right)}{b-a}\right)d\tau \\ y=\frac{\pi }{A(b-a)}{\int }_{t_0}^{t_0+1/n}\left(e^{-a(1-\tau )}-e^{-(t-\tau )}\right)d\tau \\ y=\frac{\pi }{A(b-a)}\left(\frac{\left.e^{-d-p-q}\right)}{a}\left(e^{\frac{\pi }{\pi }}-1\right)+\frac{e^{-\alpha (t-q)}}{b}\left(1-e^{\frac{a}{\pi }}\right)\right) \end{gathered}$$

Now, use Maclaurin series expansion$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\dots$ for $x=a/n$and$x=b/n$.

$$\begin{gathered} n{e}^{d/n}-n=a+\frac{a^2}{2n}+\dots n-n{e}^{-b/n} \\ =-b-\frac{b^2}{2n}+\dots \end{gathered}$$

Then,

$$\begin{gathered} y=\frac{\pi }{A(b-a)}\left(\frac{e^{-ab-b)}}{a}\left(a\right)+\frac{\left.e^{-a\left(\right(-a)}\right)}{b}(-b)\right) \\ y=\frac{\pi }{A(b-a)}\left(\frac{e^{-\alpha -\alpha -b)}}{1}-\frac{e^{-\alpha \left(\alpha -b_0\right)}}{1}\right) \\ y=\frac{n}{A(b-a)}\left(e^{-a\left(t-t_0\right)}-e^{-b\left(t-b_0\right)}\right) \end{gathered}$$

The function $f\left(t\right)=\delta \left(t-t_0\right)$.

$$\begin{gathered} g\left(t\right)=\frac{1}{A}\left(e^{-ut}-e^{-bl}\right)/(b-a)\text{and}f\left(t\right)=\delta \left(t-t_0\right) \\ g\left(t\right)=\frac{1}{A}\left(e^{-ut}-e^{-bt}\right)/(b-a) \\ G\left(p\right)=\frac{1}{A(b-a)}\left(\frac{1}{p+a}-\frac{1}{p+b}\right) \\ =\frac{1}{A}\left(\frac{1}{(p+a)(p+b)}\right) \end{gathered}$$

$$\begin{gathered} f\left(t\right)=\delta \left(t-t_0\right) \\ F\left(p\right)=e^{-t_{0}p} \end{gathered}$$

Now,

$$\begin{gathered} y=g*f \\ y={\int }_0^{t}g(t-\tau )f\left(\tau \right)d\tau \end{gathered}$$

Take Laplace transform on both sides.

role="math" localid="1664351460254" $$\begin{gathered} Y=G\left(p\right)F\left(p\right) \\ y={\int }_0^{l}g(t-\tau )f\left(\tau \right)d\tau \\ Y=G\left(p\right)F\left(p\right)=\frac{1}{A}\left(\frac{1}{(p+a)(p+b)}\right)e^{-t_{0}p} \\ =\frac{1}{A(b-a)}\left(\frac{e^{-b0}}{p+a}-\frac{e^{-to}}{p+b}\right) \end{gathered}$$

Now, apply Inverse Laplace transform.

$y=\frac{1}{A(b-a)}\left(e^{-a\left(t-t_0\right)}-e^{-b\left(l-I_0\right)}\right)$

Replace t₀with 0 in the solution of the differential equation to get:

$y=\frac{1}{\lambda }\left(e^{-at}-e^{-bt}\right)/(b-a)$

Now, take inverse Laplace transform of the transfer function in the given differential equation.

$L^{-1}\left\{\frac{1}{A(p+a)(p+b)}\right\}=\frac{1}{A}\left(e^{-at}-e^{-bl}\right)/(b-a)$