Differential equations and Laplace transforms: Laplace-transformations
Transfer and response functions
We summarize the use of the Laplace transformation by describing a general method for finding specific solutions of second-order linear equations with constant coefficients.
Consider the initial value problem \[a\cdot x''+b\cdot x'+c\cdot y = g(t)\qquad\text{ with }\qquad x(0) = x_0\quad \text{ and }\quad x'(0)=x'_0\]Let
- \(H(s) = \frac{1}{a\cdot s^2+b\cdot s+c}\) be the transfer function, and
- \(h=\mathcal{L}^{-1}(H)\) the impulse response function of the IVP.
In terms of these functions, the solution of the above IVP is the sum of the following two functions:
- the forced response: the convolution \(g\ast h\); it is the solution of the IVP \(a\cdot x''+b\cdot x'+c\cdot y = g(t)\) with \(x(0)=x'(0)=0\)
- the free response: \( \mathcal{L}^{-1}\left(H(s)\cdot\left((a\cdot s+b)\cdot x_0+a\cdot x_0'\right)\right)\); it is the solution of the IVP \(a\cdot x''+b\cdot x'+c\cdot y = 0\) with \(x(0)=x_0\) and \(x'(0)=x_0'\)
The example below shows how the above transfer and response functions help in solving a second-order linear IVP.
\[ {{d^2}\over{d t^2}} x+3 {{d}\over{d t}} x=3 t+1\qquad \text{ with }\qquad x(0)=1\ \text{ and }\ x'(0)=3\]
\(x(t)=\) \( {{\euler^ {- 3\cdot t }\cdot \left(\left(t^2+4\right)\cdot \euler^{3\cdot t}-2\right)}\over{2}} \)
We begin by determining the transfer function #H(s)# of the IVP. By definition of the transfer function, we have
\[\begin{array}{rcl}
H(s) &=&\dfrac{1}{as^2+bs+c}\\
&&\phantom{xx}\color{blue}{\text{if }a,\ b,\ c\ \text{are constants such that the ODE is of the form}}\\
&&\phantom{xx1234567891011}\color{blue}{ax''+bx'+cx = g(t)}\\
&=&\displaystyle {{1}\over{s^2+3 s}}
\end{array}\]
Thus, \(H(s) =\) \( {{1}\over{s^2+3 s}} \).
We next determine the impulse response function \(h(t)\) of this problem. By definition of the impulse response function and the Inverse Laplace transforms of rational functions, we have
\[\begin{array}{rcl}
h(t) &=&\mathcal{L}^{-1}(H(s))(t)\\
&=&\displaystyle \mathcal{L}^{-1}\left({{1}\over{s^2+3 s}}\right)(t)\\
&&\phantom{xx}\color{blue}{\text{expression found for }H(s)\text{ filled in}}\\
&=&\displaystyle {{1}\over{3}}-{{\euler^ {- 3\cdot t }}\over{3}}\\
&&\phantom{xx}\color{blue}{\text{the inverse Laplace transform of a rational function}}\\
\end{array}\]
Thus, \(h(t) =\) \( {{1}\over{3}}-{{\euler^ {- 3\cdot t }}\over{3}} \).
The Laplace transform of the free reponse is
\[H(s)\cdot\left((a\cdot s+b)\cdot x_0+a\cdot x_0'\right) = {{s+6}\over{s^2+3\cdot s}}\]
The free reponse itself is the inverse Laplace transform of this function:
\[ \mathcal{L}^{-1}\left(H(s)\cdot\left((a\cdot s+b)\cdot x_0+a\cdot x_0'\right)\right) = 2-\euler^ {- 3\cdot t }\]
In order to determine the forced response, we compute the following convolution product:
\[\begin{array}{rcl} (g\ast h)(t) &=&\displaystyle \int_0^t h(\tau)\cdot g(t-\tau)\,\dd \tau\\
&=&\displaystyle \int_0^t -{{\euler^ {- 3\cdot \tau }\cdot \left(\left(3\cdot \tau-3\cdot t-1\right)\cdot \euler^{3\cdot \tau}-3\cdot \tau+3\cdot t+1\right)}\over{3}}\,\dd \tau\\
&=&\displaystyle \left[ -{{\euler^ {- 3\cdot \tau }\cdot \left(\left(3\cdot \tau^2+\left(-6\cdot t-2\right)\cdot \tau\right)\cdot \euler^{3\cdot \tau}+2\cdot \tau-2\cdot t\right)}\over{6}}\right]_{\tau=0}^{\tau=t}\\ &=&\displaystyle{{-3\cdot t^2-\left(-6\cdot t-2\right)\cdot t}\over{6}}-{{t}\over{3}}\\ &=&\displaystyle {{t^2}\over{2}}
\end{array}\]
Adding the free and forced response, we find the solution of the IVP:
\[\begin{array}{rcl} x(t)& =& \displaystyle 2-\euler^ {- 3\cdot t }+{{t^2}\over{2}}\\ &=& \displaystyle {{\euler^ {- 3\cdot t }\cdot \left(\left(t^2+4\right)\cdot \euler^{3\cdot t}-2\right)}\over{2}}\end{array}\]
Or visit omptest.org if jou are taking an OMPT exam.