Q7.3 - 29E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The transfer function of a linear system is defined as the ratio of the Laplace transform of the output function y(t) to the Laplace transform of the input function g(t), when all initial conditions are zero. If a linear system is governed by the differential equation

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{t}\mathbf{>}\mathbf{0}$

use the linearity property of the Laplace transform and Theorem 5 on page363 on the Laplace transform of higher-order derivatives to determine the transfer function $\mathbf{H}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{Y}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{/}\mathbf{G}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$of this system.

Step-by-Step Solution

Answer

The value of transfer function $H\left(s\right)=Y\left(s\right)/G\left(s\right)$ of this system is $\frac{1}{s^2+6s+10}$.

Define Laplace transform

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. Laplace transform $\mathcal{L}$of a function f(t) is defined as:

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^{\frac{<}{>}}{e}^{-st}f\left(t\right)dt$

In words, we can describe this expression as the Laplace transform of f(t) equals function F of s, that is, $\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$.

Find the transfer function

Consider the differential equation $y\text{'}\text{'}\left(t\right)+6y\text{'}\left(t\right)+10y\left(t\right)=g\left(t\right)$.

 

Rewrite the equation as:

$g\left(t\right)=y\text{'}\text{'}\left(t\right)+6y\text{'}\left(t\right)+10y\left(t\right)$

Find the Laplace transform of $g\left(t\right)=y\text{'}\text{'}\left(t\right)+6y\text{'}\left(t\right)+10y\left(t\right)$ using $\mathcal{L}\left\{f\left(t\right)\right\}=F\left(s\right)$ as:

$\begin{array}{rcl}\mathcal{L}\left\{g\left(t\right)\right\}\left(s\right)& =& \mathcal{L}\left\{y\text{'}\text{'}\left(t\right)+6y\text{'}\left(t\right)+10y\left(t\right)\right\}\left(s\right)\\ G\left(s\right)& =& \mathcal{L}\left\{y\text{'}\text{'}\left(t\right)\right\}\left(s\right)+6\mathcal{L}\left\{y\text{'}\left(t\right)\right\}\left(s\right)+10\mathcal{L}\left\{y\right(t\left)\right\}\left(s\right)\\ G\left(s\right)& =& \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+6\left[sY\right(s)-y(0\left)\right]+10Y\left(s\right)\\ G\left(s\right)& =& s^{2}Y\left(s\right)+6sY\left(s\right)+10Y\left(s\right)\end{array}$

Simplify the equation as:

$G\left(s\right)=\left(s^2+6s+10\right)Y\left(s\right)$

Since $G\left(s\right)=\left(s^2+6s+10\right)Y\left(s\right)$ and transfer function is $H\left(s\right)=Y\left(s\right)/G\left(s\right)$,

$H\left(s\right)=\frac{1}{s^2+6s+10}$.

Hence, the value of transfer function $H\left(s\right)=Y\left(s\right)/G\left(s\right)$ is $\frac{1}{s^2+6s+10}$.