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Answers without the blur. Sign up and see all textbooks for free! Q7.3 - 29E

Expert-verified Found in: Page 365 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # 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.

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

See the step by step solution

## 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:

${\mathcal{L}}{\left\{}{f}{\left(}{t}{\right)}{\right\}}{=}{{\int }}_{{0}}^{\frac{<}{>}}{{e}}^{-st}{f}{\left(}{t}{\right)}{d}{t}$

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

## 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:

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

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 $\frac{1}{{s}^{2}+6s+10}$. ### Want to see more solutions like these? 