# Evaluating Continuous Transfer Functions

The transfer function can be evaluated using different inputs. We commonly use the impulse, step and sinusoidal input functions.$$\require{AMSsymbols}\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \def\fourier{\lfz{\mathcal{F}}} \def\ztransform{\lfz{\mathcal{Z}}} \require{cancel} \newcommand\ccancel[2][black] {\color{#1}{\cancel{\color{black}{#2}}}}$$

Let $$H$$ be a stable system with transfer function $$H(s)$$, input signal $$x(t)$$, and output $$y(t)$$. In this, “stable” implies that the poles are in left half of $$s$$-plane.

## Impulse Response

Earlier, we derived the Laplace transform for the impulse function as
$$\delta(t)\laplace\Delta(s)=1$$

Substituting this input function
$$Y(s)=H(s)\,\Delta(s)$$

The response to an impulse input function, is the transfer function $$H(s)$$ itself
$$\shaded{Y(s)=H(s)}$$

## Unit Step Response

The page Laplace Transforms gives the Laplace transform for the unit step function as
$$\gamma(t)\laplace\Gamma(s)\frac{1}{s}$$

Substitute the input function
$$Y(s)=H(s)\,\Gamma(s)$$
The response to an unit step input function follows as
$$\shaded{Y(s)=\frac{H(s)}{s}}$$

## Frequency Response

The frequency response is defined as the steady state response of system to a sinusoidal input.

Given sinusoidal input and transfer function
\left\{\begin{align} x(t)&=\sin(\omega t)\gamma(t) &\mathrm{input\ signal}\nonumber\\ H(s)&=|H(s)|\,e^{j\angle H(s)} &\mathrm{transfer\ function}\nonumber \end{align}\right.

Find the output signal $$y(t)$$, using the Laplace transform of the sinodial input function
\left.\begin{align} X(s)&=\frac{\omega}{s^2+\omega^2} \nonumber\\[8mu] Y(s)&=H(s)\,X(s)\nonumber \end{align} \right\}

Substitute $$X(s)$$ in $$Y(s)$$
\begin{align} Y(s)&=H(s)\,\frac{\omega}{s^2+\omega^2}\nonumber\\ &=H(s)\,\frac{\omega}{(s+j\omega)(s-j\omega)} \end{align}

According to Heaviside, this can be expressed as partial fractions [swarthmore, MIT-cu], where the term $$C_h(s)$$ represents the transient response resulting from $$H(s)$$. This term is independent of $$j\omega$$, and dies out for $$t\to\infty$$.
$$Y(s)=\frac{\omega}{(s+j\omega)(s-j\omega)}H(s)=\underbrace{\frac{c_0}{s+j\omega}+\frac{c_1}{s-j\omega}}_{Y_{ss}(s)=\mathrm{steady\ state\ response}}+\underbrace{C_h(s)}_\mathrm{transient\ response}\label{eq:partialfractions}$$

Find $$c_0$$
\begin{align} &H(s)\,\frac{\omega\cancel{(s+j\omega)}}{\cancel{(s+j\omega)}(s-j\omega)}=\frac{c_0\cancel{(s+j\omega)}}{\cancel{s+j\omega}}+\frac{c_1(s+j\omega)}{s-j\omega}+C_h(s)(s+j\omega) \nonumber\\ \Rightarrow\,& \left.H(s)\,\frac{\omega}{s-j\omega}=c_0+c_1\frac{s+j\omega}{s-j\omega}+C_h(s)(s+j\omega)\right|_{s=-j\omega} \nonumber\\ \Rightarrow\,& H(-j\omega)\,\frac{\omega}{-j\omega-j\omega}=c_0+c_1\frac{\cancelto{0}{-j\omega+j\omega}}{s-j\omega}+C_h(j\omega)(\cancelto{0}{-j\omega+j\omega}) \nonumber\\ \Rightarrow\,& c_0=H(-j\omega)\,\frac{\cancel{\omega}}{-2j\cancel{\omega}} =\frac{H(-j\omega)}{-2j} \end{align}

Similarly, find $$c_1$$
\begin{align} &H(s)\,\frac{\omega\cancel{(s-j\omega)}}{(s+j\omega)\cancel{(s-j\omega)}}=\frac{c_0(s-j\omega)}{s+j\omega}+\frac{c_1\cancel{(s-j\omega)}}{\cancel{s-j\omega}}+C_h(s)(s-j\omega) \nonumber\\ \Rightarrow\,& \left.H(s)\,\frac{\omega}{s+j\omega}=c_0\frac{s-j\omega}{s-j\omega}+c_1+C_h(s)(s-j\omega)\right|_{s=j\omega} \nonumber\\ \Rightarrow\,& H(j\omega)\,\frac{\omega}{j\omega+j\omega}=c_0\frac{\cancelto{0}{j\omega-j\omega}}{s-j\omega}+c_1+C_h(j\omega)(\cancelto{0}{j\omega-j\omega}) \nonumber\\ \Rightarrow\,& c_1=H(j\omega)\,\frac{\cancel{\omega}}{2j\cancel{\omega}} =\frac{H(j\omega)}{2j} \end{align}

Inverse Laplace transform of $$\eqref{eq:partialfractions}$$ back to the time domain
$$y(t)=\underbrace{c_0e^{-j\omega t}+c_1e^{j\omega t}}_{y_{ss}(t)=\mathrm{steady\ state\ response}}+\underbrace{\cancelto{0\mathrm{\ as\ }t \to\infty}{\mathfrak{L}^{-1}\left\{C_h(s)\right\}}}_{\mathrm{transient\ response}}$$

Substitute $$c_0$$ and $$c_1$$ in $$y_{ss}(t)$$
\begin{align} y_{ss}(t)&=\frac{H(-j\omega)}{-2j}e^{-j\omega t}+\frac{H(j\omega)}{2j}e^{j\omega t}\nonumber\\ &=\frac{H(j\omega)e^{j\omega t}-H(-j\omega)e^{-j\omega t}}{2j}\label{eq:yss} \end{align}

Based on Euler’s formula we can express $$H(s)$$ in polar coordinates

\left\{ \begin{align} H(s)&=|H(s)|\,e^{j\angle H(s)}\nonumber\\ |H(j\omega)| &= K \frac{\prod_{i=1}^m \sqrt{\left(\Re\{{z_i}\}\right)^2+\left(\omega+\Im\{z_i\}\right)^2}}{\prod_{i=1}^n \sqrt{\left(\Re\{{p_i}\}\right)^2+\left(\omega+\Im\{p_i\}\right)^2}} \nonumber\\ \angle{H(s)} &= \sum_{i=1}^m\mathrm{atan2}\left(\omega+\Im\{z_i\}, \Re\{{z_i}\}\right) -\sum_{i=1}^n\mathrm{atan2}\left(\omega+\Im\{p_i\}, \Re\{{p_i}\}\right)\nonumber\\ \end{align} \right.\label{eq:euler}

Substitute the polar representation of $$H(s)$$, in $$y_{ss}$$
\begin{align} y_{ss}(t) &=|H(j\omega)|\left( \frac{e^{j\angle H(j\omega)}e^{j\omega t}-e^{j\angle H(-j\omega)}e^{-j\omega t}}{2j} \right) \nonumber\\ &=|H(j\omega)|\left( \frac{e^{j\left(\angle H(j\omega)+\omega t\right)}-e^{j\left(\angle H(-j\omega)-\omega t\right)}}{2j} \right) \nonumber\\ &=|H(j\omega)|\, \underbrace{\frac{e^{j(\omega t + \angle{H(j\omega)})}-e^{-j(\omega t + \angle H(j\omega))}}{2j}}_{\sin(\omega t+\angle H(j\omega))} \end{align}

In this, we recognize the Laplace transfer for a sinodial function
$$\sin(\omega t)\gamma(t) \laplace \frac{\omega}{s^2+\omega^2}$$

The frequency response follows as
$$\shaded{ y_{ss}(t)=|H(j\omega)|\,\sin(\omega t+\angle H(j\omega))\,\gamma(t)}$$

In other words, for a linear system a sinusoidal input generates a sinusoidal output with the same frequency, but different amplitude and phase.