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# Evaluating Transfer Functions

The transfer function can be evaluated using different inputs.  We commonly use the impulse, step and sinusoidal input functions.  

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

### Impulse response

The page Laplace Transforms gives the Laplace transform for the impulse function as

$$\Delta(s)=1$$

Substituting this input function

\begin{aligned} Y(s)&=H(s)\,\Delta(s) \end{aligned}

The response to an impulse input function, is the transfer function $$H(s)$$ itself

\def\doubleunderline#1{\underline{\underline{#1}}} \doubleunderline{ \begin{aligned} Y(s)=H(s) \end{aligned} }\tag{1}

### Unit step response

The page Laplace Transforms gives the Laplace transform for the unit step function as

\begin{aligned} \Gamma(s)\,&=\frac{1}{s} \end{aligned}

Substitute the input function

\begin{aligned} Y(s)&=H(s)\,\Gamma(s) \end{aligned}

The response to an unit step input function follows as

\def\doubleunderline#1{\underline{\underline{#1}}} \doubleunderline{ \begin{aligned} Y(s)=\frac{H(s)}{s} \end{aligned} }\tag{2}

### Frequency Response

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

Given sinusoidal  input and transfer function

\begin{aligned} u(t)&=\sin(\omega t)\gamma(t) & \mathrm{input\ signal}\\ H(s)&=|H(s)|\,e^{j\angle H(s)} & \mathrm{transfer\ function} \end{aligned}

Find the output signal $$y(t)$$, using the Laplace transform of the sinodial input function

\left. \begin{aligned} U(s)&=\frac{\omega}{s^2+\omega^2} \\ Y(s)&=H(s)\,U(s) \end{aligned} \right\}

Substitute $$U(s)$$ in $$Y(s)$$

\begin{aligned} Y(s)&=H(s)\,\frac{\omega}{s^2+\omega^2}\\ &=H(s)\,\frac{\omega}{(s+j\omega)(s-j\omega)} \end{aligned}

According to Heaviside, this can be expressed as partial fractions [swarthmoreMIT-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 $$t\to\infty$$.

$$Y(s)=\frac{\omega}{(s+j\omega)(s-j\omega)}H(s)=\underbrace{\frac{c_0}{s+j\omega}+\frac{c_1}{s-j\omega}}_{Y_{ss}(s)=\mathrm{steady\ state\ response}}+\underbrace{C_h(s)}_\mathrm{transient\ response}\label{eq:partialfractions}\tag{3}$$

Find $$c_0$$

\require{cancel} \begin{aligned} &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) \\ \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} \\ \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}) \\ \Rightarrow\,& c_0=H(-j\omega)\,\frac{\cancel{\omega}}{-2j\cancel{\omega}} =\frac{H(-j\omega)}{-2j} \end{aligned}

Similarly, find $$c_1$$

\require{cancel} \begin{aligned} &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) \\ \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} \\ \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}) \\ \Rightarrow\,& c_1=H(j\omega)\,\frac{\cancel{\omega}}{2j\cancel{\omega}} =\frac{H(j\omega)}{2j} \end{aligned}

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{aligned} y_{ss}(t)&=\frac{H(-j\omega)}{-2j}e^{-j\omega t}+\frac{H(j\omega)}{2j}e^{j\omega t}\\ &=\frac{H(j\omega)e^{j\omega t}-H(-j\omega)e^{-j\omega t}}{2j} \end{aligned}\label{eq:yss}\tag{4}

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

\left\{ \begin{aligned} H(s)&=|H(s)|\,e^{j\angle H(s)}\\ |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}} \\ \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)\\ \end{aligned} \right.\label{eq:euler}\tag{5}

Substitute the polar representation of $$H(s)$$, equation $$\eqref{eq:euler}$$, in $$y_{ss}$$, equation $$\eqref{eq:yss}$$

\begin{aligned} 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) \\ &=|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) \\ &=|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{aligned}

In this, we recognize the Laplace transfer for a sinodial function

\begin{aligned} \sin(\omega t)\gamma(t) \overset{\mathfrak{L}}\longleftrightarrow \frac{\omega}{s^2+\omega^2} \end{aligned}

The frequency response follows as

\def\doubleunderline#1{\underline{\underline{#1}}} \doubleunderline{ \begin{aligned} y_{ss}(t)=|H(j\omega)|\,\sin(\omega t+\angle H(j\omega))\,\gamma(t) \end{aligned} }\tag{6}

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