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.

Transfer Function in s-domain

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.

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Suggested next reading is Impedance.

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