.

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.

TransferFunction

 

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.

 

Next

Suggested next reading is Impedance.

 

Coert Vonk

Coert Vonk

Independent Firmware Engineer at Los Altos, CA
Welcome to the things that I couldn’t find.This blog shares some of the notes that I took while deep diving into various fields.Many such endeavors were triggered by curious inquiries from students. Even though the notes often cover a broader area, the key goal is to help the them adopt, flourish and inspire them to invent new technology.
Coert Vonk

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