System response

We will consider first-order and second-order system

first-order systems, time constant

"\\usepackage{tikz}\n\\usetikzlibrary{positioning}\n\n\\begin{document}\n\\begin{tikzpicture}[auto, node distance=2cm, >=latex]\n\n% Nodes\n\\node[draw, rectangle, minimum width=3cm, minimum height=1.5cm] (block) {$\\frac{a}{s + a}$};\n\\node[left=1.5cm of block] (input) {$X(s) = \\frac{1}{s}$};\n\\node[right=1.5cm of block] (output) {$Y(s)$};\n\\node[above=0.5cm of block] (G) {$G(s)$};\n\n% Arrows\n\\draw[->] (input) -- (block);\n\\draw[->] (block) -- (output);\n\n\\end{tikzpicture}\n\\end{document}"

source code

Output of a general first-order system is

Y(s) = X(s)G(s) = \frac{a}{s(s+a)}

thus the time domain output is $y(t) = 1 - e^{-at}$

time constant

usually, $t=\frac{1}{a}$ , and $y(t) = 0.63$ , hence 63.2% to find the rise time.

rise time $T_r$

$T_r$ , time for the waveform to go from 0.1 to 0.8 of its final value

for first order: $T_r = \frac{2.2}{a}$

settling time $T_s$

$T_s$ , time for response to reach and stay with 2% of its final value

for first order: $T_s = \frac{4}{a}$

general order system:

G(s) = \frac{b}{s^{2}+as +b}

Thus the pole for this system:

s_{1},s_{2}= \frac{-a + \sqrt{a^2 - 4b}}{2}

happens when $a=0$

The transfer function is $G(s)=\frac{b}{s^{2}+b}$ , and poles will only have imaginary $\pm jw$

$w_n = \sqrt{b}$ is the frequency of oscillation of this system.

in a sense, this is the undamped case:

complex poles has real part $\sigma = -\frac{a}{2}$

definition

damping ratio is defined as:
$\zeta = \frac{\text{exponential decay frequency}}{\text{natural frequency}} = \frac{|\sigma|}{w_n}$

So that $a = 2 \zeta w_n$

\begin{aligned} G(s) &= \frac{w_n^2}{s^2 + 2 \zeta w_n s + w_n^2} \\[12pt] s_{1},s_{2} &= - \zeta w_n \pm w_n \sqrt{\zeta^2 - 1} \end{aligned}

Condition	Poles	pole type	Damping Ratio ( $\zeta$ )	Natural Response $c(t)$
Undamped	$\pm j \omega_n$	imaginary	$\zeta = 0$	$A \cos (\omega_n t - \varphi)$
Underdamped	$\omega_d \pm j \omega_d$	complex	$0 < \zeta < 1$	$A e^{(-\sigma_d)t} \cos (\omega_d t - \varphi)$ where $w_d = w_n \sqrt{1- \zeta^2}$
critically damped	$\sigma_1$	real	$\zeta = 1$	$K t e^{\sigma_1 t}$
overdamped	$\sigma_1 \quad \sigma_2$	real	$\zeta > 1$	$K (e^{\sigma_1 t} + e^{\sigma_2 t})$

Transfer function $C(s)$ is given by

C(s) = \frac{w_n^2}{s(s^2 + 2 \zeta w_n s + w_n^2)}

response in time-domain via inverse Laplace transform:

c(t) = 1 - \frac{1}{\sqrt{1- \zeta^2}} e^{- \zeta w_n t} \cos (\sqrt{1- \zeta^2}\omega_n t + \varphi)

where $\varphi = \tan^{-1} (\frac{\zeta}{\sqrt{1-\zeta^2}})$

time required to reach the first or maximum peak

T_p = \frac{\pi}{\omega_n \sqrt{1-\zeta^2}}

\%OS = e^{\zeta \pi / \sqrt{1-\zeta^2}} \times 100 \%

or in terms of damping ratio $\zeta$ :

\zeta = \frac{-\ln \frac{\text{\%OS}}{100}}{\sqrt{\pi^2 + \ln^2(\frac{\text{\%OS}}{100})}}

\begin{aligned} G(s) &= \frac{w_n^2}{s^2 + 2 \zeta w_n s + w_n^2} \\[12pt] s_{1},s_{2} &= - \zeta w_n \pm w_n \sqrt{\zeta^2 - 1} \\[8pt] T_p &= \frac{\pi}{\omega_n \sqrt{1-\zeta^2}} \\ T_s &\cong \frac{4}{\zeta \omega_n} \end{aligned}

figure1: poles of second-order underdamped system

"\\usepackage{tikz}\n\\usepackage{pgfplots}\n\\pgfplotsset{compat=1.16}\n\n\\begin{document}\n\\begin{tikzpicture}\n \\begin{axis}[\n width=12cm, height=8cm,\n xlabel={$t$ (time)},\n ylabel={Amplitude},\n grid=major,\n legend style={at={(0.5,1.1)}, anchor=north, legend columns=-1},\n xmin=0, xmax=10,\n ymin=-1.2, ymax=1.2\n ]\n % First Transfer Function (Sinusoidal)\n \\addplot[blue, thick, samples=100, domain=0:10]\n {exp(-0.1*x)*sin(deg(2*pi*0.5*x))};\n\n % Second Transfer Function (Envelope with Different Frequency)\n \\addplot[red, dashed, thick, samples=100, domain=0:10]\n {exp(-0.1*x)*sin(deg(2*pi*0.8*x))};\n\n % Envelope (Exponential Decay)\n \\addplot[black, dotted, thick, samples=100, domain=0:10]\n {exp(-0.1*x)};\n\n \\addplot[black, dotted, thick, samples=100, domain=0:10]\n {-exp(-0.1*x)};\n \\end{axis}\n\\end{tikzpicture}\n\\end{document}"

source code

"\\usepackage{tikz}\n\\usepackage{pgfplots}\n\\pgfplotsset{compat=1.16}\n\n\\begin{document}\n\\begin{tikzpicture}\n \\begin{axis}[\n width=12cm, height=8cm,\n xlabel={$t$ (time)},\n ylabel={Amplitude},\n grid=major,\n legend style={at={(0.5,1.1)}, anchor=north, legend columns=-1},\n xmin=0, xmax=10,\n ymin=-3, ymax=3\n ]\n % First Transfer Function (Higher Amplitude)\n \\addplot[blue, thick, samples=100, domain=0:10]\n {2*sin(deg(2*pi*0.5*x))};\n\n % Second Transfer Function (Lower Amplitude)\n \\addplot[red, dashed, thick, samples=100, domain=0:10]\n {1*sin(deg(2*pi*0.5*x))};\n \\end{axis}\n\\end{tikzpicture}\n\\end{document}"

source code

"\\usepackage{tikz}\n\\usepackage{pgfplots}\n\\pgfplotsset{compat=1.16}\n\n\\begin{document}\n\\begin{tikzpicture}\n \\begin{axis}[\n width=12cm, height=8cm,\n xlabel={Time $t$},\n ylabel={Response},\n grid=major,\n legend style={at={(0.5,1.1)}, anchor=north, legend columns=-1},\n xmin=0, xmax=10,\n ymin=0, ymax=2\n ]\n % First System Response\n \\addplot[blue, thick, samples=200, domain=0:10]\n {1 - exp(-0.5*x)*(cos(deg(2*pi*0.5*x)) + 0.1*sin(deg(2*pi*0.5*x)))};\n\n % Second System Response (Same Overshoot, Different Natural Frequency)\n \\addplot[red, dashed, thick, samples=200, domain=0:10]\n {1 - exp(-1*x)*(cos(deg(2*pi*1*x)) + 0.05*sin(deg(2*pi*1*x)))};\n\n % Reference Line for Steady-State Response\n \\addplot[black, dotted, thick] coordinates {(0,1) (10,1)};\n \\end{axis}\n\\end{tikzpicture}\n\\end{document}"

source code

We will consider first-order and second-order system