Complex poles

Coert Vonk

Shows the math of a underdamped RLC low pass filter. Visualizes the poles in the Laplace domain. The step and frequency response. Part of the article RLC Low-pass Filter.\(\)

Complex Poles (underdamped case)

For complex conjugate poles, the transfer function can be written as below. Given that \(\zeta\lt1\), the argument of the square root in the poles will be negative. Multiply this argument with \(-j^2\) to highlight the imaginary part apart.

$$ H(s)=K\frac{1}{(s-p)(s-p^\ast)}\label{eq:case3a_transferpoles} $$

Where

$$ \begin{align} \mathrm{where\ }K&=\frac{1}{LC},\nonumber\\ p,p^\ast&=\omega_n\left(-\zeta \pm \sqrt{-j^2\left(\zeta^2-1\right)}\right) =\omega_n\left(-\zeta \pm j\sqrt{1-\zeta^2}\right),\nonumber\\ \omega_n&=\sqrt{\frac{1}{LC}},\nonumber\\ \zeta&=\frac{R}{2}\sqrt{\frac{C}{L}}\nonumber \end{align} \label{eq:case3a_transferpoleswhere} $$

Split the conjugate poles in their real and imaginary parts by defining the poles from equation \(\eqref{eq:case3a_transferpoles}\) as \(p,\,p^*\equiv -\sigma\pm j\omega_d\)

$$ p,\,p^\ast=\omega_n\left(-\zeta \pm j\sqrt{1-\zeta^2}\right) \equiv -\sigma\pm j\omega_d \label{eq:sigmaomegad} $$

So that

$$ \begin{eqnarray} \left\{ \begin{aligned} \sigma&=\omega_n\zeta,&\text{attenuation}\nonumber\\ \omega_d&=\omega_n\sqrt{1-\zeta^2},&\text{damped natural frequency}\nonumber \end{aligned} \right. \end{eqnarray} \label{eq:sigmaomegad_} $$

This equation indicates that the conjugate poles \(p, p^*\) lay in the left half of the \(s\)-plane. The length of the line segment from the origin to pole \(p\) represents the natural frequency \(\omega_n\) and the angle of the imaginary axis with that line is \(\arcsin\) of the attenuation \(\zeta\). [MIT-me]

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\(s\)-plane for underdamped case

Unit Step Response

Multiplication of the Laplace transform of the unit step function, \(\Gamma(s)\), with the transfer function \(\eqref{eq:case3a_transferpoles}\) gives the unit step response \(Y(s)\).

$$ \begin{align} Y(s)&=\Gamma(s)\cdot H(s)\nonumber \\ &= \frac{1}{s}\cdot K\frac{1}{(s-p)(s-p^\ast)}\nonumber \\ &= K\frac{1}{s(s-p)(s-p^\ast)} \end{align} \label{eq:case3a_multiplication} $$

Using Heaviside, split this equation partial fractions that can be found in the Laplace Transform table

$$ Y(s)=K\frac{1}{s(s-p)(s-p^\ast)} \equiv \frac{c_0}{s}+\frac{c_1}{s-p}+\frac{c_2}{s-p^\ast} \label{eq:case3a_heaviside} $$

Substitute \(K=p\cdot p^\ast\) in the pole equations and use Heaviside’s cover up method to find the constants \(c_{0,1,2}\).

$$ \begin{gather} \left\{ \begin{array}{cclclcc} c_0 &=& \left.\frac{p\,p^*}{(s-p)(s-p^*)}\right|_{s=0} &=& 1 \\ c_1 &=& \left.\frac{p\,p^*}{s(s-p^*)}\right|_{s=p} &=& \frac{p^*}{p-p^*} \\ c_2 &=& \left.\frac{p\,p^*}{s(s-p)}\right|_{s=p^*} &=& \frac{p}{p^*-p} &\equiv& {c_1}^* \end{array} \right. \label{eq:case3a_constants} \end{gather} $$

The constants \(c_1\) and \(c_2\) are complex conjugates of each other since they are equivalent except for the sign on the imaginary part. To highlight this, substitute the values for the poles from \(\eqref{eq:sigmaomegad}\) and write these constants in polar notation

$$ \begin{align} c_1&=\frac{p^*}{p-p^\ast} =\frac{-\sigma-j\omega_d}{(-\sigma+j\omega_d)-(-\sigma-j\omega_d)}\nonumber\\ &=\frac{-\sigma-j\omega_d}{j2\omega_d} =\frac{j(\sigma+j\omega_d)}{j(-j2\omega_d)} =\frac{j\sigma-\omega_d}{2\omega_d}\nonumber\\ &=\frac{1}{2\omega_d}\left(-\omega_d+j\sigma\right)\overset{Euler}\Rightarrow\nonumber\\ &=\frac{|p|}{2\omega_d}\,e^{j\varphi}, \mathrm{\ where\ } \left\{\begin{aligned} |p|&=\sqrt{{\omega_d}^2+\sigma^2}\mathrm{\ and\ }\nonumber\\ \varphi&=\arctan\left(\frac{\sigma}{-\omega_d}\right)+\pi \end{aligned}\right.\label{eq:case3a_c2polar} \\ c_2&={c_1}^\ast=\frac{|p|}{2\omega_d}\,e^{-j\varphi}\label{eq:case3a_c3polar} \end{align} $$

The unit step response \(y(t)\) follows from the inverse Laplace transform of \(\eqref{eq:case3a_heaviside}\), substituting \(c_{0,1,2}\) from \(\eqref{eq:case3a_constants}\), \(\eqref{eq:case3a_c2polar}\) and \(\eqref{eq:case3a_c3polar}\).

$$ \begin{align} y(t)&= \mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\mathcal{L}^{-1}\left\{\frac{c_1}{s-p}\right\}+\mathcal{L}^{-1}\left\{\frac{c_2}{s-p^*}\right\} & t\geq0\nonumber\\ &=c_0+c_1e^{pt}+c_2e^{p^*t} \overset{\mathrm{subst\ consts}}\Rightarrow &t\geq0\nonumber \\ &=1+\frac{|p|}{2\omega_d}\,e^{j\varphi}e^{(-\sigma+j\omega_d)t}+\frac{|p|}{2\omega_d}\,e^{-j\varphi}e^{(-\sigma-j\omega_d)t} &t\geq0\nonumber \\ &=1+\frac{|p|}{\omega_d}\ e^{-\sigma t}\ \underbrace{\frac{ e^{j(\omega_d t+\varphi)}+e^{-j(\omega_d t+\varphi)}}{2}}_{\text{Euler identity for cosine}} &t\geq0 \end{align} $$

Apply the Euler identify for cosine, and reference \(|p|\) and \(\varphi\) from equation \(\eqref{eq:case3a_c2polar}\) and \(\eqref{eq:case3a_c3polar}\), \(\sigma\) and \(\omega_d\) from equation \(\eqref{eq:sigmaomegad}\) and \(\zeta\) and \(\omega_n\) from \(\eqref{eq:case3a_transferpoles}\)

$$ \shaded{ y(t) = \left( 1+\frac{|p|}{\omega_d}e^{-\sigma t}\cos(\omega_d t+\varphi) \right)\gamma(t) } \label{eq:case3a_usr} $$

Where

$$ \begin{aligned} |p|&=\sqrt{{\omega_d}^2+\sigma^2}\\ \omega_d&=\omega_n\sqrt{1-\zeta^2}\\ \sigma&=\omega_n\zeta\\ \omega_n&=\sqrt{\frac{1}{LC}}\\ \zeta&=\frac{R}{2}\sqrt{\frac{C}{L}}\\ \varphi&=\pi-\arctan\frac{\sigma}{\omega_d} \end{aligned} \label{eq:case3a_usr_where} $$

The graph shows the response for different values of \(R\). This underdamped circuit oscillates, with the amplitude exceeding that of the input (\(1\)).

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Step response for underdamped case

For the extreme case, where \(R=0\), the response becomes \(\left(1-cos(\omega_n t)\right)\gamma(t)\), oscillating with an amplitude reaching twice the input (\(1\)).

Frequency Response

The frequency response \(y_{ss}(t)\) is defined as the steady state response to a sinusoidal input signal

$$ u(t)=\sin(\omega t)\,\gamma(t) $$

We can rewrite the transfer functionby substituting the poles from \(\eqref{eq:sigmaomegad}\)

$$ \begin{align} H(s)&=K\frac{1}{(s-p)(s-p^\ast)} =K\frac{1}{s^2-s(p+p^\ast)+p\,p^\ast}\overset{\mathrm{subst\ }p,p\ast}\Rightarrow\nonumber\\ &=K\frac{1}{s^2-s( \omega_n\left(-\zeta + j\sqrt{1-\zeta^2}\right) + \omega_n\left(-\zeta – j\sqrt{1-\zeta^2}\right) )+p\,p^\ast}\overset{\mathrm{subst\ }pp^\ast={\omega_n}^2}\Rightarrow\nonumber\\ &=K\frac{1}{s^2-s\omega_n\left( \left(-\zeta \cancel{+j\sqrt{1-\zeta^2}}\right) + \left(-\zeta \cancel{-j\sqrt{1-\zeta^2}}\right) \right)+{\omega_n}^2}\nonumber\\ &=K\frac{1}{s^2+2\zeta\omega_n s +{\omega_n}^2}\label{eq:case3a_newhs} \end{align} $$

In Evaluating Transfer Functions, we derived the frequency response as

$$ y_{ss}(t)=|H(j\omega)|\,\sin(\omega t+\angle H(j\omega))\,\gamma(t) $$

This transfer function with the poles at \(p\) and \(p^\ast\), evaluated for \(s=j\omega\) can be visualized with vectors from the poles to \(j\omega\).

z-plane complex poles jw
Transfer function evaluated at \(s=j\omega\) for underdamped case1a_constants

Substitute \(s=j\omega\) into the transfer function \(\eqref{eq:case3a_newhs}\)

$$ \begin{align} H(j\omega) &= K\frac{1}{(j\omega)^2+2\zeta\omega_n (j\omega) +{\omega_n}^2} \nonumber \\ &= K\frac{1}{({\omega_n}^2-\omega^2)+j(2\zeta\omega_n \omega)} \end{align} $$

The transfer function can be expressed in polar form using Euler’s formula as

$$ \begin{gather} \left\{ \begin{aligned} H(s) &= |H(s)|\ e^{j\angle{H(s)}}\\\\ |H(s)| &= K\frac{1}{\sqrt{({\omega_n}^2-\omega^2)^2+(2\zeta\omega_n \omega)^2}}\\ \angle{H(s)}&=-\mathrm{atan2}\left(2\zeta\omega_n \omega,\ {\omega_n}^2-\omega^2\right)\\ \end{aligned} \right. \end{gather} \label{eq:case3b_polar} $$

The graph shows the magnitude of the output for different values of \(R\). The magnitude of the frequency response demonstrates resonant behavior. Note the voltage amplification around the natural frequency \(\omega_n\) .

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Bode magnitude for underdamped case

The corresponding Nyquist plot shows that the system gets less stable as the resistor value decreases

own work; requires svg enabled browser
Nyquist plot for underdamped case

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