# Complex poles

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]

### 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 ($$1$$).

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 $$j\omega$$.

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 $$\omega_n$$ .

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