Underdamped RLC low pass filter

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]

$s$-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 ($1$).

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 $j\omega$.

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 $\omega_n$ .

Bode magnitude for underdamped case

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

Nyquist plot for underdamped case

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