Coinciting real poles

Shows the math of a critically-damped RLC low pass filter. Visualizes the poles in the Laplace domain. Visualizes the step and frequency response. Part of the article RLC Low-pass Filter.

Coinciting Real Poles (critically-damped case)

In the critically-damped case, the two poles from the transfer polynominal coincite on the negative real axis.

Substitute $$\zeta=1$$ and $$\omega_n$$ in the equation for the poles.

$$\begin{array}{cr} p = -\frac{R}{2L} = \sqrt\frac{1}{LC}, & R=2\sqrt\frac{L}{C}\\ \end{array} \label{eq:case2a_p}$$

This double pole $$p\lt0$$ is on the left real axis, as visualized in the $$s$$-plane

Unit Step Response

Multiplying the Laplace transform of the unit step function, $$\Gamma(s)$$, with the transfer polynominal 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)^2}, & K=\frac{1}{LC}, & \nonumber\\ &=K\frac{1}{s(s-p)^2} \end{align} \label{eq:case2a_multiplication}

Split up this complicated fraction into forms that are in the Laplace Transform table. According to Heaviside, this can be expressed as partial fractions. Note the factor $$\frac{c_2}{s-p}$$. [swarthmore, MIT-cu]

\begin{align} Y(s) &= K\frac{1}{s(s-p)^2} \nonumber \\ &\equiv \frac{c_0}{s}+\frac{c_1}{\left(s-p\right)^2}+\frac{c_2}{s-p} \label{eq:case2a_heaviside} \end{align}

Substitute $$K=p^2$$ in the pole equations and use Heaviside’s cover up method to find the first two constants $$c_{0,1}$$.

$$\begin{array}{ll} c_0=\left.\frac{p^2}{(s-p)^2}\right|_{s=0}&=\frac{p^2}{p^2}&=1\\ c_1=\left.\frac{p^2}{s}\right|_{s=p}&=\frac{p^2}{p}&=p\\ \end{array} \label{eq:case2a_constants1}$$

Given $$c_0$$ and $$c_1$$, constant $$c_2$$ can be found by substituting any numerical value (other than $$0$$ or $$p$$) in equation $$\eqref{eq:case2a_heaviside}$$. In this case, we substitute $$s=-p$$ [MIT-ex4]

\begin{align} &\frac{p^2}{s(s-p)^2} = \frac{1}{s}+\frac{p}{\left(s-p\right)^2}+\frac{c_2}{s-p}\left.\right|_{s=-p}\nonumber\\ \Rightarrow\ &-\frac{p^2}{4p^3} = -\frac{1}{p}+\frac{p}{4p^2}-\frac{c_2}{2p}\nonumber\nonumber\\ \Rightarrow\ &c_2 = -1\label{eq:case2a_constants2} \end{align}

The unit step response $$y(t)$$ follows from the inverse Laplace transform of $$\eqref{eq:case2a_heaviside}$$

\begin{align} y(t)&= \mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\mathcal{L}^{-1}\left\{\frac{c_1}{(s-p)^2}\right\}+\mathcal{L}^{-1}\left\{\frac{c_2}{s-p}\right\} & t\geq0\nonumber\\ &=c_0+c_1te^{pt}+c_2e^{pt} & t\geq0\nonumber \\ &=c_0+(c_1t+c_2)e^{pt} & t\geq0 \\ \end{align}

Substituting the constants $$\eqref{eq:case2a_constants1}$$ and $$\eqref{eq:case2a_constants2}$$ yields

$$\shaded{ y(t)=\left(1+(pt-1)e^{pt}\right)\,\gamma(t) }, \quad \mathrm{where\ }p=-\sqrt\frac{1}{LC}\\$$

As shown in the graph below, this unit step response is a relatively fast rising exponential curve, demonstrating the shortest possible rise time without overshoot.

Frequency Response

The frequency response $$y_{ss}(t)$$ is defined as the steady state response to a sinusoidal input signal $$u(t)=\sin(\omega t)\,\gamma(t)$$.

In Evaluating Transfer Functions, we have proven that

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

The transfer function $$H(s)$$ for this RLC Filter is given by the transfer polynominal and the poles given by $$\eqref{eq:case2a_p}$$

$$H(s)=K\frac{1}{(s-p)^2},\ K=\frac{1}{LC},\ p =\sqrt{\frac{1}{LC}} \label{eq:case2b_splane}$$

Based on Euler’s formula, we can express $$H(s)$$ in polar coordinates

\begin{gather} \left\{ \begin{aligned} H(s) &= |H(s)|\ e^{j\angle{H(s)}}\nonumber\\ |H(s)| &= K \frac{\prod_{i=1}^m\left|(s-z_i)\right|}{\prod_{i=1}^n\left|(s-p_i)\right|} \nonumber \\ &= K \frac{1}{\left|s-p\right|\,\left|s-p\right|}\nonumber\\ \angle{H(s)}&=\sum_{i=1}^m\angle(s-z_i)-\sum_{i=1}^n\angle(s-p_i) \nonumber \\ &= -\left(\angle(s-p)+\angle(s-p) \right)\nonumber \end{aligned} \right. \end{gather}

This transfer function with the double poles at $$p$$, evaluated for $$s=j\omega$$ can be visualized with vectors from the poles to $$j\omega$$.

The square of the length of the vector corresponds to $$|(H(j\omega)|$$, and minus twice the angles with the real axis corresponds to phase shift $$\angle H(j\omega)$$.

\begin{gather} \left\{ \begin{aligned} |H(j\omega)| &=K \frac{1}{\left|j\omega-p\right|\,\left|j\omega-p\right|}= K \frac{1}{\sqrt{\omega^2+{p}^2}\sqrt{\omega^2+{p}^2}}\nonumber\\ &=K\frac{1}{\omega^2+p^2}\nonumber\\ \angle{H(j\omega)}&=-\angle(j\omega-p)-\angle(j\omega-p)=-2\angle(j\omega-p)\nonumber\\ &=-2\cdot \mathrm{atan2}(\omega,-p) =-2\cdot\arctan\frac{\omega}{-p}\nonumber\\ &=2\arctan\frac{\omega}{p},\ p\lt0\land p\in\mathbb{R}\nonumber \end{aligned} \right. \label{eq:case2_polar1} \end{gather}

The output signal $$y_{ss}(t)$$ for a sinusoidal input signal $$\sin(\omega t)\,\gamma(t)$$ for $$\zeta=1$$ follows as

\begin{gather} \left\{ \begin{aligned} y_{ss}(t)&= |H(j\omega)|\,\sin(\omega t+\angle H(j\omega))\,\gamma(t)\nonumber\\ |H(j\omega)| &=K \frac{1}{\omega^2+p^2}\nonumber\\ \angle{H(j\omega)} &=2\arctan\frac{\omega}{p},\quad p\lt0\land p\in\mathbb{R}\nonumber\\ K&=\frac{1}{LC}\nonumber\\ p &= \sqrt{\frac{1}{LC}}\nonumber \end{aligned} \right. \label{eq:case2_frequencyresponse} \end{gather}

The magnitude of the transfer function $$\eqref{eq:case2_frequencyresponse}$$ expressed on a logarithmic scale

\begin{align} p,\,p^* &=\omega_n\left(-\zeta \pm \sqrt{-j^2\left(\zeta^2-1\right)}\right) \nonumber \\ &=\omega_n\left(-\zeta \pm j\sqrt{1-\zeta^2}\right)\\ \end{align} \label{eq:case2b_polar}

The magnitude of the frequency response has a relatively steep -40 dB/decade drop-off at $$\omega_n$$ without signs of resonance.

The corresponding Nyquist plot