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

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\(s\)-plane for critically-damped case

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.

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Step response for critically-damped case

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 \(j\omega\).

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Transfer function evaluated at \(s=j\omega\) for critically-damped case

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.

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Bode magnitude for critically-damped case

The corresponding Nyquist plot

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Nyquist plot for critically-damped case

Continue reading about Complex poles

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