# Two different real poles

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

## Two Different Real Poles (overdamped case)

The two poles from the transfer polynominal are on separate locations on the negative real axis.

$$\begin{array}{ccc} p_{1,2} = -\frac{R}{2L} \pm \sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}},&{R>2\sqrt\frac{L}{C}} \end{array}$$

Note that $$p_1\lt p_2\lt0$$ and $$|p_1|>|p_2|$$, as visualized in the $$s$$-plane

### Unit Step Response

The unit step response shows how the system reacts to the input going from $$0$$ to $$1$$ volt at time $$t=0$$. This input is called the Unit Step Function, here represented by $$u(t)=\gamma(t)$$. The unit step response gives an impression of the system behavior in the time domain.

Multiplying the Laplace transform of the unit step function, $$\Gamma(s)$$, with the transfer function (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_1)(s-p_2)}\nonumber \\ &= K\frac{1}{s(s-p_1)(s-p_2)} \end{align} \label{eq:case1a_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 that we need to set up a partial fraction for each descending power of the denominator. [swarthmore, MIT-cu]

\begin{align} Y(s) &= K\frac{1}{s(s-p_1)(s-p_2)} \nonumber \\ &\equiv \frac{c_0}{s}+\frac{c_1}{s-p_1}+\frac{c_2}{s-p_2} \label{eq:case1a_heaviside} \end{align}

Substitute $$K=p_1 p_2$$ from the pole equations and use Heaviside’s cover up method to find the constants $$c_{0,1,2}$$.

$$\left. \begin{array}{cccc} c_0 &=& \left.\frac{p_1p_2}{(s-p_1)(s-p_2)}\right|_{s=0} &=& 1 \\ c_1 &=& \left.\frac{p_1p_2}{s(s-p_2)}\right|_{s=p1} &=& \frac{p_2}{p1-p_2} \\ c_2 &=& \left.\frac{p_1p_2}{s(s-p_1)}\right|_{s=p2} &=& \frac{p_1}{p_2-p_1} \end{array} \right\} \label{eq:case1a_constants}$$

The unit step response $$y(t)$$ follows from the inverse Laplace transform of $$\eqref{eq:case1a_heaviside}$$ and substituting the constants $$\eqref{eq:case1a_constants}$$

\begin{align} y(t)&=\mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\mathcal{L}^{-1}\left\{\frac{c_1}{s-p_1}\right\}+\mathcal{L}^{-1}\left\{\frac{c_2}{s-p_2}\right\} & t\geq0 \nonumber \\ &=c_0+c_1e^{p_1t}+c_2e^{p_2t} & t\geq0 \end{align}

Substituting the constants $$\eqref{eq:case1a_constants}$$ gives the unit step response

$$\shaded{ y(t)=\left(1+\frac{p_2}{p_1-p_2}e^{p_1t}+\frac{p_1}{p_2-p_1}e^{p_2t}\right)\gamma(t) }$$
where
$$p_{1,2} = -\frac{R}{2L} \pm \sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}} \label{eq:case1a_usr}$$

As shown in the graph, the unit step response is a relatively slow decaying exponential curve (with $$p_1\lt p_2\lt 0$$). The figure was generated using the source code listed in the appendix.

### Frequency Response

The frequency response $$y_{ss}(t)$$ is defined as the steady state response to a sinusoidal input signal $$u(t)=\sin(\omega t)\,\gamma(t)$$. It describes how well the filter can distinguish between different frequencies.

In Evaluating Transfer Functions, we have proven that

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

The transfer function $$H(s)$$ for this RLC Filter is given by transfer polynominal.

$$H(s)=K\frac{1}{(s-p_1)(s-p_2)}$$
where
$$p_{1,2} = -\frac{R}{2L} \pm \sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}} \nonumber$$

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)}}\\\\ |H(s)| &= K \frac{\prod_{i=1}^m\left|(s-z_i)\right|}{\prod_{i=1}^n\left|(s-p_i)\right|} \\ &= K \frac{1}{\left|s-p_1\right|\,\left|s-p_2\right|} \\ \angle{H(s)}&=\sum_{i=1}^m\angle(s-z_i)-\sum_{i=1}^n\angle(s-p_i) \\ &= -\left(\angle(s-p_1)+\angle(s-p_2) \right) \end{aligned} \right. \end{gather}

This transfer function with poles at $$p_1$$ and $$p_2$$, evaluated for $$s=j\omega$$ can be visualized with vectors from the poles to $$j\omega$$.

The product of the length of the vector corresponds to $$|(H(j\omega)|$$, and minus the sum of 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_2\right|\,\left|j\omega-p_2\right|} \\ &= K \frac{1}{\sqrt{\omega^2+{p_1}^2}\sqrt{\omega^2+{p_2}^2}} \nonumber \\ \angle{H(j\omega)}&=-\angle(j\omega-p_1)-\angle(j\omega-p_2)\nonumber\\ &= -\mathrm{atan2}(\omega,-p_1)-\mathrm{atan2}(\omega,-p_2)\nonumber\\ &= -\arctan\frac{\omega}{-p_1}-\arctan\frac{\omega}{-p_2}\nonumber\\ &= \arctan\frac{\omega}{p_1}+\arctan\frac{\omega}{p_2},\ p\lt0\land p\in\mathbb{R}\nonumber \end{aligned} \right. \end{gather} \label{eq:polar1}

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}{\sqrt{\omega^2+{p_1}^2}\sqrt{\omega^2+{p_2}^2}}\nonumber\\ \angle{H(j\omega)} &=\arctan\frac{\omega}{p_1}+\arctan\frac{\omega}{p_2},\ p\lt0\land p\in\mathbb{R}\nonumber\\ K&=\frac{1}{LC}\nonumber\\ p_{1,2} &= -\frac{R}{2L} \pm \sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}}\nonumber \end{aligned} \right. \label{eq:frequencyresponse} \end{gather}

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

\begin{align} |H_{dB}(\omega)| =& \,20\log\left(K\right) \nonumber \\ & -20\log\sqrt{\omega^2+{p_1}^2} \nonumber \\ & -20\log\sqrt{\omega^2+{p_2}^2}, & p_{1,2}\in\mathbb{R}\\ \end{align}

The Bode plot shows the magnitude of the frequency response has a relatively shallow -20 dB/decade drop-off between $$|p_2| \lt \omega \lt |p_1|$$.

The corresponding Nyquist plot shows a very stable system