# Frequency response

Derives the frequency response of RC low-pass filter using the Laplace transform. Part of a series about the properties of the RC low-pass filter.

## 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

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

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

$$H(s )= K\frac{1}{s-p},\ K = \frac{1}{RC},\ p = -\frac{1}{RC} \label{eq:splane}$$

The system behavior at $$\omega=0$$ and at $$\omega\rightarrow \infty$$ indicates that this is a low pass filter.

\begin{align} \lim_{s \rightarrow j0} |H(s)| &=-p\frac{1}{0-p} =1 \\ \lim_{s \rightarrow j\infty} |H(s)| &=-p\frac{1}{\infty-p} =0 \end{align}

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

\left\{ \begin{align} 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|} =K \frac{1}{\left|s-p\right|}\nonumber \\ \angle{H(s)}&=\sum_{i=1}^m\angle(s-z_i)-\sum_{i=1}^n\angle(s-p_i) =-\angle(s-p)\nonumber \end{align} \right.

This transfer function with pole $$p$$, evaluated for $$s=j\omega$$ can be visualized with a vector from the pole to $$j\omega$$.

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

\left\{ \begin{align} |H(j\omega)| &=K \frac{1}{\left|j\omega-p\right|}= K \frac{1}{\sqrt{\omega^2+p^2}}\nonumber \\ \angle{H(j\omega)}&=-\angle(j\omega-p)=\mathrm{atan2}(\omega,-p)\nonumber\\ &=-\arctan\frac{\omega}{-p},\ p\lt 0\land p\in\mathbb{R}\nonumber \end{align} \right. \label{eq:polar1}

Substitute $$p=-\frac{1}{RC}$$ and $$K=\frac{1}{RC}$$

$$\left\{ \begin{eqnarray} |H(j\omega)| &=&\frac{1}{RC} \frac{1}{\sqrt{\omega^2+\left(\frac{1}{RC}\right)^2}}\nonumber \\ \angle{H(j\omega)}&=&-\arctan\frac{\omega}{\frac{1}{RC}}\nonumber \end{eqnarray} \right. \label{eq:eq101}$$

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

\shaded{ \begin{aligned} y_{ss}(t)&=|H(j\omega)|\,\sin(\omega t+\angle H(j\omega))\,\gamma(t)\nonumber \\ |H(j\omega)| &=\frac{1}{\sqrt{(1+\omega RC)^2}}\\ \angle{H(j\omega)}&=-\arctan(\omega RC)\nonumber \end{aligned} } \label{eq:frequencyresponse}

This frequency response for different frequencies can be visualized in a Bode plot or a Nyquist diagram. Each of these are a topic of the remaining sections.

### Effect on Input with Harmonics

As a side step, we examine the effect of the filter on a square wave input signal. The Fourier series of the square wave shows that it consists of a base frequency and odd harmonics.

$$x(t)=\frac{4}{\pi}\sum_{n=1,3,\dots }^\infty \frac{\sin\left(n\omega t\right) }{n}$$

Substituting $$C=470\ \mathrm{nF}$$, $$R=100\ \Omega$$ and 20 kHz in $$\eqref{eq:frequencyresponse}$$, gives the output signal $$y_{ss}(t)$$

\left\{ \begin{align} y_{ss}(t)&=\frac{4}{\pi}\sum_{n=1,3,\dots }^\infty |H(jn\omega)| \frac{\sin\left(n\omega t+\angle H(jn\omega)\right) }{n}\nonumber\\ |H(jn\omega)| &=\frac{1}{\sqrt{(1+n\omega RC)^2}}\nonumber\\ \angle{H(jn\omega)}&=-\arctan(n\omega RC)\nonumber\\ RC&=47\cdot 10^{-6}\nonumber\\ \omega&=2\pi20\cdot 10^{3}\nonumber \end{align} \right.

UNFINISHED

### Bode plot

A Bode plots frequency as the horizontal axis and usually consists of two separate plots to that show the magnitude and phase of the frequency response $$y_{tt}$$. Since the range of magnitudes may also be large, the amplitude scale is usually expressed in decibels $$20\log_{10}\left|H(j\omega)\right|$$ . The frequency axis uses a logarithmic scale as well.

\begin{align} |H(j\omega)|\ &= \frac{1}{\sqrt{1+(\omega RC)^2}} \\ |H_{dB}(j\omega)|\ &= -20\log\sqrt{1+ (\omega RC)^2} \label{eq:polar2} \\ \angle{H(j\omega)}\ &=-\arctan\left(\omega RC\right) \end{align}

The magnitude of the frequency response has a relatively shallow drop-off.

The phase shift depends on the frequency, causing signals composed of multiple frequencies to be distorted.

Angular frequency $$\omega_c=|p|$$, is is known as the cutoff, break, -3dB or half-power frequency because the magnitude of the transfer function $$\eqref{eq:polar2}$$ equals $$1/\sqrt{2}$$

\begin{align} |H(j\omega_c)|\ &=\dfrac{1}{\sqrt{1+(\omega_c RC)^2}}=\frac{1}{\sqrt{2}},\quad\omega_c=\frac{1}{RC} \\ |H_{dB}(j\omega_c)|\ &= 20\log\frac{1}{\sqrt{2}} \approx-3\rm{\ dB} \end{align}

The attenuation slope is calculated by first expressing the magnitude relative to the cutoff angular frequency $$\omega_c$$

$$\begin{array}{llr} &|H(j\omega)| = \frac{1}{\sqrt{1+\left(\frac{\omega}{\omega_c}\right)^2}}\\ \Rightarrow&|H(j\omega)| =\frac{\omega}{\omega_c} & \forall\ {\omega\gg\omega_c} \end{array}$$

The rate of change of attenuation is usually expressed in dB/decade, where an decade is a factor of 10 in frequency, $$\omega_2=10\omega_1$$

\begin{gather}\begin{aligned} |H_{dB}(j\omega_2)|-|H_{dB}(\omega_1)|\ &= -20\log\left(\frac{10\omega_1}{\omega_c}\right)+20\log\left(\frac{\omega_1}{\omega_c}\right) \\ &=20\log\left(\frac{\omega_1}{\omega_c}\frac{\omega_c}{10\omega_1}\right) \\ &=-20 \mathrm{\ dB/decade} \end{aligned}\end{gather}

This single-pole filter gives has a relatively shallow -20 dB/decade drop-off.

In general, the cutoff frequency is equal to the radial distance of the poles or zeros from the origin of the $$s$$-plane. For information on sketching the Bode magnitude plot from the poles and zeros, refer to Understanding Poles and Zeros [MIT 3.1].

### Nyquist plot

The Nyquist plots display both amplitude and phase angle on a single plot, using the angular frequency as the parameter. It helps visualize if a system is stable or unstable.

Starting with the transfer function transfer polynominal

$$\begin{gather}\begin{array}{ccc} H(s) = K\frac{1}{s-p}, &K = \frac{1}{RC}, &p = -\frac{1}{RC} \end{array}\end{gather}$$

Evaluate at $$s=j\omega$$ and split into real and imaginary parts

\begin{gather} \begin{aligned} H(j\omega)\ &=K\frac{1}{j\omega-p} \\ &=K\frac{1}{j\omega-p}\frac{j\omega+p}{j\omega+p} \\ &=K\frac{j\omega+p}{-\omega^2-p^2} \\ &=K\frac{-j\omega-p}{\omega^2+p^2} \end{aligned} \\ \Rightarrow \left\{ \begin{aligned} \Re\left\{{H(j\omega)}\right\}=&K\frac{-p}{p^2+\omega^2} \\ \Im\left\{{H(j\omega)}\right\}=&K\frac{-\omega}{p^2+\omega^2} \\ \end{aligned} \right. \end{gather}

Plot the frequency transfer function for $$-\infty\lt\omega\gt\infty$$, indicating an increase of frequency using an arrow. A dashed line is used for negative frequencies. (The plot was generated using the GNU/Octave as shown in Appendix A.)

From the plot we see that for $$\omega=0$$ the gain is 1, and for $$\omega\to\infty$$ the gain becomes 0. The high frequency portion of the plot approaches the origin at an angle of -90 degrees. For more information on Nyquist refer to Determining Stability using the Nyquist Plot [swarthmore].

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## One Reply to “Frequency response”

1. edwardjs says:

Hi, The tutorial will be very useful if you correct several mistakes, like in Eq. 5 (magnitude).

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