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

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

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

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

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 \(\angle H(j\omega)\).

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

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

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.

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

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### 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.

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 \(1/\sqrt{2}\)

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

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

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

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

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].

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