ft4.md

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Unit 4.4: Introduction to Filters

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Colophon

An annotatable worksheet for this presentation is available as {ref}ws9.

• The source code for this page is fourier_transform/4/ft4.md.

• You can view the notes for this presentation as a webpage ({ref}ft4).

• This page is downloadable as a PDF file.

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Scope and Background Reading

This section is Based on the section Filtering from Chapter 5 of Benoit Boulet, Fundamentals of Signals and Systems{cite}boulet from the Recommended Reading List.

This material is an introduction to analogue filters. You will find much more in-depth coverage on Pages 11-1—11-48 of {cite}karris.

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Agenda

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• Frequency Selective Filters

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• Ideal low-pass filter

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• Butterworth low-pass filter

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• High-pass filter

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• Bandpass filter

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Introduction

• Filter design is an important application of the Fourier transform
• Filtering is a rich topic often taught in graduate courses so we give only an introduction.
• Our introduction will illustrate the usefulness of the frequency domain viewpoint.
• We will explore how filters can shape the spectrum of a signal.

Other applications of the Fourier transform are sampling theory (introduced next week) and modulation.

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Frequency Selective Filters

An ideal frequency-selective filter is a system that let's the frequency components of a signal through undistorted while frequency components at other frequency are completely cut off.

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• The range of frequencies which are let through belong to the pass Band
• The range of frequencies which are cut-off by the filter are called the stopband
• A typical scenario where filtering is needed is when noise $n(t)$ is added to a signal $x(t)$ but that signal has most of its energy outside the bandwidth of a signal.

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Typical filtering problem

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Signal

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Out-of Bandwidth Noise

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Signal plus Noise

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Results of filtering

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Motivating example

See the video and script on Canvas Week 6.

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Ideal Low-Pass Filter (LPF)

An ideal low pass filter cuts-off frequencies higher than its cut-off frequency, $\omega_c$.

$$H_{\rm{lp}}(\omega ) = \left{ {\begin{array}{*{20}{c}} {1,}&{|\omega |{\kern 1pt} , < {\omega _c}}\\ {0,}&{|\omega |{\kern 1pt} , \ge {\omega _c}} \end{array}} \right.$$

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Frequency response of an ideal LPF

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Impulse response of an ideal LPF

$$h_{\rm{lp}}(t) = \frac{\omega _c}{\pi }{\mathop{\rm sinc}\nolimits} \left( \frac{\omega _c}{\pi }t \right)$$

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Filtering is Convolution

The output of an LTI system with impulse response

$$h(t) \Leftrightarrow H(\omega)$$

subject to an input signal

$$x(t) \Leftrightarrow X(\omega)$$

is given by

$$y(t) = h(t)*x(t) \Leftrightarrow Y(\omega) = H(\omega)X(\omega)$$

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Issues with the "ideal" filter

This is the step response:

(reproduced from {cite}boulet Fig. 5.23 p. 205)

Ripples in the impulse resonse would be undesireable, and because the impulse response is non-causal it cannot actually be implemented.

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Butterworth low-pass filter

N-th Order Butterworth Filter

$$\left| H_B(\omega ) \right| = \frac{1}{\left(1 + \left(\frac{\omega }{\omega _c}\right)^{2N}\right)^{\frac{1}{2}}}$$

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Remarks

• DC gain is

  $$|H_B(j0)|=1$$

• Attenuation at the cut-off frequency is

  $$|H_B(j\omega_c)|=1/\sqrt{2}$$

  for any $N$

More about the Butterworth filter: Wikipedia Article

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Example 5: Second-order BW Filter

The second-order butterworth Filter is defined by is Characteristic Equation (CE):

$$p(s) = s^2 + \omega_c\sqrt{2}s+\omega_c^2 = 0^*$$

Calculate the roots of $p(s)$ (the poles of the filter transfer function) in both Cartesian and polar form.

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Note: This has the same characteristic as a control system with damping ratio $\zeta = 1/\sqrt{2}$ and $\omega_n = \omega_c$!

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Example 6

Derive the differential equation relating the input $x(t)$ to output $y(t)$ of the 2nd-Order Butterworth Low-Pass Filter with cutoff frequency $\omega_c$.

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Example 7

Determine the frequency response $H_B(\omega)=Y(\omega)/X(\omega)$

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Magnitude of frequency response of a 2nd-order Butterworth Filter

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wc = 100;

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Transfer function

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H = tf(wc^2,[1, wc*sqrt(2), wc^2])

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Poles of $H(s)$

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[poles] = pole(H)

Natural frequency $\omega_n$ and damping ratio $\zeta$ of the poles

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[mag,phase] = damp(H)

Phase of the poles

phase = angle(poles)*180/pi % degrees

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Magnitude frequency response

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slideshow:
  slide_type: subslide
---
w = -400:400;
mHlp = 1./(sqrt(1 + (w./wc).^4));
plot(w,mHlp)
grid
ylabel('|H_B(j\omega)|')
title('Magnitude Frequency Response for 2nd-Order LP Butterworth Filter (\omega_c = 100 rad/s)')
xlabel('Radian Frequency \omega [rad/s]')
text(100,0.1,'\omega_c')
text(-100,0.1,'-\omega_c')
hold on
plot([-400,-100,-100,100,100,400],[0,0,1,1,0,0],'r:')
hold off

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Bode plot

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---
bode(H)
grid
title('Bode-plot of Butterworth 2nd-Order Butterworth Low Pass Filter')

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Note that the attentuation of the filter is flat at 0 dB in the pass-band at frequencies below the cut-off frequency $\omega &lt; \omega_c$; has a value of $-3$ dB at the cut-off frquency $\omega = \omega_c$; and has a "roll-off" (rate of decrease) of $N\times 20$ dB/decade in the stop-band.

In this case, $N=2$, and $\omega_c = 100$ rad/s so the attenuation is -40 dB at $\omega = 10\omega_c = 1,000$ rad/s and $\omega = -80$ dB at $\omega = 100\omega_c = 10,000$ rad/s.

The phase is $0^\circ$ at $\omega = 0$; $N\times 90^\circ$ at $\omega = \infty$; and $N\times 45^\circ$ and $\omega = \omega_c$.

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Example 8

Determine the impulse and step response of a butterworth low-pass filter.

You will find this Fourier transform pair useful:

$$e^{-at}\sin\omega_0 t;u_0(t) \Leftrightarrow \frac{\omega_0}{(j\omega + a)^2+\omega_0^2}$$

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Impulse response

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slideshow:
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---
impulse(H,0.1)
grid
title('Impulse Response of 2nd-Order Butterworth Low Pass Filter')

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Step response

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slideshow:
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---
step(H,0.1)
title('Step Response of Butterworth 2nd-Order Butterworth Low Pass Filter')
grid
text(0.008,1,'s_B(t) for \omega_c = 100 rad/s')

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High-pass filter (HPF)

An ideal highpass filter cuts-off frequencies lower than its cutoff frequency, $\omega_c$.

$$H_{\rm{hp}}(\omega ) = \left{ {\begin{array}{*{20}{c}} {0,}&{|\omega |{\kern 1pt} , \le {\omega _c}}\\ {1,}&{|\omega |{\kern 1pt} , > {\omega _c}} \end{array}} \right.$$

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Frequency response of an ideal HPF

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Responses

Frequency response

$$H_{\mathrm{hp}}(\omega)=1-H_{\mathrm{lp}}(\omega)$$

Impulse response

$$h_{\mathrm{hp}}(t)=\delta(t)-h_{\mathrm{lp}}(t)$$

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Example 9

Determine the frequency response of a 2nd-order butterworth highpass filter

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Magnitude frequency response

---
slideshow:
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---
w = -400:400;
plot(w,1-mHlp)
grid
ylabel('|H_B(j\omega)|')
title('Magnitude Frequency Response for 2nd-Order HP Butterworth Filter (\omega_c = 100 rad/s)')
xlabel('Radian Frequency \omega [rad/s]')
text(100,0.9,'\omega_c')
text(-100,0.9,'-\omega_c')
hold on
plot([-400,-100,-100,100,100,400],[0,0,1,1,0,0],'r:')
hold off

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High-pass filter

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---
Hhp = 1 - H
bode(Hhp)
grid
title('Bode-plot of Butterworth 2nd-Order Butterworth High Pass Filter')

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Band-pass filter (BPF)

An ideal bandpass filter cuts-off frequencies lower than its first cutoff frequency $\omega_{c1}$, and higher than its second cutoff frequency $\omega_{c2}$.

$$H_{\rm{bp}}(\omega ) = \left{ {\begin{array}{*{20}{c}} {1,}&\omega _{c1} < ,|\omega |, < \omega _{c2}\\ {0,}&\rm{otherwise} \end{array}} \right.$$

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Frequency response of an ideal BPF

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Bandpass filter design

A bandpass filter can be obtained by multiplying the frequency responses of a lowpass filter by a highpass filter.

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$$H_{\mathrm{bp}}(\omega) = H_{\mathrm{hp}}(\omega)H_{\mathrm{lp}}(\omega)$$

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• The highpass filter should have cut-off frequency of $\omega_{c1}$

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• The lowpass filter should have cut-off frequency of $\omega_{c2}$

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Summary

• Frequency-Selective Filters
• Ideal low-pass filter
• Butterworth low-pass filter
• High-pass filter
• Bandpass filter

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Solutions

Handwritten

Solutions to Examples 5-9 are captured as a PenCast in filters.pdf.

MATLAB

To generate all the plots shown in this presentation, you can use butter2_ex.mlx