To accompany Unit 5.4: Models of Discrete-Time Systems#

Colophon#

This worksheet can be downloaded as a PDF file. We will step through this worksheet in class.

An annotatable copy of the notes for this presentation will be distributed before the second class meeting as Worksheet 17 in the Week 9: Classroom Activities section of the Canvas site. I will also distribute a copy to your personal Worksheets section of the OneNote Class Notebook so that you can add your own notes using OneNote.

You are expected to have at least watched the video presentation of Models of Discrete-Time Systems of the notes before coming to class. If you haven’t watch it afterwards!

After class, the lecture recording and the annotated version of the worksheets will be made available through Canvas.

Agenda#

• Discrete Time Systems (Notes)

• Transfer Functions in the Z-Domain (Notes)

• Modelling digital systems in MATLAB/Simulink

• Converting Continuous Time Systems to Discrete Time Systems

• Example: Digital Butterworth Filter

Discrete Time Systems#

In the lecture that introduced the z-transform we talked about the representation of a discrete-time (DT) system by the model shown below:

In this session, we want to explore the contents of the central block.

clear all
cd matlab
pwd
format compact

ans =

    '/Users/eechris/code/src/github.com/cpjobling/eg-247-textbook/dt_systems/4/matlab'

Nz = [1 1 0];

Dz = [1 -0.5 0.125];

[r,p,k] = residue(Nz,Dz)

Hz = tf(Nz,Dz,-1)
hn = impulse(Hz, 15);

stem([0:15], hn)
grid
title('Example 5 - Part 2')
xlabel('n')
ylabel('Impulse response h[n]')

impulse(Hz,15)
grid
title('Example 5 - Part 2 - As Analogue Signal')
xlabel('nTs [s]')
ylabel('Impulse response h(t)')

open dtm_ex1_3

Modelling DT systems in MATLAB and Simulink#

We will consider some examples in class

Ts = 1;
z = tf('z', Ts);

Hz = (z^2 + z)/(z^2 - 0.5 * z + 0.125)

step(Hz)
grid
title('Example 1 - Part 3 - As Analogue Signal')
xlabel('nTs [s]')
ylabel('Step response y(t)')
axis([0,15,0,3.5])

dtm

Converting Continuous Time Systems to Discrete Time Systems#

help c2d

Example: Digital Butterworth Filter#

• Design a 2nd-order butterworth low-pass anti-aliasing filter with transfer function $H(s)$ for use in sampling music.

• The cut-off frequency ${\omega }_c=20$ kHz and the filter should have an attenuation of at least $-80$ dB in the stop band.

• Choose a suitable sampling frequency for the audio signal and give the transfer function $H(z)$ and an algorithm to implement $h[n]$

Solution#

See digi_butter.mlx.

First determine the cut-off frequency ${\omega }_c$

$${\omega }_c=2\pi f_c=2\times \pi \times 20\times {10}^3\mathrm{rad}/\mathrm{s}$$

wc = 2*pi*20e3

wc =
   1.2566e+05

$${\omega }_c=125.66\times {10}^3\mathrm{rad}/\mathrm{s}$$

From the lecture on filters, we know the 2nd-order butterworth filter has transfer function:

$$H(s)=\frac{Y(s)}{U(s)}=\frac{{\omega }_c^2}{s^2+{\omega }_c\sqrt{2}s+{\omega }_c^2}$$

Substituting for ${\omega }_c=125.6637\times {10}^3$ this is …?

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

Hs =
 
           1.579e10
  ---------------------------
  s^2 + 1.777e05 s + 1.579e10
 
Continuous-time transfer function.

$$H(s)=\frac{15.79\times {10}^9}{s^2+177.7\times {10}^{3}s+15.79\times {10}^9}$$

Bode plot#

MATLAB:

bode(Hs,{1e4,1e8})
grid

Sampling Frequency#

From the bode diagram, the frequency roll-off is -40 dB/decade for frequencies $\omega \gg {\omega }_c$. So, $|H(j\omega )|=-80$ dB is approximately 2 decades above ${\omega }_c$.

w_stop = 100*wc

w_stop =
   1.2566e+07

To avoid aliasing, we should choose a sampling frequency twice this = ?

${\omega }_s=2\times 12.6\times {10}^6$ rad/s.

ws = 2 * w_stop

ws =
   2.5133e+07

So

${\omega }_s=25.133\times {10}^6$ rad/s.

Sampling frequency ($f_s$) in Hz = ?

$$f_s={\omega }_s/(2\pi )\mathrm{Hz}$$

fs = ws/(2*pi)

fs =
     4000000

Sampling time $T_s=?$

$T_s=1/fs\mathrm{s}$

Ts = 1/fs

Ts =
   2.5000e-07

Digital Butterworth#

zero-order-hold equivalent

Hz = c2d(Hs, Ts)

Hz =
 
  0.0004862 z + 0.0004791
  -----------------------
  z^2 - 1.956 z + 0.9565
 
Sample time: 2.5e-07 seconds
Discrete-time transfer function.

Step response#

step(Hz)

Algorithm#

From final result:

$$H(z)=\frac{Y(z)}{U(z)}=\frac{486.2\times {10}^{-6}z+479.1\times {10}^{-6}}{z^2-1.956z+0.9665}$$

Dividing top and bottom by $z^2$ …

$$H(z)=\frac{Y(z)}{U(z)}=\frac{486.2\times {10}^{-6}{z}^{-1}+479.1\times {10}^{-6}{z}^{-2}}{1-1.956z^{-1}+0.9665z^{-2}}$$

expanding out …

$$\begin{array}{r}\begin{array}{l}Y(z)-1.956z^{-1}Y(z)+0.9665z^{-2}Y(z)=\\ 486.2\times {10}^{-6}{z}^{-1}U(z)+479.1\times {10}^{-6}{z}^{-2}U(z)\end{array}\end{array}$$

Inverse z-transform gives …

$$\begin{array}{r}\begin{array}{l}y[n]-1.956y[n-1]+0.9665y[n-2]=\\ 486.2\times {10}^{-6}u[n-1]+479.1\times {10}^{-6}u[n-2]\end{array}\end{array}$$

in algorithmic form (compute $y[n]$ from past values of $u$ and $y$) …

$$\begin{array}{r}\begin{array}{l}y[n]=1.956y[n-1]-0.9665y[n-2]+486.2\times {10}^{-6}u[n-1]+...\\ 479.1\times {10}^{-6}u[n-2]\end{array}\end{array}$$

Block Diagram of the digital BW filter#

As Simulink Model#

digifilter.slx

open digifilter

Convert to code#

To implement:

$$y[n]=1.956y[n-1]-0.9665y[n-2]+486.2\times {10}^{-6}u[n-1]+479.1\times {10}^{-6}u[n-2]$$

/* Initialize */
Ts = 0.25e-06; /* more probably some fraction of clock speed */
ynm1 = 0; ynm2 = 0; unm1 = 0; unm2 = 0;
while (true) {
    un = read_adc;
    yn = 1.956*ynm1 - 0.9665*ynm2 + 479.1e-6*unm1 + 476.5e-6*unm2;
    write_dac(yn);
    /* store past values */
    ynm2 = ynm1; ynm1 = yn;
    unm2 = unm1; unm1 = un;
    wait(Ts);
}