Lab 12: Poles and Zeros

Table of Contents

1. Introduction
2. Procedure
3. Feedback

1. Introduction

This exercise shows how to using MATLAB on find transfer functions of differential equation systems to discover the various properties related to the linear systems, and generate plots to demonstrate these properties.

2. Procedure

2.1. Analysis of differential equations

\frac{d^{2} y}{d t^{2}} + 4 \frac{d^{} y}{d t^{}} + 3 y (t) = \frac{d^{} x}{d t^{}} + 3 x (t)

The transfer function is H = tf([1,3],[1,4,3]):

H (s) = \frac{s + 3}{s^{2} + 4 s + 3}

Poles are -3, -1, zero is -3.

Figure 1. Pole-zero map of H = tf([1,3],[1,4,3])

The impluse response converges to zero so the system is asymptotically stable.

Figure 2. Impluse response of H = tf([1,3],[1,4,3])

\frac{d^{3} y}{d t^{3}} + 3 \frac{d^{2} y}{d t^{2}} + \frac{d^{} y}{d t^{}} - 5 y (t) = \frac{d^{2} x}{d t^{2}} - 7 \frac{d^{} x}{d t^{}} + 12 x (t)

The transfer function is H = tf([1,-7,12],[1,3,1,-5]):

Poles are -2+j, -2-j, 1, zeros are 4, 3.

Figure 3. Pole-zero map of H = tf([1,-7,12],[1,3,1,-5])

The impluse response diverges so the system is unstable.

Figure 4. Impluse response of H = tf([1,-7,12],[1,3,1,-5])

\frac{d^{4} y}{d t^{4}} + 5 \frac{d^{3} y}{d t^{3}} + 9 \frac{d^{2} y}{d t^{2}} + 5 \frac{d^{} y}{d t^{}} = \frac{d^{2} x}{d t^{2}} - 10 \frac{d^{} x}{d t^{}} + 21 x (t)

The transfer function is H = tf([1 -10 21],[1 5 9 5 0]):

Poles are 0, -2+j, -2-j, -1, zeros are 7, 3.

Figure 5. Pole-zero map of H = tf([1 -10 21],[1 5 9 5 0])

The impluse response converges to a non-zero number so the system is stable.

The curve converges to a positive number

Figure 6. Impluse response of H = tf([1 -10 21],[1 5 9 5 0])

2.1.1. Design a stable transfer function

The transfer function is H = tf([1 -6 -16], [1 10 21 0]):

Poles are 0, -3, -7, zeros are -2, 8.

Figure 7. Pole-zero map of H = tf([1 -6 -16], [1 10 21 0])

The impluse response is stable because poles are real numbers and includes a zero.

Figure 8. Impluse response of H = tf([1 -6 -16], [1 10 21 0])

2.1.2. Design an asymptotically stable transfer function

The transfer function is H = tf([1 -7 10], [1 3 12 10]):

Poles are -1-3j, -1+3j, -1, zeros are 2, 5.

Figure 9. Pole-zero map of H = tf([1 -7 10], [1 3 12 10])

The impluse response is asymptotically stable because poles contains a pair of counjugate numbers.

Figure 10. Impluse response of H = tf([1 -7 10], [1 3 12 10])

2.1.3. Design an unstable transfer function

The transfer function is H = tf([1 0 -1], [1 3 -4 -12]):

Poles are 2, -3, -2, zeros are -1, 1.

Figure 11. Pole-zero map of H = tf([1 0 -1], [1 3 -4 -12])

The impluse response is unstable because poles contains a positive number.

Figure 12. Impluse response of H = tf([1 0 -1], [1 3 -4 -12])

2.2. RLC circuit

H (s) = \frac{\frac{1}{s C}}{R + s L + \frac{1}{s C}} = \frac{1}{s^{2} L C + s R C + 1}

Equation 1. Transfer function of the RLC circuit

For R = 40; L = 3*10^-3; C = 5*10^-6;:

Poles show signs of asymptotically stable

Figure 13. Pole-zero map of the system

The impluse response is asymptotically stable, as the poles contains a pair of counjugate complex numbers.

Figure 14. Impluse response of the system

Figure 15. Step response of the system

Setting R = 1, 2, 3, 10, we can discover that as resistance increases, the oscillation drastics reduces, and the pole magnitude is always the same, only the angle varies.

2.3. Pole-zero cancelation

For H = tf([1 -7 12], [1 1 0 -2]), we could find that it has poles 1, -1+j, -1-j, zeros 4, 3.

Figure 16. Impluse response of H = tf([1 -7 12], [1 1 0 -2])

H (s) = \frac{(s - 4) (s - 3)}{(s + 1 - j) (s + 1 + j) (s - 1)}

Equation 2. Factorized

H
= tf([1 -7 12], [1 1 0 -2])

\begin{aligned} H (s) & = \frac{(s - 4) (s - 3) (s - 1)}{(s + 1 - j) (s + 1 + j) (s - 1)} \\ = \frac{(s - 4) (s - 3)}{(s + 1 - j) (s + 1 + j)} \\ = \frac{s^{2} - 7 s + 12}{s^{2} + 2 s + 2} \end{aligned}

Equation 3. Cancelation

Figure 17. The pole-zero canceled impluse response

3. Feedback

This lab is a very intensive exercise. While I do apprciate that MATLAB is the most popular toolbox for engineers, I believe that using a CAS would probably be more effective for this type of task.

The tf used in the exercise needs the Control System Toolbox add-on to work, and there are other add-ons has the function of the same name but for different purpose. I think this needs a clearification.