# Scilab/C4/Control-systems/English

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Title of script: Advanced Control Systems

Author: Manas, Shamika

Keywords: control, continuous time, response

Visual Cue
Narration
Slide 1 Dear Friends,

Welcome to the spoken tutorial on “Advanced Control of Continuous Time systems

Slide 2,3-Learning Objective Slide At the end of this tutorial, you will learn how to:

1.Define a continuous time system: second and higher order

2.Plot response to step and sine inputs

3.Do a Bode plot

4.Study numer and denom Scilab functions

5. Plot poles and zeros of a system

Slide 4-System Requirement slide To record this tutorial, I am using Ubuntu 12.04 as the operating system with Scilab 5.3.3 version
Slide 5- Prerequisite slide Before practising this tutorial, a learner should have basic knowledge of Scilab and control systems.

For scilab, please refer to the Scilab tutorials available on the Spoken Tutorial website.

Slide 6 In this tutorial, I will describe how to define second-order linear system.

So, first we have to define complex domain variable s.

Switch to the Scilab Console Window and type:

s = poly(0, ’s’)

Let us switch to the Scilab Console Window.

Here type:

s equal to poly open paranthesis zero comma open single quote s close single quote close paranthesis

and press Enter.

Display the output polynomial The output is s.

On the console window type:

s = %s

and press Enter.

There is another way to define 's' as continuous time complex variable

On the Console window type:

s equal to percentage s

and press Enter.

Slide 7 Let us study the syslin Scilab command

Use the Scilab function ’syslin’ to define the continuous time system

G of s is equal to 2 over 9 plus 2 s plus s square

Use csim with step option to obtain the step response and then plot the step response.

Switch to the Scilab Console Window and type:

sysG = syslin(’c’,2/(sˆ2+2*s+9))

Let us switch to the Scilab Console Window.

Here type:

sys capital G equal to syslin open paranthesis open single quote c close single quote comma two divided by open paranthesis s square plus two asterik s plus nine close paranthesis close paranthesis

Here c is used as we are defining a continuous time system

Press Enter

Display the output generated The output is linear second order system represented by

2 over 9 plus 2 s plus s square

Type:

t=0:0.1:10;

Press Enter.

Then type

t equal to zero colon zero point one colon ten semi colon

Press Enter.

Then type

y1 = csim(’step’, t, sysG);

Press Enter.

Then type

y one is equal to c sim open paranthesis open single quote step close single quote comma t comma sys capital G close the paranthesis semicolon

Press Enter.

Then type

plot(t, y1);

Press Enter.

Then type

plot open paranthesis t comma y one close paranthesis semicolon

Press Enter.

Display the output generated The output will display the step response of the given second order system.
Slide 8 Let us study the Second Order system response for sine input.

Sine inputs can easily be given as inputs to a second order system to a continuous time system.

Switch to the Scilab Console Window and type this on your Scilab Console

u2=sin(t);

Press Enter.

Let us switch to the Scilab Console Window.

Type

U two is equal to sine open paranthesis t close paranthesis semi colon

Press Enter.

Type y2 = csim(u2, t, sysG);

Press Enter.

Then type

y two is equal to csim open paranthesis u two comma t comma sys capital G close the bracket semicolon

Press Enter.

Here we are using sysG, the continuous time second order system we had defined earlier.

Type plot(t, [u2; y2])

Press Enter.

Then type

plot open paranthesis t comma open square bracket u two semicolon y two close square bracket close paranthesis

Make sure that you place a Semicolon between u2 and y2 because u2 and y2 are row vectors of the same size

Press Enter.

This plot shows the response of the system to a step input and sine input. It is called the response plot.

Slide 9, 10 Response Plot plots both the input and the output on the same graph.

As expected,

• the output is also a sine wave, and
• there is a phase lag between the input and output
• amplitude is different for the input and the output as it is being passed through a transfer function.
• This is a typical under-damped example

Slide 11 Let us plot bode plot of 2 over 9 plus 2 s plus s square

Please note command 'f r e q' is a Scilab command for frequency response.

Do not use f r e q as a variable !!

Switch to the Scilab console and type

fr = [0.01:0.1:10]; // Hertz

Press Enter.

Open the Scilab Console and type

f r is equal to open square bracket zero point zero one colon zero point one colon ten close square bracket semicolon.

Press Enter.

The frequency is in Hertz.

Type bode(sysG, fr) and press Enter. Then type

bode open paranthesis sys capital G comma fr close paranthesis

and press Enter.

The bode plot is shown

Slide 12 Let us define another system

We have an over-damped system p equal to s square plus nine s plus nine

Let us plot step response for this system

Switch to the Scilab console and type

p=s^2 +9*s+9

Press Enter.

Switch to Scilab console

Type this on your Scilab Console

p is equal to s square plus nine asterik s plus nine

and then press Enter.

Type sys2 = syslin('c', 9/p)

Press Enter.

Then type this on your Scilab Console

sys two is equal to syslin open paranthesis open single quote c close single quote comma nine divided by p close paranthesis

and press Enter.

Then type

t equal to zero colon zero point one colon ten semi colon

Press Enter.

y is equal to c sim open paranthesis open single quote step close single quote comma t comma sys two close the paranthesis semicolon

Press enter

Then type plot open paranthesis t comma y close paranthesis.

Press enter

The response plot for over damped system is shown.

roots(p)

and press Enter.

To find the roots of p type this on your on Scilab console.

Roots of p

and press Enter.

These roots are the poles of the system sys two

Display the output The roots or poles of the system are shown
Slide 13, 14 Please plot Step response for this system

along similar lines as for over damped system.

G of s is equal to 2 over 9 plus 6 s plus s square which is a critically damped system

Then G of s is equal to two over 9 plus s square which is an undamped system

G of s is equal to 2 over 9 minus 6 s plus s square which is an unstable system

Check response to sinusoidal inputs for all the cases and plot bode plot too.

Switch to the Scilab Console Window and type this on your Scilab Console

--> sys3 = syslin(’c’,s+6,sˆ2+6*s+19) and press Enter

Alternatively:

--> g = (s+6)/(sˆ2+6*s+19) and press Enter

Then type this on your Scilab Console

--> sys4 = syslin(’c’,g) and press Enter

Switch to Scilab console.

For a general transfer function, the numerator and denominator can be specified separately. Let me show you how.

Type this on your Scilab Console

sys three is equal to syslin open paranthesis open single quote c close single quote comma s plus six comma s square plus six asterik s plus nineteen close paranthesis

Press Enter

Another way of defining a system is to type

g is equal to open paranthesis s plus six close paranthesis divided by open paranthesis s square plus six asterik s plus nineteen close paranthesis

Press enter

Then type this on your Scilab Console

sys four is equal to syslin open paranthesis open single quote c close single quote comma g close paranthesis

Press enter

Both ways, we get the same output

six plus s over 19 plus six s plus s square

Slide 15,16 The variable ’sys’ is of type ’rational’.

Its numerator and denominator can be extracted by various ways.

Sys of two , numer of sys or numer of g gives the numerator

The denominator can be calculated using sys(3) or denom of sys functions.

Slide 17 The poles and zeros of the system can be plotted using p l z r function.

The syntax is p l z r of sys

The plot shows x for poles and circles for zeros.

Switch to Scilab and type this on your Scilab Console

--> sys3(2) and press Enter

Type

numer(sys3)

Switch to Scilab console.

Type this on your Scilab Console

sys three open paranthesis two close paranthesis

Press enter

This gives the numerator of the rational function ’sys three’ that is 6 + s

Otherwise you can type

numer open paranthesis sys three close paranthesis

The numerator of sys three is shown

To get the denominator type

sys three open paranthesis three close paranthesis. Press enter

The denominator of the function is shown.

You can also type denom open paranthesis sys three close paranthesis. Press enter

Then type p l z r open paranthesis sys three close paranthesis. Press enter

Display output The output graph plots the poles and zeros.

It shows cross and circle for poles and zeros of the system respectively

It is plotted on the complex plane.

Slide 18 In this tutorial we have learnt how to:
• Define a system by its transfer function.
• Plot step and sinusoidal responses.
• Extract poles and zeros of a transfer function.

Show Slide 19

Title: About the Spoken Tutorial Project

• It summarises the Spoken Tutorial project
• If you do not have good bandwidth, you can download and watch it

* Watch the video available at the following link
• It summarises the Spoken Tutorial project
• If you do not have good bandwidth, you can download and watch it

Show Slide 20

Title: Spoken Tutorial Workshops

The Spoken Tutorial Project Team

• Conducts workshops using spoken tutorials
• Gives certificates for those who pass an online test
• For more details, please write to contact@spoken-tutorial.org

The Spoken Tutorial Project Team
• Conducts workshops using spoken tutorials
• Gives certificates for those who pass an online test
• For more details, please write to contact at spoken hyphen tutorial dot org

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Title: Acknowledgement 21

• Spoken Tutorial Project is a part of the Talk to a Teacher project
• It is supported by the National Mission on Education through ICT, MHRD, Government of India