Scilab/C4/Controlsystems/English
Title of script: Advanced Control Systems
Author: Manas, Shamika
Keywords: control, continuous time, response



Slide 1  Dear friends, welcome to the spoken tutorial on “Advanced Control of Continuous Time systems” 
Slide 2,3Learning Objective Slide  At the end of this tutorial, you will learn how to:

Slide 4System Requirement slide  To record this tutorial, I am using

Slide 5 Prerequisite slide  Before practising this tutorial, a learner should have basic knowledge of Scilab and control systems.

Slide 6  In this tutorial, I will describe how to define secondorder linear system.

Switch to the Scilab Console Window and type:
s = poly(0, ’s’) 
Let us switch to the Scilab console window.
Here type:

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.

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:

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 semicolon Press Enter. 
Then type
y1 = csim(’step’, t, sysG); Press Enter. 
Then type

Then type
plot(t, y1); Press Enter. 
Then type
plot open paranthesis t comma y one close paranthesis semicolon

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
Press Enter. 
Let us switch to the Scilab console window.
U two is equal to sine open paranthesis t close paranthesis semicolon

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

Then type
y two is equal to c sim open paranthesis u two comma t comma sys capital G close the bracket semicolon

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

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

Slide 10 

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

Type bode(sysG, fr) and press Enter.  Then type
bode open paranthesis sys capital G comma fr close paranthesis and press Enter.

Slide 12  Let us define another system.
We have an overdamped 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.

Type sys2 = syslin('c', 9/p)
Press Enter. 
Then type this on your console.
sys two is equal to syslin open paranthesis open single quote c close single quote comma nine divided by p close paranthesis
t equal to zero colon zero point one colon ten semicolon 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.

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.

Switch to the Scilab Console Window and type this on your Scilab Console
Alternatively: Type this on your Console
Then type this on your Scilab Console
> sys4 = syslin(’c’,g) and press Enter

Switch to Scilab console.
Press Enter

Both ways, we get the same output;
six plus s over 19 plus six s plus s square  
Slide 15  The variable ’sys’ is of type ’rational’.
Its numerator and denominator can be extracted by various ways. 
Slide 16 
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
> sys3(2) and press Enter

Switch to Scilab console.
sys three open paranthesis two close paranthesis
This gives the numerator of the rational function 'sys three' that is 6 + s 
Type numer(sys3) >> Press Enter

Otherwise, you can type
numer open paranthesis sys three close paranthesis The numerator of sys three is shown. 
Type sys3(3) >> Press Enter

To get the denominator, type
sys three open paranthesis three close paranthesis. Press Enter.

Type
denom(sys3) >> Press Enter. 
You can also type denom open paranthesis sys three close paranthesis.
Press Enter. 
Type
plzr(sys3) >> 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:

Show Slide 19
Title: About the Spoken Tutorial Project

* Watch the video available at the following link

Show Slide 20
Title: Spoken Tutorial Workshops The Spoken Tutorial Project Team

The Spoken Tutorial Project Team

Show Slide
Title: Acknowledgement 21

* Spoken Tutorial Project is a part of the Talk to a Teacher project

On previous slide  This is Ashwini Patil signing off. Thank you for joining. 