Scilab/C4/Control-systems/English
Title of script: Advanced Control Systems
Author: Manas, Shamika
Keywords: control, continuous time, response
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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:
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Slide 4-System Requirement slide | To record this tutorial, I am using
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Slide 5- Prerequisite slide | Before practising this tutorial, a learner should have basic knowledge of Scilab and control systems.
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Slide 6 | In this tutorial, I will describe how to define second-order linear system.
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Switch to the Scilab Console Window and type:
s = poly(0, ’s’) |
Let us switch to the Scilab console window.
Here type:
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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.
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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:
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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
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Then type
plot(t, y1); Press Enter. |
Then type
plot open paranthesis t comma y one close paranthesis semicolon
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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
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Type y2 = csim(u2, t, sysG);
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Then type
y two is equal to c sim open paranthesis u two comma t comma sys capital G close the bracket semicolon
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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
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Slide 9 | Response Plot plots both the input and the output on the same graph.
As expected,
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Slide 10 |
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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.
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Type bode(sysG, fr) and press Enter. | Then type
bode open paranthesis sys capital G comma fr close paranthesis and press Enter.
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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.
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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.
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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.
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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
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Switch to Scilab console.
Press Enter
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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
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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
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Otherwise, you can type
numer open paranthesis sys three close paranthesis The numerator of sys three is shown. |
Type sys3(3) >> Press Enter
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To get the denominator, type
sys three open paranthesis three close paranthesis. Press Enter.
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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:
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Show Slide 19
Title: About the Spoken Tutorial Project
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* Watch the video available at the following link
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Show Slide 20
Title: Spoken Tutorial Workshops The Spoken Tutorial Project Team
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The Spoken Tutorial Project Team
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Show Slide
Title: Acknowledgement 21
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* Spoken Tutorial Project is a part of the Talk to a Teacher project
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On previous slide | This is Ashwini Patil signing off. Thank you for joining. |