Scilab/C4/Control-systems/English

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 and 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 c sim 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: Type this on your Console --> 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.
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On previous slide	This is Ashwini Patil signing off. Thank you for joining.

Contributors and Content Editors

Lavitha Pereira, Nancyvarkey