Scilab/C4/Control-systems/English-timed
From Script | Spoken-Tutorial
Time | Narration |
00:01 | Dear Friends, |
00:02 | |
00:09 | At the end of this tutorial, you will learn how to |
00:12 | Define a continuous time system: second and higher order
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00:17 | Plot response to step and sine inputs
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00:20 | Do a Bode plot |
00:22 | Study numer and denom Scilab functions
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00:26 | Plot poles and zeros of a system |
00:30 | To record this tutorial, I am using |
00:33 | Ubuntu 12.04 as the operating system with |
00:36 | Scilab 5.3.3 version |
00:40 | Before practising this tutorial, a learner should have basic knowledge of Scilab and control systems. |
00:48 | For Scilab, please refer to the Scilab tutorials available on the Spoken Tutorial website. |
00:55 | In this tutorial, I will describe how to define second-order linear system. |
01:02 | So, first we have to define complex domain variable 's'. |
01:08 | Let us switch to the Scilab console window. |
01:11 | Here type s equal to poly open paranthesis zero comma open single quote s close single quote close paranthesis , press Enter.
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01:25 | The output is 's'.
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01:27 | There is another way to define 's' as continuous time complex variable. |
01:32 | On the console window, type:
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01:35 | s equal to percentage s, press Enter.
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01:41 | Let us study the syslin Scilab command. |
01:44 | Use the Scilab function ’syslin’ to define the continuous time system.
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01:51 | G of s is equal to 2 over 9 plus 2 s plus s square
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01:58 | Use csim with step option, to obtain the step response and then plot the step response.
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02:06 | Let us switch to the Scilab console window. |
02:09 | Here type: sys capital G equal to syslin open paranthesis open single quote c close single quote comma two divide by open paranthesis s square plus two asterik s plus nine close paranthesis close paranthesis |
02:32 | Here c is used as we are defining a continuous time system. |
02:38 | Press Enter |
02:40 | The output is linear second order system represented by
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02:44 | 2 over 9 plus 2 s plus s square
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02:49 | Then type t equal to zero colon zero point one colon ten semicolon
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02:57 | Press Enter.
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02:59 | 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 |
03:15 | Press Enter.
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03:17 | Then type plot open paranthesis t comma y one close paranthesis semicolon |
03:24 | Press Enter. |
03:26 | The output will display the step response of the given second order system. |
03:33 | Let us study the Second Order system response for sine input.
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03:39 | Sine inputs can easily be given as inputs to a second order system to a continuous time system. |
03:47 | Let us switch to the Scilab console window.
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03:51 | Type U two is equal to sine open paranthesis t close paranthesis semicolon |
03:59 | Press Enter.
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04:01 | 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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04:15 | Press Enter |
04:17 | Here we are using sysG, the continuous time second order system we had defined earlier.
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04:25 | 'Then type plot open paranthesis t comma open square bracket u two semicolon y two close square bracket close paranthesis
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04:39 | Make sure that you place a semicolon between u2 and y2 because u2 and y2 are row vectors of the same size.
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04:50 | 'Press Enter.
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04:52 | This plot shows the response of the system to a step input and sine input. It is called the response plot. |
05:01 | Response Plot plots both the input and the output on the same graph. |
05:06 | As expected, the output is also a sine wave, and
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05:11 | there is a phase lag between the input and output |
05:15 | Amplitude is different for the input and the output, as it is being passed through a transfer function.
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05:23 | This is a typical under-damped example.
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05:26 | Let us plot bode plot of 2 over 9 plus 2 s plus s square |
05:32 | Please note command 'f r e q' is a Scilab command for frequency response. |
05:39 | Do not use f r e q as a variable !!
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05:44 | Open the Scilab console and type |
05:47 | f r is equal to open square bracket zero point zero one colon zero point one colon ten close square bracket semicolon.
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06:00 | Press Enter. |
06:03 | The frequency is in Hertz. |
06:06 | Then type bode open paranthesis sys capital G comma fr close paranthesis
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06:15 | and press Enter.'
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06:17 | The bode plot is shown |
06:20 | Let us define another system.
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06:23 | We have an over-damped system p equal to s square plus nine s plus nine
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06:32 | Let us plot step response for this system.
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06:36 | Switch to Scilab console. |
06:38 | Type this on your console. |
06:40 | p is equal to s square plus nine asterik s plus nine |
06:47 | and then press Enter. |
06:49 | Then type this on your console. |
06:51 | sys two is equal to syslin open paranthesis open single quote c close single quote comma nine divided by p close paranthesis |
07:04 | and press Enter. |
07:07 | Then type t equal to zero colon zero point one colon ten semicolon |
07:14 | Press Enter.
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07:17 | y is equal to c sim open paranthesis open single quote step close single quote comma t comma sys two close the paranthesis semicolon |
07:31 | Press Enter. |
07:33 | Then type plot open paranthesis t comma y close paranthesis |
07:39 | Press Enter. |
07:41 | The response plot for over damped system is shown. |
07:46 | To find the roots of p type this on your console - |
07:49 | Roots of p and press Enter.
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07:54 | These roots are the poles of the system sys two |
07:59 | The roots or poles of the system are shown. |
08:02 | Please plot Step response for this system along similar lines, as for over damped system. |
08:11 | G of s is equal to 2 over 9 plus 6 s plus s square which is a critically damped system |
08:20 | Then G of s is equal to two over 9 plus s square which is an undamped system |
08:28 | G of s is equal to 2 over 9 minus 6 s plus s square which is an unstable system |
08:36 | Check response to sinusoidal inputs for all the cases and plot bode plot too. |
08:45 | Switch to Scilab console. ; |
08:48 | For a general transfer function, the numerator and denominator can be specified separately. |
08:55 | Let me show you how.
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08:57 | Type on console |
08:59 | 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 |
09:19 | Press Enter
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09:21 | Another way of defining a system, is to type |
09:24 | 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
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09:40 | Press Enter.
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09:42 | Then type this on your console |
09:44 | sys four is equal to syslin open paranthesis open single quote c close single quote comma g close paranthesis |
09:55 | Press Enter. |
09:58 | Both ways, we get the same output;
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10:01 | six plus s over 19 plus six s plus s square
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10:07 | The variable ’sys’ is of type ’rational’. |
10:10 | Its numerator and denominator can be extracted by various ways. |
10:16 | Sys of two , numer of sys or numer of g gives the numerator.
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10:22 | The denominator can be calculated using sys(3) or denom of sys functions.
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10:30 | The poles and zeros of the system can be plotted using p l z r function. |
10:37 | The syntax is p l z r of sys
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10:41 | The plot shows x for poles and circles for zeros.
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10:46 | Switch to Scilab console. |
10:48 | Type this on your Scilab console. |
10:50 | sys three open paranthesis two close paranthesis
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10:55 | Press Enter. |
10:56 | This gives the numerator of the rational function 'sys three' that is 6 + s
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11:03 | Otherwise, you can type
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11:05 | numer open paranthesis sys three close paranthesis. |
11:11 | Press Enter |
11:13 | The numerator of system three is shown. |
11:17 | To get the denominator, type |
11:19 | sys three open paranthesis three close paranthesis. Press Enter.
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11:26 | The denominator of the function is shown. |
11:30 | You can also type denom open paranthesis sys three close paranthesis. |
11:36 | Press Enter.
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11:38 | Then type p l z r open paranthesis sys three close paranthesis. |
11:44 | Press Enter. |
11:47 | The output graph plots the poles and zeros. |
11:50 | It shows cross and circle' for poles and zeros of the system respectively.
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11:58 | It is plotted on the complex plane.
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12:01 | In this tutorial, we have learnt how to: |
12:03 | Define a system by its transfer function. |
12:08 | Plot step and sinusoidal responses. |
12:11 | Extract poles and zeros of a transfer function.
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12:15 | Watch the video available at the following link |
12:19 | It summarises the Spoken Tutorial project
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12:22 | If you do not have good bandwidth, you can download and watch it |
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12:43 | Spoken Tutorial Project is a part of the Talk to a Teacher project
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12:47 | It is supported by the National Mission on Eduction through ICT, MHRD, Government of India. |
12:55 | More information on this mission is available at spoken-tutorial.org/NMEICT-Intro |
13:06 | This is Ashwini Patil signing off. |
13:08 | Thank you for joining Good Bye. |