Difference between revisions of "Scilab/C4/ODE-Applications/English-timed"

Latest revision as of 11:11, 10 March 2017

Time	Narration
00:01	Dear Friends, Welcome to the Spoken Tutorial on Solving ODEs using Scilab ode function
00:09	At the end of this tutorial, you will learn how to:
00:12	Use Scilab ode function
00:15	Solve typical examples of ODEs and
00:18	Plot the solution.
00:21	The typical examples will be:
00:24	Motion of simple pendulum
00:26	Van der Pol equation
00:28	and Lorenz system.
00:30	To record this tutorial, I am using
00:33	Ubuntu 12.04 as the operating system
00:36	and Scilab 5.3.3 version.
00:40	To practice this tutorial, a learner should have basic knowledge of Scilab
00:45	and should know how to solve ODEs.
00:48	To learn Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website.
00:56	The ode function is an ordinary differential equation solver.
01:01	The syntax is y equal to ode within parenthesis y zero, t zero, t and f.
01:10	Here y zero is the initial condition of the ODEs,
01:15	t zero is the initial time,
01:17	t is the time range,
01:19	and f is the function.
01:22	Consider the motion of simple pendulum.
01:25	Let theta t be the angle made by the pendulum with the vertical at time t.
01:33	We are given the initial conditions-
01:36	theta of zero is equal to pi by four and
01:39	theta dash of zero is equal to zero.
01:44	Then the position of the pendulum is given by:
01:47	theta double dash t minus g by l into sin of theta t equal to zero.
01:56	Here g equal to 9.8 m per second square is the acceleration due to gravity and
02:03	l equal to zero point five meter is the length of the pendulum.
02:11	For the given initial conditions, we have to solve the ODE within the time range zero less than equal to t less than equal to five.
02:22	We also have to plot the solution.
02:25	Let us look at the code for solving this problem.
02:29	Open Pendulum dot sci on Scilab editor.
02:34	The first line of the code defines the initial conditions of the ODE.
02:40	Then we define the intial time value. And we provide the time range.
02:46	Next, we convert the given equation to a system of first order ODEs.
02:52	We substitute the values of g and l .
02:56	Here we take y to be the given variable theta and y dash to be theta dash.
03:03	Then we call the ode function with arguments y zero, t zero, t and the function Pendulum.
03:12	The solution to the equation will be a matrix with two rows.
03:17	The first row will contain the values of y in the given time range.
03:21	The second row will contain the values of y dash within the time range.
03:27	Hence we plot both the rows with respect to time.
03:31	Save and execute the file Pendulum dot sci.
03:37	The plot shows how the values of y and y dash vary with time.
03:44	Switch to Scilab console.
03:47	If you want to see the values of y, type y on the console and press Enter.
03:54	The values of y and y dash are displayed.
03:58	Let us solve Van der Pol equation using the ode function.
04:03	We are given the equation -
04:06	v double dash of t plus epsilon into v of t square minus one into v dash of t plus v of t equal to zero.
04:20	The initial conditions are v of two equal to one and v dash of two equal to zero.
04:28	Assume epsilon is equal to zero point eight nine seven.
04:32	We have to find the solution within the time range two less than t less than ten and then plot the solution.
04:42	Let us look at the code for Van der Pol equation.
04:47	Switch to Scilab editor and open Vander pol dot sci.
04:53	We define the initial conditions of the ODEs and time and then define the time range.
05:01	Since the inital time value is given as two, we start the time range at two.
05:07	Then we define the function Vander pol and construct a system of first order ODEs.
05:15	We substitute the value of epsilon with zero point eight nine seven.
05:21	Here y refers to the voltage v.
05:25	Then we call ode function and solve the system of equations.
05:30	Finally we plot y and y dash versus t.
05:35	Save and execute the file Vander pol dot sci.
05:41	The plot showing voltage versus time is shown.
05:45	Let's move onto Lorenz system of equations.
05:50	The Lorenz system of equations is given by:
05:53	x one dash equal to sigma into x two minus x one,
06:00	x two dash equal to one plus r minus x three into x one minus x two and
06:08	x three dash equal to x one into x two minus b into x three.
06:16	The initial conditions are x one zero equal to minus ten, x two zero equal to ten and x three zero equal to twenty five.
06:29	Let sigma be equal to ten, r be equal to twenty eight and b equal to eight by three.
06:37	Switch to Scilab editor and open Lorenz dot sci.
06:44	We start by defining the initial conditions of the ODEs.
06:48	Since there are three different ODEs, there are three initial conditions.
06:54	Then we define the inital time condition and next the time range.
07:00	We define the function Lorenz and then define the given constants sigma, r and b.
07:08	Then we define the first order ODEs.
07:12	Then we call the ode function to solve the Lorenz system of equations.
07:18	We equate the solution to x.
07:21	Then we plot x one, x two and x three versus time.
07:28	Save and execute the file Lorenz dot sci.
07:33	The plot of x one, x two and x three versus time is shown.
07:39	Let us summarize this tutorial.
07:41	In this tutorial we have learnt to develop Scilab code to solve an ODE using Scilab ode function.
07:50	Then we have learnt to plot the solution.
07:53	Watch the video available at the link shown below.
07:56	It summarizes the Spoken Tutorial project.
07:59	If you do not have good bandwidth, you can download and watch it.
08:04	The spoken tutorial project Team:
08:06	Conducts workshops using spoken tutorials.
08:09	Gives certificates to those who pass an online test.
08:13	For more details, please write to contact@spoken-tutorial.org.
08:20	Spoken Tutorial Project is a part of the Talk to a Teacher project.
08:23	It is supported by the National Mission on Eduction through ICT, MHRD, Government of India.
08:31	More information on this mission is available at the link shown below.
08:36	This is Ashwini Patil, signing off.
08:38	Thank you for joining.

Contributors and Content Editors

Gaurav, PoojaMoolya, Sandhya.np14

Difference between revisions of "Scilab/C4/ODE-Applications/English-timed"

Latest revision as of 11:11, 10 March 2017

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@@ Line 7: / Line 7: @@
 |-
 | 00:01
-|Dear Friends,
+|Dear Friends,  Welcome to the Spoken Tutorial on '''Solving ODEs using Scilab ode function'''
-|-
-| 00:02
-| Welcome to the Spoken Tutorial on '''Solving ODEs using Scilab ode Function'''
 |-
@@ Line 19: / Line 15: @@
 |-
 |00:12
-|'''Use Scilab ode function'''
+| Use Scilab ode function
 |-
 |00:15
-|Solve typical examples of '''ODEs''' and
+| Solve typical examples of '''ODEs''' and
 |-
 | 00:18
-|Plot the solution.
+| Plot the solution.
 |-
@@ Line 35: / Line 31: @@
 |-
 | 00:24
-|* Motion of '''simple pendulum'''
+| Motion of '''simple pendulum'''
 |-
 | 00:26
-|* '''Van der Pol equation'''
+| '''Van der Pol equation'''
 |-
 |00:28
-|* and ''' Lorenz system'''.
+| and ''' Lorenz system'''.
 |-
@@ Line 62: / Line 58: @@
 |-
 |00:45
 |and should know how to solve '''ODEs.'''
 |-
 |00:48
 |To learn '''Scilab,''' please refer to the relevant tutorials available on the '''Spoken Tutorial''' website.
 |-
 | 00:56
+|The '''ode''' function is an '''ordinary differential equation solver'''.
-|The '''ode''' function is an '''ordinary differential equation''' solver.
 |-
 | 01:01
 ||The syntax is '''y equal to ode''' within parenthesis '''y zero, t zero, t''' and '''f'''.
 |-
 |01:10
 || Here '''y zero''' is the initial condition of the '''ODEs''',
 |-
 |01:15
 | '''t zero''' is the '''initial time''',
@@ Line 100: / Line 86: @@
 |-
 |01:19
 |and '''f''' is the '''function'''.
 |-
 |01:22
 ||Consider the motion of '''simple pendulum.'''
@@ Line 116: / Line 98: @@
 |-
 | 01:33
 |We are given the initial conditions-
 |-
 |01:36
+|'''theta of zero''' is equal to '''pi by four''' and
-|'''theta zero''' is equal to '''pi by four''' and
 |-
 |01:39
 | '''theta dash of zero''' is equal to '''zero.'''
 |-
 |01:44
+| Then the position of the '''pendulum''' is given  by:
-| Then the position of the '''pendulum''' is given  by:
 |-
 | 01:47
 || '''theta double dash t minus g by l into sin of theta t equal to zero.'''
@@ Line 149: / Line 123: @@
 |-
 |02:03
-|''' l equal to zero point five meter''' is the '''length''' of the '''pendulum.'''
+|''' l equal to zero point five meter''' is the length of the '''pendulum.'''
 |-
@@ Line 165: / Line 139: @@
 |-
 | 02:29
-|Open '''pendulum dot sci''' on '''Scilab editor.'''
+|Open '''Pendulum dot sci''' on '''Scilab editor.'''
 |-
@@ Line 180: / Line 154: @@
 |-
 | 02:52
 |We substitute the values of '''g''' and '''l''' .
 |-
 | 02:56
 ||Here we take '''y''' to be the given '''variable theta''' and '''y dash''' to be '''theta dash.'''
 |-
 |03:03
+|Then we call the '''ode''' function with arguments '''y zero, t zero, t''' and the function '''Pendulum.'''
-|Then we call the '''ode function''' with '''arguments y zero, t zero, t''' and the '''function Pendulum.'''
 |-
 | 03:12
-| The solution to the '''equation''' will be a '''matrix''' with two '''rows.'''
+| The solution to the equation will be a '''matrix''' with two '''rows.'''
 |-
 | 03:17
 | The first '''row''' will contain the values of '''y''' in the given '''time range.'''
 |-
 | 03:21
 | The second '''row''' will contain the values of '''y dash ''' within the '''time range. '''
 |-
 |03:27
 |Hence we plot both the '''rows''' with respect to '''time. '''
 |-
 |03:31
 |Save and execute the file '''Pendulum dot sci'''.
 |-
 | 03:37
 |The plot shows how the values of '''y''' and '''y dash''' vary with '''time. '''
 |-
 | 03:44
 |Switch to '''Scilab console'''.
 |-
 | 03:47
 | If you want to see the values of '''y,''' type '''y''' on the '''console''' and press '''Enter.'''
 |-
 | 03:54
 |The values of '''y''' and '''y dash''' are displayed.
 |-
 | 03:58
+|Let us solve '''Van der Pol equation''' using the '''ode''' function.
-|Let us solve '''Van der Pol equation''' using the '''ode function.'''
 |-
 | 04:03
 |We are given the '''equation '''-
 |-
 | 04:06
 |'''v double dash of t plus epsilon into v of t square minus one into v dash of t plus v of t equal to zero.'''
 |-
 | 04:20
+|The initial conditions are '''v of two equal to one''' and '''v dash of two equal to zero. '''
-|The initial '''conditions''' are '''v of two equal to one''' and '''v dash of two equal to zero. '''
 |-
 | 04:28
 |Assume '''epsilon is equal to zero point eight nine seven. '''
 |-
 |04:32
 |We have to find the solution within the '''time range two less than t less than ten''' and then '''plot'''  the solution.
 |-
 | 04:42
 |Let us look at the code for '''Van der Pol equation. '''
 |-
 | 04:47
+| Switch to '''Scilab editor''' and open '''Vander pol dot sci.'''
-| Switch to '''Scilab editor''' and open '''vander pol dot sci.'''
 |-
 | 04:53
 | We define the initial conditions of the '''ODEs''' and '''time''' and then define the '''time range. '''
 |-
 | 05:01
 |Since the '''inital time value''' is given as '''two''', we start the time range at two.
 |-
 | 05:07
-| Then we define the '''function vander pol''' and construct a system of '''first order ODEs.'''
+| Then we define the '''function Vander pol''' and construct a system of '''first order ODEs.'''
 |-
@@ Line 317: / Line 249: @@
 |-
 | 05:21
 |Here '''y''' refers to the '''voltage v.'''
 |-
 | 05:25
+| Then we call '''ode''' function and solve the system of '''equations.'''
-| Then we call '''ode function''' and solve the system of '''equations.'''
 |-
 | 05:30
 | Finally we plot '''y''' and '''y dash versus t.'''
 |-
 | 05:35
+| Save and execute the file '''Vander pol dot sci.'''
-| Save and execute the file '''vander pol dot sci.'''
 |-
 | 05:41
 |The '''plot''' showing '''voltage versus time''' is shown.
 |-
 | 05:45
 |Let's move onto '''Lorenz system of equations.'''
 |-
 | 05:50
 |The '''Lorenz system of equations''' is given by:
 |-
 | 05:53
 |'''x one dash equal to sigma into x two minus x one''',
 |-
 | 06:00
 |'''x two dash equal to one plus r minus x three into x one minus x two''' and
 |-
 | 06:08
 |'''x three dash equal to x one into x two minus b into x three.'''
 |-
 | 06:16
+|The initial conditions are '''x one zero  equal to minus ten''',  '''x two zero equal to ten''' and '''x three zero equal to twenty five.'''
-|The initial conditions are '''x one zero''' equal to '''minus ten,  x two zero''' equal to '''ten''' and '''x three zero''' equal to '''twenty five.'''
 |-
 | 06:29
+|Let '''sigma''' be equal to '''ten,  r''' be equal to '''twenty eight''' and '''b''' equal to '''eight by three.'''
-|Let '''sigma''' be equal to '''ten,  r''' be equal to '''twenty eight and b''' equal to '''eight by three.'''
 |-
 | 06:37
 |Switch to '''Scilab editor''' and open '''Lorenz dot sci'''.
 |-
 | 06:44
 |We start by defining the initial conditions of the '''ODEs.'''
 |-
 | 06:48
 |Since there are three different '''ODEs,''' there are three initial conditions.
 |-
 | 06:54
 |Then we define the '''inital time''' condition and next the '''time range.'''
 |-
 | 07:00
+|We define the '''function Lorenz''' and then define the given constants '''sigma, r''' and '''b.'''
-|We define the '''function Lorenz''' and then define the given constants '''sigma, r and b.'''
 |-
 | 07:08
 |Then we define the '''first order ODEs.'''
 |-
 | 07:12
+|Then we call the '''ode''' function to solve the '''Lorenz system of equations.'''
-|Then we call the '''ode function''' to solve the '''Lorenz system of equations.'''
 |-
 | 07:18
 |We equate the solution to '''x.'''
 |-
 | 07:21
 |Then we '''plot x one, x two''' and '''x three versus time.'''
 |-
 | 07:28
 |Save and execute the file '''Lorenz dot sci.'''
 |-
 | 07:33
 |The '''plot''' of '''x one, x two''' and '''x three versus time''' is shown.
 |-
 | 07:39
 |Let us summarize this tutorial.
 |-
 | 07:41
 |In this tutorial we have learnt to develop '''Scilab code''' to solve an '''ODE''' using '''Scilab ode function'''.
 |-
 | 07:50
 |Then we have learnt to '''plot''' the solution.
@@ Line 476: / Line 357: @@
 |-
 | 07:56
 | It summarizes the Spoken Tutorial project.
 |-
 |07:59
 ||If you do not have good bandwidth, you can download and watch it.
 |-
 |08:04
 ||The spoken tutorial project Team:
 |-
 |08:06
 ||Conducts workshops using spoken tutorials.
 |-
 |08:09
 ||Gives certificates to those who pass an online test.
 |-
 |08:13
 ||For more details, please write to contact@spoken-tutorial.org.
 |-
 |08:20
 |Spoken Tutorial Project is a part of the Talk to a Teacher project.
 |-
 | 08:23
 | It is supported by the National Mission on Eduction through ICT, MHRD, Government of India.
 |-
 | 08:31
 |More information on this mission is available at the link shown below.
 |-
 | 08:36
 |This is Ashwini Patil, signing off.
 |-
 |08:38
 | Thank you for joining.