An Introduction to Xcos - INFLIBNET Centrecontent.inflibnet.ac.in/data-server/eacharya-documents/... · 2017-04-20 · 1 Xos/Scicos: Scilab connected object simulator !visual editor

Ordinary Differential Equations Why Scilab/Xcos? How do I construct dynamic models? More examples

An Introduction to Xcos

A. B. Raju Ph.D

Professor and Head,Electrical and Electronics Engineering Department,

B. V. B. College of Engineering and Technology,HUBLI-580 031, KARNATAKA

[email protected]

28th September 2010

A. B. Raju BVBCET, Hubli



Outline of Presentation

1 Ordinary Differential Equations

2 Why Scilab/Xcos?

3 How do I construct dynamic models?

4 More examples




Ordinary Differential Equations

These equations are of great importance in Science andEngineering

Many physical laws and relations appear mathematically inthe form of differential equations

A physical law involving a rate of change of function, such asvelocity or acceleration–it leads to a differential equation




Example (1): Experiments show that a radioactive substancedecomposes at a rate proportional to the amount present. Startingwith a given amount of substance, say, 2 gms, at a certain time,say, t=0, what can be said about the amount available at a latertime?

dy

dt= ky

where k is a constant, whose value depends on the radioactivesubstance.




Example (2): The tank shown below contains 200 gal of water inwhich 40 lb of salt are dissolved. Five gal of water, each containing2 lb of dissolved salt, run into the tank per minute, and themixture, kept uniform by stirring, run out at the same rate. Findthe amount of salt y(t) in the tank at any time t.

dy

dt= inflow rate− outflow rate

= 10− 0.025y




Example (3): A body slides on a surface, it experiences frictionforce F. Experiments show that |F | = µ|N |. Assume that the bodyweighs 45 nt (about 10 lb), µ = 0.20, α = 300, the slide is 10 mtrlong, initial velocity is zero. Find the velocity of the body at theend of the slide.

W

α

Ν

y

x

(W/g )*a

F

WNv(t)

s(t)

Equating horizontal and vertical forces, we get

N = W cosα

F = µN

= µW cosα

0 = W sinα− µW cosα− W

g

dx

dtA. B. Raju BVBCET, Hubli



Example (4): Find the transfer function, X(s)/F(s) for the systemgiven below.

M

K

D

x(t)

f(t)

Equating horizontal forces, we get

f(t) = Md2x(t)

dt2+D

dx(t)

dt+Kx(t)




Example (5): RL circuit

R

V

L

i

−

+

The corresponding differential equation is given by:

di

dt=

1

L(V − iR)




How do I solve such differential equations?

1 Get all the data required for the solution (viz., all constants,initial conditions, time-step etc)

2 Select a suitable numerical integration algorithm among manyavailable methods

3 Write a program to implement selected algorithm using anyprograming languages (viz., C, C++, Fortran, Scilab, Matlabetc)

4 Collect data required for visualization

5 Interpret results so obtained








































Let us see the implementation of aforementioned steps in action:Run rlckt1.sce from Scilab




Why Scilab/Xcos?

1 Xos/Scicos: Scilab connected object simulator → visual editor

2 Scilab package for modelling and simulation of dynamicsystems

3 Dynamic systems can include continuous or discretesub-systems

4 It has a friendly GUI for editing models by interconnectingXcos blocks




Why Scilab/Xcos?








Why Scilab/Xcos?








Why Scilab/Xcos?








How do I construct dynamic models?

1 Starting Xcos with an empty diagram

2 Opening one or more palettes

3 Copying blocks of interest from the palettes into the diagram

4 Setting parameters of the blocks to desired values

5 Connecting the blocks’ input and output ports

6 Compiling and simulating the diagram

7 Renaming and saving the diagram






































































Xcos implementation RL circuit will be




Xcos response of this circuit for R=10Ω, L=100mH and a stepinput voltage of 50 V is applied

0

1

2

3

4

5

6

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

Time (s)

Current (A)

Graphic 1




RLC circuit with step input voltage

V

R L

C

i

Vc−

+

The corresponding differential equations are given by:

di

dt=

1

L(V − iR− vc)

dvcdt

=i

C




Xcos block diagram is




Xcos response for R=2Ω, L=100mH, C=1000µF and a step inputvoltage of 50 V is applied

-5

-4

-3

-2

-1

0

1

2

3

4

5

0.0 0.1 0.2 0.3 0.4 0.5

Time (s)

Current (A)

Graphic 1

-20

0

20

40

60

80

100

0.0 0.1 0.2 0.3 0.4 0.5

Time (s)

Voltage (V)

Graphic 2




A Variable Frequency Oscillator

d2y1dt2

= −ω2y1

Converting this equation into two first-order differential equations

y2 =

(1

ω

)dy1dt

dy2dt

= −ωy1




Xcos implementation isOD

-10

-8

-6

-4

-2

0

2

4

6

8

10

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

Time (s)

y

Graphic 1

-10

-8

-6

-4

-2

0

2

4

6

8

10

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

Time (s)

y

Graphic 2




Separately excited/permanent magnet dc motor

Motor parameters:

Rated power = 2 kWRated armature voltage = 125 VRated armature current = 16 ARated speed = 1750 rpmRa = 0.24 Ω, La = 18 mH,Kaφ = 0.6699, J = 0.5 kgm2,Tl = 0.01 + 3.189× 10−4 × ω2

m

DC Motor defining equations are:

Va = Raia + Ladiadt

+Kaφω

Kaφia = Jdω

dt+Bω + Tl







Xcos response is

0

50

100

150

200

250

300

350

400

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Time (s)

Current (A)

Graphic 1

0

20

40

60

80

100

120

140

160

180

200

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Time (s)

Speed (rps)

Graphic 2




DC motor closed loop speed control system




Xcos response is

0

5

10

15

20

25

30

35

40

45

50

0 2 4 6 8 10 12 14 16 18 20

Time (s)

Current (A)

Graphic 1

0

50

100

150

0 2 4 6 8 10 12 14 16 18 20

Time (s)

y

Graphic 2




Thank YouAny Questions?



Documents

An Introduction to Xcos - INFLIBNET Centrecontent.inflibnet.ac.in/data-server/eacharya-documents/... · 2017-04-20 · 1 Xos/Scicos: Scilab connected object simulator !visual editor