Nonlinear Second Order Models 1-Mathematical Modeling and Simulation-Lecture Slides, Slides of Mathematical Modeling and Simulation

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Nonlinear Second Order Models

Mathematical Modeling

& Simulation

Consider a model of two-dimensional dynamical system that obeys the following ordinary differential equations (ODEs):

f ( x , y )

dt

dy

 g(x,y) dt

dx 

It is an autonomous system as it does not explicitly involve the independent variable t. If such a system involves a nonlinear term such as x^2 (t), x(t)y(t), sinx(t), exp(ay) etc., then the system is said to be nonlinear one. Recall the chain rule for differentials: (dy/dx)(dx/dt) is equal to dy/dt. It means that

g(x,y)

f(x,y)

dx / dt

dy / dt

dx

dy  

Non-Linear Models

 The phase space is a coordinate space with x, vx, y, vy,.. ., where v is velocity or any other canonical state variable.  Phase space can have fixed points (x, v) such that they satisfy the model (say Eq. as f(x, v) = 0 and g(x, v) = 0). This corresponds to a steady state.  The set of points in the phase space are identified as orbit or trajectory****.  If the set of points in the simulation repeat itself after some time (T), then the orbit is said to be periodic , that is x(t + T) = x(t).  The orbit of mass-spring system in a friction free environment is an ellipse or a circle in phase space.

Visualization in Phase Space

 A closed curve is called a limit cycle in phase space

towards which an orbit evolves as time goes to large

values.

 When all the neighboring trajectories are going

towards the limit cycle it is called a stable or

attracting cycle , otherwise it is an unstable or

repelli ng one.

Visualization in Phase Space

Physical Pendulum

w g L

g m s

x

L

g

dt

d x

sin

2

2

2

2

This equation can be simplified by dividing by mL. Also, choose L = 9.8 m ,

x

dt

d x

2 sin

2

Because this model has sinx rather than x on the right-hand side, it forms a second-order non-linear differential equation.

2 2 sin^0 ;^2 /^1.^0

2

     g L 

dt

d





Major features of the mathematical model:

Here Dependent and independent variable are y and t;

Order: 2; Linearity: It is a nonlinear equation as it has no product terms. Homogeneity: It is homogeneous equation with no force term Conditions: Initial conditions are given. Coefficients: There are constant coefficients. Model Equation Type: It is a single ordinary differential eq. based model.

Example 1: Undamped System of Simple Pendulum

θ

m

L

You may also write a program in language of your choice to implement the algorithm.

OR you may use MATLAB-SIMULINK to simulate this model.

Use this for h = 0.01, x = 0 at t = 0 , and v = 1.0 at t = 0 ;

here the domain is [0, 25], and the range is [-2, 2].

Also try a different set of initial conditions; say x 0 = 1 and v 0 = - 0..

mass = 1 and L = 9.8, x(0) = 10 and v(0) = 0.

Example 1: Undamped System of simple pendulum

-4 0 50 100 150 200 250 300

0

1

2

3

4

angular displacement,



time, t

undamped motion of pendulum

initial conditions : and ' = 0.

XY Graph

sin TrigonometricFunction simout To Workspace

simout To Workspace (^1) s Integrator

(^1) s Integrator

- Gain

-4 0 50 100 150 200 250 300

0

1

2

3

4

angular velocity, d

dt



time, t

undamped motion of pendulum

initial conditions : and ' = 0.

2 sin^0 ;^2 /^1.^0

2 ddtx  x  as   g L 

mass = 1 and L = 9.8,

Example 1: Undamped System of simple pendulum

XY Graph

sin TrigonometricFunction simout To Workspace

simout To Workspace (^1) s Integrator

(^1) s Integrator

- Gain

-3-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

0

1

2

3

d

/dt

time, t

Phase portrait for given initial conditions:

x(0) = -2.0 and dx/dt(0) = 0.

2 sin^0 ;^2 /^1.^0

2 ddtx  x  as   g L 

mass = 1 and L = 9.8,

Example 1: Undamped System of simple pendulum

XY Graph

sin TrigonometricFunction simout To Workspace

simout To Workspace (^1) s Integrator

(^1) s Integrator

- Gain

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

0

1

2

d

/dt

time, t

Phase portrait for given initial conditions:

x(0) = -3.0 and dx/dt(0) = 0.

2 sin^0 ;^2 /^1.^0

2 ddtx  x  as   g L 

Example 1: Undamped System of simple pendulum

2 sin^0 ;

2  x  dt

d x Phase portrait for various initial conditions:

 Let us allow velocity range to be from -4 to 4 in steps of 0.

and domain is given by [ -3p, 3p ].

 We can modify the initial conditions to get a set of

concentric ellipses centered on points (2nπ, 0).

 The point in phase space (0, 0) phase space (0, 0) and the set

of points ( 2nπ, 0), where n is an integer are points of stable

equilibrium for the simple pendulum; the pendulum is

hanging down from the pivot.

 The set of points [(2n - 1)π, 0] are points of unstable

equilibrium; the pendulum is " standing " on its pivot.

Phase Space: a Simple Pendulum