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# SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS - PowerPoint PPT Presentation

SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS. Example:. Mathematical model of a mechanical system is defined as a system of differential equations as follows:. where f is input, x 1 are x 2 outputs. At t=0 x 1 =2 and x 2 =-1. Find the eigenvalues of the system.

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Example:

Mathematical model of a mechanical system is defined as a system of differential equations as follows:

where f is input, x1 are x2 outputs.

At t=0 x1=2 and x2=-1.

• Find the eigenvalues of the system.

• If f is a step input having magnitude of 3, find x1(t).

• If f is a step input having magnitude of 3, find x2(t).

• Find the response of x1 due to the initial conditions.

• Find the response of x2 due to the initial conditions.

• How do you obtain [sI-A]-1 with MATLAB?

State Variables Form

A

B

D(s)

Let us obtain the State Variables Form so as to 1st order derivative terms are left-hand side and non-derivative terms are on the right-hand side.

or

Initial Conditions

Solution due to the input Particular Solution

Solution due to the initial conditions

Homogeneous Solution

General Solution

clc;clear;

num=[4.5 67.5];

den=[1 15 -280 0];

[r,p,k]=residue(num,den)

a) Eigenvalues are roots of the polynomial D(s) or eigenvalues of the matrix A.

b) x1(t) due to the forcing

System is instable because of the positive root.

Laplace transform of x2p

clc;clear;

num=[6 174];

den=[1 15 -280 0];

[r,p,k]=residue(num,den)

c) x2(t) due to input

clc;clear;

num=[2 -25];

den=[1 15 -280];

[r,p,k]=residue(num,den)

clc;clear;

num=[-1 4];

den=[1 15 -280];

[r,p,k]= residue(num,den)

f) [sI-A]-1 with Matlab.

clc;clear;

syms s;

i1=eye(2)

A=[-20 15;12 5];

a1=inv(s*i1-A)

pretty(a1)

d) x1 due to the initial conditions.

e) x2 due to the initial conditions

V2(t)

2

t (s)

Example:

• Mathematical model of a system is given below. Where V(t) is input, q1(t) and q2(t) are outputs.

• Write the equations in the form of state variables.

• Write Matlab code to obtain eigenvalues of the system.

• Write Matlab code to obtain matrix [sI-A]-1.

• Results of (b) and (c) which are obtained by computer are as follows:

At t=0

and V(t) is a step input having magnitude of 2.

Find the Laplace transform of due to the initial conditions.

e) Find the Laplace transform of q1 due to the input.

State variables

B

A

a) State variables are q1, q2 and .

System of differential equations is arranged so as to 1st order derivative terms are left-hand side and non-derivative terms are on the right-hand side.

b) Matlab code which gives the eigenvalues of the system.

A=[-1.5 1.5 0;0 0 1;3.75 -3.75 0]; eig(A)

c) Matlab code which produces [sI-A]-1

clc;clear

A=[-1.5 1.5 0;0 0 1;3.75 -3.75 0];

syms s;

i1=eye(3);

sia=inv(s*i1-A);

pretty(sia)

Example:Write the equation of motion of the mechanical system given below in the State Variables Form. Force applied on the system is F(t)=100 u(t) (a step input having magnitude 100 Newtons) and at t=0 x0=0.05 m and dx/dt=0. Find x(t) and v(t).

State variables are x and v=dx/dt .

m=20 kg

c=40 Ns/m

k=5000 N/m

Matlab program to obtain eigenvalues:

>>a=[0 1;-250 -2];eig(a)

Applying Laplace transform and arranging,

Solution due to the initial conditions

Solution due to the input

clc;clear;

syms s;

A=[0 1;-250 -2];

i1=eye(2); %unit matix with dimension 2x2

siA=s*i1-A;

x0=[0.05;0]; %Initial conditions

B=[0;0.05];

Fs=100/s;

X=inv(siA)*x0+inv(siA)*B*Fs;

pretty(X)

For x(t) ;

clc;clear;

num=[0.05 0.1 5];

den=[1 2 250 0];

[r,p,k]=residue(num,den)

Initial value, x0

For v(t)

clc;clear;

num=[-7.5];

den=[1 2 250];

[r,p,k]=residue(num,den)

State variables

Example: Mathematical model of a mechanical system having two degrees of freedom is given below. If F(t) is a step input having magnitude 50 Newtons, find the Laplace transforms of x and θ.

R=0.2 m

m=10 kg

k=2000 N/m

c=20 Ns/m

clc;clear

A=[0 0 1 0;0 0 0 1;-400 80 0 0;2000 -600 0 -2];

syms s;

eig(A)

i1=eye(4);

sia=inv(s*i1-A);

pretty(sia)

System is stable since real parts of all eigenvalues are negative.

Eigenvalues:

If the initial conditions are zero, only the solution due to the input exists.