This example is from the previous slides. Here we can find the state space representationR=1;L=.25; C=4; n=1; d=[L*C R*C 1]; s=tf(n,d)

1-------------s^2 + 4 s + 1n=sqrt(1/(L*C))= 50=(R/2)*sqrt(C/L)z1(s)+z2(s)t=0:20;y=1+.0774*exp(-3.732*t)-1.0774*exp(-.2679*t);

A=[-R/L -1/L;1/C 0]; eig(A)

ans =Eigenvalues of A matrix and poles of transfer function are same-3.7321 -0.2679

State Equations

dif.mfunction c= dif(t,x);A=[-4 -4; 0.25 0];B=[4;0];c=A*x+B;x0=([0;.0]); ts=[0 20]; [t,x]=ode23('dif',ts,x0);

u=1This in an m file named difCopy this in command window

Plot(t,x)

plot(x(:,1),x(:,2))

Simple Pendulum

Simple Pendulum

function c= pend(t,x);

c=[x(2);-sin(x(1))];ts=[0 20]; x0=[.5;-.5];[t,x]=ode23('pend', ts,x0); plot(t,x) plot(x(:,1),x(:,2))

Assume m=g=1

Solving state equations

Converting a Transfer Function to State SpaceA set of state variables is called phase variables, where each subsequent state variable is defined to be the derivative of the previous state variable. Consider the differential equation

Choosing the state variables

Converting a transfer function with constant term in numeratorStep 1. Find the associated differential equation.

Step 2. Select the state variables. Choosing the state variables as successive derivatives.

Converting a transfer function with polynomial in numeratorStep1. Decomposing a transfer function.

Step 2. Converting the transfer function with constant term in numerator.

Step 3. Inverse Laplace transform.

Converting from state space to a transfer functionGiven the state and output equations

Take the Laplace transform assuming zero initial conditions:

(1)

(2)

Solving for in Eq. (1),

or (3)Substitutin Eq. (3) into Eq. (2) yields

The transfer function is

Ex. Find the transfer from the state-space representationSolution:

Types of Eqbm. pointsNodeeig(A) -1 -2

Types of Eqbm. pointsNodeeig(A) -1 -2

eig(A)

-0.5000 + 0.8660i -0.5000 - 0.8660i

eig(A) 0 + 1.0000i 0 - 1.0000i

eig(A)

-0.5000 + 0.8660i -0.5000 - 0.8660i

eig(A) 0 + 1.0000i 0 - 1.0000i

Types of equilibrium points of second order linear systems

Prey predator model lions and gazelles, birds and insects, pandas and eucalyptus trees, herbivore-plant,parasitoid-host,Venus fly traps and flies. x1 is the prey populationX2 predator population

the predator species is totally dependent on a single prey species as its only food supply,

the prey species has an unlimited food supply, and

there is no threat to the prey other than the specific predator. . AssumptionsA good model must be simple enough to be mathematically tractable, but complex enough to represent a system realistically.

In the model the parameter a is the growth rate of species (the prey) in the absence of interaction with species (the predators). If there were no predators, the prey species grows exponentially. But there are predators, which must account for a negative component in the prey growth rate. The rate at which predators encounter prey is jointly proportional to the sizes of the two populations. A fixed proportion of encounters lead to the death of the prey. Thus the negative component of the prey growth rate is proportional to the product of the population sizes, i.e., The parameter d is the death (or emigration) rate of species in the absence of interaction with species. If there were no food supply, the population would die out at a rate proportional to its size, i.e. we would findthe term c denotes the net rate growth of the predator population in response to the size of the prey population.