using gradient with ode45

i have this code, where gradient(u,t) should be the derivative of u:

function [xdot]=pid_practica9_ejercicio3_ec_diferencial_prueba2(t,x)

Vp=5;
u=Vp*sin(2*pi*t)+5;

xdot = [
        x(2, :);
        1.776*0.05252*20*gradient(u,t)-10*x(1, :)-7*x(2, :);
       ];

%[t,x]=ode45('pid_practica9_ejercicio3_ec_diferencial_prueba2',[0,10],[0,0])
% plot(t,x)

but in the solution i get only zeros (and they shouldn't be)

is it possible to work with a derivative in the definition of a diferential equation like i do?

[madhan ravi]

Can you post the question to solve?

[jose luis guillan suarez]

this is the equation:

Vp=5;

u=Vp*sin(2*pi*t)+5;

x''=-7x'-10x+1.776*0.05252*20*u'

Answer 1

[Walter Roberson]

The t and x values passed into your function will be purely numeric, with t being a scalar and x being a vector the length of your initial conditions (so a vector of length 2 in this case.)

You calculate u from the scalar t value, and you pass the scalar u and scalar t into gradient -- the numeric gradient routine. The numeric gradient() with respect to scalar F and scalar H is always 0.

[jose luis guillan suarez]

but t is a vector from 0 to 10:

[t,x]=ode45('pid_practica9_ejercicio3_ec_diferencial_prueba2', [0,10],[0,0])

[Walter Roberson]

When you use ode45, the [0 10] tells it that it needs to integrate from t = 0 to t = 10. It will use variable time steps to do so, and it will report at whatever times it wants as long as the first is 0 and the last is 10. If you had specified something like linspace(0,10,101) then it would report with respect to 0:.1:10 but it would still calculate for whatever times it wants.

ode45 is not a fixed timestep solver.

ode45 calls the function you provide with a scalar time, and with a vector of boundary conditions that is the same length as your initial condition. It does not pass all of the times in a single call, because the boundary conditions (x) evolve with time.

[jose luis guillan suarez]

i deduce that is not possible to work with the derivative of the input (u')

[Walter Roberson]

> diff(Vp*sin(2*Pi*t)+5, t);

                      2 Vp Pi cos(2 Pi t)

so if you want u' at time t, then use

2 * pi * Vp * cos(2 * pi * t)

[jose luis guillan suarez]

and how can i do if the input is a squared waveform like this?:

u=5*square(2*pi*t)+5;

[Walter Roberson]

The derivative is 0 except at the integers, and it is undefined at the integers because square() is discontinuous. There are no rising or falling edges for a mathematical square wave.

There are approximations to square waves for which it is meaningful to ask about the derivative. http://mathworld.wolfram.com/FourierSeriesSquareWave.html

[jose luis guillan suarez]

my problem is that i'm trying to represent this system with differential equations and with ode45, where the derivative of u is needed (if im not commiting a mistake), and it is perfectly representable in matlab using the blocks (in the 's' domain) and with the input as a step input (which is very similar to a squared waveform)

[jose luis guillan suarez]

i simulated this system with a step input with this instructions:

>> num1=20

num1 =

      20

>> den1=[1 7 10]

den1 =

       1     7    10

>> sys1=tf(num1,den1)

sys1 =

          20
    --------------
    s^2 + 7 s + 10

Continuous-time transfer function.

>> num2=[1 0]

num2 =

       1     0

>> den2=1

den2 =

       1

>> sys2=tf(num2,den2)

sys2 =

    s

Continuous-time transfer function.

>> sys=series(sys1,sys2)

sys =

         20 s
    --------------
    s^2 + 7 s + 10

Continuous-time transfer function.

>> step(sys)

and i obtain this response:

but it seems to be impossible to simulate it with diferential equations