using gradient with ode45

2018 7 6
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更新时间：2018 7 19
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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?

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madhan ravi
madhan ravi 2018-7-7
Can you post the question to solve?
jose luis guillan suarez
编辑：Walter Roberson 2018-7-7
this is the equation:
Vp=5;
u=Vp*sin(2*pi*t)+5;
x''=-7x'-10x+1.776*0.05252*20*u'

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回答（1 个）

Walter Roberson
Walter Roberson 2018-7-7

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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.

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but t is a vector from 0 to 10:
[t,x]=ode45('pid_practica9_ejercicio3_ec_diferencial_prueba2', [0,10],[0,0])
Walter Roberson
Walter Roberson 2018-7-9
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.
i deduce that is not possible to work with the derivative of the input (u')
Walter Roberson
Walter Roberson 2018-7-10
> 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)
and how can i do if the input is a squared waveform like this?:
u=5*square(2*pi*t)+5;
Walter Roberson
Walter Roberson 2018-7-11
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
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)
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

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