using gradient with ode45

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jose luis guillan suarez 2018-7-6

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/409195-using-gradient-with-ode45

评论： jose luis guillan suarez 2018-7-19

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 2018-7-7

Can you post the question to solve?

jose luis guillan suarez 2018-7-7

编辑：Walter Roberson 2018-7-7

在 MATLAB Online 中打开

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 2018-7-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/409195-using-gradient-with-ode45#answer_327911

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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jose luis guillan suarez 2018-7-12

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 2018-7-19

在 MATLAB Online 中打开

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)