Evaluate gradient function in the for loop.

is a function dependent on time. It is n the function and need to be differentiated to get and the differentiation should pass through the for loop. But for loop picks one value at time t and differentiation failed. Code attached.

function S = Get_Vel(t)
    ts=0.0001;
    x(t)=cos(2*pi*t)
    y(t)=sin(2*pi*t)
    vx(t)=gradient(x,ts)
    vy(t)=gradient(y,ts)
    S=[vx;vy]
end
function A = Compute(t,ts,~,~,~) 
S=Get_Vel(t)
end
function solve = solver(F,t0,tf,y0,~)
    for t=t0:ts:tf-ts
        A =F(t,ts,~,~)
    end
    solve =A
end
%% MAIN 
Result=solver(@Compute,t0,tf,y0,~)

But, since solver used a for loop gradient failed.

Any help is apperciated.

Thank you

[Sara Boznik]

What happens if you write A(t)?

[HN]

Always zero. Gradient is not calculated.

[Sara Boznik]

It looks like you have only constant.

[HN]

No, x and y changes over time. but gradient is not because of lost previous information while passing through for loop.

Thanks

Answer 1

[KSSV]

ts=0.0001;
x(t)=cos(2*pi*t)    % index of x is t, it cannot be, it shows error
y(t)=sin(2*pi*t)
vx(t)=gradient(x,ts)  % index cannot be fraction and to use gradient you need to have x as vector
vy(t)=gradient(y,ts)
S=[vx;vy]

You may rather use:

ts=0.0001 ; 
vx = sin(2*pi*t) ; 
vy = cos(2*pi*t) ; 
S=[vx;vy]

[HN]

Actually x is different from what is shown in the code. It is a bit complicated to simply differentiate it. I just tried to show my problem. if any other tip maybe,

Thank you

[KSSV]

USe the vector version, don't call the function for each step. You proceed like this:

t0 = 0 ; t1 = 10 ; 
dt = 0.0001 ;
t=t0:dt:t1 ;

x = cos(2*pi*t)
y = sin(2*pi*t)
vx = gradient(x,t)
vy = gradient(y,t)
S=[vx;vy]

[HN]

Well, how bout the for loop ? The function solver has for loop and the vector must pass through that.

How it can be handled ?

Thanks

[HN]

Is there any technique to save previous value of S outside the function and calculate the gradient from that ?

[KSSV]

What is F in the solver?

[HN]

F is RK4 integrator, to inegrate some function that is made from S and other variables .

Thank you

[KSSV]

F is a function, it can be evaluated at once. Loop is not required.

[HN]

Can you brief me a bit mote, Dr,

I didnot get it very well. If needed I can post the code here.

[KSSV]

It is suggested to post the code here..so that if not me others also can help you.

[HN]

KSSV ,

Here is the program. I tried to use persistant for t but the dimension exceeded. If t= not a time vector, S become zero all the time.

% time function is to create a time vector   
function time(t)    
    persistent n
    n=t;
    if isempty(n)
        n = 0;
    end
    n = n+1
end
function [Pose, S] = get_St3PhRS(t) 
         ts=1/1000;
         tt=time(t);
         rp=1000;   % Radius of the base plate in mm 
         th=-0.2*cos(2*pi*tt);
         psi=0.2*sin(2*pi*tt);
         z=707.1068;
         phi=atan2(sin(psi)*sin(th),(cos(psi)+cos(th)));
         T=Rot('y',th)*Rot('x',psi)*Rot('z',phi)
         x=1/2*rp*(-cos(phi)*cos(psi)+cos(phi)*cos(th)+sin(phi)*sin(psi)*sin(th));
         y=-rp*cos(psi)*sin(phi);
         Pose=[x;y;z;th;psi;phi]
         
         % derivative of moving plate rotation angles
          vx=gradient(x,ts);
          vy=gradient(y,ts);
          vz=gradient(z,ts);
          dth=gradient(th,ts);  %(2*pi*sin(2*pi*t))/5;
          dpsi=gradient(psi,ts); %(2*pi*cos(2*pi*t))/5;          
          dphi=gradient(phi,ts);%(cos(psi)*sigma1*dpsi+cos(th)*dpsi-cos(psi)*sin(psi)*sin(th)*dth)/(-sigma2*sigma1+sigma2+2*cos(psi)*cos(th)+2*sigma1);
          JT=[0,  cos(th), cos(psi)*sin(th);1,  0,  -sin(psi);0, -sin(th), cos(psi)*cos(th)]; 
          dTh=[dth; dpsi;dphi]
          w=JT*dTh;
          S=[vx;vy;vz;w(1);w(2);w(3)];
end 
function A  = InverseVelocityPhRs20200820(T,th,~,q)
         :
         :
         :
         S=get_St3PhRS(T)  % Here is where the function is called 
         :
         :
         A=~~; 
end 
function [thout,Stm_out,MM_out,Pout,G_out,FRK_OUT,R_out,q_out,Pd_out] = RK4_RhPRS(F,t0,h,tfinal,y0,p0,q0)
   % ODE4  Classical Runge-Kutta ODE solver.
      th = y0;
      R = eye(3);
      P =  p0;
      q =  q0;
      q_out=q;
      thout = th;
      Pout = P;
      Pd=[-9.9667;0;707.1068];
      Pd_out=Pd;
      Stm=[0;0;0;1.1080;0; 0.8593];
      Stm_out = Stm;
      MM_out = [];
      G_out = [];      
      R_out=R;
      FRK=[0;0;0;1.1080;0; 0.8593];
      FRK_OUT = FRK;
      for t = t0 : h : tfinal-h
         [k1,~,~,P1,~,~,~,q1] = F(t,th,P,q); % Error in RK4_RhPRS (line 20) : Runge-Kutta
         :
         :
      end 
end 
% Main function call 
[J,S,M,P,G,F,R,q_out,Pd]= RK4_RhPRS(@InverseVelocityPhRs20200820,0,ts,1,th,P,q);
%% Error 
Error using time
Too many output arguments.
Error in get_St3PhRS (line 3)
         tt=time(t); % t contains all t values but 
Error in InverseVelocityPhRs20200820 (line 19)
         S=get_St3PhRS(T)
Error in RK4_RhPRS (line 20)
         [k1,~,~,P1,~,~,~,q1] = F(t,th,P,q);
Error in Main3PhRS_20200820 (line 103)
[J,S,M,P,G,F,R,q_out,Pd]= RK4_RhPRS(@InverseVelocityPhRs20200820,0,ts,1,th,P,q);