Evaluate gradient function in the for loop.

4 次查看（过去 30 天）
HN
HN 2020-8-14
编辑： HN 2020-8-22
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
  4 个评论
显示 2更早的评论
HN
HN 2020-8-14
No, x and y changes over time. but gradient is not because of lost previous information while passing through for loop.
Thanks

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

KSSV
KSSV 2020-8-14
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]
  10 个评论
显示 8更早的评论
KSSV
KSSV 2020-8-14
It is suggested to post the code here..so that if not me others also can help you.
HN
HN 2020-8-22
编辑：HN 2020-8-22
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);

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Get Started with Embedded Coder Support Package for STMicroelectronics STM32 Processors 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by