Evaluate gradient function in the for loop.

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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
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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

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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]
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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);

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