How to convert the attached equations in Matlab?

I have two equations which I have attached in the attachment here. I want to convert them in Matlab code. But how? I tried but in vain. I don't know why it cannot be coded? The required desired vector u is given belwo:

u=[1 2 0.1 0.1 3 4 30 40 50 60];% u=[a1 a2 r1 r2 f1 f2 theta1 theta2 phi1 phi2];
fmax=10;
m=1:5;
n=m;

As you can see the values of ap in given equations have 1st two values inside u namely 1 and 2. Likewise, the values of rp in given equations have 3rd and 4th values inside u. Similarly, the fp in given equations have values 3 and 4 inside u. Likewise, the values of thetap are 30 and 40 inside u and the values of phip are 50 and 60 inside u.

I want to find the values of xo and yo in MATLAB?

Answer 1

[Torsten]

I'm not sure if you mean m.^2.*f.^2./16.*r or m.^2.*f.^2./(16.*r) in your equations.

I assumed that your equations.png is correct.

u=[1 2 0.1 0.1 3 4 30 40 50 60];
a = u(1:2);
r = u(3:4);
f = u(5:6);
theta = u(7:8);
phi = u(9:10);
fmax=10;
m=(1:5).';
xo = sum(a.*exp(-1i*((pi/fmax).*(-m.*f/2).*sind(theta).*cosd(phi)+m.^2.*f.^2./16.*r).*(1-sind(theta).^2.*cosd(phi).^2)),2)

xo = 
   2.9880 + 0.2678i
   2.9975 + 0.0768i
   2.9297 - 0.5680i
   2.4816 - 1.5650i
   1.3000 - 2.4949i

yo = sum(a.*exp(-1i*((pi/fmax).*(-m.*f/2).*sind(theta).*sind(phi)+m.^2.*f.^2./16.*r).*(1-sind(theta).^2.*sind(phi).^2)),2)

yo = 
   2.9647 + 0.4489i
   2.9507 + 0.5263i
   2.9889 + 0.2358i
   2.9693 - 0.4278i
   2.6463 - 1.4077i

[Sadiq Akbar]

@Torsten

Thanks a lot for your kind response. I am surprised how you people do it because it's not very easy task. But in my given equations, the 2nd equation has n instead of m. However, now I can replace it by n. However, thanks again.

[Torsten]

u=[1 2 0.1 0.1 3 4 30 40 50 60];
a = u(1:2);
r = u(3:4);
f = u(5:6);
theta = u(7:8);
phi = u(9:10);
fmax=10;
m=(1:5).';
n=m;
xo = sum(a.*exp(-1i*((pi/fmax).*(-m.*f/2).*sind(theta).*cosd(phi)+m.^2.*f.^2./16.*r).*(1-sind(theta).^2.*cosd(phi).^2)),2)
yo = sum(a.*exp(-1i*((pi/fmax).*(-n.*f/2).*sind(theta).*sind(phi)+n.^2.*f.^2./16.*r).*(1-sind(theta).^2.*sind(phi).^2)),2)

And check whether m.^2.*f.^2./16.*r and n.^2.*f.^2./16.*r or m.^2.*f.^2./(16.*r) and n.^2.*f.^2./(16.*r) is correct.

[Sadiq Akbar]

@Torsten

Sorry for disturbing you but there is again a problem. The problem is that there are two vectors like u. I mean

u=[1 2 0.1 0.1 3 4 30 40 50 60];
b=[10 20 0.11 0.12 13 14 33 45 56 69];

You have done it for u only like this:

a = u(1:2);
r = u(3:4);
f = u(5:6);
theta = u(7:8);
phi = u(9:10);
fmax=10;
m=(1:5).';

But how will we do it for both the vectors simultaneously? and then how will we find the MSE? I mean how will we do it when we put vector u in the given equations and vector b in the same two equations but the subscript "o" in x and y is replaced by e i.e., xo and yo by xe and ye respectively. Then how will we find the MSE beteeen the xo and xe and likewise the MSE between yo and ye?

Regards,

[Sadiq Akbar]

@Torsten

I did it like this:

clear;clc
u=[1 2 0.1 0.1 3 4 30 40 50 60];b=u;
a1 = u(1:2);
r1 = u(3:4);
f1 = u(5:6);
theta1 = u(7:8);
phi1 = u(9:10);
fmax1=10;
m=(1:5).';
n=m;
% for b
a2 = b(1:2);
r2 = b(3:4);
f2 = b(5:6);
theta2 = b(7:8);
phi2 = b(9:10);
fmax2=10;
m=(1:5).';
n=m;

xo = sum(a1.*exp(-1i*((pi/fmax1).*(-m.*f1/2).*sind(theta1).*cosd(phi1)+m.^2.*f1.^2./16.*r1).*(1-sind(theta1).^2.*cosd(phi1).^2)),2)
yo = sum(a1.*exp(-1i*((pi/fmax1).*(-n.*f1/2).*sind(theta1).*sind(phi1)+n.^2.*f1.^2./16.*r1).*(1-sind(theta1).^2.*sind(phi1).^2)),2)


xe = sum(a2.*exp(-1i*((pi/fmax2).*(-m.*f2/2).*sind(theta2).*cosd(phi2)+m.^2.*f2.^2./16.*r2).*(1-sind(theta2).^2.*cosd(phi2).^2)),2)
ye = sum(a2.*exp(-1i*((pi/fmax2).*(-n.*f2/2).*sind(theta2).*sind(phi2)+n.^2.*f2.^2./16.*r2).*(1-sind(theta2).^2.*sind(phi2).^2)),2)

%%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%%%
%e=norm(xo-xe).^2/(M);
errx=norm(xo-xe).^2/(m);
erry=norm(yo-ye).^2/(n);
err=errx+erry

But in this the final err is a row vector instead of a single zero value. I don't know why?

[Sadiq Akbar]

@Torsten

It is (m^2*f^2/16*r) and (n^2*f^2/16*r).

[Torsten]

But in this the final err is a row vector instead of a single zero value. I don't know why?

m and n are 5x1 vectors. Dividing a scalar by a 5x1 vector gives a 1x5 vector:

x = 0;
y = (1:5).';
x/y

ans = 1×5

0 0 0 0 0

[Sadiq Akbar]

@Torsten

Thanks a lot for your guiadance.