MATLAB: Solving Mathematical Relation Having If and Else Statement in Matlab

if statementmatrix

Hi,

Here I am writing a mathematical relation which is needed to be solved using Matlab:

N=4;
j=sqrt(-1);
Sx=[27 0 4 19;
  19 -8 -10 16;
  20 -4 -7 9;
  27 0 2 24];
Sy=[22 4 -7 -25;
  7 2 -8 -12;
  -8 -5 3 5;
  -23 -9 5 13];
P(k,l)={0; if k,l=0 and k,l=N/2
      {((exp(-j*2*pi*k)/N)-1)*((exp(-j*2*pi*l)/N)+1)*DFT(Sx(k,l)+
      ((exp(-j*2*pi*l)/N)-1)*((exp(-j*2*pi*k)/N)+1)*DFT(Sy(k,l)}*
      {8*((sin(k*pi/N).^2)*(cos(l*pi/N).^2)+(sin(l*pi/N).^2)*(cos(k*pi/N).^2)}.^(-1), else
        where k and l varies from 1 to 4 each

********** Can anyone please help me in solving the above problem with the help of Matlab? Thanking You!

Best Answer

Well you are practically there already. Just put that code in a loop over k and l (l is a bad variable name choice). Inside the loop have the if statements where you assign P(k,l) to different things depending on what the values of k and l are.

Related Solutions

MATLAB: Decreasing three loops to two loops with sum

----------- Updated answer ------------------------------

Vectorize the k-loop and use sum() to define theta2.

thetao=1;
alfa=1;
l=1;
theta2(1:7,1:21)=0;
T=[0.01, 0.1, 0.2, 0.5, 1, 2, 5];
x=0:0.05:l;
n=0:50;
for t=1:7
    for j=1:length(x)
        theta1 = sin((2*n+1)*pi*x(j)/(2*l)).* exp(-(2*n+1).^ 2*pi^2*alfa*T(t)/(4*l^2))./((2*n+1)*pi);
        theta2 = sum(theta1);
        theta(t,j)=thetao-4*thetao*theta2;
    end
end

To demonstrate that this produces the same values at the original version, run the block and above and then two two blocks below.

% Save the theta values above and then clear all variables 
theta_version2 = theta; 
clearvars -except theta_version2
% Original version
thetao=1;
alfa=1;
l=1;
theta2(1:7,1:21)=0;
T=[0.01, 0.1, 0.2, 0.5, 1, 2, 5];
x=0:0.05:l;
n=0:50;
for t=1:7
    for j=1:length(x)
        for k=1:length(n)
           
           theta1(t,j)=(sin((2*n(k)+1)*pi*x(j)/(2*l))*exp(-(2*n(k)+1).^2*pi^2*alfa*T(t)/(4*l^2))/((2*n(k)+1)*pi));
           theta2(t,j)=theta2(t,j)+theta1(t,j);
           theta(t,j)=thetao-4*thetao*theta2(t,j);
            
       end
    end
end

Now compare the results

max(abs(theta - theta_version2),[],'all')
% ans =
%    4.4409e-16

The max difference in theta between the original and new versions is 4.4409e-16. This is due to the precision in floating point calculations explained [here].

MATLAB: Simple fft code problem

%k and m start at 0

 x=rand(1,8);
for k=0:7 
    for m=0:7
       l(m+1)=x(m+1)*exp(-i*k*m*pi/4);
    end
    X(k+1)=sum(l);
end

%or

R=exp(-i*2*pi/8)
k=0:7;
XX=exp(-i*pi/4).^(k'*k)*x'

Remark: for big array, you must use FFT algorithm

Best Answer

Related Solutions

MATLAB: Decreasing three loops to two loops with sum

MATLAB: Simple fft code problem

Related Question