MATLAB: How to code for an equation for numbers of input

matlab function

Hi,

I am new in Matlab.

Would like to know how to code for this problem.

I have this equation

Qc = 2*N*((alpha*I0*Tc)-(0.5*I0*I0*(rho/G))-(k*G*(Th-Tc)))

And i want to compute this equation for I0 from 1 to 10 and Tc from -60 to 240.

I try coding it this way.

    //
  N= 96;
  G = 0.072;
  alpha = 0.000203;
  rho = 0.001095;
  k = 0.01438;
  Th = 323;
  for I0 = 0.0
      for Tc = -60.0
          Qc = 2*N*((alpha*I0*Tc)-(0.5*I0*I0*(rho/G))-(k*G*(Th-Tc)))
          I0 = I0+1
          Tc = Qc+30
          if I0>=11 && Tc >=270
          end
      end
  end

But, it only give me 1 answer instead of I need the Qc to give me the output in terms of I(1:10) and Tc (-60:240)

Please help

Best Answer

N = 96;
G = 0.072;
alpha = 0.000203;
rho = 0.001095;
k = 0.01438;
Th = 323;
[I0, Tc] = ndgrid(1:10, -60:240);
Qc = 2*N*((alpha .* I0 .* Tc) - (0.5 .* I0 .* I0 .* (rho./G)) - (k .* G .* (Th-Tc)));
surf(I0, Tc, Qc, 'edgecolor', 'none')

Related Solutions

MATLAB: How to speed up matrix operations with lots of scalar constants involved

If you have a modern Matlab version >= 2016b, use the auto expanding instead of an explicit repmat:

% Ipc = repmat(I(:,jt2),[1,npmax(jz)-1]);
Ipc = I(:,jt2);
... CnE(:,1:npmax(jz)-1,jl)) .* Ipc * dt / hb

I guess, that dt and hb are scalars (please do not let the readers guess!). Then remember, that Matlab evaluate the code from left to right:

Y = X * dt / hb

is equivalent to:

Temp = X * dt;
Y    = Temp / hb;

These are 2 matrix operations. But you can combine the scalar operations:

Y = X * (dt / hb)

This is 1 matrix operation and a cheap scalar division.

As Matt J has explained already, operating on subarrays as in |dCnE(:,2:npmax(jz)) is expensive. Better crop the data, such that you can omit the ugly indexing.

I cannot run it by my own, because you did not provide the input values, but compare it optically:

Ipc=repmat(I(:,jt2),[1,npmax(jz)-1]);
dCnE(:,2:npmax(jz))=(sigma0*CnZ(:,2:npmax(jz),jl)-sigma*CnE(:,2:npmax(jz),jl)+ ...
      vs2*sigma*CnE(:,1:npmax(jz)-1,jl)).*Ipc*dt/hb+CnB(:,2:npmax(jz),jl)*(1-exp(-dt/tau2))...
      +vs2*(beta0*CnZ(:,1:npmax(jz)-1,jl).*(Ipc.^2))*dt/(2*hb)-CnE(:,2:npmax(jz),jl)/tau1*dt;

or:

c1 = 1 - exp(-dt / tau2);   % Calculate constants outside the loops!
Ipc  = I(:,jt2);
dCnE = (sigma0 * CnZ(:, :, jl) - sigma * CnE(:, :, jl) + ...
       vs2 * sigma * CnE(:, :, jl)) .* (Ipc * (dt / hb)) + ...
       CnB(:,:,jl) * c1 + ...
       (vs2 * beta0 * dt / (2*hb)) * CnZ(:, :, jl) .* (Ipc .^ 2) - ...
       CnE(:, :, jl) / (tau1 * dt);

This can be simplified a little bit for the CnE term:

dCnE = (sigma0 * CnZ(:, :, jl) + ...
       CnE(:, :, jl)) .* (Ipc * ((vs2 * sigma - sigma) * dt / hb)) + ...
       CnB(:,:,jl) * c1 + ...
       (vs2 * beta0 * dt / (2*hb)) * CnZ(:, :, jl) .* (Ipc .^ 2) - ...
       CnE(:, :, jl) / (tau1 * dt);

Methods:

Combine the scalar operations to reduce the number of matrix operations.
Move constants out of the loop.
Crop the data outside the loops instead of applying an a:b indexing in each iteration.
Use auto-expanding instead of inflating by repmat. bsxfun is sometimes faster or slower than auto-expanding - try it.
Avoid repeated calculations. In your case e.g. Ipc(:,1)*dt/hb is computed several times. Using this avoids this:

C1 = I(:, jt2) * (dt/hb);

The latter includes another idea: You inflate I(:,jt2) at first by Ipc = repmat(...), and waste time with cropping out a subvector Ipc(:,1) later on. Better create the subvector once only: Ipc = I(:,jt2).

Finally: Replace

        non0=Cn0(:,jl);
        non0(non0<0)=0;
        Cn0(:,jl)=non0;

by:

        Cn0 = max(Cn0, 0);

and move it outside the loops, because you do not use the cleaned values inside the loop.

MATLAB: Unable to perform assignment because the indices on the left side are not compatible with the size of the right side

Pre allocate B1x as cell array instead of numeric array to solve the problem.

close
clear
clc
%%


mu0 = 4*pi*1e-7; % Vs/Am
M0 = 1e3; % A/m
maxnum = 31;
rho1_min = 0;
rho1_max = 0.25;
xlimit = [-1, 1];
ylimit = xlimit;
zlimit = xlimit;
x = linspace(min(xlimit), max(xlimit), maxnum);
y = linspace(min(ylimit), max(ylimit), maxnum);
z = linspace(min(zlimit), max(zlimit), maxnum);
[Xg, Yg, Zg] = ndgrid(x, y, z);
rho = sqrt(Xg.^2 + Yg.^2 + Zg.^2);
phi = angle(Xg + 1i*Yg);
theta = angle(Zg + 1i*sqrt(Xg.^2 + Yg.^2));
%%
RHO = sqrt(x.^2 + y.^2 + z.^2);
THETA = linspace(0, pi, 31);                          % Trapz

PHI = linspace(0, 2*pi, 31);                          % Trapz
%%
% Pre-allocate F_x and B1x as cell array
F_x = cell(numel(RHO), numel(THETA), numel(PHI));
B1x = cell(numel(RHO), numel(THETA), numel(PHI));
for ii = 1:numel(RHO)
    for jj = 1:numel(THETA)
        for kk = 1:numel(PHI)
            F_x{ii, jj, kk} = (RHO(ii)>= rho1_max) .* 2/3*M0*mu0 .* sin(theta) .* (RHO(ii) .* (sin(THETA(jj)) .* cos(theta) .* cos(PHI(kk)-phi) - cos(THETA(jj)) .* sin(theta)) ./ ...
                (RHO(ii).^2 + rho1_max.^2 - 2.*RHO(ii) .* rho1_max .* (sin(THETA(jj)) .* sin(theta) .* cos(PHI(kk)-phi) + cos(THETA(jj)).* cos(theta))).^3/2) .* rho1_max.^2 .* sin(theta);
            B1x{ii, jj, kk} = squeeze(-trapz(PHI,trapz(THETA,F_x{ii,jj,kk},2)));
        end
    end
end

Best Answer

Related Solutions

MATLAB: How to speed up matrix operations with lots of scalar constants involved

MATLAB: Unable to perform assignment because the indices on the left side are not compatible with the size of the right side

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