How to fix matrix dimensions

Given the specific values of y, sigmay, and sigmaz (see code), I must calculate them and plot each line on a graph

close all;
clear all;
clc

Q = 1;
y = [10, 20, 30, 40, 50, 65, 80, 100, 200, 300, 400, 500, 650, 800, 1000];
sigmay = [27, 62, 115, 165, 210];
sigmaz = [14, 38, 105, 250 450];
u = 2;

for j = 1:y
    for k = 1:sigmay
        for l = 1:sigmaz
            C = (Q./pi.*sigmay.*sigmaz.*u).*exp(-0.5.*(y./sigmay).^2); %% this is where the error is at%%
        end
    end
end

Warning: Colon operands must be real scalars. This warning will become an error in a future release.

Warning: Colon operands must be real scalars. This warning will become an error in a future release.

Warning: Colon operands must be real scalars. This warning will become an error in a future release.

Arrays have incompatible sizes for this operation.

plot(y,C)
title('Excercise 1')
xlabel('distance (m)')
ylabel('concentrations')
grid on

As you can see, I tried using the ./ or .* but, it didn't work.

[Stephen23]

You seem to be mixing up loops with vectorized code. How about:

Q = 1;
y = [10, 20, 30, 40, 50, 65, 80, 100, 200, 300, 400, 500, 650, 800, 1000];
sigmay = [27, 62, 115, 165, 210];
sigmaz = [14, 38, 105, 250 450];
u = 2;
C = (Q./pi.*sigmay.*sigmaz.*u).*exp(-0.5.*(y(:)./sigmay).^2)

C = 15x5

1.0e+04 * 0.0225 0.1480 0.7658 2.6212 6.0092 0.0183 0.1424 0.7572 2.6068 5.9888 0.0130 0.1334 0.7430 2.5830 5.9550 0.0080 0.1218 0.7236 2.5500 5.9079 0.0043 0.1084 0.6994 2.5082 5.8479 0.0013 0.0866 0.6552 2.4300 5.7347 0.0003 0.0652 0.6035 2.3348 5.5950 0.0000 0.0408 0.5267 2.1855 5.3712 0.0000 0.0008 0.1694 1.2597 3.8225 0.0000 0.0000 0.0256 0.5029 2.1685

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[Jonathon Klepatzki]

I originally had just the equation C listed then tried it with a for loop. I just tried your method and it worked great. However, concentration values are way too large. They actually need to be at an order of 10^-6 or lower. Any suggestions?

[Stephen23]

"Any suggestions?"

Do you have a reference for the formula?

[Jonathon Klepatzki]

Just this, C = (Q/pi*sigmaz*sigmay*u)*exp(-1/2(y/sigmay)^2)

Answer 1

[the cyclist]

In this expression

Q/pi*sigmaz*sigmay*u

only pi is going to be in the denominator.

Since your expression bears some resemblance to the PDF of a normal distribution, I expect that at least sigmaz and sigmay are also supposed to be in the denominator. Probably u as well.

I expect that you actually need

Q/(pi*sigmaz*sigmay*u)

which would give

Q = 1;
y = [10, 20, 30, 40, 50, 65, 80, 100, 200, 300, 400, 500, 650, 800, 1000];
sigmay = [27, 62, 115, 165, 210];
sigmaz = [14, 38, 105, 250 450];
u = 2;
C = (Q./(pi.*sigmay.*sigmaz.*u)).*exp(-0.5.*(y(:)./sigmay).^2)

C = 15x5

1.0e-03 * 0.3931 0.0667 0.0131 0.0039 0.0017 0.3200 0.0641 0.0130 0.0038 0.0017 0.2271 0.0601 0.0127 0.0038 0.0017 0.1405 0.0549 0.0124 0.0037 0.0017 0.0758 0.0488 0.0120 0.0037 0.0016 0.0232 0.0390 0.0112 0.0036 0.0016 0.0052 0.0294 0.0103 0.0034 0.0016 0.0004 0.0184 0.0090 0.0032 0.0015 0.0000 0.0004 0.0029 0.0019 0.0011 0.0000 0.0000 0.0004 0.0007 0.0006

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This looks to be in the ballpark of what you expected.

[Jonathon Klepatzki]

That actually fixed it. It's been one of those days. Thank you @Stephen23@the cyclist