mean square error (MSE) between two graphs

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adriane duarte
adriane duarte am 1 Nov. 2022
Kommentiert: Michael am 8 Nov. 2022
Hello guys.
I want to calculate the MSE between two curves that I drew but I don't know how to do it. Plot information has different dimensions.
I put the script with the plot of the figures.
The figures start (X,Y) at the zero point.
% practical values
V_red = [0,1,1.62,1.65,1.66,1.67,1.68,1.69,1.69,1.70,1.71,1.71,1.72,1.72,1.73,1.73,1.73,1.74,1.74,1.74,1.75,1.75,1.77,1.78,1.80,1.82,1.84,1.86,1.90,1.91,1.93,1.96,1.98,2.01,2.03,2.05,2.06,2.07,2.12,2.19,2.20,2.21,2.28,2.41,2.52,3.01,3.55]; % volts
I_red_amp = [0,0,0.00005,0.0001,0.00015,0.0002,0.00025,0.0003,0.00035,0.0004,0.00045,0.0005,0.00055,0.0006,0.00065,0.0007,0.00075,0.00080,0.00085,0.0009,0.00095,0.00102,0.00131,0.00152,0.00206,0.00253,0.00324,0.00406,0.00561,0.00631,0.00711,0.00842,0.00967,0.01158,0.01245,0.01367,0.01440,0.01530,0.01851,0.02311,0.02411,0.02630,0.03125,0.04556,0.059,0.1281,0.2485]; % ampere
% LED information
Vt_red = 3.55; % led voltage
Is_red = 0.2485; % LED current
K = 0.5; % can have multiple values of K
imax_red = 0.3; % maximum LED current
SNRdB = -10:2:25;
for sk = 1:length(SNRdB)
for tk = 1:max_run
ss = -10:0.001:24;
iLED_red = Is_red.*(exp(ss./Vt_red) - 1);
xLED_red = iLED_red./((1 + (iLED_red./imax_red).^(2.*K)).^(1./(2.*K)));
% plot figures
hold on;
xlabel('entry (V)');
hold on;
ylabel('exit (A)');
hold on;
grid on;
the resulting graph is this

Antworten (1)

Michael am 1 Nov. 2022
Interp them so they are on the same grid and then do the regulare MSE calc?
xLED_vermelho_K_INTERP = interp1(ss,xLED_vermelho_K,V_vermelho)
n = numel(I_vermelho_amp);
MSE = 1/n * sum((xLED_vermelho_K_INTERP - I_vermelho_amp).^2)
  2 Kommentare
adriane duarte
adriane duarte am 8 Nov. 2022
I was thinking of doing it like this, but as the vectors have different sizes I didn't know how to apply to calculate the error.
Michael am 8 Nov. 2022
Did my answer work for you?

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