Please help me understand whether I have set up a loop and performed an RMSE (Root Mean Square Error) calculation correctly

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Ann St 2020-8-10

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编辑： Ann St 2020-8-11

I am returning to Matlab after such a long break that I'm essentially a newbie. I'd be very grateful if someone would please help me figure out if I have done what I intend to do!

I have a column of data (avgW) from measuring something over time.
I have an equation (second line of my code below) to calculate the same thing I measured (wCalc). In the equation, the parameters, a, b, c, d, e, and f are all constants. The parameter "columnOfData" is a column of distinct values.
My goal is to find the value of "x" in the equation. What I am trying to do below is first plug in a test value for x (1 - 10000). Then figure out which value (between 1 and 10000) gives me the smallest difference between wCalc and avgW.

I think my method so far would work if "columnOfData" were a constant, but it isn't and I am lost. I'm not even sure what I'm going to end up with when my current calculation finishes. Would anyone happen to know how to find and plot and record the lowest value of RMSE for each value of the column/vector, columnOfData?

Please feel free to comment if there are more appropriate tags for this question. Thank you.

for x = [1:1:10000]
    wCalc = ( ( a ./ ( ( columnOfData ./ x ) - b ) ) - c - d - e ) .* f; 
    RMSE(x) = sqrt( sum( ( wCalc - avgW).^2) ./ g );
end

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hosein Javan 2020-8-10

if RMSE is calculated for each "x", then plot(x,RMSE) would work. and the lowest value is found by "[i,m]=min(RMSE)".

if columnOfData changes, then add another for loop to calculate for each Data set.

if the goal is to find the error of RMS value calculation, why not using integral and calculate the convergence error?

Ann St 2020-8-10

@hosein Javan, I don't know how to use the integral method you mentioned (I'm pretty new to error analysis) but will look it up. I'll think about your loop suggestion and see if I can come up with a solution. Thank you.

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Answer 1

Jon 2020-8-10

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编辑：Jon 2020-8-10

在 MATLAB Online 中打开

You could do something like this.

I just made up some numbers to try the code.

Also, obviously variable names like columnOfData are not very meaningful and I wouldn't recommend them but I didn't want to depart to far from your starting point.

I removed some of the ./ and just used / where the operation involved a vector and a scalar so element by element calculations were not needed. Would work with ./ but makes it less clear what you are doing. I also defined x separately from the index in the loop, in case you wanted to use other x values and also so tha you could have an index for recording the RMSE values.

With some more work, you could probably fully vectorize this (eliminate the loop) but at least this should get you started.

% define constants
a = 2.2
b = 3.4
c = 8.1
d = 10.4
e = 0.6
f = 22.4
% define measured value 
avgW = [8;7.2;44;3.12] 
% define column of data
columnOfData = [3;9;8.3;2]
% number of elements in avgW (and column of data)
g = length(avgW)
x = 1:10000;
for k = 1:10000
    wCalc = ( ( a ./ ( ( columnOfData / x(k) ) - b ) ) - c - d - e ) * f; % a vector
    RMSE(k) = sqrt( sum( ( wCalc - avgW).^2) / g );
end
% find optimal value, and it's index in the array x
[RMSEOpt,idx] = min(RMSE)
xOpt = x(idx)

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Jon 2020-8-11

If you are stuck, and you could always attach your code, or a relevant fragment of it. If at this point your original question is answered, when you have a chance, please accept this answer so others will know it has been answered.

Ann St 2020-8-11

编辑：Ann St 2020-8-11

Thank you very much for all your help, Jon. I would not have been able to figure this out otherwise. I had to add another step because I kept getting a "nan" result (even though I thought I had elimated nans from my inputs), but it is finally working. Your solution is much more convenient than my original idea.

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Answer 2

hosein Javan 2020-8-11

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在 MATLAB Online 中打开

is this way more efficient?

freq = 1;
T = 1/freq;
w = 2*pi*freq;
a = [1 1/2]; % harmonic amplitudes
y = @(t) a(1)*sin(w*t) + a(2)*sin(2*w*t); % define a periodic function with 2 harmonics
format long
RT = 1e-5; % relative tolerance of integral
RMS_cal = sqrt(freq*integral(@(t)y(t).^2,0,T, 'RelTol', RT )) % calculated RMS value using integral
RMS_exact = sqrt(sum(a.^2))/sqrt(2) % exact value of RMS

RMS_cal =

0.790569415042095

RMS_exact =

0.790569415042095

if your signal's frequency is known, you can find the exact rms value and compare it to your integral.

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Ann St 2020-8-11

Thanks, hosein. I'm not sure if harmonic analysis would be applicable to my problem, but I will think it over once I'm done with testing another possible solution.

hosein Javan 2020-8-11

you're welcome. the harmonics are used only to compute the exact RMS value only for comparison. the integral part is what we must use. if you have your signal function you don't need harmonics at all however it could be easily derived.

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