How to do a double fourier transform or double summation equation in Matlab?

Hi Community,

Iam very confused to this equation, because it should be written in a double fourier transform which re very difficult for me.... SO here is the equation i should solve in my Thesis :

T is hour of the years (it can be replaced with a datenum, right?) and M is month number. I just want to know how this difficult equation should be written as a code in Matlab... Im very confused.... I think it would be written in a combination of symsum, syms, and etc, but i still doubt how to write it correctly in Matlab.... WOuld anyone lend me a hand for this equation, please... Thank you so much, everyone... /.\ /.\ /.\

[Hiro Yoshino]

This is not a Fourier transform but a Fourier series expansion.

Although I am unfamiliar with this type, all you need is determine A_n^m term upto 6. You may want to make a reference to this for example. Find orthogonal functions and multiply them to the given equatio and you can obtain A_n^m terms.

[Tyann Hardyn]

@Hiro Im sorry, Sir. But the m actually not an exponent number just like An^m... In other ways, the equation is like this :

Iam just wonder, Is the equation use for loop or not, Sir? Thank you very much

Answer 1

[Walter Roberson]

T cannot be replaced by a datetime. datetime objects cannot be multiplied by anything.

At one level, T could potentially be replaced by a duration() object, as durations can be multiplied... but you cannot take cos() of a duration.

syms T M
syms A [6 6]
syms alpha [1 6]
syms beta [1 6]
n = (1:6).';
m = 1:6;
Sq_est = sum(sum(A .* cos(m .* T + alpha) .* cos(n .* M + beta)))

Sq_est = 

[Tyann Hardyn]

Alright, i ll test this code, Sir... This is the exact code, right?

syms T M
syms A [6 6]
syms alpha [1 6]
syms beta [1 6]
n = (1:6).';
m = 1:6;
Sq_est = sum(sum(A .* cos(m .* T + alpha) .* cos(n .* M + beta)))

Hmmm... So what should i replace the variable T with? Actually, it should be replaced with a Hour of Years... But what is Hour of the years actually meant....

[Tyann Hardyn]

@Walter Roberson Oh, i know it now...

The T = Hours Of Year is actually can be determined by this :

( [datenum of hour time in T (ex. datenum of 2022-04-01 02:29:00) - datenum of hour time in the very FIRST year hour time (ex. datenum of 2022-01-01 00:29:00)] / [datenum of hour time the very LAST year hour time (ex. datenum of 2022-12-31 23:29:00) - datenum of hour time in the very FIRST year hour time (ex. datenum of 2022-01-01 00:29:00)] ) * 360 degree

[Walter Roberson]

Your calculations are off by 1 hour.

Suppose that your "year" was 1 day long. 23:29 - 00:29 is 23 hours, but we know that a complete day is 24 hours.

Take datetime of the time in question, subtract datetime of the very beginning of the year, getting a duration object. Take hours() of that. Divide by the number of hours in the year.

But I wonder about that 360 degrees. Are you working with solar years or sidereal years?

[Tyann Hardyn]

@Walter Roberson Im working with solar years that count 365.25 or 366 days in a year, Sir...

I wonder, is the code of yours didnt use a "for" loop or something? Is that code can just be inputed by the variable in which T (Hours Of Year) and M (Month Number) and A as Data Series? Thank you so much...

[Tyann Hardyn]

@Walter Roberson Im sorry, Sir. Maybe the equation should be like this one :

Iam just wonder, Is the equation use for loop or not, Sir? Thank you very much....

[Walter Roberson]

The sq_est I showed does not need a loop even for multiple T and M, if you use matlabFunction() or subs() the right way. You would need a loop to process different A, alpha, or beta.

[Tyann Hardyn]

Alright, Sir.. So the input of T variable is hour of the year, sir? and the M is month number (ex. January = 1, February = 2, etc), and the A is the data in T and M?

[Tyann Hardyn]

@Walter Roberson Would u like to tell me the version of loop code from your code before, Sir?