Re: [ferret_users] Re: FFTA

[Ryo Furue <furue@xxxxxxxxxx>]

Dear Yadidya and Saurabh,

On Mon, Jun 4, 2018 at 2:06 PM, saurabh rathore <rohitsrb2020@xxxxxxxxx> wrote:

Fast Fourier Transform i.e. ffta yields the same unit of the variable as it is the amplitude of the variable for various frequencies. so on y-axis it is celcius to best of my knowledge.

Without knowing Ferret's implementation of FFT, I think Saurabh's answer is most likely to be correct.  Here I would like to add a bit of background.

Normally "Fourier transform" (time -> frequency) is defined as

F(ω) = integral f(t) exp( i ω t) dt 

(give or take a factor of 1/2π or 1/(2π)^{1/2}).  Therefore, the units of the Fourier transform of temperature f(t) will be degrees × seconds. (Recall exp is dimensionless.) F(ω) is sometimes called the "spectral density".

On computers, we often talk about FFT, which is a "Fourier series expansion":

f(t) = sum{n = 0 to infinity} of A(n) exp(2 π i n t / T)

for 0 ≤ t < T.  We call these A(0), A(1), A(2), . . .  Fourier coefficients.  Because exp is dimensionless, A's and f(t) have the same units.

This distinction is sometimes important.  For example, if you compare your spectral density to those in the literature, you need to make sure you use the same units and you often need to convert your Fourier coefficients A(n) to spectral density A(n) Δt.

(Bonus exercise: Write down Parseval's relations for Fourier transform and Fourier series. Then you'll see the importance of the units.)

Best regards,

Ryo