Topic

FFT and IFFT

Ce sujet contient 4 réponses, 2 participants et a été mis à jour pour la dernière fois par paulwarshawsky@yahoo.com, le il y a 20 années et 5 mois.

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01.12.2004 à 08:20 #34162

paulwarshawsky@yahoo.com
Participant

I’m experiencing some strange behaviour, and I wonder if someone could explain it to me.

1) Take a signal with only real values (the ultrasonic burst used in “The first document”, for example)
2) Create a formula FFT(Burst)
3) Evaluate the formula – you now have a complex signal
4) Do an inverse FFT on that signal IFFT(Formula)

This should return the original signal, but does not! Due to rounding errors in the calculations I would expect a small imaginary component to the results, but the magnitude of the imaginary component is not small at all!

What am I missing here?

Paul Warshawsky

01.12.2004 à 21:05 #34159

Bernhard Kantz
Participant

Since the input signal has only real values you must use [b]IRFFT-function[/b] instead of the [b]IFFT-function[/b].
IFFT Function calculates the complex time signal as the inverse transform of a complex Fourier spectrum.
IRFFT Function calculates the real time signal as the inverse transform of a complex Fourier spectrum.

Please look at:

FFT Function (FPScript)
IFFT Function (FPScript)
IRFFT Function (FPScript)
Fourier Transform Analysis Object

Support@weisang.com

02.12.2004 à 02:06 #34160

paulwarshawsky@yahoo.com
Participant

Thank you for the response, and I shall definitely use the IRFFT function. Nonetheless, IFFT(FFT(F(x))) should be the same as F(x). The imaginary component should be near 0. Using the IRFFT function will fix my application, but I would suggest someone look into the IFFT algorhythm to make sure it is correct.

Paul

03.12.2004 à 04:32 #34161

paulwarshawsky@yahoo.com
Participant

Ahh… I understand why this doesn’t work. When you do an FFT on a signal that is entirely real, because of the symetry of the FFT result, only the first half of the spectrum is returned (the remainder being the negative complement of the first half.) Thus you have to tell the reverse FFT algorithm whether you have the entire spectrum or just the first half.

Paul

03.12.2004 à 18:09 #34158

paulwarshawsky@yahoo.com
Participant

I’m experiencing some strange behaviour, and I wonder if someone could explain it to me.

1) Take a signal with only real values (the ultrasonic burst used in “The first document”, for example)
2) Create a formula FFT(Burst)
3) Evaluate the formula – you now have a complex signal
4) Do an inverse FFT on that signal IFFT(Formula)

This should return the original signal, but does not! Due to rounding errors in the calculations I would expect a small imaginary component to the results, but the magnitude of the imaginary component is not small at all!

What am I missing here?

Paul Warshawsky
Auteur

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