## the unit after taking FFT transform

## the unit after taking FFT transform

(OP)

what is the unit of the data after taking FFT transform?

say, if we have a group of acceleration data, usually the acceleration is in m/s^2 or g, what is the unit after the FFT transform and take the magnitude?

many people just say it is Frequency Response Function(FRF) and label Magnitude for the Y x-axis, what really is it?

Thanks

say, if we have a group of acceleration data, usually the acceleration is in m/s^2 or g, what is the unit after the FFT transform and take the magnitude?

many people just say it is Frequency Response Function(FRF) and label Magnitude for the Y x-axis, what really is it?

Thanks

## RE: the unit after taking FFT transform

A spectrum display is a plot where the X axis represents vibration frequency and the Y axis represents amplitude. If the data is taken in acceleration, the Y axis may be displayed in acceleration or integrated to velocity or double integrated to displacement.

## RE: the unit after taking FFT transform

But....

I do think that under some circumstances it is reasonable to consider the units of the continous fourier transform (and by extension the FFT spectral components) to be g-sec.

Consider Parseval's relation:

Integral(x^2(t)dt)=Integral(X^2(w)dw)

where x(t) is time and X(w) is frequency domain.

The left side will have units of g^2*sec

The right side if we treat X as g's would have units of g^2/sec.

But if we treat X as g-sec, the right side will be (g-sec)^2/sec = g^2*sec.

The discrete fourier transform has a summed Parseval's equation which does not create the same motivation for g-sec. But if we are interpretting our FFT as representative of a continuous spectrum... g-sec might be reasonable. Am I off base here?

## RE: the unit after taking FFT transform

say 1*sin(2*pi*t) + 2*cos(6*pi*t)

for a number of points such that the signal is periodic, say 32 points at a time interval of 3-(3/32) seconds for the above example, and run it through the FFT algorithm. This allows you to check both the scaling of the FFT algorithm and the ordering of the positive, negative, nyquist and dc spectral lines.

M

## RE: the unit after taking FFT transform

All our FFTs have units up the side, they wouldn't be much use for engineering if they didn't.

The actual units of the sine and cosine functions in a hand calculated fourier analysis are the same as the original units of measure... so I'm inclined to think that they are in g, in this case (obviously so for tonal components if you think about it).

Since 75% of my career has been based on assuming that an FFT is a reasonable approximation to a proper fourier series, I'm going to stic with that!

Cheers

Greg Locock

## RE: the unit after taking FFT transform

Mikey - if I understand you correctly the N will carry with it some units of time? I suspect you're right but I can't connect the dots. Can you explain a little more how that would come about?

Greg - If you assume the FFT is the same as Fourier series (special case of Fourier transform).... then what about the continuous Parseval's equation? It won't hold if you assume X(w) and x(t) have the same units.

## RE: the unit after taking FFT transform

However, on rereading it, what it seems to be saying is that the energy of the spectrum is the same as the energy of the time history, which is a damn good place to start.

I agree with your reasoning, but I don't like it very much!

On further examination, you are definitely calculating a Real energy in the time domain, but is integral(((X(w))^2)dw) always Real? I don't think so, always, for instance if X(w)=j then you would get a very silly result, but the Iin the time domain it is a sensible if strange waveform.

Puxxled? I am.Puzzled as well.

Cheers

Greg Locock

## RE: the unit after taking FFT transform

## RE: the unit after taking FFT transform

multiplied by timeI think MikeyP was onto something. IF I consider the sample frequency to be constant, then multiplying by N has the same proportional effect for discrete transforms as multiplying by T has for continuous transforms. That multiplication (or division...I've lost track) might remove the time units.