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I know that frequency is not the same as the rate of change of amplitude, but why?

Surely a high frequency is when your amplitude changes rapidly. Does an FFT perform something similar to differentiation of a function?

Furthermore, what else is needed other than my amplitude values to plug into an FFT to get frequencies?enter image description here

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    $\begingroup$ It's not very clear what you are asking. Obviously by rate of change of amplitude you mean derivative of your signal? FT is not performing anything similar to signal differentiation. Most simplistic explanation would be that its fitting lot's of complex sinusoids to your signal and calculating the correlation with it. Although you can calculate the FT of derivative of your signal very easily. Just use the relation: $\mathcal{F}[\frac{d}{dt}f(t)]=-i2\pi f F(f) $, where $F(f)=\mathcal{F}[f(t)] $. $\endgroup$ – jojek Jul 1 '14 at 16:24
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Fourier transforms and differentiation are indeed very different, but your intuition is still not entirely wrong. Here's how to make sense of it.

Let's assume you have a signal that is essentially sine-shaped locally, i.e. we can write it as $s(t) = a \sin(2 \pi f t)$ for a times $t$ close to some time $t_0$. In other words: The signal oscillates like a sine and has a sufficiently slowly changing envelope and instantaneous frequency.

The time derivative of this signal is $s'(t) = 2 a \pi f \cos(2 \pi f t)$. This signal is also oscillating with the same frequency and its amplitude is scaled by $2\pi f$ compared to the original signal. So the frequency is encoded in the derivative.

There are a few different methods for decoding the frequency from this derivative, and I'll describe the one that is probably closest to your intuition. We have to get rid of the oscillating part. So let's take first the magnitude of both signals and afterwards apply a low pass filter $H_l$ that removes the residual oscillation and leaves you with just the average value:

$$H_l(|s(t)|) = a \frac{2}{\pi}$$ $$H_l(|s'(t)|) = 2 a \pi f \frac{2}{\pi}$$

( The 2/$\pi$ comes from averaging over $|\sin(t)|$. ) With this, we can find the quotient of the two results and have

$$\frac{H_l(|s'(t)|)}{H_l(|s(t)|)}=2\pi f$$

which demonstrates that the derivative of the signal in fact yields a very accurate approximation of the frequency under the assumptions we made.

For more general signals this method calculates not the instantaneous frequency but the so called spectral centroid frequency.

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  • $\begingroup$ Thanks, but what is $2 \pi ft$? $\endgroup$ – Cobbles Jul 1 '14 at 18:13
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    $\begingroup$ $\pi$ is 3.141..., $f$ is frequency and $t$ is time. Together they form the argument of a sine that oscillates with frequency $f$. $\endgroup$ – Jazzmaniac Jul 1 '14 at 19:43
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but why?

Some counter examples showing why it would be hard to try to define frequency from rate of change:

  • Rate of change scales with amplitude (multiply a signal by 1000, the rate of change will be multiplied by 1000 too) ; while frequency should be amplitude independent.
  • A sine wave has constant frequency but its rate of change varies over the course of the waveform (and gets null at minima maxima).
  • Consider a 10 Hz sine wave, of amplitude 1V, to which are added spiky bursts of noise of amplitude 1µV with an attack time of 1ns. The rate of change will become extremely high at these points - but the noise is absolutely irrelevant in determining the dominant frequency of the signal.

Does an FFT perform something similar to differentiation of a function?

No.

what else is needed other than my amplitude values to plug into an FFT to get frequencies?

FFT does not give frequencies. It gives complex amplitudes for a set of equally spaced frequencies.

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  • $\begingroup$ Thank you for explaining that! What do you mean by "it gives complex amplitudes for a set of equally spaced frequencies"? $\endgroup$ – Cobbles Jul 1 '14 at 18:11

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