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I'm studying electrical engineering and use correlation, Fourier transform and Laplace transform a lot. I know how and when to use them, however, the interpretation I've seen in the lectures still leave me a bit hungry.

Fourier series is easily interpreted as how much a harmonic contributes to a periodic signal. However, I feel that the way Fourier transform were taught, fails to give a similar simple interpretation.

The focus of these transformations in my lectures is mainly on the transformation of known functions or using the properties (e.g. $d/dt => s$), without much interpretation for transformation a randomly chosen signal. In labs however we focus on FT of measures signals of sensors. We do not know the function beforehand.

Let's assume well behaving (so realistic) real signals $f(t)$ and $g(t)$ here (which we for instance measure). Now I understand that the cross-correlation

$R(\tau)=\int_{-\infty}^{\infty} f(t)g(t+\tau)dt$

is a measure of similarity of functions $f(t)$ and $g(t)$ over a time shift $\tau$.

Can you then say that

$F(s)=\int_{0}^{\infty} f(t)e^{-st}dt = \int_{0}^{\infty} f(t)e^{-\sigma t}[cos(\omega t)+isin(\omega t)]dt $

Is then a measure for similarity between $f(t)$ and the decaying exponential $e^{-st} = e^{-(\sigma +i\omega)t} = e^{-\sigma t}[cos(\omega t)) +i sin(\omega t)]$ that starts at $t=0$? The real part of $F(s)$ expresses the similarity to a decaying cosine and the imaginary part to a decaying sine.

Similarly

$F(\omega)=\int_{-\infty}^{\infty} f(t)e^{-\omega t}dt = \int_{-\infty}^{\infty} f(t)(cos(\omega t)+isin(\omega t))dt $

expresses in its real part the similarity of $cos(\omega t)$ and in the imaginary part $sin(\omega t)$. This then easily explains that some function that is highly similar (but not exactly) to $cos(\omega_{0} t)$ has relative high magnitude for F($\omega_{0}$) in $\omega_{0}$ but also for a band around $\omega_{0}$, as the integral is not exactly zero (which I see in my measurement labs)

Is this interpretation of measures of similarity correct to some degree (especially in the context of measuring unknown signals)?

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  • $\begingroup$ Your intuition about the Laplace transform and Fourier series as measures of similarity is basically on the right track. Could you elaborate on what makes you think the Fourier transform cannot be interpreted the same way? $\endgroup$ – Oberdada May 16 '16 at 18:42
  • $\begingroup$ The part about Fourier series was a bit poorly states. It belongs in the beginning of the question. When FS is taught, the interpretation of a sum of harmonics is always given. While the focus of FT (and LT) lays on transforms of known functions and on properties, leaving out a similar interpretation. From the cross-correlation I then try give an interpretation to FT and LT myself, to increase my understanding of FT and LT. Editted question to make it more clear. $\endgroup$ – pivu0 May 16 '16 at 18:59
  • $\begingroup$ I think this is a great way to think about both transforms. I could not find more material along this line of thinking elsewhere on the internet. For $\tau = 0$ and $g(t) = e^{-st}$ I think $F(s) = R(0)$ since the integral is 0 from $-\infty$ to $0$. When $s = i\omega$ i.e. purely imaginary, this - the fourier transform makes sense to me as it now measures the correlation of f(t) with the sinusoid of frequency $\omega$ giving us a measure for its presence. But I can't see how the decaying exponential helps, i.e. what the laplace transform is doing... $\endgroup$ – Aditya Aug 19 '18 at 12:11
  • $\begingroup$ @Aditya, Similar to fitting a frequency with the FT, I think the Laplace Transform fits a envelope of exponential decay or growth $e^{-st}$ if $s=\sigma$ which can be useful to measure for instance damping. I have however never seen this applied. $\endgroup$ – pivu0 Aug 20 '18 at 7:50
  • $\begingroup$ @pivu0 Right I see your point, the Laplace transform fits a larger class of functions and this makes sense, Laplace transform is broader and works for functions which don't have a FT because it can handle the function as long as it of "exponential order". I hope someone who is more experienced answers this question. I'll be sure to try and answer this comprehensively once I feel confident in the topics involved. We need to atleast see how far this intuition can be carried and where it breaks down. $\endgroup$ – Aditya Aug 20 '18 at 8:19

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