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Does the Kramer-Kronig relations apply to this example $f(t) =\left(1-t^2\right)^4\cdot\theta(1-t^2)$?

with $\theta(t)$ is the Heaviside step function.

I made a detailed related question here with full explanations, where I got no answers, but the main doubt could be solved just by knowing if the KK-relation conditions are fulfilled or not by this example.

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    $\begingroup$ I would suggest to add the definition for $\theta(\cdot)$ for completeness $\endgroup$ Mar 20, 2023 at 14:19
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    $\begingroup$ @LaurentDuval done. $\endgroup$
    – Joako
    Mar 20, 2023 at 16:21
  • $\begingroup$ Hmm, the KK relations may be what I was looking for... $\endgroup$ Mar 21, 2023 at 14:52

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The function $f(t)$ is real-valued and even, and so is its Fourier transform $F(\omega)$. Clearly, the real and imaginary parts (the latter being zero) of $F(\omega)$ are not related via the Hilbert transform. But this is also not to be expected, because $f(t)$ is not causal. However, since $f(t)$ vanishes for $|t|>1$, we can shift it such that the resulting function is causal:

$$g(t)=f(t-1)$$

The Fourier transform of $g(t)$, expressed in terms of $F(\omega)$, is

$$G(\omega)=F(\omega)e^{-j\omega}=F(\omega)\cos(\omega)-jF(\omega)\sin(\omega)$$

The real and imaginary parts of $G(\omega)$ satisfy the well-known Hilbert transform relations due to the causality of $g(t)$. This relationship can be shown using Bedrosian's theorem.


For general non-causal right-sided sequences we can also derive equations relating the real and imaginary parts of their Fourier transform. Let $f(t)=0$ for $t<-T$, $T>0$. Consequently,

$$f(t)=f(t)u(t+T)\tag{1}$$

where $u(t)$ denotes the unit step function. Taking the Fourier transform of $(1)$ gives

$$F(\omega)=\frac{1}{2\pi}F(\omega)\star U(\omega)e^{j\omega T}\tag{2}$$

where $\star$ denotes convolution, and $F(\omega)$ and $U(\omega)$ are the Fourier transforms of $f(t)$ and $u(t)$, respectively. With

$$U(\omega)=\pi\delta(\omega)+\frac{1}{j\omega}\tag{3}$$

Equation $(2)$ becomes

$$F(\omega)=\frac12 F(\omega)+\frac{1}{2\pi}F(\omega)\star \frac{e^{j\omega T}}{j\omega}\tag{4}$$

which is equivalent to

$$F(\omega)=F(\omega)\star \frac{\sin\omega T-j\cos\omega T}{\pi\omega}\tag{5}$$

Splitting $(5)$ into real and imaginary parts, and with $F(\omega)=F_R(\omega)+jF_I(\omega)$ we obtain

$$\begin{align}F_R(\omega)&=F_R(\omega)\star\frac{\sin\omega T}{\pi\omega}+F_I(\omega)\star\frac{\cos\omega T}{\pi\omega}\\F_I(\omega)&=F_I(\omega)\star\frac{\sin\omega T}{\pi\omega}-F_R(\omega)\star\frac{\cos\omega T}{\pi\omega}\end{align}\tag{6}$$

For $T=0$, i.e., for causal $f(t)$, Equation $(6)$ simplifies to the well-known Hilbert transform relationships between real and imaginary parts of $F(\omega)$:

$$\begin{align}F_R(\omega)&=F_I(\omega)\star\frac{1}{\pi\omega}=\mathcal{H}\big\{F_I(\omega)\big\}\\F_I(\omega)&=-F_R(\omega)\star\frac{1}{\pi\omega}=-\mathcal{H}\big\{F_R(\omega)\big\}\end{align}\tag{7}$$

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  • $\begingroup$ Thanks you for the answer. Could you extend it with these: $\endgroup$
    – Joako
    Mar 21, 2023 at 16:08
  • $\begingroup$ (i) Why $f(t)$ is not causal? it got finite starting and ending times... is due it have values $f(t)\neq 0\text{ for some }t<0$? It looks arbitrary since I could place $t_0$ at any point just with a translation that doesn't change the "form" of the signal, so I expect to got the same frequency components... Why this "logic" is mistaken? $\endgroup$
    – Joako
    Mar 21, 2023 at 16:12
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    $\begingroup$ Causality requires $f(t)=0$ for $t<0$. If $f(t)=0$ for $t<t_0$ with some $t_0\neq 0$, you can also derive a relationship between the real and imaginary parts of $F(\omega)$ but it's not the standard form of the Hilbert transform. $\endgroup$
    – Matt L.
    Mar 21, 2023 at 16:54
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    $\begingroup$ @Joako: A causal function has indeed a complex-valued Fourier transform, and its real and imaginary parts are related via the Hilbert transform. $\endgroup$
    – Matt L.
    Mar 21, 2023 at 17:02
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    $\begingroup$ @Joako: I don't have a reference for that, I'll add the math to my answer. I'll also provide an answer to your question on the math site. It might take a few days though. $\endgroup$
    – Matt L.
    Mar 22, 2023 at 10:18

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