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

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.

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

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}

• Thanks you for the answer. Could you extend it with these: Mar 21 at 16:08
• (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? Mar 21 at 16:12
• 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. Mar 21 at 16:54
• @Joako: A causal function has indeed a complex-valued Fourier transform, and its real and imaginary parts are related via the Hilbert transform. Mar 21 at 17:02
• @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. Mar 22 at 10:18