# Trying to understand the Kramer-Kronig relations with example $f(t) =\left(1-t^2\right)^4\cdot\theta(1-t^2)$

Trying to understand the Kramer-Kronig relations with example $$f(t) =\left(1-t^2\right)^4\cdot\theta(1-t^2)$$

Introduction

Let think in the function:

$$f(t) =\left(1-t^2\right)^4\cdot\theta(1-t^2)\equiv \left(\frac{1-t^2+|1-t^2|}{2}\right)^4\tag{Eq. 1}\label{Eq. 1}$$

which is a real-valued square-integrable even function, with $$\theta(t)$$ the unitary step function.

At can be seen in Wolfram-Alpha, the Fourier Transform is given by: $$\begin{array}{r c l} F(w) & = & \int\limits_{-\infty}^{\infty} f(t)\,e^{-iwt}\,dt \\ & = & \int\limits_{-1}^{1} (1-t^2)^4\,e^{-iwt}\,dt \\ & = & \displaystyle{\frac{768}{w^9}\cdot \Big(5\ w\ (2\ w^2-21)\cos(w)+(w^4-45\ w^2+105) \sin(w)\Big)} \tag{Eq. 2}\label{Eq. 2} \\ \end{array}$$

which is reviewed in detail in this question since shows an horrid spike when plotted, but it end to be an artifact due current software, since it is continuous at the origin as is show in this answer by @CalvinKhor.

As expected, since $$f(t)$$ is real-valued and an even function, its Fourier transform $$F(w)$$ is also real-valued and even function (see Wikipedia Table line $$111$$).

Main part

In the Wikipedia page for the Kramers-Kronig relations is stated that if a function $$x(t)$$ have an initial time $$t_0$$ such as $$x(t) = 0\ \forall\ t, and without loss of generality using the assumption that $$t_0=0$$, then if the Fourier Transform $$X(w) \in \mathbb{C}$$ could be split as $$X(w) = U(w)+iV(w)$$ with both $$U(w),\ V(w) \in \mathbb{R}$$ real-valued, then it is possible to find under some mild-assumptions that:

$$\begin{array}{r c l} U(w) & = & \frac{1}{\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}\ d\xi \\ V(w) & = & -\frac{1}{\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{U(\xi)}{\xi-w}\ d\xi\tag{Eq. 3}\label{Eq. 3} \end{array}$$

which is traditionally stated as the property a causal function should fulfill.

And here is where I got into a problem: thinking that $$f(t)$$ it is a real-valued causal function, in principle it should be fulfilling \eqref{Eq. 3}, but since its Fourier transform its real valued, the split $$F(w) = U(w)+iV(w)$$ will lead: $$\begin{array}{r c l} U(w) & = & \frac{768}{w^9}\cdot \Big(5\ w\ (2\ w^2-21)\cos(w)+(w^4-45\ w^2+105) \sin(w)\Big) \\ V(w) & = & 0 \tag{Eq. 4}\label{Eq. 4} \end{array}$$

And I don't know how it could be possible then to fulfill with these values the relations of \eqref{Eq. 3}: $$\begin{array}{r c l} U(w) & = & \frac{1}{\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{\textbf{0}}{\xi-w}\ d\xi \overset{?!}{\equiv} \frac{768}{w^9}\cdot \Big(5\ w\ (2\ w^2-21)\cos(w)+(w^4-45\ w^2+105) \sin(w)\Big)\\ V(w) & = & -\frac{1}{\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{\frac{768}{\xi^9}\cdot \Big(5\ \xi\ (2\ \xi^2-21)\cos(\xi)+(\xi^4-45\ \xi^2+105) \sin(\xi)\Big)}{\xi-w}\ d\xi \overset{?!}{\equiv} 0 \tag{Eq. 5}\label{Eq. 5} \end{array}$$

and even it could be possible of having a zero-valued integral, in \eqref{Eq. 5} showing a function as result of integrating a zero makes no sense for me, so surely I have something mistakenly understood, or otherwise there some assumptions of the Kramer-Kronig relation that $$f(t)$$ is not fulfilling, and I don't know which it is (at least I checked that $$F(w)$$ decrease much master than $$1/|w|$$, other assumptions I don't really understand them so I cannot check them).

Hope you could explain with detail and sources what is happening here.

Reading again the Wikipedia page for the Kramers-Kronig relations I think the problem could be in the requirement "Suppose this function is analytic in the closed upper half-plane of $$w$$", which could be requiring than the Fourier Transform must have a non-zero imaginary part $$V(w) \neq 0$$ from which I think I have understood from the page for analytic signal.

But this makes the problem even worst: with the following procedure I would try to make appear the Kramers-Kronig relations under a few assumptions and using the real/imaginary split in general form, and I think without loss of generality it means that by replacing the terms by zero on the imaginary part of the spectrum it should still be holding as true, but as it could be seen is not what end by happening, and I don't know why is that so:

Let be $$x(t)$$ a square-integrable real-valued continuous function with Fourier Transform $$X(w)$$ such as $$X(w)=U(w)+iV(w)$$ with $$U(w)$$ and $$V(w)$$ real-valued functions in the angular frequency variable $$w$$ ($$X(w)\in\mathbb{C};\,U(w),V(w) \in\,\mathbb{R}$$). Since $$x(t)$$ is real-valued then $$X(-w) = \overline{X(w)}$$ his complex conjugate.

Let also $$x(t)$$ be a causal function, so, there exist an initial time $$t_0$$ where $$x(t)=0,\ \forall t. For simplicity, but without loss of generality, I will assume that $$t_0=0$$. With this, it will be equivalent to represent $$x(t)$$ as $$x(t) = x(t)\cdot\theta(t)$$ with $$\theta(t)$$ the unitary step function. This assumption have a serious consequence: since $$x(t)$$ is not extended from $$t \to \pm \infty$$, then it cannot be band-limited, as is stated in Wikipedia, so the domain on the frequencies must go from $$w\in \ (-\infty,\ \infty)$$.

Also having an initial time can be described through the Kramer-Kronig relation, which are used as a characteristic a causal function must fulfill, obtained by applying the Fourier Transform the last equation:

$$\begin{array}{r c l} X(w) & = & \mathbb{F}\left\{x(t)\cdot\theta(t) \right\}(w) \\ & = & \frac{1}{2\pi} X(w)\circledast\mathbb{F}\left\{\theta(t) \right\}(w) \\ & = & \frac{1}{2\pi} X(w)\circledast\left[\pi\delta(w) - \mathit{P\!V}\frac{i}{w} \right](w)\\ & = & \frac{X(0)}{2}-\frac{1}{2\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{iX(\xi)}{\xi-w}\ d\xi \tag{Eq. 6}\label{Eq. 6} \end{array}$$ where $$\circledast$$ is the Convolution operator, and $$\mathit{P\!V}$$ means the Principal Value which if there exist a singularity at point "$$\epsilon$$", it can be calculated as $$\mathit{P\!V}\int\limits_{-\infty}^{\infty}f(x)\ dx =\lim\limits_{c\rightarrow 0^+}\left[\int\limits_{-\infty}^{\epsilon-c}f(x)\ dx +\int\limits_{\epsilon+c}^{\infty}f(x)\ dx \right]$$.

By splitting $$X(w) = U(w)+iV(w)$$ and using the property for real-valued functions one gets that: $$\begin{array}{r c l} X(-w) & = & \overline{X(w)}\\ \Rightarrow U(-w) + iV(-w) & = & U(w)-iV(w) \\ \Rightarrow U(-w) & = & U(w)\,\,\text{even function}\\ V(-w) & = & -V(w)\,\,\text{odd function}\\ \Rightarrow V(-0) = -V(0) & \Rightarrow & V(0) = 0 \end{array}$$

This split into \eqref{Eq. 6} leads to: $$\begin{array}{r c l} U(w)+iV(w) & = & \frac{U(0)}{2}+\frac{i\require{cancel}\cancel{V(0)}}{2}-\frac{1}{2\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{iU(\xi)}{\xi-w}\ d\xi +\frac{1}{2\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}\ d\xi \\ \text{pairing real and imaginary terms}\,\Rightarrow U(w) & = & \frac{U(0)}{2} +\frac{1}{2\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}\ d\xi \\ V(w) & = & -\frac{1}{2\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{U(\xi)}{\xi-w}\ d\xi\tag{Eq. 7}\label{Eq. 7} \end{array}$$

Which are "almost" the relation shown in Wikipedia page for the Kramer-Kronig relation (show on \eqref{Eq. 8}): there is a constant $$U(0)$$ and a $$1/2$$ term don't fit: I don't know exactly why is that, and this is not the formal way to find the relations, but since it don't change why I want to ask, and I found more intuitive this procedure, please keep it mind. If you know how to get the following:

$$\begin{array}{r c l} U(w) & = & \frac{1}{\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}\ d\xi \\ V(w) & = & -\frac{1}{\pi}\mathit{P\!V}\int\limits_{-\infty}^{\infty} \frac{U(\xi)}{\xi-w}\ d\xi\tag{Eq. 8}\label{Eq. 8} \end{array}$$

from \eqref{Eq. 7} I will love to know how to do it: I think is related with the fact that the Fourier Transform just go from $$t=0\to\infty$$ as it kind of fit with the definition of the analytic signal, but to be honest I don't know in the formal definitions how they make appear the DC component $$U(0)$$ from an even function $$V(w)$$.

Also, as you can see from \eqref{Eq. 7} by replacing just $$V(w)=0$$ the equalities do not hold true, which means that the Kramer-Kronig relations, the only tool for causality analysis I know from signal analysis, don't hold for any squared-integrable real-valued causal continuous function $$x(t)$$, which don't seen right to me since physical signals in time should fulfill it (as the particular example of the question $$f(t)$$).

As can you see I am quite lost, so please elaborate about why this is isn't working.

The full conditions shown in Wikipedia for the Kramer-Kronig relation says: "Suppose this function is analytic in the closed upper half-plane of $$w$$ and vanishes faster than $$1/|w|$$ as $$|w|\to \infty$$. Slightly weaker conditions are also possible."

The closed upper half-plane is defined in Wikipedia as "the union of the upper half-plane $$\mathcal{H}\equiv \{x+iy\,|\,y>0;\,x,\ y\in\mathbb{R}\}$$ and the real axis. It is the closure of the upper half-plane."

So, since $$f(t)$$ lives in the real axis as also do $$F(w)$$, which I believe it is indeed analytic in $$w$$, and also if I plot the Fourier Transform of Eq. 2 jointly with the function $$1/|w|$$ it shows indeed a decay faster than the $$1/|w|$$, so in principle I think its fulfilling the conditions, as can be seen here

So $$f(t)$$ should be fulfilling the conditions of the Kramers-Kronig relations, so I don't know why the previous equations aren't fulfilled. Hope you explain in detail if I have misunderstood some of the conditions.

• Remember that the function should decay faster than 1/z in th upper half plane. I think your example is not a good example. Commented Dec 17, 2022 at 2:36
• @CraigThone Thanks for commenting. I am confused exactly with the meaning about the "upper half plane". So far, if I plot the Fourier Transform of Eq. 2 with the function $1/|w|$ it shows indeed a decay faster than the $1/|w|$, so in principle I think its fulfilling the condition, as can be seen here... Could you elaborate what I am interpreting mistakenly in an answer? Commented Jan 1, 2023 at 22:06
• btw it's "without loss of generality" Commented Jan 4, 2023 at 23:06
• @FShrike thanks, I have corrected it now. Commented Jan 5, 2023 at 2:49
• Shouldn't there be 2 in the denominator inside the brackets of eq. 1? Commented Jan 6, 2023 at 1:41

The Kramer-Kronig relations are not satisfied for your example because the given function is not causal, i.e., $$f(t)=0$$ for $$t<0$$ is not satisfied. Due to the finite support of $$f(t)$$, the function $$g(t)=f(t-1)$$ does satisfy the Kramer-Kronig relations.