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Can anyone please help me with this problem. I must let you know from the beginning that it's not an easy one.

"Two functions are given: $u, y \in L^{2}(-\infty,\infty), y(t)=\frac{u(t)}{u(t)+b}$ , with $|u(t)|\leq c<b$. Knowing that the bandwidth of $u$ is $\nu$, i.e., $(\mathcal{F}u)(s)=0 , \forall s\in \mathbb{R} \backslash [-\nu,\nu]$, how can one approximate the bandwidth of $y(t)$ as accurate as possible?"

Formula for Fourier transform used: $(\mathcal{F}u)(s)=\int_{-\infty}^{\infty}u(t) e^{-2\pi i st}dt $.

PS: I know that literally the bandwidth of $y$ is infinite, but the spectrum (absolute value of Fourier Transform) clearly decreases very fast to infinity. So instead of the normal bandwidth, another definition of bandwidth can be used (you can use whatever definition you find suitable). The one that I propose is the FOBE bandwidth, which is defined by $\nu_\alpha\in \mathbb{R}$, such that:

$\int_{-\nu_\alpha}^{\nu_\alpha}(\mathcal{F}y)^2(s)ds \geq \alpha\int_{-\infty}^{\infty}(\mathcal{F}y)^2(s)ds = \alpha||y||_2^2$, with $\alpha$ a constant of choice close to 1 (e.g. $\alpha=0.99).$

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2 Answers 2

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This may be too crude for what you need, but you could expand y in a geometric series: $$y=\sum_{n=1}^\infty (-1)^{n+1}({u\over b})^n.$$ Clearly every finite truncation of the series has finite bandwidth. Then you can combine this with an estimate of the error of the series.

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  • $\begingroup$ This is not crude at all, especially if you combine it with the observation that the FOBE bandwidth of $u^n$ grows like $\nu\sqrt n$ rather than $\nu n$ due to the central limit theorem. $\endgroup$
    – fedja
    Commented Nov 7, 2015 at 11:41
  • $\begingroup$ Thanks Michael, in the end that's how I've done it, and the proof was included in my paper: ieeexplore.ieee.org/abstract/document/7226558 $\endgroup$
    – Dorian Florescu
    Commented Oct 27, 2017 at 18:16
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I haven't found an answer but I looks like the solution would be easier to find if your signal u(t) is defined on a finite time domain,i.e., $L^2(0,T)$ so you can find an upper bound for your Fourier transform

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  • $\begingroup$ I can use it both ways, to be honest, as long as the final solution works. $\endgroup$
    – Dorian Florescu
    Commented Sep 17, 2013 at 13:09

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