3
$\begingroup$

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. $\alpha=0.99).$

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Nov 7 '15 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 Oct 27 '17 at 18:16
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I can use it both ways, to be honest, as long as the final solution works. $\endgroup$ – Dorian Florescu Sep 17 '13 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.