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For this question purposes a function, $ f \left( t \right) $ is said to hold the Partition of Unity if:

$$ \forall t, \ \sum_{n \in \mathbb{Z}} f \left ( t - n \right ) = 1 $$

Using the Poisson Summation Formula the Partition of Unity is equivalent of $ F \left( 2 \pi k \right ) = \delta \left ( k \right ) $ where $ F \left ( \omega \right ) $ is the Fourier Transform of $ f \left ( t \right ) $.

Looking at $$ \frac{ \operatorname{sinc}^{p} \left ( pt \right ) }{\int_{-\infty}^{\infty} \operatorname{sinc}^{p} \left ( pt \right ) } $$ Does it holds the Partition of Unity for $ p \geq 2 $?

At first I thought it does, since the Fourier Transform equivalent property will hold.
Yet it seems I'm missing something.
Any ideas?

Thank You.

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For any finite-energy signal $x(t)$, the function $$y(t) = \sum_{n=-\infty}^\infty x(t-nT), -\infty < t < \infty$$ is a periodic function of $t$ with period $T$, that is, $y(t-T) = y(t), -\infty < t < \infty$. Note that $T$ is not necessarily the least period of $y(t)$. Now, a periodic function can be expressed as a Fourier series $$y(t) = \sum_{m=-\infty}^\infty c_m \exp(j2\pi mt/T)$$ where $$\begin{align} c_m &= \frac{1}{T}\int_0^T y(t)\exp(-j2\pi mt/T)\,\mathrm dt\\ &= \frac{1}{T}\int_0^T \sum_{n=-\infty}^\infty x(t-nT)\exp(-j2\pi mt/T)\,\mathrm dt\\ &= \frac{1}{T}\sum_{n=-\infty}^\infty \int_0^T x(t-nT)\exp(-j2\pi mt/T)\,\mathrm dt\\ &= \frac{1}{T}\sum_{n=-\infty}^\infty \int_{-nT}^{-(n-1)T} x(s)\exp(-j2\pi ms/T)\,\mathrm ds\\ &= \frac{1}{T}\int_{-\infty}^{\infty} x(s)\exp(-j2\pi ms/T)\,\mathrm ds\\ &= \frac{1}{T}X\left(\frac{m}{T}\right) \end{align}$$ where $X(f)$ is the Fourier transform of $x(t)$. Now, for the case $T = 1$, if $y(t) = 1$ for all $t$ as the OP desires, then it must be that $c_0 = 1$ and $c_m = 0$ for all $m\neq 0$, and therefore the Fourier transform of $x(t)$ must have the property that $$X(m) = \begin{cases} 1, & m = 0,\\ 0, &m \neq 0,\end{cases}\tag{1}$$ that is, $X(f)$ is what is sometimes referred to as a Nyquist pulse (in the frequency domain). We are familiar with Nyquist pulses in the time domain, with $\operatorname{sinc}(t) = \frac{\sin(\pi t)}{\pi t}$ being the most familiar one, but there are many others, including as a special case $[\operatorname{sinc}(t)]^k$. Applying duality, $X(f) = [\operatorname{sinc}(f)]^k$ is a Nyquist pulse in the frequency domain, and its inverse Fourier transform $x(t)$ is a signal with the property that $$\sum_{n=-\infty}^\infty x(t-nT) = 1, -\infty < t < \infty.$$ Note that $x(t)$ is the $k$-fold convolution of $\operatorname{rect}(t)$ with itself. For example, with $k = 2$, $$x(t) = \begin{cases}1 - |t|, &-1 < t < 1,\\ 0, & \text{otherwise.}\end{cases}$$ What about the other way around? Well, $\operatorname{sinc}(t)$ has Fourier transform $\operatorname{rect}(f)$ which has the Nyquist pulse property $(1)$ in the frequency domain as does $[\operatorname{sinc}(t)]^2$ whose Fourier transform $$X(f) = \begin{cases}1 - |f|, &-1 < f < 1,\\ 0, & \text{otherwise,}\end{cases}$$ also satisfies $(1)$. However, for $k > 2$, the Fourier transform of $[\operatorname{sinc}(t)]^k$ does not have the Nyquist pulse property $(1)$ because the three-fold (or more generally $k$-fold convolution of $\operatorname{rect}(f)$ with itself does not satisfy $(1)$: the $k$-fold convolution "bulges" outside the interval $(-1,1)$. See for, example, the Irwin-Hall distribution for details of the exact values of the convolution results.

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  • $\begingroup$ I see what you meant. I'm after the proof that these family of Sinc's is "Nyquist Pulse". $\endgroup$ – Royi Jul 17 '13 at 14:52
  • $\begingroup$ @Drazick Unfortunately the family does not have the right properties except when $k$ (or $p$ as you called it) has value $1$ or $2$. I have added this information to the end of my answer. $\endgroup$ – Dilip Sarwate Jul 17 '13 at 17:56
  • $\begingroup$ I think it doesn't hold for p=2 as well. The scaling at time works twice (Scaling and the Power). I was given to prove they do hold it and I think the question was wrong. $\endgroup$ – Royi Jul 17 '13 at 18:27
  • $\begingroup$ I just gave an argument above that it does hold for $p=2$ but not for $p > 2$. Feel free to edit my calculations to correct what you claim is a mistake in my answer, or down-vote it if you prefer. $\endgroup$ – Dilip Sarwate Jul 17 '13 at 18:58
  • $\begingroup$ The normalized sinc^2(2t) doesn't hold, I think. The convolution extends its support as you pointed. You didn't take into account the scaling account which stretches it further (form -2 to 2 in you example). Am I wrong? $\endgroup$ – Royi Jul 17 '13 at 20:06

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