0
$\begingroup$

Consider a band limited function $f(x)$ in $L_2(\mathbb{R})$ with frequency support in $[-B,B]$ . For small delays $\delta$ it is straightforward to show that $f(x)$ and $f(x+\delta)$ are not orthogonal in the sense of the $L_2(\mathbb{R})$ inner product. Here, the delay $\delta$ must be small relative to $\frac{1}{B}$.

In addition, Poisson Summation can be used to show that the sequences of samples $f[n], f[n+\delta], n\in \mathbb{Z}$ are not orthogonal in the sense of $L_2$ sequences.

However, when we pass to finite sums it now becomes possible that the samples $f[n], f[n+\delta], n=0,\dots,N$ are in fact orthogonal in the sense of finite real vectors. In particular, if $N=1$ and either $f(0)$ or $f(\delta)$ are $0$, then the sequences are orthogonal. Similar constructions can be found for $N=2$, but the general case becomes less tractable.

Can we find conditions on the spectrum of $f$ and the number of samples $N$ that result in orthogonal sequences?

My initial thoughts: The difference, I believe, between the finite case and the two previous cases is the time-limiting causing an expansion in frequency domain. What was once a band limited function now had a spectrum that is spread over a much larger range. Can we use this fact along with the periodic summation of the spectrum induced by summation to find forms for $\hat{f}$ that result in the desired orthogonality? What is the required relationship between $N$, $B$, and $\delta$ to be able to find such a spectrum?

$\endgroup$
3
  • 1
    $\begingroup$ Have you looked into MIMO? This seems to be along the lines of FDMA, OFDM, circulating waveforms, etc. $\endgroup$ Commented Nov 9 at 22:51
  • 1
    $\begingroup$ Another example: sine and cosine are orthogonal and delayed versions of each other. This works for both infinite length continuous signals and also finite sequences if the N contains an integer number of periods. $\endgroup$ Commented Nov 10 at 12:20
  • 1
    $\begingroup$ If this discrete-time function is a Maximum Length Sequence, then shifts of any amount other than an integer number of periods, will have an inner product that is virtually zero. $\endgroup$ Commented Nov 10 at 22:20

0

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.