# Another mathy question from the Signal Processing community…

This is the question as posed to the DSP folks. Electrical engineers like to use particular definitions of the Fourier Transform (using Hz, rather than angular frequency) and the sinc function.

Out of consideration to you kind folks, I will try to eliminate a few symbols, like I will, without loss of generality, set the sample rate, $f_\text{s}$, to 1 and I try to use angular frequency.

So let $$0 < \omega_0 < \pi$$

and for any real $W$ such that

$$\omega_0 < W < 2\pi - \omega_0$$

$$\cos(\omega_0 t + \phi) = \sum\limits_{n=-\infty}^{\infty} \cos\left(\omega_0 n + \phi \right) \, \frac{\sin\big( W(t - n) \big)}{\pi(t - n)}$$

without the use of the Fourier Transform.

I s'pose one of the " Whittaker–Nyquist–Kotelnikov–Shannon" folks did this, but I can't see exactly how this gets extracted out of the Poisson summation formula.

I spent all of my rep on a bounty of a previous question. Sorry, I don't have much rep to spend here.

well, 21 hours left to get the bounty!! don't let it go to waste.

UPDATE: the 100 rep bounty has expired. so i guess it's wasted.

• Just out of interest, why is it important to you that it is proved without the fourier transform? – user159517 Jul 18 '17 at 8:00
• And what about fourier series? – user159517 Jul 18 '17 at 8:01
• geez, i know sorta how it's done with the Fourier Transform. (see the link that takes it to the DSP SE). i just thought that with a zillion different values for $W$, the sum of the sinc functions (of width $W$) still adds up, no matter what $W$ is, as long as it's in that range. – robert bristow-johnson Jul 18 '17 at 8:13
• @robertbristow-johnson Anything you don't understand with my answer ? – reuns Jul 18 '17 at 9:42
• an error was pointed out to me in the formation of the question at the DSP.SE . that required i change $W$ to $\pi$ in the denominator of the $\frac{\sin(x)}{x}$ factor in this question. – robert bristow-johnson Jul 19 '17 at 8:35

• Let $$f(x) = \frac{\sin(\pi x)}{\pi x} = \int_{-1/2}^{1/2} e^{2 i\pi \xi x}d\xi, \qquad F(x) = \int_{-\infty}^x f(y)dy, \quad F(-\infty) = 0, \ F(+\infty) = 1$$ If $g$ is bounded and $C^1$ then $$\lim_{m \to \infty}\int_{k-1/2}^{k+1/2} m f(mx) g(x)dx=\lim_{m \to \infty} F(mx) g(x)|_{k-1/2}^{k+1/2}- \int_{k-1/2}^{k+1/2} F(mx) g'(x)dx \\ = 1_{k> -1/2} g(k+1/2)-1_{k> 1/2}g(k-1/2)-\int_{k-1/2}^{k+1/2}1_{x > 0} g'(x)dx= g(0) 1_{|k| < 1/2}$$ (ie. $m f(mx) \to \delta(x)$ in the sense of distributions)

• Then see the Dirichlet kernel

$$\sum_{n=-\infty}^\infty e^{2i \pi k n}f(x-n) =\lim_{m \to \infty} \sum_{n=-m}^m e^{2i \pi k n} f(x-n) =\lim_{m \to \infty}\sum_{n=-m}^m \int_{-1/2}^{1/2} e^{2i\pi( \xi x+(k-\xi) n)}d\xi \\= \lim_{m \to \infty} \int_{-1/2}^{1/2} e^{2i\pi \xi x} \frac{\sin(2\pi(m+1/2) (k-\xi))}{\sin(\pi (k-\xi))}d\xi \\=\lim_{m \to \infty} \int_{-k-1/2}^{-k+1/2} e^{2 i\pi (\xi+k) x} \frac{\pi \xi}{\sin(\pi \xi)}m f(m\xi) d\xi =e^{2i \pi k x} 1_{|k| < 1/2}$$

With a subtelty if $|k| = 1/2$, in that case we need a factor $1/2$.

The whole proves the Fourier series, the Fourier transform as well as the Shannon sampling theorem !

• i just wanna confirm a notation that appears to be used by the math community, but is unfamiliar to me. is this true?: $$1_{|k|<\tfrac12} = \Pi(k)$$ where $\Pi(u)$ (sometimes "$\operatorname{rect}(u)$") is the rectangular function $$\Pi(u) \triangleq \begin{cases} 1 \qquad & \text{ if } |u| < \tfrac12 \\ 0 \qquad & \text{ if } |u| \ge \tfrac12 \\ \end{cases}$$ – robert bristow-johnson Jul 18 '17 at 21:45
• and $$1_{x \ge 0} = H(x)$$ is the Heaviside unit step function $$H(u) \triangleq \begin{cases} 1 \qquad & \text{ if } u \ge 0 \\ 0 \qquad & \text{ if } u < 0 \\ \end{cases}$$ if we don't worry too much about the value at the discontinuity? – robert bristow-johnson Jul 18 '17 at 21:45
• reuns, i had a mistake in the question that had to be corrected. see my comment above. does that change anything in your proof? – robert bristow-johnson Jul 19 '17 at 8:36