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I am trying to use DTFT (as asked in a problem) to find the following sum

$$\sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)$$ for real $\alpha_1>0$ and $\alpha_2<1$.

I tried few approaches, but none seems to work. In the following, $x[n]$ is the discrete-time signal (sampled signal) and $X(\omega)$ the continuous-frequency transformed variable (frequency response).


Approach 1

\begin{align} \sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)&=\sum_{n=-\infty }^{\infty }\frac{1}{\alpha_1\alpha_2 n^2}\sin(n\alpha_1)\sin(n\alpha_2)\\ &=\sum_{n=-\infty }^{\infty }\frac{1}{\alpha_1\alpha_2 n^2}\frac{1}{2}\left [ \cos(\alpha_1-\alpha_2)n-\cos(\alpha_1+\alpha_2)n \right ]\\ &=\frac{1}{2\alpha_1\alpha_2}\sum_{n=-\infty }^{\infty }\frac{1}{n^2}\cos(\alpha_1-\alpha_2)n-\frac{1}{2\alpha_1\alpha_2}\sum_{n=-\infty }^{\infty }\frac{1}{n^2}\cos(\alpha_1+\alpha_2)n \end{align}

Then I thought about using this property $$\sum_{n=-\infty }^{\infty }x[n]\leftrightarrow X(0)$$

And I don't think the DTFT of both terms is easy.

Approach 2

\begin{align} DTFT\left \{ \sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2) \right \}&=\sum_{m=-\infty}^{\infty}\left \{ \sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2) \right \} e^{-i\omega m}\\ &=\sum_{n=-\infty}^{\infty} \sum_{m=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)e^{-i\omega m} \end{align}

And then I need to find the DTFT of $\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)$ using this property $$x_1[n]x_2[n]\leftrightarrow \frac{1}{2\pi}X_1(\omega)\ast X_2(\omega)$$ where $\ast$ denotes the convolution operator. The final step would be to sum this result over all $n$, which doesn't seem to help (unless the result has a well-known sum).


Any hints would be appreciated.

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Assuming $\alpha,\beta> 0$, in order to compute: $$\sum_{n\in\mathbb{Z}}\text{sinc}(n\alpha)\,\text{sinc}(n\beta)=1+\frac{2}{\alpha \beta}\sum_{n\geq 1}\frac{\sin(n\alpha)\sin(n\beta)}{n^2}\tag{1}$$ it is enough to find a closed formula for: $$ g(\gamma)=\sum_{n\geq 1}\frac{\cos(n\gamma)}{n^2},\qquad \gamma\geq 0.\tag{2}$$ It is straightforward to notice that $g(\gamma)$ is a continuous, $2\pi$-periodic function, whose value in zero equals $\frac{\pi^2}{6}$ and whose value in $\pi$ equals $-\frac{\pi^2}{12}$. Since $g'(\gamma)$ is a well-known Fourier series (the Fourier series of the sawtooth-wave), it follows that $g(\gamma)$ is the periodic continuation of the function: $$ h(\gamma)=\frac{\pi^2}{6}-\frac{\gamma(2\pi-\gamma)}{4},\qquad \gamma\in[0,2\pi]\tag{3}$$ hence we have, by further assuming $\alpha+\beta<2\pi$: $$\begin{eqnarray*}\sum_{n\in\mathbb{Z}}\text{sinc}(n\alpha)\text{sinc}(n\beta)&=&1+\frac{1}{\alpha\beta}\left(g(|\alpha-\beta|)-g(\alpha+\beta)\right)\\&=&1+\frac{1}{4\alpha\beta}\left[(\alpha-\beta)^2-(\alpha+\beta)^2+2\pi\left(\alpha+\beta-|\alpha-\beta|\right)\right]\\&=&1+\frac{1}{4\alpha\beta}\left[-4\alpha\beta+4\pi\min(\alpha,\beta)\right]\\&=&{\frac{\pi}{\alpha\beta}\min(\alpha,\beta)}=\color{red}{\frac{\pi}{\max(\alpha,\beta)}}.\tag{4}\end{eqnarray*}$$

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I think you already have the answer.

using these two properties:

Property 1. $$x_1[n]x_2[n]\leftrightarrow \frac{1}{2\pi}X_1(\omega)\ast X_2(\omega)$$

Property 2. $$ \sum_{n=-\infty }^{\infty }x[n]\leftrightarrow X(0)$$

Step 1. Using property 1.

find the DTFT of $y[n]=\text{sinc}[n\alpha_1]\text{sinc}[n\alpha_2]$, which is a periodic convolution of two rectangle function in frequency domain. Now you have a function of $\omega$, $Y(\omega)$

Step2. Using property 2.

set $\omega$ to $0$ in your previous answer $Y(\omega)$

$$\sum_{n=-\infty }^{\infty }y[n]\leftrightarrow Y(0)$$

Then you have the answer

In detail :

$\text{sinc}[\alpha_1 n]\leftrightarrow X_1(\omega)=\begin{cases} \frac{1}{\alpha_1} & \text{for} & |\omega|<\pi\alpha_1 \\ 0 &\text{for} & \pi\alpha_1<|\omega|\leq \pi\\ \end{cases}$

Similarly

$\text{sinc}[\alpha_2 n]\leftrightarrow X_2(\omega)=\begin{cases} \frac{1}{\alpha_2} & \text{for} & |\omega|<\pi\alpha_2 \\ 0 &\text{for} & \pi\alpha_2<|\omega|\leq \pi\\ \end{cases}$

and also both $X_1(\omega)$and$X_2(\omega)$ repeats in every $2\pi$ interval.

by property 1, $\text{sinc}[\alpha_1n]\text{sinc}[\alpha_2n]\leftrightarrow \frac{1}{2\pi}X_1(\omega)\ast X_2(\omega)=\frac{1}{2\pi}\int^{\pi}_{-\pi}X_1(\tau)X_2(\omega-\tau)d\tau=Y(\omega)$

Since the final solution is $Y(0)=\frac{1}{2\pi}\int^{\pi}_{-\pi}X_1(\tau)X_2(-\tau)d\tau$ ,the integration is simply calculating the overlapping area of these two rectangular function in $[-\pi,\pi]$, then the answer is $$2*\frac{\pi \text{min}(\alpha_1,\alpha_2)}{2\pi\alpha_1\alpha_2}=\frac{\text{min}(\alpha_1,\alpha_2)}{\alpha_1\alpha_2}$$ (assuming $\alpha_1,\alpha_2>0$ and $\alpha_1,\alpha_2<1$)

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  • $\begingroup$ So, what is the value of the original series? The approach, in principle, is working, but details about the computation are important, too. $\endgroup$ – Jack D'Aurizio Sep 13 '15 at 15:12
  • $\begingroup$ The detail calculation is appended. $\endgroup$ – Allen Kuo Sep 13 '15 at 17:14
  • $\begingroup$ Way better, but please check your normalizazion constants and consider that $$\sum_{n\in\mathbb{Z}}\text{sinc}(2n)\text{sinc}(3n)=\frac{\pi}{3}.$$ $\endgroup$ – Jack D'Aurizio Sep 13 '15 at 17:21
  • $\begingroup$ thanks for reminding me about the normalization, I indeed make a mistake there. also, seems like we are using different definition for sinc function (saw on the wikipedia), I am using the convention in digital signal processing. $\endgroup$ – Allen Kuo Sep 13 '15 at 17:53
  • $\begingroup$ All right, that explains the difference between our answers, I assumed $\text{sinc}(x)=\frac{\sin(x)}{x}$ for $x>0$. Good work, (+1). $\endgroup$ – Jack D'Aurizio Sep 13 '15 at 22:53

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