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.


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 $\alpha+\beta<2\pi$: $$\begin{eqnarray*}\sum_{n\in\mathbb{Z}}\text{sinc}(n\alpha)\text{sinc}(n\beta)&=&1+\frac{1}{\alpha\beta}\left(g(|\alpha-\beta|)-g(\alpha+\beta)\right)\\&=&1+\frac{1}{4\alpha\beta}\left[(\alpha-\beta)^2-(\alpha+\beta)^2+2\pi\left(\alpha+\beta-|\alpha-\beta|\right)\right]\\&=&1+\frac{1}{4\alpha\beta}\left[-4\alpha\beta+4\pi\min(\alpha,\beta)\right]\\&=&{\frac{\pi}{\alpha\beta}\min(\alpha,\beta)}=\color{red}{\frac{\pi}{\max(\alpha,\beta)}}.\tag{4}\end{eqnarray*}$$


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 :

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


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

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 $\alpha_1,\alpha_2<1$)

  • $\begingroup$ 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

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.