0

Prove that, for any real numbers $\lambda$ and $\nu$, one has $$\int_{-\infty}^\infty f(\tau) {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau=\frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}{F(\omega_1+\omega_2)e^{j(\omega_1\lambda+\omega_2\nu)}d\omega_1d\omega_2}$$ where $F(\omega)$ is the Fourier transform of $f(t)$.

Afterward, calculate: $$\int_{-\infty}^\infty f(\tau) {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau$$ for $f(t)=\sin(\omega_0t)$.

In particular, I do not know how to prove the first formula. Suggestions are welcome!

Lely
  • 1,057

1 Answers1

1

The inverse Fourier transform of $a(\tau)=\text{sinc}(\tau-\lambda)$ is given by a constant times $e^{-\lambda i \xi}\cdot \mathbb{1}_{\xi\in(-1,1)} $. If we set $b(\tau)=\text{sinc}(\tau-\nu)$, from $\mathcal{F}(a\cdot b)=\mathcal{F}(a)*\mathcal{F}(b)$ we get that the inverse Fourier transform of $a(\tau)\cdot b(\tau)$ is given by a constant times: $$ e^{-\frac{\nu+\lambda}{2}i\xi}\cdot\sin\left(1-\frac{\xi}{2}\,\text{sign}(\xi)\right)\cdot \mathbb{1}_{\xi\in(-2,2)}.$$ By rearranging a bit and exploiting: $$ \int_{\mathbb{R}} f(\tau)\cdot a(\tau)b(\tau)\,d\tau = \int_{\mathbb{R}}\mathcal{F}(f)(\xi)\cdot \mathcal{F}^{-1}(a\cdot b)(\xi)\,d\xi $$ the claim follows. I leave the second part to you: the Fourier transform of $\sin(\omega_0 \tau)$ is just the difference of two Dirac delta function multiplied by a constant.

Jack D'Aurizio
  • 353,855
  • Thank you @Jack! Could you add some other steps? Excuse me but I'm not expert on Fourier Analysis. For $\omega_0\in[0,2\pi]$, I found this: $$\int_{-\infty}^{\infty}{\sin(\omega_0\tau) \operatorname{sinc}(\tau-\lambda) \operatorname{sinc}(\tau-\nu)d\tau }=\frac{1}{2\pi(\nu-\lambda)}\left{\sin[\pi(\nu-\lambda)+\omega_0\lambda ]-\sin[\pi(\nu-\lambda)-\omega_0\nu ]\right},$$ instead, for $\omega_0\geq 2\pi$, result is $0$. Is it true? – Lely Sep 10 '15 at 08:19
  • @Lely: I am not 100% sure about the constants, but it should be right. – Jack D'Aurizio Sep 10 '15 at 12:38