- #1
mnb96
- 715
- 5
Hello,
I'd like to prove the orthogonality of two "shifted" Sinc functions, but I can't find the mistake.
Here is my attempt:
[tex]\int_{-\infty}^{+\infty}sinc(x)sinc(x-x_0)dx[/tex]
Observing this quantity can be obtained by evaluating the Fourier transform at zero, we have:
[tex]\mathcal{F}\{ sinc(x)sinc(x-x_0) \}(0)[/tex]
and using the convolution theorem and the shift theorem (for the second sinc), we get:
[tex](rect(\omega)\otimes e^{-i2\pi\omega x_0}rect(\omega))(0) = [/tex]
[tex]= \int_{-1/2}^{+1/2} e^{-i2\pi\omega x_0}d\omega = [/tex]
[tex]= \left[ \frac{sin(2\pi\omega x_0)}{2\pi x_0} \right]_{-1/2}^{1/2} = [/tex]
[tex]= \frac{sin(\pi x_0)}{\pi x_0} = [/tex]
[tex]= sinc(x_0) [/tex]
Now, this quantity is 0 iff [itex]x_0[/itex] is a non-zero integer!
Is this the correct result?
Aren't two sinc functions supposed to be orthogonal for any [itex]x_0[/itex] real?
I'd like to prove the orthogonality of two "shifted" Sinc functions, but I can't find the mistake.
Here is my attempt:
[tex]\int_{-\infty}^{+\infty}sinc(x)sinc(x-x_0)dx[/tex]
Observing this quantity can be obtained by evaluating the Fourier transform at zero, we have:
[tex]\mathcal{F}\{ sinc(x)sinc(x-x_0) \}(0)[/tex]
and using the convolution theorem and the shift theorem (for the second sinc), we get:
[tex](rect(\omega)\otimes e^{-i2\pi\omega x_0}rect(\omega))(0) = [/tex]
[tex]= \int_{-1/2}^{+1/2} e^{-i2\pi\omega x_0}d\omega = [/tex]
[tex]= \left[ \frac{sin(2\pi\omega x_0)}{2\pi x_0} \right]_{-1/2}^{1/2} = [/tex]
[tex]= \frac{sin(\pi x_0)}{\pi x_0} = [/tex]
[tex]= sinc(x_0) [/tex]
Now, this quantity is 0 iff [itex]x_0[/itex] is a non-zero integer!
Is this the correct result?
Aren't two sinc functions supposed to be orthogonal for any [itex]x_0[/itex] real?