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I have a vector $x = (..., x_{-1}, x_0, x_1, ...)$ and a vector $w = (..., 0, 0, 1, 1, .. , 1, 1, 0, 0, ..)$ (with $2M + 1$ components equals to 1) such that

$y = x \cdot w = (0, 0, .., x_{-M},x_{-M + 1}, .., x_{M-1}, x_M, 0, 0, ..) $ (product cordinate by cordinate, ie, $y_k = x_kw_k$ for all $k \in \mathbb{Z}$).

All these vectors are in $L^2(\mathbb{Z})$.

I can prove that the discret time Fourier transform of $y$, named, $Y(f)$, is $Y(f) = X(f) * W(f)$ where $*$ denotes the convolution

and I can show too that $W(f) = \frac{\sin{((2M + 1)\pi f)}}{\sin{(\pi f)}}$.

Now, I need to prove that when $M$ goes to infinity, $Y(f)$ converges to $X(f)$.

The book gives the follow hint:

$$\lim_{R \rightarrow \infty} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\sin(R\pi x) \phi(x)}{\sin(\pi x)} dx = \phi(0) $$

But I don't know how to use it. If I made $\phi(f) = X(f)$ and put $Y(f)$ into the integral I don't arrive in any place.

I hope you could help me.

Thank you.

Rodolpho
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1 Answers1

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The circular convolution $(X * W)(f)$ is defined by $\displaystyle{\int_{-\frac{1}{2}}^\frac{1}{2} X(f - g)W(g)dg}$, so, you can solve it using the hint if you do the follow:

Define $R = 2M + 1$ and $\phi(g) = X(f - g)$, where $f$ is any constant in $[-\frac{1}{2}, \frac{1}{2}]$, so

$$\lim_{R \rightarrow \infty} Y(f) = \lim_{R \rightarrow \infty} \int_{-\frac{1}{2}}^{\frac{1}{2}} \phi(g) W(g)dg = \lim_{R \rightarrow \infty} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\sin(R\pi g) \phi(g)}{\sin(\pi g)} dg = \phi(0) $$

but $\phi(0)$ is equals to $X(f)$, then

$$\lim_{R \rightarrow \infty} Y(f) = X(f)$$

and it's done.

Hilder Vitor Lima Pereira
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