Finding an ideal low pass filter convolution kernel

Question

Let $f \in L^2[-\pi,\pi] $ and let: $$f = \sum_{k=-\infty}^{\infty}\hat{f}(k)e^{ikx}$$ the Fourier expansion of $f$. I want to find a convoultion kernel $g_N$ so that:

$$\frac{1}{2\pi}h(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}g_N(x-t)f(t)dt = \sum_{k=-N}^{N}\hat{f}(k)e^{ikx}$$

i.e. an ideal low pass filter. By the convolution theorem:

$$\frac{1}{2\pi}\hat{h}(k)=\hat{f}(k)\hat{g_N}(k)= \left\{ \begin{array}{ll} \hat{f}(k) & \mbox{if $k \leq N$};\\ 0 & \mbox{if $x > N$}.\end{array} \right. $$

So we get:

$$\hat{g_N}(k)= \left\{ \begin{array}{ll} 1 & \mbox{if $k \leq N$};\\ 0 & \mbox{if $x > N$}.\end{array} \right. $$

And

$$g_N(x) = \sum_{k=-N}^{N}e^{ikx}$$

But how do I transform $g_N$ to a "real" form?

Thanks.

Answer 1

How about this: since

$e^{ikx} + e^{-ikx} = (\cos kx + i\sin kx) + (\cos kx - i\sin kx) = 2\cos kx, \tag{1}$

when we write $g_N(x)$ in the form

$g_N(x) = 1 + \sum_{k = 1}^{k = N} (e^{ikx} + e^{-ikx}) \tag{2}$

we immediately see that

$g_N(x) = 1 + 2\sum_{k = 1}^{k = N} \cos kx, \tag{3}$

we see that $g_N(x)$ is in a real form as is, as real as real can be!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!