# Fourier Series and Filter in Function

Let $L=1, A=1$ and $f\in L^2([-L/2, L/2])$, with Fourier series $$f^{t}=\sum_{n=-K}^{K}a_n \exp(2j\pi xn/L),$$ truncated at $K$. Has this function, $f^{t}$, any relation with Fourier Inverse Transform of $\hat{W}(w)*\hat{f}(w)$?.

I begin calculated $f^t$ and Inverse Fourier of $\hat{W}(w)*\hat{f}(w)$ I get: $$\sum_{n=-K}^{K}\dfrac{A}{2\pi j n}(1-\exp(-\pi j n x))\exp(2\pi j n x)$$

and

$$\int_{-K}^{K}\dfrac{A}{2\pi j w}(1-\exp(-\pi j w x))\exp(2\pi j w x)dx$$ respectively, then I think that relation is when $K\rightarrow \infty$. The first and second expression are equals, Is true this?

pdta: $W(x) = \dfrac{\sin(2\pi Kx)}{\pi x}\;$ and $f(x) = 0,$ if $-L/2 \leq x < 0$; $f(x) = A,$ if $0 \leq x < L/2$
What's $A$ doing there? And which Fourier series is $f^t$ the truncated version of? You didn't assume that $f$ is periodic, so $f$ itself has no Fourier series, truncated or not. In the current formulation it's not apparent what connection, if any, should obtain between $f$ and $f^t$. – joriki Oct 30 '12 at 7:15
The edit doesn't resolve the questions I raised. $f$ is now defined on $[-L/2,L/2]$. Perhaps you meant $f\in L^1([-L/2,L/2])$ instead of $f\in L^1(\mathbb R)$? – joriki Oct 30 '12 at 15:36