3
$\begingroup$

let $f(x)$ show that $$f(x)\approx\sum_{i=0}^{2n-1}f(t_{j})L_{j}(t)$$

where $$t_{j}=\dfrac{\pi}{n}j$$ $$L_{j}(t)=\dfrac{1}{2n}\left[1+2\displaystyle\sum_{m=1}^{n-1}\cos{m(t-t_{j})}+\cos{n(t-t_{j})}\right], j=0,1,2,\cdots,2n-1$$

It is say that “Trigonometric interpolation”,But I find sometimes, I can't find this and can't anywhere have solution,But this is from china book,I think this have solution,Have you someone can help me,Thank you very much.

is from:enter image description here

and the background is this:Howuse this $R_{l}=\frac{1}{n}\left(\frac{(-1)^l}{2n}+\sum\limits_{m=1}^{n-1}\frac{1}{m}\cos{\frac{ml\pi}{n}}\right)$ and MATLAB get this four fig?

$\endgroup$
2
+200
$\begingroup$

This is based on the Discrete Fourier Transform for a function on the sample points $t_j$ for $j\in\{0, \ldots, 2n-1\}$. Let $e_m(t_j) = \exp(imt_j)$ for $m\in \mathbb{Z}$. Note that $e_{m+2n}=e_m$ on the sample points $t_j$. For a function $f$ the DFT $\hat{f}$ is then expressed by

$$\hat{f}(m) = \sum_{j=0}^{2n-1} f(t_j) e_{-m}(t_j)$$

and the inversion is given by

$$\begin{eqnarray} 2n f(t_k) &=& \sum_{m=0}^{2n-1}\hat{f}(m)e_m(t_k)\\ &=& \sum_{j=0}^{2n-1}f(t_j)\sum_{m=0}^{2n-1}e_{-m}(t_j)e_m(t_k)\\ &=& \sum_{j=0}^{2n-1}f(t_j)\sum_{m=0}^{2n-1}e_m(t_k-t_j)\\ &=& \sum_{j=0}^{2n-1}f(t_j)\left(1+\cos n(t_k-t_j) + \sum_{m=1}^{n-1}e_m(t_k-t_j) + \sum_{m=1}^{n-1}e_{2n - m}(t_k-t_j)\right)\\ &=& \sum_{j=0}^{2n-1}f(t_j)\left(1+\cos n(t_k-t_j) + 2\sum_{m=1}^{n-1}\cos m(t_k-t_j)\right) \end{eqnarray}$$

The interpolation simply replaces $t_k$ by an arbitrary $x$ to get a periodic real function with period $2\pi$. This interpolation uses the low frequency information from the sampled signal.

$\endgroup$
  • $\begingroup$ It's very very nice.Thank you +1 $\endgroup$ – math110 Oct 15 '13 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.