z变换求响应
一个数列

求证

痴心男http://im.cx/n posted @ 2008年11月25日 03:41 in 未分类 , 1204 阅读

证明：

\[ \sum\limits_{l = 0}^{N - 1} {\frac{{\sin ^2 \pi \left( {l + \varepsilon } \right)}}{{N^2 *\sin ^2 \frac{{\pi \left( {l + \varepsilon } \right)}}{N}}}} = 1 \]

其中N是2的ｍ次幂，l是整数，0<ε<0.5



证明:






以此类推，对于N=2m，则有：



用归纳法，对N=2

\[ \begin{array}{l} A\left( {N = 2} \right) = \sum\limits_{l = 0}^1 {\cos ^2 \frac{{\pi \left( {l + \varepsilon } \right)}}{2}} \\ = \cos ^2 \frac{{\pi \varepsilon }}{2} + \cos ^2 \frac{{\pi \left( {1 + \varepsilon } \right)}}{2} \\ = \cos ^2 \frac{{\pi \varepsilon }}{2} + sin^2 \frac{{\pi \varepsilon }}{2} \\ = 1 \\ \end{array} \]

对N=4

\[ \begin{array}{l} A(N = 4) = \sum\limits_{l = 0}^3 {\cos ^2 \frac{{\pi \left( {l + \varepsilon } \right)}}{4}} \cos ^2 \frac{{\pi \left( {l + \varepsilon } \right)}}{2} \\ = \underbrace {\cos ^2 \frac{{\pi \varepsilon }}{4}\cos ^2 \frac{{\pi \varepsilon }}{2}}_{even} + \underbrace {\cos ^2 \frac{{\pi \left( {1 + \varepsilon } \right)}}{4}\cos ^2 \frac{{\pi \left( {1 + \varepsilon } \right)}}{2}}_{odd} \\ + \underbrace {\cos ^2 \frac{{\pi \left( {2 + \varepsilon } \right)}}{4}\cos ^2 \frac{{\pi \left( {2 + \varepsilon } \right)}}{2}}_{even} + \underbrace {\cos ^2 \frac{{\pi \left( {3 + \varepsilon } \right)}}{4}\cos ^2 \frac{{\pi \left( {3 + \varepsilon } \right)}}{2}}_{odd} \\ = \underbrace {\cos ^2 \frac{{\pi \varepsilon }}{2}\cos ^2 \frac{{\pi \varepsilon }}{2} + sin^2 \frac{{\pi \varepsilon }}{2}\cos ^2 \frac{{\pi \varepsilon }}{2}}_{even} \\ + \underbrace {\cos ^2 \frac{{\pi \left( {1 + \varepsilon } \right)}}{4}\cos ^2 \frac{{\pi \left( {1 + \varepsilon } \right)}}{2} + sin^2 \frac{{\pi \left( {1 + \varepsilon } \right)}}{4}\cos ^2 \frac{{\pi \left( {1 + \varepsilon } \right)}}{2}}_{odd} \\ = \underbrace {\cos ^2 \frac{{\pi \varepsilon }}{2}}_{even} + \underbrace {\cos ^2 \frac{{\pi \left( {1 + \varepsilon } \right)}}{2}}_{odd} \\ = A(N = 2) \\ = 1 \\ \end{array} \]

以此类推就可以得证。

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