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 , 936 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

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

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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