统计信号处理基础 - 估计与检测理论 估计部分习题3.7公式推导
- 题目
- 证明
- 结论
- 得证
题目
相信学习信号检测与估计的童鞋们肯定看到过Steven M.Kay大牛的书,非常厚的一本,不得不说,人家的书就是写得好,浅显易懂(当然是要从头把基础的东西都掌握了),在估计部分第三章中例题3.4中遇到了下面这个公式,在习题3.7中要求证明,首先看题目
1N∑n=0N−1cos(4πf0n+2ϕ)≈0\frac{1}{N}\sum\limits_{n = 0}^{N - 1} {\cos (4\pi {f_0}n + 2\phi ) \approx 0}N1n=0∑N−1cos(4πf0n+2ϕ)≈0
要使上式成立,f0{f_0}f0必须满足以下条件f0≠0,f0≠1/2{f_0} \ne 0,\quad {f_0} \ne 1/2f0=0,f0=1/2
证明
令α=4πf0,β=2ϕ\alpha = 4\pi {f_0},\quad \beta = 2\phiα=4πf0,β=2ϕ
则1N∑n=0N−1cos(αn+β)=1NRe(∑nej(αn+β))=1NRe(ejβ⋅1−ejαN1−ejα)=1NRe(ejβ⋅ejαN/2ejα/2⋅e−jαN/2−ejαN/2e−jα/2−ejα/2)=1NRe(ejβ⋅ejαN/2ejα/2⋅e−jαN/2−ejαN/2e−jα/2−ejα/2)=1NRe(ejβ⋅ejαN−12⋅sin(αN/2)sin(α/2))=sin(αN/2)Nsin(α/2)⋅cos(αN−12+β)\begin{array}{l} \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {\cos (\alpha n + \beta )} = \frac{1}{N}{\mathop{\rm Re}\nolimits} \left( {\sum\limits_n {{e^{j(\alpha n + \beta )}}} } \right)\\ = \frac{1}{N}{\mathop{\rm Re}\nolimits} \left( {{e^{j\beta }} \cdot \frac{{1 - {e^{j\alpha N}}}}{{1 - {e^{j\alpha }}}}} \right)\\ = \frac{1}{N}{\mathop{\rm Re}\nolimits} \left( {{e^{j\beta }} \cdot \frac{{{e^{j\alpha N/2}}}}{{{e^{j\alpha /2}}}} \cdot \frac{{{e^{ - j\alpha N/2}} - {e^{j\alpha N/2}}}}{{{e^{ - j\alpha /2}} - {e^{j\alpha /2}}}}} \right)\\ = \frac{1}{N}{\mathop{\rm Re}\nolimits} \left( {{e^{j\beta }} \cdot \frac{{{e^{j\alpha N/2}}}}{{{e^{j\alpha /2}}}} \cdot \frac{{{e^{ - j\alpha N/2}} - {e^{j\alpha N/2}}}}{{{e^{ - j\alpha /2}} - {e^{j\alpha /2}}}}} \right)\\ = \frac{1}{N}{\mathop{\rm Re}\nolimits} \left( {{e^{j\beta }} \cdot {e^{j\alpha \frac{{N - 1}}{2}}} \cdot \frac{{\sin (\alpha N/2)}}{{\sin (\alpha /2)}}} \right)\\ = \frac{{\sin (\alpha N/2)}}{{N\sin (\alpha /2)}} \cdot \cos \left( {\alpha \frac{{N - 1}}{2} + \beta } \right) \end{array}N1n=0∑N−1cos(αn+β)=N1Re(n∑ej(αn+β))=N1Re(ejβ⋅1−ejα1−ejαN)=N1Re(ejβ⋅ejα/2ejαN/2⋅e−jα/2−ejα/2e−jαN/2−ejαN/2)=N1Re(ejβ⋅ejα/2ejαN/2⋅e−jα/2−ejα/2e−jαN/2−ejαN/2)=N1Re(ejβ⋅ejα2N−1⋅sin(α/2)sin(αN/2))=Nsin(α/2)sin(αN/2)⋅cos(α2N−1+β)
结论
当α≠0,α≠2π时,即f0≠0,f0≠1/2时,原式约为0。\alpha \ne 0,\;\alpha \ne 2\pi时,即 {f_0} \ne 0,\quad {f_0} \ne 1/2时,原式约为0。α=0,α=2π时,即f0=0,f0=1/2时,原式约为0。