傅立叶变换与小波变换

傅立叶变换与小波变换
Fourier Transform and Wavelet Transform
------ 再谈积分变换
1. 窗式傅里叶变换
1.1 Fourier Transform 局域化特性分析 傅里叶变换（FT）在平稳信号分析和处理中有着突出的贡献， 利用它可以把复杂的时间信号和空间信号转换到频率域中， 然后用频 谱特性去分析和表示时域信号的特性。 FT PAIR
f (t ) ⇔ F (u )
F (ω ) =
+∞
−∞
∫
f ( t ) exp [ − jω t ] dt ,
+∞
ω∈R
1 f (t ) = 2π
−∞
∫
F (ω ) e x p
[ jω t ] d ω
频域过程 F (ω ) 的任一频率组成部分的值： 由时域过程 f (t ) 在 （－∞，∞）上决定的。 过程 f (t ) 在任一时刻的状态：由 F (ω ) 在整个频域（－∞，∞） 的量决定。
f (t ) 和 F (ω ) 彼此间的刻画是“全局性”的，不能反映各自局部
区域上的特征。也就是说，人们虽然从傅立叶变换能清楚地看到一个 信息包含的每一个频率的多少， 但很难看出不同信号的发射时间和发 射的延续时间，缺少时间信息使得傅立叶分析变得脆弱而容易失误。
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傅立叶变换与小波变换
伊利诺依斯大学教授 Y. Meyer 曾说： “若你记录 1 小时长的信息 而在最后 5 分钟出错， 这一错误就会毁了整个傅立叶变换。 相位的错 误是灾难性的， 如果在相位上哪怕犯了一个错误， 你最后就会发现你 所干的事与最初的信号无关了。 ” 实际上，对常见的不平稳信号，如语音信号、音乐信号、探地信 号、核探测的脉冲信号、以及核医学的图像信号等，人们需要了解某 些局部时段上所对应的主要频率特性是什么， 也需要了解某些频率的 信息出现在哪些时段上，也就是需要了解时－频局部化要求。对于这 种时－频局部化要求，傅立叶变换是无能为力的。 Examples. A stationary signal
x ( t ) = cos ( 2 • π • 10 • t ) + cos ( 2 • π • 25 • t ) + cos ( 2 • π • 50 • t ) + cos ( 2 • π • 100 • t )
is a stationary signal, because it has frequencies of 10, 25, 50, and 100 Hz at any given time instant.
Fig. 1
a stationary signal
And the following is its FT:
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傅立叶变换与小波变换
Fig.2
The next figure shows A signal with four different frequency components at four different time intervals, hence a non-stationary signal. The interval 0 to 300 ms has a 100 Hz sinusoid, the interval 300 to 600 ms has a 50 Hz sinusoid, the interval 600 to 800 ms has a 25 Hz sinusoid, and finally the interval 800 to 1000 ms has a 10 Hz sinusoid.
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傅立叶变换与小波变换
Fig.3
And the following is its FT:
Fig. 4
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傅立叶变换与小波变换
For the “chirp signal” (唧声信号,), the frequency components change continuously.
Fig. 5 Chirp signal
The similarity between these two spectrum (in Figs. 2and 4) should be apparent. Both of them show four spectral components at exactly the same frequencies, i.e., at 10, 25, 50, and 100 Hz. Other than the ripples, and the difference in amplitude (which can always be normalized), the two spectrums are almost identical, although the corresponding time-domain signals are not even close to each other. Both of the signals involves the same frequency components, but the first one has these frequencies at all times, the second one has these frequencies at different intervals. So, how come the spectrums of two entirely different signals look very much alike? Recall that the FT gives the spectral content of the signal, but it gives no information regarding where in time those spectral components appear . Therefore, FT is not a suitable technique for non-stationary signal, with one exception:
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