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*# <math>x_n(t)</math>'s have zero mean for all <math>t</math>'s
*# <math>x_n(t)</math>'s are mutually independent for all <math>t</math>'s and <math>\tau</math>'s
:then:
::<math>E[x_m(t+\tau/2)x_n^*(t-\tau/2)] = E[x_m(t+\tau/2)]E[x_n^*(t-\tau/2)] = 0</math>
* if <math>m \neq n</math>, then
::<math>E[W_h(t,f)] = \sum_{n=1}^k E[W_{x_n}(t,f)]</math> ::<math>E[A_h(\eta,\tau)] = \sum_{n=1}^k E[A_{x_n}(\eta,\tau)]</math>
=== Short-time Fourier transform ===
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