19Zoom - MAE 591 RANDOM DATA Zoom Transform Let y (t ) be...

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MAE 591 RANDOM DATA Zoom Transform Let y t ( ) be the bandpass filtered record of the original data record x t ( ) , i.e. yt xt f B ff B otherwise () () ; ; = −≤≤+ 00 22 0 and vt , which is the modulated record of yte jf t = 2 1 π y t ( ) with the modulating frequency , then we have f 1 11 ( ) 2 1 ( ) ( ) TT t j f f t t Vf y te e d t y d t Yf f ππ π −− + == = ∫∫ + Gf Gf f vv yy ( ) = + 1 If B 10 2 =− , then Gf f B B otherwise yy xx ; = 0 0 0 with f f f B c s =≤+ 0 and f fB N + 2 0 with Gff f otherwise vv xx ; ; = +≤ 1 0 0 B f f B c s
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