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Signal Processing

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  • $\begingroup$ I've read these pages in Widrow, there is no direct explanations why 1 sample shift is equal to $-\pi/2$.. It is so if sample rate is $2\omega_0$ but it isn't told to us. But I think it isn't too important for this case. In the scheme above you have only one tap and it exactly shifts the signal for $\pi/2$, so maybe author means "lag of 1" is this tap. It's only my suggestion. $\endgroup$ Commented Jan 12, 2015 at 20:35