Hi, I read a paragraph of "Theory and application of digital signal processing" of L. R. Rabiner on page 70. It says: For v(n), a Hiltert transformed signal, its Fourier transform V(e^ (jw)) has the property V(e^(jw))=0 pi< w <2*pi (2.187) Clearly v(n) is a complex signal since the Fourier transforms of real signals have the property V*(e^(-jw))=V(e^(jw)) (2.188) which would imply V(e^(jw))=0 if v(n) were real. I don't understand the above would sentence. (2.188) is a fact for real signal, right? Why does it get the following: V(e^(jw))=0 Thanks.

# Any one can explain the Hilbert transform related question for me?

Started by ●April 29, 2009

Reply by ●April 29, 20092009-04-29

On Apr 29, 10:57�am, fl <rxjw...@gmail.com> wrote:> Hi, > I read a paragraph of "Theory and application of digital signal > processing" of L. R. Rabiner on page 70. It says: > > For v(n), a Hiltert transformed signal, its Fourier transform V(e^ > (jw)) has the property > > V(e^(jw))=0 � � � � �pi< w <2*pi � � � �(2.187) > > Clearly v(n) is a complex signal since the Fourier transforms of real > signals have the property > > V*(e^(-jw))=V(e^(jw)) � � � � � � � � � � (2.188) > > which would imply V(e^(jw))=0 if v(n) were real. > > I don't understand the above would sentence. (2.188) is a fact for > real signal, right? Why does it get the following: > > �V(e^(jw))=0 >first of all, V(e^(jw)) is periodic with period 2*pi. you know why that is, right? if v[n] was real, how do you satisfy V(e^(jw)) = 0 pi< w <2*pi *and* V*(e^(-jw)) = V(e^(jw)) ? you could have V(e^(j0)) = somthing non-zero. same with V(e^(j*pi)). but how could both equations be true for other values of w. r b-j

Reply by ●May 2, 20092009-05-02

On Wed, 29 Apr 2009 07:57:05 -0700 (PDT), fl <rxjwg98@gmail.com> wrote:>Hi, >I read a paragraph of "Theory and application of digital signal >processing" of L. R. Rabiner on page 70. It says: > > > > >For v(n), a Hiltert transformed signal, its Fourier transform V(e^ >(jw)) has the property > >V(e^(jw))=0 pi< w <2*pi (2.187) > >Clearly v(n) is a complex signal since the Fourier transforms of real >signals have the property > >V*(e^(-jw))=V(e^(jw)) (2.188) > >which would imply V(e^(jw))=0 if v(n) were real. > >I don't understand the above would sentence. (2.188) is a fact for >real signal, right? Why does it get the following: > > > V(e^(jw))=0 > >Thanks.Hello fl, I think you're justified in being puzzled by those equations. First the authors say: V(e^(jw))=0 pi< w <2*pi (2.187) OK, fine, ...that covers the negative freq range of V(e^(jw)), the spectrum of a complex-valued v(n) time sequence. Then they introduce a spectrum that they call V(e^(-jw)) in a conjugated form. Well, that also looks like an expression for a spectrum over a negative freq range. I'll bet their explanation confuses many readers. Perhaps after the sentence, "... which would imply V(e^(jw)) = 0 if v(n) were real." they should have added the sentence: "Thus, the only way to simultaneously satisfy Equations (2.187) and (2.188) is for v(n) to be complex-valued." Good Luck, [-Rick-]