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

Started by fl April 29, 2009
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.
On Apr 29, 10:57&#2013266080;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 &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;pi< w <2*pi &#2013266080; &#2013266080; &#2013266080; &#2013266080;(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)) &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; (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: > > &#2013266080;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
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-]