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Re: [Fwd: Re: Re: problem in normalization factor in low pass filter implementation]

Started by Logeshwaran Vijayan July 17, 2002
Hi Premkiran Mannava,

The lms algorithm for adaptive filtering is not that complex to
implement. just get hold of this book -

Adaptive filtering by Simon Haykin

hope this helps.

logesh. > Premkiran Mannava-
>
> Thanks for your inquiry. I don't think I can help with this. Please reply to
the
> group.
>
> Jeff Brower
> DSP sw/hw engineer
> Signalogic
>
> -------- Original Message --------
> Subject: Re: [matlab] Re: problem in normalization factor in low pass filter
> implementation
> Date: Wed, 17 Jul 2002 12:13:53 +0100 (BST)
> From: premkiran mannava <>
> To: Jeff Brower < >
> Hello sir,
>
> Could you help me out with a matlab program which is written for widro hoff
lms
> algorithm for adaptive filtering? It is needed as a part for my work and it is
> urgent. if possible write the code itself
>
> kiran >
> Jeff Brower <> wrote:
>
> Henry-
>
> You're not reading carefully. I said all of the "x[n]" components, like
> this:
>
> y(n) = 2*y(n-1) - y(n-2) + x(n)/8 - x(n-3)/4 + x(n-6)/8
>
> Jeff Brower
> DSP sw/hw engineer
> Signalogic > i_am_henry wrote:
> >
> > Hi Jeff,
> >
> > Thank you for you help first.
> >
> > Yes, I have applied the gain/normalization factor (1/8) to all the
> > input components according to the equation.
> >
> > However, for
> > y(n) = 2*y(n-1) - y(n-2) + x(n) - 2*x(n-3) + x(n-6)
> > the filter response tranfer function is
> > Yn 1 - 2*z^-3 + z^-6
> > ---- = -------------------- .............. (1)
> > Xn 1 - 2*z^-1 + z^-2
> >
> >
> > For
> > y(n) = 1/8 * (2*y(n-1) - y(n-2) + x(n) - 2*x(n-3) + x(n-6))
> > the filter response tranfer function is
> > Yn 1 - 2*z^-3 + z^-6
> > ---- = -------------------- .............. (2)
> > Xn 8 - 2*z^-1 + z^-2
> >
> > The filter frequency respose of Equ. (2) is different from (1).
> >
> > Regards
> > Henry Chang
> > Assistant Computer Officer
> > Chinese Univerisity of HK
> >
> > --- In matlab@y..., Jeff Brower wrote:
> > > Henry-
> > >
> > > Are you sure the gain factor (1/8) is not supposed to be applied to
> > the x[n]
> > > components only? What happens if you try that?
> > >
> > > Typically the gain/normalization factor is intended to insure unity
> > gain (or desired
> > > gain) in the passband, but the frequency response of the filter
> > (shape) should remain
> > > unchanged.
> > >
> > > Jeff Brower
> > > DSP sw/hw engineer
> > > Signalogic
> > >
> > >
> > > i_am_henry wrote:
> > > >
> > > > Dear all,
> > > >
> > > > Currently, I am working on implementing recursive low pass
> > filter
> > > > with design equation extracted from manual
> > > > y(n) = 2*y(n-1) - y(n-2) + x(n) - 2*x(n-3) + x(n-6)
> > > >
> > > > The frequency response of the filter is good.
> > > >
> > > > However, the manual also mention, in actual implementation,
> > there
> > > > is a normalization factor of 1/8 based upon the gain of the
> > filter.
> > > > That is, the equation become:
> > > > y(n) = 1/8 * (2*y(n-1) - y(n-2) + x(n) - 2*x(n-3) + x(n-6))
> > > >
> > > > But the frequency response of this filter is bad.
> > > >
> > > > Is it some common practice to design a filter with desired good
> > > > frequency response but later add some normalization factor to
> > > > normalize the signal? What is the advantage to do so?
> > > >
> > > > Thank you for your help.
> > > >
> > > > Henry >
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