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group delay of FIR filter

Hi all,

Consider a FIR filter with coefficients [1 2 3 2 1]

I know the group delay is 2, (5-1)/2. Can anyone tell how to deduce the
group delay of 2 without going through the d(theta)/d(omega) mathematics?
Is there an easy way to find the group delay?


cfy30
0
cfy301 (90)
3/21/2009 7:02:09 PM
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cfy30 wrote:
> Hi all,
> 
> Consider a FIR filter with coefficients [1 2 3 2 1]
> 
> I know the group delay is 2, (5-1)/2. Can anyone tell how to deduce the
> group delay of 2 without going through the d(theta)/d(omega) mathematics?
> Is there an easy way to find the group delay?

You just found it. What more do you want? (5-1)/2 bypasses 
d(theta)/d(omega).

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
0
jya (12871)
3/21/2009 7:26:19 PM
On 21 Mar, 20:02, "cfy30" <cf...@yahoo.com> wrote:
> Hi all,
>
> Consider a FIR filter with coefficients [1 2 3 2 1]
>
> I know the group delay is 2, (5-1)/2. Can anyone tell how to deduce the
> group delay of 2 without going through the d(theta)/d(omega) mathematics?
> Is there an easy way to find the group delay?

Well, there is the difference between *deriving*
the group delay (that is, finding the formula),
and *computing* the group delay.

You can do the computation in two ways:

- Observe that the filter's impulse response
  is symmetric, so the filter is linear phase
  and thus the group delay is constant (N-1)/2 = 2.
  I can't imagine anything easier than that.
- A somewhat more involved way is to compute the
  spectrum, unwrap the phase, and find the gradient
  of the phase vs w.

Rune
0
allnor (8509)
3/21/2009 7:31:23 PM
On 21 Mrz., 20:02, "cfy30" <cf...@yahoo.com> wrote:
> Consider a FIR filter with coefficients [1 2 3 2 1]
>
> I know the group delay is 2, (5-1)/2. Can anyone tell how to deduce the
> group delay of 2 without going through the d(theta)/d(omega) mathematics?
> Is there an easy way to find the group delay?

For this kind of filter you mentioned (symmetric + odd number of taps)
it is easy.  It's always group delay = (number of taps-1)/2.

Let H(z) be the transfer function of such a filter, let k=(number of
taps-1)/2, let G(z)=z^k. Note: The filter that corresponds to G(z) has
a group delay of -k.  Let's compose both filters:

Now, F(z) := H(z)*G(z) is of the form

  F(z) = a_0 + a_1 * (z+z^-1) + a_2 * (z^2+z^-2)
       + ... + a_k * (z^k z^-k)

Observe that  z^p+z^-p  is real for |z|=1.  That makes F(z) real for |
z|=1.  So, the phase response of F doesn't change with the frequency.
So, the group delay of F is zero.  Since F=H*G has a group delay of
zero and G has a group delay of -k and group delay of F = group delay
of H + group delay of G, H has a group delay of k.

q.e.d.

You can prove this similarily for all remaining combinations of
symmetric/antisymmetric + even/odd.


Cheers!
SG
0
s.gesemann (692)
3/21/2009 7:50:58 PM
On Sat, 21 Mar 2009 14:02:09 -0500, "cfy30" <cfy30@yahoo.com> wrote:

>Hi all,
>
>Consider a FIR filter with coefficients [1 2 3 2 1]
>
>I know the group delay is 2, (5-1)/2. Can anyone tell how to deduce the
>group delay of 2 without going through the d(theta)/d(omega) mathematics?
>Is there an easy way to find the group delay?
>
>
>cfy30

In a very basic sense, the amount of delay to the filter's peak
coefficient is the group delay.   You've picked a symmetric filter so
that the assumption that the peak is in the middle holds.

If you want an analytical reason for that, the coefficient set is the
impulse response, and the location of the peak coefficient is usually
taken to be the location of "the impulse" in the output stream.   So,
the delay to the peak coefficient is, in a basic sense, the group
delay.

Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
0
eric.jacobsen (2636)
3/21/2009 9:39:39 PM
Thanks all!

Hi SG,

Your proof is insightful! Simple and elegant! 

Hi Eric,

"The amount of delay to the filter's peak coefficient is the group delay."
Finding group delay from the impulse response is new to me. Does it apply
to all filters? 

Thanks,
cfy30


>On Sat, 21 Mar 2009 14:02:09 -0500, "cfy30" <cfy30@yahoo.com> wrote:
>
>>Hi all,
>>
>>Consider a FIR filter with coefficients [1 2 3 2 1]
>>
>>I know the group delay is 2, (5-1)/2. Can anyone tell how to deduce the
>>group delay of 2 without going through the d(theta)/d(omega)
mathematics?
>>Is there an easy way to find the group delay?
>>
>>
>>cfy30
>
>In a very basic sense, the amount of delay to the filter's peak
>coefficient is the group delay.   You've picked a symmetric filter so
>that the assumption that the peak is in the middle holds.
>
>If you want an analytical reason for that, the coefficient set is the
>impulse response, and the location of the peak coefficient is usually
>taken to be the location of "the impulse" in the output stream.   So,
>the delay to the peak coefficient is, in a basic sense, the group
>delay.
>
>Eric Jacobsen
>Minister of Algorithms
>Abineau Communications
>http://www.ericjacobsen.org
>
>Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
>
0
cfy301 (90)
3/22/2009 4:16:39 AM
cfy30 wrote:
> Thanks all!
> 
> Hi SG,
> 
> Your proof is insightful! Simple and elegant! 
> 
> Hi Eric,
> 
> "The amount of delay to the filter's peak coefficient is the group delay."
> Finding group delay from the impulse response is new to me. Does it apply
> to all filters? 

No. it applies only to filters with constant group delay.

   ...

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
0
jya (12871)
3/22/2009 3:42:35 PM
On Sun, 22 Mar 2009 11:42:35 -0400, Jerry Avins <jya@ieee.org> wrote:

>cfy30 wrote:
>> Thanks all!
>> 
>> Hi SG,
>> 
>> Your proof is insightful! Simple and elegant! 
>> 
>> Hi Eric,
>> 
>> "The amount of delay to the filter's peak coefficient is the group delay."
>> Finding group delay from the impulse response is new to me. Does it apply
>> to all filters? 
>
>No. it applies only to filters with constant group delay.

Correct, and in cases where the delay in the passband is reasonably
constant.  Another caveat is that the amount of energy in the peak, or
the size of the peak coefficient relative to the rest, gives a hint to
how well the generalization might hold.



Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
0
eric.jacobsen (2636)
3/22/2009 5:26:54 PM
On Sat, 21 Mar 2009 14:02:09 -0500, cfy30 wrote:

> Hi all,
> 
> Consider a FIR filter with coefficients [1 2 3 2 1]
> 
> I know the group delay is 2, (5-1)/2. Can anyone tell how to deduce the
> group delay of 2 without going through the d(theta)/d(omega)
> mathematics? Is there an easy way to find the group delay?
> 
> 
> cfy30

For a symmetric filter it is easy.

First, consider the World's Simplest Filter.  It has coefficient [1], 
meaning that it's input/output relationship is y_n = x_n, and all it does 
is passes the input to the output identically.  Clearly it does nothing 
at all because it isn't really even a filter.

Now consider the World's Second Most Simple Filter.  It has coefficients 
[0 0 ... 1], meaning it's input/output relationship is y_n = x_{n-m}, and 
all it does is delay the input by m samples.

Now consider a filter whose coefficients are perfectly symmetrical around 
zero delay, i.e. y_n = a_0 x_n + a_1(x_{n-1} + x_{n+1}) + ...

This filter has a serious problem in that it must look into the future 
(the x_{n+1} and other future-looking terms), but it has exactly zero 
delay.  You can see* that it must not have any delay because anything 
that happens in the past to delay things is balanced by the exact same 
thing happening in the future to "advance" things.  So there can be 
attenuation or gain, but no delay or advance.

A "normal", i.e. realizable, FIR filter just combines the impossible 
'perfect' symmetrical filter with the World's Second Most Simple Filter.  
The phase delays add, so you just get 0 + m samples.  Now toss in the 
(unproven) assertion that it still works for a half-sample delay, and 
you're done.

* You asked for no math, so -- no math.  Instead, I give you hand-
waving.  If you want math, start with the property of the Fourier 
transform that states that the Fourier transform of a real-valued, even-
symmetric function must be real-valued and even-symmetric, and see how 
many steps you need to prove zero phase shift over all frequencies.

-- 
http://www.wescottdesign.com
0
tim177 (4433)
3/22/2009 6:03:33 PM
On Sat, 21 Mar 2009 14:02:09 -0500, "cfy30" <cfy30@yahoo.com> wrote:

>Hi all,
>
>Consider a FIR filter with coefficients [1 2 3 2 1]
>
>I know the group delay is 2, (5-1)/2. Can anyone tell how to deduce the
>group delay of 2 without going through the d(theta)/d(omega) mathematics?
>Is there an easy way to find the group delay?
>
>
>cfy30

Hi cfy30,
  your question has, what I consider to be, a fairly 
deep (significant underlying) meaning.
I like Tim's "hand waving" explanation.

Sometimes in their algebraic derivations of z-domain 
transfer functions, some authors arrive at a transfer 
function equation of some network whose impulse response 
begins before an impulse input is applied.  They call such 
networks "acausal".  Then they say something 
like: "To make the network causal, we add a delay to the 
network."  That kind of explanation always irritates 
me because how can they add a delay to a network when that 
network cannot exist in our physical universe?  Ha ha.   

By the way, you asked for ways to compute "group delay of 
without going through the d(theta)/d(omega) mathematics".

You might have a look at:
   
http://www.dsprelated.com/showarticle/69.php

The above scheme avoids the "d(theta)/d(omega) mathematics" 
but still has plenty of other mathematics.

See Ya',
[-Rick-]
0
Rick
3/24/2009 12:33:11 PM
On 24 Mrz., 13:33, Rick Lyons <R.Lyons@_BOGUS_ieee.org> wrote:
> Sometimes in their algebraic derivations of z-domain
> transfer functions, some authors arrive at a transfer
> function equation of some network whose impulse response
> begins before an impulse input is applied. =A0They call such
> networks "acausal". =A0Then they say something
> like: "To make the network causal, we add a delay to the
> network."

I actually did the reverse: I started with a causal FIR filter (H(z))
and made it acausal (F(z), "positive powers of z =3D future") to show
that the acausal version has a group delay of zero.

> That kind of explanation always irritates
> me because how can they add a delay to a network when that
> network cannot exist in our physical universe? =A0Ha ha. =A0

An acausal FIR filter needs to look a couple of samples into the
future. If you combine such an acausal filter with an appropriate
delay you won't need to predict the future anymore.

Cheers!
SG
0
s.gesemann (692)
3/24/2009 1:13:09 PM
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delay filter
hallo to everyone i'm trying to implement a digital filter which is a pure delay. is this code right? h=fir1(delay*2, 0.999, blackman(2*delay+1)); I use this code because a fir filter give to me a delay equal to half of filter order. fir1 function seems to not accept 1 as second parameter. why? Is there any other simpler code ? (sorry for my english) On Oct 6, 6:36=A0am, "aenima1891" <aenima1...@gmail.com> wrote: > hallo to everyone > i'm trying to implement a digital filter which is a pure delay. > is this code right? > > h=3Dfir1(delay*2, 0.999, blackman(2*delay+1)); > > I use this code because a fir filter give to me a delay equal to half > of filter order. > fir1 function seems to not accept 1 as second parameter. why? > > Is there any other simpler code ? > > (sorry for my english) How about using a simple delay line? John >On Oct 6, 6:36=A0am, "aenima1891" <aenima1...@gmail.com> wrote: >> hallo to everyone >> i'm trying to implement a digital filter which is a pure delay. >> is this code right? >> >> h=3Dfir1(delay*2, 0.999, blackman(2*delay+1)); >> >> I use this code because a fir filter give to me a delay equal to half >> of filter order. >> fir1 function seems to not accept 1 as second parameter. why? >> >> Is there any other simpler code ? >> >> (sorry for my english) > >How about using a si...

What is the delay of this filter,
Dear all this is my homework, I should determine the delay of this symmetric FIR filter whose length is 3. y[n]=x[n]+x[n-1]+x[n-2]. The answer in the manual solutions is 1. I am confused. In my view, clearly h[0]=1, h(1)=1, h(2)=0, otherwise =zero and it's delay must be two. I guess that the answer is 1 because the filter is symmetric so that h[k]=h[M-k-1] (M=3 is the length of the filter). So delay is only (M-1)/2 if M is odd. Using this thinking, I can find the same answers as in the solution manuals. But indeed I am not satisfied with this explaination and would like to ask you al...

Filter a group
I have a report that uses a query as the recordsource. One field in the query is called Header which can be either Approved, Received, Not Taken, Closed Out. On the report footer - I total each of these by txtTotalApproves = Sum(IIF(Header="Approved",1,0)). Then I reference these in the report header as a summary page at the TOP of the report. So this works - but I want the detail lines of a group to only include header types of Closed Out or Not Taken. I know I could create a subreport, but can a group be filtered somehow? I tried entering an expression in the grouping : IIF(Header="Not Taken" or Header="Closed Out",Header), but that still gave me all status. Thanks, Brian In Report Design Mode, modify the "Record Source" property and then click on the elipses "...". In there, modify the query to filter your data as required. Dominic Vella "BerkshireGuy" <berkshireguy2005-commerical@yahoo.com> wrote in message news:a2f6daa6-340a-4720-9c05-dd8b0d075dcf@m34g2000hsb.googlegroups.com... >I have a report that uses a query as the recordsource. > > One field in the query is called Header which can be either Approved, > Received, Not Taken, Closed Out. > > On the report footer - I total each of these by > > txtTotalApproves = Sum(IIF(Header="Approved",1,0)). Then I reference > these in the report header as a summary page at the TOP of the > report. > > So...

Quadratic phase response all padd filter using FIR filter
Hi, I am trying to find a way to implement a FIR filter which can give allpass filter with quadratic phase response. The transfer function I am trying to implement is : H = exp( j * alpha * w^2 ). Can some one suggest what is best filter design approach that can be used to find the filter cofficients. Most of the filter design methods concentrate on the amplitude response and linear phase. Thanks sunil skhatana wrote: > Hi, > > I am trying to find a way to implement a FIR filter which can give > allpass filter with quadratic phase response. > > The transfer function I am trying to implement is : > > H = exp( j * alpha * w^2 ). > > Can some one suggest what is best filter design approach that can be used > to find the filter cofficients. Most of the filter design methods > concentrate on the amplitude response and linear phase. You probably want a Hilbert transformer. The length of your filter will depend on the bandwidth and your tolerance for in-band ripple. Jerry -- Engineering is the art of making what you want from things you can get. ����������������������������������������������������������������������� On Mon, 08 Jun 2009 14:27:17 -0500, skhatana wrote: > Hi, > > I am trying to find a way to implement a FIR filter which can give > allpass filter with quadratic phase response. > > The transfer function I am trying to implement is : > > H = exp( j * alpha * w^2 ). > > Can some...

Reg:Design of Butterworth approximated filter using FIR filter design technique
I need to design a butterworth approximated low pass, high pass, band pass, band stop filter characteristics.This i did using DFT technique which is a very heavy algorithm.So My question is how to design the FIR for my requirement.So can anybody help in how to proceed... Thank you :-) With Regards, SRS On Sep 5, 9:56 am, SRS <sitaram...@gmail.com> wrote: > I need to design a butterworth approximated low pass, high pass, band > pass, band stop filter characteristics.This i did using DFT technique > which is a very heavy algorithm.So My question is how to design the > FIR for my requirement.So can anybody help in how to proceed... > Thank you :-) > > With Regards, > SRS A recent article: Butterworth Window for Power Spectral Density Estimation Tae Hyun Yoon, and Eon Kyeong Joo, ETRI Journal, vol.31, no.3, June 2009, pp. 292-297. available from: http://etrij.etri.re.kr/Cyber/servlet/BrowseAbstract?vol=31&num=3&pg=292 gives an approximate Butterworth window approach that can be applied to FIR filters through the window based FIR design approach. Dale B. Dalrymple On 9/5/2009 9:56 AM, SRS wrote: > I need to design a butterworth approximated low pass, high pass, band > pass, band stop filter characteristics.This i did using DFT technique > which is a very heavy algorithm.So My question is how to design the > FIR for my requirement.So can anybody help in how to proceed... > Thank you :-) > > With Regards, ...

How to Calculate Wiener Filter Coefficients for a 8 Tap FIR Filter from 2 Input Wave-Files
Hello World, i work with a dsp and programmed a adaptiv filter for it. i would like to compare the results from the LMS on the DSP, with the optimal filter coefficients from the wiener-filter. the optimal filter weights = wiener solution you get if you use this w = p * R^-1 in words optimal filter coefficient vector = crosscorrelationvektor * inverse autocorrelation everything on the dsp works fine but i lack the mathematics to calculate the wiener optimal filter settings for a given FIR Filter length of n=8. Goal I would like to get 8 nummbers = that would be the filter coefficients ...

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