Droup delay in FIR implementation

Hi,

I want to know more about group delay in FIR filter hardware.

In fir filter hardware, there are three main components.
1. delay elements
2. Multiplier
3. Adder

For every input sample, output takes same amount of time as there is fix
delay for each.
Then where the group delay comes into effect?

Plz help me to clarify this.

Dhaval

On Sat, 25 Dec 2010 08:14:29 -0600, Dhaval_DSP wrote:

> Hi,
> 
> I want to know more about group delay in FIR filter hardware.
> 
> In fir filter hardware, there are three main components. 1. delay
> elements
> 2. Multiplier
> 3. Adder
> 
> For every input sample, output takes same amount of time as there is fix
> delay for each.
> Then where the group delay comes into effect?
> 
> Plz help me to clarify this.

Group delay is the effective delay experienced by a signal at a 
particular frequency or around a particular frequency.  Note the word 
"effective".

A FIR filter that is symmetric around the center tap will have a group 
delay equal to half the length of the filter.  Filters that are not 
symmetric around some point in the time response, whether FIR or IIR, do 
not have group delays that are constant with frequency.

The delays involved in the individual tap calculations don't have much to 
do with the overall group delay, unless you are using a short filter with 
a long pipeline.

(I was going to refer you to the Wikipedia article on group delay, but I 
didn't like it -- it may still help, though: http://en.wikipedia.org/wiki/
Group_delay).

-- 
http://www.wescottdesign.com

On Sat, 25 Dec 2010 08:14:29 -0600, "Dhaval_DSP"
<shah.dhavalshah@n_o_s_p_a_m.gmail.com> wrote:

>Hi,
>
>I want to know more about group delay in FIR filter hardware.
>
>In fir filter hardware, there are three main components.
>1. delay elements
>2. Multiplier
>3. Adder
>
>For every input sample, output takes same amount of time as there is fix
>delay for each.
>Then where the group delay comes into effect?
>
>Plz help me to clarify this.
>
>Dhaval

Hi Dhaval,
  The group delay (time delay measured in "samples") 
through a FIR filter (that has symmetrical coefficients) 
is equal to the number of delay elements divided 
by two.  You can prove this to yourself by way of 
software modeling.

Create a lowpass symmetrical-coefficient lowpass 
FIR filter, and apply a low-frequency sinewave 
sequence to the filter's input.  Now align a plot 
of the filter's input sequence with a plot of 
the filter's output sequence.

Do you see the delay between the output and the 
input sequences?  If your lowpass filter had 
14 delay elements (15 symmetrical coefficients) 
then the output will be a delayed from the input 
by 14/2 = 7 samples.

Coefficient symmetry is *KEY* here.

[-Rick-]

On 12/25/2010 6:14 AM, Dhaval_DSP wrote:
> Hi,
>
> I want to know more about group delay in FIR filter hardware.
>
> In fir filter hardware, there are three main components.
> 1. delay elements
> 2. Multiplier
> 3. Adder
>
> For every input sample, output takes same amount of time as there is fix
> delay for each.
> Then where the group delay comes into effect?
>
> Plz help me to clarify this.
>
> Dhaval
>
>

Do be careful to not expect a tone burst to necessarily have the same 
"delay" as the group delay at the same frequency as the tone.

That's because the group delay at a particular frequency is generally 
measured in the *steady state* response at that frequency and, as such, 
can be expressed in terms of phase.
Whereas, the delay of a burst may be measured differently.

In truth, I don't know how to relate the two just offhand...

Fred

On Dec 27, 8:19&#4294967295;pm, Fred Marshall <fmarshall_xremove_the...@xacm.org>
wrote:
>
> Do be careful to not expect a tone burst to necessarily have the same
> "delay" as the group delay at the same frequency as the tone.
>
> That's because the group delay at a particular frequency is generally
> measured in the *steady state* response at that frequency and, as such,
> can be expressed in terms of phase.
> Whereas, the delay of a burst may be measured differently.
>

this is curious to me, Fred.  isn't that precisely what "group delay"
is?  it's the delay of the envelope applied to a sinusoidal tone,
assuming that the envelope is bandlimited to a frequency much less
than that of the tone.  group delay *is* the delay of the envelope of
a tone burst.  that's how it got the name, no?

r b-j

On 12/27/2010 10:13 PM, robert bristow-johnson wrote:
> On Dec 27, 8:19 pm, Fred Marshall<fmarshall_xremove_the...@xacm.org>
> wrote:
>>
>> Do be careful to not expect a tone burst to necessarily have the same
>> "delay" as the group delay at the same frequency as the tone.
>>
>> That's because the group delay at a particular frequency is generally
>> measured in the *steady state* response at that frequency and, as such,
>> can be expressed in terms of phase.
>> Whereas, the delay of a burst may be measured differently.
>>
>
> this is curious to me, Fred.  isn't that precisely what "group delay"
> is?  it's the delay of the envelope applied to a sinusoidal tone,
> assuming that the envelope is bandlimited to a frequency much less
> than that of the tone.  group delay *is* the delay of the envelope of
> a tone burst.  that's how it got the name, no?
>
> r b-j

r b-j,

Well, I tried to describe the caution.  And I did say that the 
measurement might be different.

Consider this:

The group delay is directly related to the phase measurement at any 
frequency.  The magnitude and phase measurements are *steady state* 
measurements - i.e. for an infinitely long sinusoid in terms of a 
Fourier Transform or for a sample in a DFT - suggesting a harmonic of a 
periodic waveform.  Not stuff we need argue about.

So, if one applies a tone burst, and if it's long enough to reach steady 
state then I guess the phase will be the phase mentioned above I do 
believe.  And, the group delay *as defined* will be related or equal to 
that phase shift.  Likely equal.

But, what I don't know is what the rise time of a tone burst would be, 
how one would like to define "rise time" and I don't know how that 
relates to the steady state phase shift when the output of a system has 
stabilized.

So, no, the definition of group delay isn't about the envelope.  But, 
I'm willing to learn that they may be nicely related or even equal.

I hope that's clearer.  I just simply don't know.  So, I offered a 
"caution".  Surely the possible variations of definition of "delay" is 
enough to justify that much.

Fred

On Dec 28, 7:13&#4294967295;am, robert bristow-johnson <r...@audioimagination.com>
wrote:
> On Dec 27, 8:19&#4294967295;pm, Fred Marshall <fmarshall_xremove_the...@xacm.org>
> wrote:
>
>
>
> > Do be careful to not expect a tone burst to necessarily have the same
> > "delay" as the group delay at the same frequency as the tone.
>
> > That's because the group delay at a particular frequency is generally
> > measured in the *steady state* response at that frequency and, as such,
> > can be expressed in terms of phase.
> > Whereas, the delay of a burst may be measured differently.
>
> this is curious to me, Fred. &#4294967295;isn't that precisely what "group delay"
> is? &#4294967295;it's the delay of the envelope applied to a sinusoidal tone,
> assuming that the envelope is bandlimited to a frequency much less
> than that of the tone. &#4294967295;group delay *is* the delay of the envelope of
> a tone burst. &#4294967295;that's how it got the name, no?

Don't know about the etymology, but these things are hard to really
understand. The problem is that the group and phase delays can change
over a bandwidth: Some parts of the signal experience one delay;
other parts some other delay. If the spectrum energy distribution is
continuous between the two frequencies where delays differ (as in a
broadband pulse) the system is dispersive and any attempt at
intuitive
reasoning about delays is prone to error and confusion.

There was a paper by Johan Leander in JASA about these things in 1995
or 1996. Read it, if you can find it.

Rune

On Dec 28, 8:44&#4294967295;pm, Rune Allnor <all...@tele.ntnu.no> wrote:
> ...

>
> There was a paper by Johan Leander in JASA about these things in 1995
> or 1996. Read it, if you can find it.
>
> Rune

On the relation between the wavefront speed and the group velocity
concept
Johan L. Leander
J. Acoust. Soc. Am. vol100(6) p3503 (1996) December

Is that the one?

Dale B. Dalrymple

On Dec 29, 9:25&#4294967295;am, dbd <d...@ieee.org> wrote:
> On Dec 28, 8:44&#4294967295;pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
> > ...
>
> > There was a paper by Johan Leander in JASA about these things in 1995
> > or 1996. Read it, if you can find it.
>
> > Rune
>
> On the relation between the wavefront speed and the group velocity
> concept
> Johan L. Leander
> J. Acoust. Soc. Am. vol100(6) p3503 (1996) December
>
> Is that the one?

Yep.

Rune

>Hi,
>
>I want to know more about group delay in FIR filter hardware.
>
>In fir filter hardware, there are three main components.
>1. delay elements
>2. Multiplier
>3. Adder
>
>For every input sample, output takes same amount of time as there is fix
>delay for each.
>Then where the group delay comes into effect?
>
>Plz help me to clarify this.
>
>Dhaval
>
>
>
There is a simple method.
http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-341-discrete-time-signal-processing-fall-2005/lecture-notes/lec02.pdf
You may use a network analyzer test the time delay,just FIR's group delay
(data rate) add your pipelines time(system clk rate).