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 all for another one

so plz help

Thank you.

"VijaKhara" <VijaKhara@gmail.com> wrote in message 
news:1132097613.156441.252760@z14g2000cwz.googlegroups.com...
> 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 all for another one
>
> so plz help
>
> Thank you.

First, h[0]=1, h(1)=1, h(2)=1
(and *not* h[0]=1, h(1)=1, h(2)=0)
Which I take to mean this is the unit sample response with the unit sample 
at t=0.
Accordingly, if the input is [1 0 0 0 0 0 0 ......]
and the output is [1 1 1 0 0 0 0 0 .....]
then the delay is at the center of the unit sample response and is
(M-1)/2 = 2/2 = 1

Fred 

VijaKhara wrote:
> 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 all for another one
> 
> so plz help
> 
> Thank you.

Consider that the first input x[0] is not a delay at all, and hence y[0] 
is not a delayed output.  Thus the delay will be the middle of the 
remaining filter taps, of which there are now 2.  2/2 = 1, and hence the 
filter has a delay length of 1.
For any odd FIR filter, the delay is (N-1)/2 samples, where N is the 
number of taps.

"VijaKhara" <VijaKhara@gmail.com> wrote in message 
news:1132097613.156441.252760@z14g2000cwz.googlegroups.com...
> 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.

Well, you've got to think about what you mean when you talk about the delay 
of the filter.  For most inputs, the output isn't just a delayed version of 
that input. If you input, say:

-1,1,0,0,0,...

you get:

-1,0,0,1,0,0,0

Looking at the two, it becomes uncomfortable to pick any number -- 0, 3, 
1.5, or whatever, and say that was the delay through the system.

There are inputs, however, for which the output *is* simply a delayed and 
scaled version of the input, and when we talk about the delay of a filter, 
we are talking about the amount by which these inputs are delayed.

Sinusoids, in particular, will be simply delayed and scaled by any (LTI) 
filter.  For linear phase filters, all sinusoids are delayed by the same 
amount, so sums of sinusoids that are scaled by the same factor will pass 
through the filter unchanged with that same delay.  That is the delay 
through the filter, and if you test your FIR with any sinusoid, you will 
find that it comes out delayed by 1 sample.

Filters that are not linear phase will delay sinusoids of different 
frequencies by differing amounts.  For these filters, it makes no sense at 
all to talk about the delay though the filter as a whole.  Instead, we 
measure the "group delay" at particular frequencies.   It is called the 
"group delay" because a waveform made by summing a group of sinusoids 
closely clustered around a single frequency will pass through the filter 
pretty much unchanged, and will be delayed by the group delay at that 
frequency.

--
Matt

"Bevan Weiss" <kaizen__@NOSPAM.hotmail.com> wrote in message 
news:Qdwef.1558$vH5.89934@news.xtra.co.nz...
> VijaKhara wrote:
>> 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 all for another one
>>
>> so plz help
>>
>> Thank you.
>
> Consider that the first input x[0] is not a delay at all, and hence y[0] 
> is not a delayed output.  Thus the delay will be the middle of the 
> remaining filter taps, of which there are now 2.  2/2 = 1, and hence the 
> filter has a delay length of 1.
> For any odd FIR filter, the delay is (N-1)/2 samples, where N is the 
> number of taps.

I think you mean for any odd-length *symmetrical* FIR filter......

Fred 

Bevan Weiss wrote:

   ...

> For any odd FIR filter, the delay is (N-1)/2 samples, where N is the 
> number of taps.

For any even FIR filter, the delay is (N-1)/2 samples, where N is the 
number of taps. Combine those statements and you get "For any FIR 
filter, the delay is (N-1)/2 samples, where N is the number of taps."

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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Thank members, I got it already, thank god the group is sooo...
helpful.

Vijay

Fred Marshall wrote:

>
> I think you mean for any odd-length *symmetrical* FIR filter......
> 

or *antisymmetric* FIR filter :-)

Ciao,

Peter K.

Using your logic wouldn't the middle of the remaining filter taps( at 1
and 2) be 1.5? Not right of course.

Dirk

dbell wrote:
> Using your logic wouldn't the middle of the remaining filter taps( at 1
> and 2) be 1.5? Not right of course.
> 
> Dirk

We're talking about delays not offsets from the initial inputs.
So that 1 and 2 you're talking about would be 1 delay and then another 
delay, hence two delays.  The output response would occur in the middle 
of this and hence give 1 delay for the total filter response.