# DSP Riddle: Infinitely Narrow Notch Filter

Started by May 19, 2006
```Jani Huhtanen wrote:
> Randy Yates wrote:
>
>
>>I think you did.
>>
>>--Randy
>
>
> I second that. By the analogy, infinitely thin but infinitely wide hamburger
> should indeed exist (at least in theory ;)) as well as there should exist a
> DC level or a sine wave. Although this has little to do with the riddle at
> hand.
>
I never seen an infinitely wide hamburger, but I've seen a close
approximation to an infinitely wide hamburger consumer. :-)

Steve
```
```robert bristow-johnson wrote:
> Andor wrote:
> >
> > is it possible to construct a causal and realizable n-th order IIR
> > filter, where n > 1 is an integer, which acts as an infinitely narrow
> > notch filter?
...
>  from a Lebesgue measure POV, there is no difference betwee the
> specification you made and one where H(w) = 1 for all w.

Shuch a brilliantly constructed riddle. Such a concise solution.
These kinds of things take all the fun out of solving puzzles.

Oh well.

Rune

```
```robert bristow-johnson wrote:

>
> Andor wrote:
>>
>> is it possible to construct a causal and realizable n-th order IIR
>> filter, where n > 1 is an integer, which acts as an infinitely narrow
>> notch filter? This filter must have frequency response
>>
>> H(w) = 1, for w =/= w0
>>      = 0, for w == w0,
>>
>> for some arbitrary 0 <  w0 < pi.
>>
>> Assume for the moment that the filter can be computed with infinite
>> precision.
>
> you can get arbitrarily close for n=2.
>
>   zeros = exp(&#2013266097;j*w0)
>   poles = (1-e)*exp(&#2013266097;j*w0)
>
>         where 0 < e << 1.
>
> but, from a Lebesgue measure POV, there is no difference betwee the
> specification you made and one where H(w) = 1 for all w.
>
> r b-j

Perhaps, but in "practice" there would be a difference. H(w) = 1, for all w,
would not effect a sine wave (from +inf to inf) in anyway, where as the
original definition (H(w)=/=0, for w0) would have filtered the sine wave
out if the frequency would have been w0. I may have mistaken of course..

--
Jani Huhtanen
Tampere University of Technology, Pori
```
```I wrote:
> robert bristow-johnson wrote:
>
>>
>> Andor wrote:
>>>
>>> is it possible to construct a causal and realizable n-th order IIR
>>> filter, where n > 1 is an integer, which acts as an infinitely narrow
>>> notch filter? This filter must have frequency response
>>>
>>> H(w) = 1, for w =/= w0
>>>      = 0, for w == w0,
>>>
>>> for some arbitrary 0 <  w0 < pi.
>>>
>>> Assume for the moment that the filter can be computed with infinite
>>> precision.
>>
>> you can get arbitrarily close for n=2.
>>
>>   zeros = exp(&#2013266097;j*w0)
>>   poles = (1-e)*exp(&#2013266097;j*w0)
>>
>>         where 0 < e << 1.
>>
>> but, from a Lebesgue measure POV, there is no difference betwee the
>> specification you made and one where H(w) = 1 for all w.
>>
>> r b-j
>
> Perhaps, but in "practice" there would be a difference. H(w) = 1, for all
> w, would not effect a sine wave (from +inf to inf) in anyway, where as the
> original definition (H(w)=/=0, for w0)

And this -------------^^^^^^^^^^^^^^^^^
was meant to be (H(w) = 0, for w=w0)

> would have filtered the sine wave
> out if the frequency would have been w0. I may have mistaken of course..
>
>
>

I really have to start proof reading my posts some day.

--
Jani Huhtanen
Tampere University of Technology, Pori
```
```Sukrut wrote:

> for a notch filter,
>
> H(ejw) = 0
> H(e-jw) = 0
>
> zero at z = e+-jw
>
> pole infinitely close to the zero
> at z = a * e+-jw
>
> where a -> 1, but a =/= 1
>
> the filter transfer function would be
>
> H(z) = (z - e^jw) * (z - e^-jw)
> 	-------------------------
>        (z - a*e^jw) * (z - a*e^-jw)
>
> 	= z^2 - z * (2 cos w) + 1
> 	  --------------------------
> 	  z^2 - z * (2*a cos w) + a
>
> This is a causal filter and can be realized using a direct form
> realization if a -> 1 can be realized
>
> As the filter can be computed with infinte precision, assuming that the
> gainterm a -> 1 can be realized,
> the filter is causal as well as realizable
>
> but, a -> 1 can't be realized in reality, which is an altogether
> different question
>
> ----------------------------------------------------------------------------------------
>
> Q: Why didnt i join this group before?

One doesn't join; one just hangs out. Welcome!

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```That was my rationale as well.

And this isn't just nit-picking. I believe that this type
of analysis as applied to the riddle at hand is going
down the wrong road. (Although I can't offer anything
more, yet.)

--Randy

```
```"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:i-SdnVQ57M0DpfPZRVn-rg@centurytel.net...
> Here is a constructive method for thinking about it:
>
> I am speaking of discrete time / discrete frequency here.
> For ease of discussion, normalize as follows:
> Assume the sample interval is 1 second and the length of the filter is N
> samples/coefficients.
> The frequency interval between samples is 1/N.
>
> Now, set a pair of frequency domain samples to zero as follows:
>  choose f0 per your specification
>  choose fs for the sampling radian frequency
>  find the index "k" of the frequency sequence that corresponds to f0<fs/2:
>    the unnormalized frequency interval is fs/N
>    f0 corresponds closest to the index k=ROUND[N*f0/fs]
>  Set F(k)=0 and F(N-k+2)=0 ... (and you worry about whether this is OK for
> your index scheme starting at 1 or at zero)..
> Compute the IFFT of the new sequence.
>
> Notice that you can only define the filter response on a discrete set of
> frequencies spaced at fs/N.  So, an infinitely narrow notch done this way
> must have infinite N.  Or, if your "infinity" is sufficient with N=10,000
> so be it.
>
> Also notice that the time domain filter that results has a spike or sample
> of amplitude 1.0 at time=0 *minus* a cosine at w0 of amplitude 2/N.  It's
> that simple.
>
> So a perfect notch filter has an impulse response that is an impulse at
> time=0 minus a cosine at w0 of amplitude 2/(infinity).
> If your "infinity" is sufficient with N=10,000 then it would have
> amplitude 2/10,000.
> Clearly infinitely narrow notches take away infinitesimal energy!
>
> Interesting, eh?
>
> Fred

Gee, I'm a bit disappointed that there was no comment to this.

As before, the narrower the notch, the smaller the amplitude affect on the
filter unit sample response.

Fred

```
```Fred Marshall wrote:

>
> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
> news:i-SdnVQ57M0DpfPZRVn-rg@centurytel.net...
>> Here is a constructive method for thinking about it:
>>
>> I am speaking of discrete time / discrete frequency here.
>> For ease of discussion, normalize as follows:
>> Assume the sample interval is 1 second and the length of the filter is N
>> samples/coefficients.
>> The frequency interval between samples is 1/N.
>>
>> Now, set a pair of frequency domain samples to zero as follows:
>>  choose f0 per your specification
>>  choose fs for the sampling radian frequency
>>  find the index "k" of the frequency sequence that corresponds to
>>  f0<fs/2:
>>    the unnormalized frequency interval is fs/N
>>    f0 corresponds closest to the index k=ROUND[N*f0/fs]
>>  Set F(k)=0 and F(N-k+2)=0 ... (and you worry about whether this is OK
>>  for
>> your index scheme starting at 1 or at zero)..
>> Compute the IFFT of the new sequence.
>>
>> Notice that you can only define the filter response on a discrete set of
>> frequencies spaced at fs/N.  So, an infinitely narrow notch done this way
>> must have infinite N.  Or, if your "infinity" is sufficient with N=10,000
>> so be it.
>>
>> Also notice that the time domain filter that results has a spike or
>> sample
>> of amplitude 1.0 at time=0 *minus* a cosine at w0 of amplitude 2/N.  It's
>> that simple.
>>
>> So a perfect notch filter has an impulse response that is an impulse at
>> time=0 minus a cosine at w0 of amplitude 2/(infinity).
>> If your "infinity" is sufficient with N=10,000 then it would have
>> amplitude 2/10,000.
>> Clearly infinitely narrow notches take away infinitesimal energy!
>>
>> Interesting, eh?
>>
>> Fred
>
> Gee, I'm a bit disappointed that there was no comment to this.
>
> As before, the narrower the notch, the smaller the amplitude affect on the
> filter unit sample response.
>
> Fred

OK, a comment. ;)

Let H(w) be the frequency response of a causal notch filter with H(w) = 0,
for w = w0 and H(w) = 1 for all other w. Let S(w) be the frequency response
of a sine wave with frequency w0 (a comb of diracs, see e.g.,
http://www.cse.ucsc.edu/classes/cmpe163/Spring03/Lec23Ap.pdf).

Convolving them yields H(w)S(w)=0 for all w, right (probably not :/)? This
is because S(w)=0 for all w except w=w0 and H(w)=0 for w=w0). This
multiplication involves infinities and as such is not well defined.
However, the dirac delta should not have a finite energy after the
multiplication and thus it seems that more than infinitesimal energy is
taken.

--
Jani Huhtanen
Tampere University of Technology, Pori
```
```Randy Yates wrote:

> So an infinitely-narrow spectrum is the same as no spectrum at all?

Yes, if it's bounded (if the flat hamburger has finite area).

Martin

--
Quidquid latine scriptum sit, altum viditur.
```
```Fred Marshall wrote:

> I am speaking of discrete time / discrete frequency here.

You can't meaningfully use the phrase "infinitely narrow" if you're
working on a grid. Alternatively, you might take the discrete
spectrum to specify a continuous spectrum through circular sinc
interpolation; but prescribing a zero at one grid point (and its
images) will leave the continuous notch finitely wide.

Martin

--
Quidquid latine scriptum sit, altum viditur.
```