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Decimation vs. Collapsing and their spectral effects

Started by Martin J. Stumpf September 21, 2004
Jerry Avins wrote:

> That seems very strange to me; I would have thought it is a sinc. Can > you explain?
You can see it from a different point of view. Two (symmetric) 1s in the frequency domain correspond to a cos() function in the time domain. But the FT is going back and forward in the same way, so two (symmetric) 1s in the time domain must be a cos() in the frequency domain... Similar thinking can be done for [1,-1] and sin(), only I guess some imaginary part has to be considered. Of course the frequency calculation is left as exercise for the student... :-) bye, -- Piergiorgio Sartor
On Fri, 24 Sep 2004 09:11:24 -0500, jim <"N0sp"@m.sjedging@mwt.net>
wrote:

> >Hi Rick > My newsreader seems to have missed some posts. This is the first one I >see since my last post. (see more below) > >Rick Lyons wrote: >> >> On Thu, 23 Sep 2004 10:23:43 -0400, Jerry Avins <jya@ieee.org> wrote: >> >> >Rick Lyons wrote: > >> >> The complex freq response of >> >> 1 + cos(w) -jsin(w) >> >> can be written >> >> 1 + exp(-jw) >> >> = exp(-jw/2)[exp(jw/2) + exp(-jw/2)]. >> >> Those terms inside the brackets are equal to >> >> 2cos(w/2). >> >> So the magnitude of the complex freq response is >> >> |exp(-jw/2)[2cos(w/2)]| = 2cos(w/2). >> >> Sheece, how simple!
Hi Jim,
>> (The above is surely what Jim was saying when he >> wrote: "e^(i(x+a)) + e^(i(x-a)) = 2*cos(a)*e^ix".)
Yep, me too.
>Yes, glad you waded through the math (and didn't make me do it). > > Remember the OP's problem was with image processing and pixels, so >there's no need to be concerned with causality. Therefore one is free to >express the frequency response with reference to the center of the >filter rather than the end as you do (that's where your exp(-jw/2) comes >from). So it could be said like this: averaging 2 pixels produces the >cosine magnitude response at the point halfway in between.
Ah ha. yes.
>> *** NOW *** >> >> Someone might read our thread and say: "Why the heck >> are Jerry and Rick screwing around with this >> silly simple two-coefficient filter's magnitude >> response?" >> >> The reason is: the above sinusoidal shaped magnitude >> response *also* applies to a transversal FIR filter >> whose coefficients are >> >> h1(k) = 1,0,0,0,0,0,0,1. >> >> This h1(k) is an example of a comb filter used in >> some "frequency sampling filters". (Section 7.1 of >> my book). >> >> In addition, that sinusoidal shaped magnitude response >> *also* applies to a transversal FIR filter whose >> coefficients are >> >> h1(k) = 1,0,0,0,0,0,0,-1 > >Right, but in that case the response will be a sine function.
Yep. You're right. That's why I used the phrase "sinusoidal shpaed".
>> which is an example of the comb portion of a cascaded >> integrator-comb (CIC) filter used in many digital >> communications systems. >> >> Neat huh? > >Yes, and if you constrain your filters to purely symmetric or purely >anti-symmetric FIR (as people inevitably do in image processing), then >any such filter can be expressedas a simple sum of sines or cosines. >For example, an even length box car filter can be considered to be the >sum of the filters: > > [1,1] + [1,0,0,1] + [1,0,0,0,0,1] ..... > >And therefore the magnitude response is just the sum of the respective >cosine functions. > >-jim
Ah ha. Yes, yes. I knew that symmetrical transversal FIR filters have a freq magnitude response that are the sums of cosine functions, but I've never looked at your example (interpretation) before. (That a boxcar can be viewed as a group of parallel comb filters. Composite impulse response is the sum of individual impulse responses.) Neat Jim. Really neat. Thanks. [-Rick-]
Rick Lyons wrote:

   ...

> Neat Jim. Really neat. > Thanks. > > [-Rick-]
And thanks again from me too. You really tied things together for me, starting with your first answer to me in this thread. It got me thinking and off autopilot. For all the good it will do! Hah! Jerry -- ... they proceeded on the sound principle that the magnitude of a lie always contains a certain factor of credibility, ... and that therefor ... they more easily fall victim to a big lie than to a little one ... A. H. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Piergiorgio Sartor wrote:

> Jerry Avins wrote: > >> That seems very strange to me; I would have thought it is a sinc. Can >> you explain? > > > You can see it from a different point of view. > > Two (symmetric) 1s in the frequency domain correspond to > a cos() function in the time domain. > > But the FT is going back and forward in the same way, > so two (symmetric) 1s in the time domain must be a cos() > in the frequency domain... > > Similar thinking can be done for [1,-1] and sin(), only > I guess some imaginary part has to be considered. > > Of course the frequency calculation is left as exercise > for the student... :-) > > bye,
Another neat answer. "I love it when things come together!" Jerry -- ... they proceeded on the sound principle that the magnitude of a lie always contains a certain factor of credibility, ... and that therefor ... they more easily fall victim to a big lie than to a little one ... A. H. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;