[ CAVEAT any term enclosed by "..." is used EXTREMELY loosely ]
Given:
1. a non-periodic signal to be "characterized"
[ "analyzed" implies much too much precision ;]
2. assumption that the data of interest is the relative "strength"
of the signal in 3 to 5 non-overlapping frequency bands as
a function of time.
3. A traditional workable solution would be to do a series of
fouier/other transforms and "sum" the appropriate components.
Anonther solution would be to filter the signal and calculate
the "strength" of the passed signal, repeating for each passband
and each time window.
Constraint: minimize calculation/delay time.
reproducibility
NON-constraint that:
function have an inverse!!!!
math have any provable correlation to process variables
[ w is defined as omega of standard discourse ]
Two possible solutions:
If each passband of interest is decade(s) wide:
Multiply incoming signal by sin(wt)*sin(3wt)*sin(5wt)...*sin([NMAX]wt)
[ note multipiers of w are prime ]
[ w -> [NMAX]w would define band of interest ]
{ I hink I can take advantage of multiplication of sines having terms
a function of initial frequencies as well as summ and difference
frewuencies )
OR
If there a M bands of interest,
could the input be multiplied by M pink-noise signals.
each pink noise signal would be limited to band of interest.
--
I'm the one that theatended to do DSP in Basic [ I've done some in
Excel. I am considering COBOL ;)
OT? -- How far can linear supperposition be stretched
Started by ●September 28, 2003
Reply by ●September 28, 20032003-09-28
Richard Owlett <rowlett@atlascomm.net> wrote in message news:<vnebb61nlodgb3@corp.supernews.com>...> [ CAVEAT any term enclosed by "..." is used EXTREMELY loosely ] > > Given: > 1. a non-periodic signal to be "characterized" > [ "analyzed" implies much too much precision ;]You want just a basic "feel" of what's going on?> 2. assumption that the data of interest is the relative "strength" > of the signal in 3 to 5 non-overlapping frequency bands as > a function of time.I'm thinking filter bank here...> 3. A traditional workable solution would be to do a series of > fouier/other transforms and "sum" the appropriate components. > Anonther solution would be to filter the signal and calculate > the "strength" of the passed signal, repeating for each passband > and each time window.Sure. A Gabor plot, a Wigner distribution, wavelets...> Constraint: minimize calculation/delay time. > reproducibilityOK, the Wigner distribution is out due to delays, wavelets... I'm not sure, Gabor plots could work.> NON-constraint that: > function have an inverse!!!! > math have any provable correlation to process variablesPoint to filter banks, wavelets appear to be way over the top.> [ w is defined as omega of standard discourse ] > > Two possible solutions: > > If each passband of interest is decade(s) wide: > Multiply incoming signal by sin(wt)*sin(3wt)*sin(5wt)...*sin([NMAX]wt) > [ note multipiers of w are prime ] > [ w -> [NMAX]w would define band of interest ] > { I hink I can take advantage of multiplication of sines having terms > a function of initial frequencies as well as summ and difference > frewuencies ) > > OR > > If there a M bands of interest, > could the input be multiplied by M pink-noise signals. > each pink noise signal would be limited to band of interest.Both solutions seem to be expensive: The sin-filter would have a very long impulse response in time domain, if taken literally. You could do what you want but with a simpler filter. The white noise sequence would have to be generated and "spectrum shaped". A simple filter bank may do the work for you sufficiently well and sufficiently fast. It may be the time of day, but "filter bank" has somehow stuck in my mind.> -- > I'm the one that theatended to do DSP in Basic [ I've done some in > Excel. I am considering COBOL ;)I've seen worse. As long as you don't do DSP in real time with a slide rule using roman numerals (no, I haven't seen that, but close enough), you'll be OK. Rune
Reply by ●September 28, 20032003-09-28
"Rune Allnor" <allnor@tele.ntnu.no> wrote in message news:f56893ae.0309281531.34e1ee7f@posting.google.com...> Richard Owlett <rowlett@atlascomm.net> wrote in messagenews:<vnebb61nlodgb3@corp.supernews.com>...> > [ CAVEAT any term enclosed by "..." is used EXTREMELY loosely ] > > > > Given: > > 1. a non-periodic signal to be "characterized" > > [ "analyzed" implies much too much precision ;] > > You want just a basic "feel" of what's going on? > > > 2. assumption that the data of interest is the relative "strength" > > of the signal in 3 to 5 non-overlapping frequency bands as > > a function of time. > > I'm thinking filter bank here... > > > 3. A traditional workable solution would be to do a series of > > fouier/other transforms and "sum" the appropriate components. > > Anonther solution would be to filter the signal and calculate > > the "strength" of the passed signal, repeating for each passband > > and each time window. > > Sure. A Gabor plot, a Wigner distribution, wavelets... > > > Constraint: minimize calculation/delay time. > > reproducibility > > OK, the Wigner distribution is out due to delays, wavelets... I'm notsure,> Gabor plots could work. > > > NON-constraint that: > > function have an inverse!!!! > > math have any provable correlation to process variables > > Point to filter banks, wavelets appear to be way over the top. > > > [ w is defined as omega of standard discourse ] > > > > Two possible solutions: > > > > If each passband of interest is decade(s) wide: > > Multiply incoming signal by sin(wt)*sin(3wt)*sin(5wt)...*sin([NMAX]wt) > > [ note multipiers of w are prime ] > > [ w -> [NMAX]w would define band of interest ] > > { I hink I can take advantage of multiplication of sines having terms > > a function of initial frequencies as well as summ and difference > > frewuencies ) > > > > OR > > > > If there a M bands of interest, > > could the input be multiplied by M pink-noise signals. > > each pink noise signal would be limited to band of interest. > > Both solutions seem to be expensive: The sin-filter would have a very > long impulse response in time domain, if taken literally. You could do > what you want but with a simpler filter. The white noise sequence would > have to be generated and "spectrum shaped". A simple filter bank may do > the work for you sufficiently well and sufficiently fast. > > It may be the time of day, but "filter bank" has somehow stuck in my mind. > > > -- > > I'm the one that theatended to do DSP in Basic [ I've done some in > > Excel. I am considering COBOL ;) > > I've seen worse. As long as you don't do DSP in real time with a slide > rule using roman numerals (no, I haven't seen that, but close enough), > you'll be OK. > > RuneRichard, We know that fft/multiply/ifft is computationally faster than doing a time-domain convolution for arrays of reasonable length. So, removing the ifft is certainly even faster! So, why not do this: 1)- [design the bandpass filters of choice - assume they are FIR filters] - fft the filter impulse responses / coefficient sequences. 2)- fft a block of data - multiply by each bandpass filter's fft - compute the sum of squares of magnitudes and take sqrt of the result (This yields a scalar measure of the signal contained in each of the bands of interest). Repeat (2) and plot the results for each filter. Overlap the blocks in time if desired. So, as Rune says, this implements a filter bank of a particular design and uses an efficient compute method. Multiplying by sines or sums of sines generates but one point in a convolution. I think you *have* to do the convolution in order to get reasonable results. Fred
Reply by ●September 29, 20032003-09-29
Fred Marshall wrote:> "Rune Allnor" <allnor@tele.ntnu.no> wrote in message > news:f56893ae.0309281531.34e1ee7f@posting.google.com... > >>Richard Owlett <rowlett@atlascomm.net> wrote in message > > news:<vnebb61nlodgb3@corp.supernews.com>... > >>>[ CAVEAT any term enclosed by "..." is used EXTREMELY loosely ] >>> >>>Given: >>> 1. a non-periodic signal to be "characterized" >>> [ "analyzed" implies much too much precision ;] >> >>You want just a basic "feel" of what's going on? >> >> >>> 2. assumption that the data of interest is the relative "strength" >>> of the signal in 3 to 5 non-overlapping frequency bands as >>> a function of time. >> >>I'm thinking filter bank here... >> >> >>> 3. A traditional workable solution would be to do a series of >>> fouier/other transforms and "sum" the appropriate components. >>> Anonther solution would be to filter the signal and calculate >>> the "strength" of the passed signal, repeating for each passband >>> and each time window. >> >>Sure. A Gabor plot, a Wigner distribution, wavelets... >> >> >>>Constraint: minimize calculation/delay time. >>> reproducibility >> >>OK, the Wigner distribution is out due to delays, wavelets... I'm not > > sure, > >>Gabor plots could work. >> >> >>>NON-constraint that: >>> function have an inverse!!!! >>> math have any provable correlation to process variables >> >>Point to filter banks, wavelets appear to be way over the top. >> >> >>>[ w is defined as omega of standard discourse ] >>> >>>Two possible solutions: >>> >>>If each passband of interest is decade(s) wide: >>>Multiply incoming signal by sin(wt)*sin(3wt)*sin(5wt)...*sin([NMAX]wt) >>>[ note multipiers of w are prime ] >>>[ w -> [NMAX]w would define band of interest ] >>>{ I hink I can take advantage of multiplication of sines having terms >>> a function of initial frequencies as well as summ and difference >>> frewuencies ) >>> >>>OR >>> >>>If there a M bands of interest, >>>could the input be multiplied by M pink-noise signals. >>>each pink noise signal would be limited to band of interest. >> >>Both solutions seem to be expensive: The sin-filter would have a very >>long impulse response in time domain, if taken literally. You could do >>what you want but with a simpler filter. The white noise sequence would >>have to be generated and "spectrum shaped". A simple filter bank may do >>the work for you sufficiently well and sufficiently fast. >> >>It may be the time of day, but "filter bank" has somehow stuck in my mind. >> >> >>> -- >>>I'm the one that theatended to do DSP in Basic [ I've done some in >>>Excel. I am considering COBOL ;) >> >>I've seen worse. As long as you don't do DSP in real time with a slide >>rule using roman numerals (no, I haven't seen that, but close enough), >>you'll be OK. >> >>Rune > > > Richard, > > We know that fft/multiply/ifft is computationally faster than doing a > time-domain convolution for arrays of reasonable length. So, removing the > ifft is certainly even faster! > So, why not do this: > 1)- [design the bandpass filters of choice - assume they are FIR filters] > - fft the filter impulse responses / coefficient sequences. > > 2)- fft a block of data > - multiply by each bandpass filter's fft > - compute the sum of squares of magnitudes and take sqrt of the result > (This yields a scalar measure of the signal contained in each of the bands > of interest). > Repeat (2) and plot the results for each filter. Overlap the blocks in time > if desired. > > So, as Rune says, this implements a filter bank of a particular design and > uses an efficient compute method. > Multiplying by sines or sums of sines generates but one point in a > convolution. I think you *have* to do the convolution in order to get > reasonable results. > > Fred > >I think there are at least 2 sorces of confusion: 1. my lack of understanding of digital FFT and filters 2. my inability to communicate in writing - I'll not admit to how many times I flunked freshman 'English' composition ;} As to #1, I understood that FFT required on order of "N SQUARED" multiplication-addition operations. The approach I was considering would require exactly N multiplication-addition operations per band of interest. Fred stated "Multiplying by sines or sums of sines generates but one point in a convolution." I had assumed that. I'm trying to test a hypothosis that says only one such point is important/required/significant/???. Now we get into #2 ;[ Lets simplify/trivialize prolem by assumming: a. only one band of interest running from f1 -> f2 b. said band has a "random" set of frequencies c. amplitude vs frequency is "relatively" flat and "continuous" d. borrow concepts of matrix math 1. input time domain data is an Nx1 matrix [A] 2. my "product of sines" or "pink noise" is a 1xN matrix [B] My assertion to be tested is that the scalar (inner) product of [A] dot [B] is significant. [B] is assumed fixed over durration of experiment. -------------------- a totaly separate way to look at my problem assume a white noise source it feeds 3 perfect bandpass filters The filters do not overlap each filter supplies input to a perfect amplifier [ gain is function of time but not ftrequency ] [ time >>> 1/(any freq of interest) ] The output of each amplifier is summed equally my task might be defined as determining gain of each of the 3 amplifiers as function of time Do either of these problem definitions make sense to anyone but me?
Reply by ●September 30, 20032003-09-30
"Richard Owlett" <rowlett@atlascomm.net> wrote in message news:vnh9fbgn2mflae@corp.supernews.com...> Fred Marshall wrote: > > "Rune Allnor" <allnor@tele.ntnu.no> wrote in message > > news:f56893ae.0309281531.34e1ee7f@posting.google.com... > > > >>Richard Owlett <rowlett@atlascomm.net> wrote in message > > > > news:<vnebb61nlodgb3@corp.supernews.com>... > > > >>>[ CAVEAT any term enclosed by "..." is used EXTREMELY loosely ] > >>> > >>>Given: > >>> 1. a non-periodic signal to be "characterized" > >>> [ "analyzed" implies much too much precision ;] > >> > >>You want just a basic "feel" of what's going on? > >> > >> > >>> 2. assumption that the data of interest is the relative "strength" > >>> of the signal in 3 to 5 non-overlapping frequency bands as > >>> a function of time. > >> > >>I'm thinking filter bank here... > >> > >> > >>> 3. A traditional workable solution would be to do a series of > >>> fouier/other transforms and "sum" the appropriate components. > >>> Anonther solution would be to filter the signal and calculate > >>> the "strength" of the passed signal, repeating for each passband > >>> and each time window. > >> > >>Sure. A Gabor plot, a Wigner distribution, wavelets... > >> > >> > >>>Constraint: minimize calculation/delay time. > >>> reproducibility > >> > >>OK, the Wigner distribution is out due to delays, wavelets... I'm not > > > > sure, > > > >>Gabor plots could work. > >> > >> > >>>NON-constraint that: > >>> function have an inverse!!!! > >>> math have any provable correlation to process variables > >> > >>Point to filter banks, wavelets appear to be way over the top. > >> > >> > >>>[ w is defined as omega of standard discourse ] > >>> > >>>Two possible solutions: > >>> > >>>If each passband of interest is decade(s) wide: > >>>Multiply incoming signal by sin(wt)*sin(3wt)*sin(5wt)...*sin([NMAX]wt) > >>>[ note multipiers of w are prime ] > >>>[ w -> [NMAX]w would define band of interest ] > >>>{ I hink I can take advantage of multiplication of sines having terms > >>> a function of initial frequencies as well as summ and difference > >>> frewuencies ) > >>> > >>>OR > >>> > >>>If there a M bands of interest, > >>>could the input be multiplied by M pink-noise signals. > >>>each pink noise signal would be limited to band of interest. > >> > >>Both solutions seem to be expensive: The sin-filter would have a very > >>long impulse response in time domain, if taken literally. You could do > >>what you want but with a simpler filter. The white noise sequence would > >>have to be generated and "spectrum shaped". A simple filter bank may do > >>the work for you sufficiently well and sufficiently fast. > >> > >>It may be the time of day, but "filter bank" has somehow stuck in mymind.> >> > >> > >>> -- > >>>I'm the one that theatended to do DSP in Basic [ I've done some in > >>>Excel. I am considering COBOL ;) > >> > >>I've seen worse. As long as you don't do DSP in real time with a slide > >>rule using roman numerals (no, I haven't seen that, but close enough), > >>you'll be OK. > >> > >>Rune > > > > > > Richard, > > > > We know that fft/multiply/ifft is computationally faster than doing a > > time-domain convolution for arrays of reasonable length. So, removingthe> > ifft is certainly even faster! > > So, why not do this: > > 1)- [design the bandpass filters of choice - assume they are FIRfilters]> > - fft the filter impulse responses / coefficient sequences. > > > > 2)- fft a block of data > > - multiply by each bandpass filter's fft > > - compute the sum of squares of magnitudes and take sqrt of the result > > (This yields a scalar measure of the signal contained in each of thebands> > of interest). > > Repeat (2) and plot the results for each filter. Overlap the blocks intime> > if desired. > > > > So, as Rune says, this implements a filter bank of a particular designand> > uses an efficient compute method. > > Multiplying by sines or sums of sines generates but one point in a > > convolution. I think you *have* to do the convolution in order to get > > reasonable results. > > > > Fred > > > > > > I think there are at least 2 sorces of confusion: > 1. my lack of understanding of digital FFT and filters > 2. my inability to communicate in writing - I'll not admit to how > many times I flunked freshman 'English' composition ;} > > As to #1, I understood that FFT required on order of "N SQUARED" > multiplication-addition operations. The approach I was considering > would require exactly N multiplication-addition operations per band of > interest. > > Fred stated "Multiplying by sines or sums of sines generates but one > point in a convolution." > I had assumed that. > I'm trying to test a hypothosis that says only one such point is > important/required/significant/???. > > Now we get into #2 ;[ > > Lets simplify/trivialize prolem by assumming: > a. only one band of interest running from f1 -> f2 > b. said band has a "random" set of frequencies > c. amplitude vs frequency is "relatively" flat and "continuous" > d. borrow concepts of matrix math > 1. input time domain data is an Nx1 matrix [A] > 2. my "product of sines" or "pink noise" is a 1xN matrix [B] > > My assertion to be tested is that the scalar (inner) product of > [A] dot [B] is significant. > > [B] is assumed fixed over durration of experiment. > > -------------------- > > a totaly separate way to look at my problem > > assume a white noise source > > it feeds 3 perfect bandpass filters > > The filters do not overlap > each filter supplies input to a perfect amplifier > [ gain is function of time but not ftrequency ] > [ time >>> 1/(any freq of interest) ] > > The output of each amplifier is summed equally > > my task might be defined as determining gain of each of the 3 > amplifiers as function of time > > Do either of these problem definitions make sense to anyone but me?Richard, Consider these responses:> Fred stated "Multiplying by sines or sums of sines generates but one > point in a convolution." > I had assumed that. > I'm trying to test a hypothosis that says only one such point is > important/required/significant/???.If this were the case, what would be the value of doing a convolution or correlation? Sometimes it's helpful to use a signal that is as simple as possible. So, if the signal is a sinusoid that's to be multiplied by a sinusoid of the same frequency in this approach, I think you will find that doing the product of the two sines of the same frequency with phase shift of pi/2, no matter the amplitudes, will result in a sum of zero over the product. So, the result is dependent on the phase - not the sort of result that you want. I don't follow the description using matrix notation. I understand A but not B. I would have said: A is the 1xN input sequence and we'll use a sinusoid of arbitrary phase as a simple test case B is an 1xN sinusoid sequence C is A.xB an 1xN product sequence [a1*b1 a2*b2 .... aN*bN] D is a Nx1 vector of Ones. CxD is a scalar sum of the elements of C. It can be zero. Fred
Reply by ●September 30, 20032003-09-30
Richard Owlett <rowlett@atlascomm.net> wrote in message news:<vnh9fbgn2mflae@corp.supernews.com>...> I think there are at least 2 sorces of confusion: > 1. my lack of understanding of digital FFT and filters > 2. my inability to communicate in writing - I'll not admit to how > many times I flunked freshman 'English' composition ;} > > As to #1, I understood that FFT required on order of "N SQUARED" > multiplication-addition operations.It's a bit more complicated than that. First, you need to choose a frame length N. Once N is chosen, the FFT costs N*log(N) mul-add computations.> The approach I was considering > would require exactly N multiplication-addition operations per band of > interest.Again, it's a bit more complicated. Depending on the type of filter (FIR, IIR) and the bandwidth, you need some M mul-add operations for each sample running through the filter. Generally speaking, FIR filters cost more than FIR filters (given similar, "reasonable" specs) and narrow-band filters cost more than broad-band filters. Depending on the filter type and specs, the FFT may or may not be cheaper than straight-forward implementations.> Fred stated "Multiplying by sines or sums of sines generates but one > point in a convolution." > I had assumed that. > I'm trying to test a hypothosis that says only one such point is > important/required/significant/???. > > Now we get into #2 ;[ > > Lets simplify/trivialize prolem by assumming: > a. only one band of interest running from f1 -> f2A lot depends on how wide this frequency band is. One measure of bandwidth is f1 + f2 Q = --------- 2(f2-f1) If this Q is "large", the filter is "narrow-band" and naive implementations cost a lot.> b. said band has a "random" set of frequencies > c. amplitude vs frequency is "relatively" flat and "continuous"Depending on the Q factor, these constraints appear to point towards broad-band filters.> d. borrow concepts of matrix math > 1. input time domain data is an Nx1 matrix [A] > 2. my "product of sines" or "pink noise" is a 1xN matrix [B] > > My assertion to be tested is that the scalar (inner) product of > [A] dot [B] is significant. > > [B] is assumed fixed over durration of experiment.The question implicit in Fred's post is how often this vector product is supposed to be computed. The convolution Fred mentions, supposes that this vector product is computed between every sample. My point is that depending on the specs, you may have similar results computed for every new sample, and with relatively few computations.> -------------------- > > a totaly separate way to look at my problem > > assume a white noise source > > it feeds 3 perfect bandpass filters > > The filters do not overlap > each filter supplies input to a perfect amplifier > [ gain is function of time but not ftrequency ] > [ time >>> 1/(any freq of interest) ] > > The output of each amplifier is summed equally > > my task might be defined as determining gain of each of the 3 > amplifiers as function of time > > Do either of these problem definitions make sense to anyone but me?It does make sense to me. I've used similar methods myself some time ago, but it would help immensely if you could give a hint on how often you need the A*B vector product to be computed, and the relative bandwidth (total bandwidth divided by center frequency, 1/Q above) of your filters. All this depend on the characteristics of the signal you process, the frame length you consider for the FFT, and the computational assets at your disposal. Rune
Reply by ●October 1, 20032003-10-01
"Rune Allnor" <allnor@tele.ntnu.no> wrote in message news:f56893ae.0309300944.1277d3b@posting.google.com...> Richard Owlett <rowlett@atlascomm.net> wrote in messagenews:<vnh9fbgn2mflae@corp.supernews.com>...> > I think there are at least 2 sorces of confusion: > > 1. my lack of understanding of digital FFT and filters > > 2. my inability to communicate in writing - I'll not admit to how > > many times I flunked freshman 'English' composition ;} > > > > As to #1, I understood that FFT required on order of "N SQUARED" > > multiplication-addition operations. > > It's a bit more complicated than that. First, you need to choose a frame > length N. Once N is chosen, the FFT costs N*log(N) mul-add computations. > > > The approach I was considering > > would require exactly N multiplication-addition operations per band of > > interest. > > Again, it's a bit more complicated. Depending on the type of filter (FIR, > IIR) and the bandwidth, you need some M mul-add operations for each > sample running through the filter. Generally speaking, FIR filters cost > more than FIR filters (given similar, "reasonable" specs) and narrow-band > filters cost more than broad-band filters. > > Depending on the filter type and specs, the FFT may or may not be cheaper > than straight-forward implementations. > > > Fred stated "Multiplying by sines or sums of sines generates but one > > point in a convolution." > > I had assumed that. > > I'm trying to test a hypothosis that says only one such point is > > important/required/significant/???. > > > > Now we get into #2 ;[ > > > > Lets simplify/trivialize prolem by assumming: > > a. only one band of interest running from f1 -> f2 > > A lot depends on how wide this frequency band is. One measure of > bandwidth is > > f1 + f2 > Q = --------- > 2(f2-f1) > > If this Q is "large", the filter is "narrow-band" and naiveimplementations> cost a lot. > > > b. said band has a "random" set of frequencies > > c. amplitude vs frequency is "relatively" flat and "continuous" > > Depending on the Q factor, these constraints appear to point towards > broad-band filters. > > > d. borrow concepts of matrix math > > 1. input time domain data is an Nx1 matrix [A] > > 2. my "product of sines" or "pink noise" is a 1xN matrix [B] > > > > My assertion to be tested is that the scalar (inner) product of > > [A] dot [B] is significant. > > > > [B] is assumed fixed over durration of experiment. > > The question implicit in Fred's post is how often this vector product is > supposed to be computed. The convolution Fred mentions, supposes that this > vector product is computed between every sample. My point is thatdepending> on the specs, you may have similar results computed for every new sample, > and with relatively few computations. > > > -------------------- > > > > a totaly separate way to look at my problem > > > > assume a white noise source > > > > it feeds 3 perfect bandpass filters > > > > The filters do not overlap > > each filter supplies input to a perfect amplifier > > [ gain is function of time but not ftrequency ] > > [ time >>> 1/(any freq of interest) ] > > > > The output of each amplifier is summed equally > > > > my task might be defined as determining gain of each of the 3 > > amplifiers as function of time > > > > Do either of these problem definitions make sense to anyone but me? > > It does make sense to me. I've used similar methods myself some time ago, > but it would help immensely if you could give a hint on how often you need > the A*B vector product to be computed, and the relative bandwidth > (total bandwidth divided by center frequency, 1/Q above) of your filters. > > All this depend on the characteristics of the signal you process, the > frame length you consider for the FFT, and the computational assets at > your disposal. > > RuneRune, It doesn't make sense to me yet because I don't understand what some of the operational steps are. 1) Feed white noise into 3 filters - OK 2) Filter outputs into a "perfect" amplifier - OK. 3) Gain (of the amplifier I guess) is a function of time - ????? Why? How? 4) The output of each amplifier is summed equally - I thought the objective was to find the relative signal strength in the "bands" which I understood to be "each band". This implies the objective is to find the total signal strength across all of the bands. If so - OK. 5) "determine the gain of each of the amplifiers? See (3). We've not talked about "linear superposition" even though it's mentioned in the OP. If the idea is to actually add the outputs of the filters as seems to be implied above, then linear phase *and* of the same slope for each filter is necessary methinks. Otherwise, the composite will not be representative of the spectral pieces of the input. Fred
Reply by ●October 1, 20032003-10-01
Extremely short answers to several questions [ have already worked 40 hrs in 3 days and expect another 13 tonight ]. Maximum Q on order of 1.5 Natural choice of frame size based on non-DSP quiteria ~4400. 4096 or 2048 would probaly be fine. Eventually I suspect that I would want to 'process' 10 to 50 frames/sec. Actually part of the experiment would be to determine how few are required. Initially not a major issue as I will be operating on a stored set of data. As FFT only costs Nlog(N) rather than N^2 changing my approach would be reasonable. The description of various pink noise sources feeding a summing amp having time variable gain per channel was meant to model the system being studied. In which case the project decription would be "given a black box of such a model, experimentally determine the gain of each of the amplifiers." More Friday when I wake up ;) Thank you.
Reply by ●October 1, 20032003-10-01
Rune Allnor wrote:> [snip]> > > Sure. A Gabor plot, [snip]Never heard of one so did Google search for "Gabor plot". Not much ;) But lead to searching "Gabor expansion" whose first hit was http://www.nari.ee.ethz.ch/commth/pubs/viewpub.php?ident=ms In the abstract it stated: "... Gabor expansion, one of the major linear time-frequency signal representations. The Gabor expansion, proposed by D. Gabor in 1946, is the decomposition of a signal into a set of time-shifted and modulated versions of an elementary window function." Given that I do not actually have the foggiest notion of what it means: It seems to "sort of" sound like what I might expect the description of what I'm trying to do would appear in a thesis. Can you point me to some elementary ( even simplistic ) description of either "Gabor plot" or "Gabor expansion" on the web? [One reference mentioned that in author's opinion not much was even available in print.]
Reply by ●October 1, 20032003-10-01
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:<DODeb.43$F4.17396@feed2.centurytel.net>...> "Rune Allnor" <allnor@tele.ntnu.no> wrote in message > news:f56893ae.0309300944.1277d3b@posting.google.com... > > Richard Owlett <rowlett@atlascomm.net> wrote in message > > > Do either of these problem definitions make sense to anyone but me? > > > > It does make sense to me....> Rune, > > It doesn't make sense to me yet because I don't understand what some of the > operational steps are. > 1) Feed white noise into 3 filters - OK > 2) Filter outputs into a "perfect" amplifier - OK. > 3) Gain (of the amplifier I guess) is a function of time - ????? Why? How? > 4) The output of each amplifier is summed equally - I thought the objective > was to find the relative signal strength in the "bands" which I understood > to be "each band". This implies the objective is to find the total signal > strength across all of the bands. If so - OK. > 5) "determine the gain of each of the amplifiers? See (3). > > We've not talked about "linear superposition" even though it's mentioned in > the OP. If the idea is to actually add the outputs of the filters as seems > to be implied above, then linear phase *and* of the same slope for each > filter is necessary methinks. Otherwise, the composite will not be > representative of the spectral pieces of the input. > > FredFred, If I read the posts literally, I'll agree with you. However, I do, for one reason or another, have the impression that the OP does not have a very profound knowledge of the intricate subtleties of the various aspects of DSP, including terminology, so I try to allow, rightly or wrongly, for that. If you go back to the first post, the analysis is not required to be very sophisticated. I suspect that a filter bank consisting of relatively simple low order IIR filters could help a lot. Only basic DSP knowledge required, no fancy stuff involved, fast run-time etc. I read the post basically as questions about implementations. Whether such crude analysis provides results that are useful in any way, is a completely different issue that only the user can decide. Sure, addressing power questions and signal reconstruction and so on is difficult stuff, but these things do not appear to be very important. Rune






