Hello I have a little question concerning the discrete wavelet transform! Let's consider one dimensional signal having a frequency up to 2kHz ( we consider that the sampling frequency is then 4kHz)which is filtered with low and highpass FIR filters, let this be the first decomposition level!! After passing the signal through a lowpass filter, half of the samples can be eliminated according to the Nyquist�s rule, since the signal now has a highest frequency of 1kHz instead of 2kHz. Simply discarding every other sample will subsample the signal by two, and the signal will then have half the number of points. The scale of the signal is now doubled Here comes my question: What do we mean actually when we say that the scale is doubled, it is a bit hard for me to understand how this scale is doubled since we are taking away every other sample!! Thanks for your time!!

# 1D Discrete wavelet transform(DWT)

Started by ●November 14, 2005

Posted by ●November 15, 2005

Umutesi Faith wrote:> Hello > > I have a little question concerning the discrete wavelet transform! > Let's consider one dimensional signal having a frequency up to 2kHz ( we > consider that the sampling frequency is then 4kHz)which is filtered with > low and highpass FIR filters, let this be the first decomposition level!! > > After passing the signal through a lowpass filter, half of the samples can > be eliminated according to the Nyquist�s rule, since the signalnow has a> highest frequency of 1kHz instead of 2kHz. Simply discarding every other > sample will subsample the signal by two, and the signal will then have > half the number of points. The scale of the signal is now doubled > > Here comes my question: > What do we mean actually when we say that the scale is doubled, it is a > bit hard for me to understand how this scale is doubled since we are > taking away every other sample!! > > Thanks for your time!!Consider an analogy: Take a photograph of the ground when you're standing on it. Your photograph covers 1 m^2. Now go up x meters and take another photo and the new image cover 4 m^2. Because the resolution of the image stays the same, you have discarded every other pixel from rows and every other from the columns of the image. But now you see the ground in a bigger scale. Oh, and there's no need for wavelets to comply with Nyquist rule. And they usually don't (really short FIRs). Aliasing happens but it cancels out at the syntesis end. -- ---- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 16, 2005

Thanks for the reply! So do u mean that the further we do the decomposition the higher gets the scale! The downsampling changes the scale,right! So what is affecting the resolution then, in the DWT? Actually this Nyquist rule refers especially to the discrete signal that we are using and i think it makes sense, since we are sampling it according the Nyquist theorem!! ************************************************************>Consider an analogy: Take a photograph of the ground when you're standingon>it. Your photograph covers 1 m^2. Now go up x meters and take anotherphoto>and the new image cover 4 m^2. Because the resolution of the image stays >the same, you have discarded every other pixel from rows and every other >from the columns of the image. But now you see the ground in a bigger >scale. > >Oh, and there's no need for wavelets to comply with Nyquist rule. Andthey>usually don't (really short FIRs). Aliasing happens but it cancels outat>the syntesis end. > >-- >---- >Jani Huhtanen >Tampere University of Technology, Pori >

Posted by ●November 16, 2005

Umutesi Faith wrote:> Thanks for the reply! > So do u mean that the further we do the decomposition the higher gets the > scale! The downsampling changes the scale,right! So what is affecting the > resolution then, in the DWT?Downsampling changes the scale, right. I'm not exactly sure what you mean by the term scale. In my previous post scale meant "how much more you see" when you downsample. So in this sence scale doubles when you downsample by two. You "see" twice as long signal. But in wavelet literature the term scale often refers to the level of decomposition and there is different ways of numbering the different scales. One may start with scale 0 as the original signal and scale 1 as the first decomposition level. Or scale K as original and K-1 as first level and 0 as the final level of decomposition. Resolution clearly gets lower each time we do the downsampling. This is because we loose the details in favor of seeing more of the signal. From dyadic decomposition (the basic DWT), we get "details" from every scale (level) and if the decomposition is carried infinitely far we could reconstruct the original signal from the details.> Actually this Nyquist rule refers especially to the discrete signal that > we are using and i think it makes sense, since we are sampling it > according the Nyquist theorem!! >The downsampling by two in DWT doesn't require that the incoming coefficients (or the signal at scale j) are bandlimited to fs/4, where fs is the sampling rate before downsampling. Take for example the simplest possible wavelet transform, lazy wavelet (use monospace font): x --->|D2|--w1-->|U2|--- | | |z^-1| |z^-1| | | -->|D2|--w2-->|U2|-> + -----> x D2 is downsampling by two (ie. discard every other sample) and U2 is upsampling by two. The diagram shows wavelet transform (analysis and synthesis) with only one level. Filters are only unit-delays so it's clear that frequency content stays intact. Signals w1 and w2 are highly aliased due to the downsampling. But at the summation alias components cancel out, resulting in perfect reconstruction altough the signals were not bandlimited. -- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 17, 2005

Sorry if this post comes twice. I sent the same post yesterday but it never appeared on the comp.dsp group, so perhaps I have better luck this time. Umutesi Faith wrote:> Thanks for the reply! > So do u mean that the further we do the decomposition the higher gets the > scale! The downsampling changes the scale,right! So what is affecting the > resolution then, in the DWT?Downsampling changes the scale, right. I'm not exactly sure what you mean by the term scale. In my previous post scale meant "how much more you see" when you downsample. So in this sence scale doubles when you downsample by two. You "see" twice as long signal. But in wavelet literature the term scale often refers to the level of decomposition and there is different ways of numbering the different scales. One may start with scale 0 as the original signal and scale 1 as the first decomposition level. Or scale K as original and K-1 as first level and 0 as the final level of decomposition. Resolution clearly gets lower each time we do the downsampling. This is because we loose the details in favor of seeing more of the signal. From dyadic decomposition (the basic DWT), we get "details" from every scale (level) and if the decomposition is carried infinitely far we could reconstruct the original signal from the details.> Actually this Nyquist rule refers especially to the discrete signal that > we are using and i think it makes sense, since we are sampling it > according the Nyquist theorem!! >The downsampling by two in DWT doesn't require that the incoming coefficients (or the signal at scale j) are bandlimited to fs/4, where fs is the sampling rate before downsampling. Take for example the simplest possible wavelet transform, lazy wavelet (use monospace font): x --->|D2|--w1-->|U2|--- | | |z^-1| |z^-1| | | -->|D2|--w2-->|U2|-> + -----> x D2 is downsampling by two (ie. discard every other sample) and U2 is upsampling by two. The diagram shows wavelet transform (analysis and synthesis) with only one level. Filters are only unit-delays so it's clear that frequency content stays intact. Signals w1 and w2 are highly aliased due to the downsampling. But at the summation alias components cancel out, resulting in perfect reconstruction altough the signals were not bandlimited. -- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 20, 2005

In my previous post scale meant "how much more you see">when you downsample. So in this sence scale doubles when you downsample- - - - - - - - - - - - - - - - - - - - - - - - - - - - Exactly, this was the point i wanted to understand,and i think you explained it well but it is really hard to see it when having 1D signal for instance of 64 samples and the downsampling gives 32 samples, this means then that we see " twice" as long signal, How?? The downsampling by two in DWT doesn't require that the incoming>coefficients (or the signal at scale j) are bandlimited to fs/4, wherefs>is the sampling rate before downsampling.- - - - - - - - - - - - - - - - - - - - - As far as i know, while doing 1D DWT ,the original signal has to be sampled according the Nyquist theorem ( fs = 2fmax) - - - - -- - - - -- - - - - - - -- - - - - By the way, i would like to load the daubechies decomposition filters coefficients, i have perfomed the following command load db4 ( in matlab command windonw) db4 = Columns 1 through 7 0.1629 0.5055 0.4461 -0.0198 -0.1323 0.0218 0.0233 Column 8 -0.0075 I got the result above. Are these coefficients for lowpass or highpass filters and how to find out what are the decomposition and synthesis coefficients. Is it correct to say the the lowpass filter is actually the scaling function and the highpass filter the wavelet function?? Kiitos!

Posted by ●November 20, 2005

Umutesi Faith wrote:> > In my previous post scale meant "how much more you see" >>when you downsample. So in this sence scale doubles when you downsample > > - - - - - - - - - - - - - - - - - - - - - - - - - - - - > Exactly, this was the point i wanted to understand,and i think you > explained it well but it is really hard to see it when having 1D signal > for instance of 64 samples and the downsampling gives 32 samples, this > means then that we see " twice" as long signal, How??You see more in sence that one sample of the downsampled signal corresponds to longer part of the original signal. So when you have limited window to the signal (ie. in the case of FIR filtering) then after downsampling you "see" longer part of the signal through the same window. Of course if the window is already of the same length than the original signal, you can't "see" more.> > > The downsampling by two in DWT doesn't require that the incoming >>coefficients (or the signal at scale j) are bandlimited to fs/4, where > fs >>is the sampling rate before downsampling. > > - - - - - - - - - - - - - - - - - - - - - > As far as i know, while doing 1D DWT ,the original signal has to be > sampled according the Nyquist theorem ( fs = 2fmax) >Actually this is somewhat false (altough true in practice ;)). DWT assumes that the orignal signal is "sampled" with the scaling function and the sampled signal is just scaling coefficients (ie. level 0 decomposition of the original continuous signal). Nyquist theorem assumes that the signal is "sampled" with the sinc function (or equivalently bandlimited to fs/2 and sampled with dirac delta). By saying that signal is "sampled" with function I mean that the original signal is described as a linear combination of time sifted functions and the samples are the weigths for the functions. If signal is sampled according to Nyquist theorem then the original signal can be reconstructed perfectly from it's samples by expressing it as a linear combination of sifted sinc-functions weighted with the samples. As far as I know, this does not necessarily hold for scaling-functions. But as I said, what you say is true in practice. We don't have ADCs that sample with scaling functions. So we often just treat the samples as level 0 scaling-coefficients (which is a "wavelet-crime" according to Strang&Nguyen). But what I _originally_ meant was simply that you don't have to design wavelet or scaling filters to bandlimit the signal before downsampling.> - - - - -- - - - -- - - - - - - -- - - - - > By the way, i would like to load the daubechies decomposition filters > coefficients, i have perfomed the following command > load db4 ( in matlab command windonw) > > db4 = > > Columns 1 through 7 > > 0.1629 0.5055 0.4461 -0.0198 -0.1323 0.0218 0.0233 > > Column 8 > > -0.0075 > > I got the result above. Are these coefficients for lowpass or highpass > filters and how to find out what are the decomposition and synthesis > coefficients.Well, you can always check whether it is lowpass by freqz(db4). (It's lowpass). Let H0(z) be the z-transform of the db4. Then the highpass filter (wavelet) H1(z) = z^-N*H0(z^-1). If H0 and H1 are analysis filters then F0 and F1 are synthesis filters given by: F0(z) = H1(-z) F1(z) = -H0(-z)> Is it correct to say the the lowpass filter is actually the > scaling function and the highpass filter the wavelet function?? >Scaling and wavelet functions are continuous-time altough they may not be continuous in the sence that it may not be differentiable at _any point_. But coefficients of the lowpass filter are the coefficient which define the scaling function and coefficients of the highpass filter are the coefficient which define the wavelet function. Scaling function is defined through so called dilation equation and wavelet function is defined through wavelet equation. More info: http://www.math.hmc.edu/faculty/ward/wavelets/pdfs/m185l2.pdf> Kiitos!Ole hyv�! :) -- ---- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 22, 2005

>More info: http://www.math.hmc.edu/faculty/ward/wavelets/pdfs/m185l2.pdfI have been at this link, well i guess the calculations of these coefficients are not as easy as i thought. It shows quite simple way of getting the scaling and wavelet coefficients instead S = (h0,h1,h2,h3) W = (h3,-h2,h1,-h0) So, does it mean that when i perform load dbN in matlab command window , what i get are the coefficients of the scaling function and from this i can obtain the coefficients of the wavelet function just according to the statement above.i have checked the freq(db4) it is a lowpass filter as you said! Once we have obtained a plot of DWT of a certain signal, where actually do we see this time-frequency representation? I hope you don't get bored with my questions. you really have a good knowledge about the wavelets! Is it possible for me to come to Tampere university to learn more about the wavelets? kiitos again!

Posted by ●November 22, 2005

Umutesi Faith wrote:>>More info: http://www.math.hmc.edu/faculty/ward/wavelets/pdfs/m185l2.pdf > > I have been at this link, well i guess the calculations of these > coefficients are not as easy as i thought. It shows quite simple way of > getting the scaling and wavelet coefficients instead > > S = (h0,h1,h2,h3) > W = (h3,-h2,h1,-h0) >I assumed that you were familiar with z-transform. Sorry about that. I'll try to explain what I meant. Here are the equations again (without the typo in H1): H0(z), analysis scaling filter (ie. S) H1(z) = (-z)^-N*H0(-z^-1), analysis wavelet filter (ie. W) F0(z) = H1(-z), synthesis scaling filter F1(z) = -H0(-z), synthesis wavelet filter Now lets say that the scaling coefficients are h0,h1,h2, and h3. Then the z-transform of the filter is H0(z) = h0 + h1*z^-1 + h2*z^-2 + h3*z^-3. OK, not let's use the equations above to derive the coefficients for other filters. First analysis wavelet: In our case N is 3 so, H1(z) = (-z)^-3 * H0(-z^-1) = (-z)^-3 * (h0 - h1*z^1 + h2*z^2 - h3*z^3) = -h0*z^-3 + h1*z^-2 - h2*z^-1 + h3. If you wonder why the signs change like that, it is simply because for example (-z)^-2 = z^-2 and (-z)^-1 = -z^-1. The exponent of the z tells us the place of the coefficient (0 corresponds to first coefficient and -1 to the second). => H1 = (h3, -h2, h1, -h0). Which is exactly the same that you said.> So, does it mean that when i perform load dbN in matlab command window , > what i get are the coefficients of the scaling function and from this i > can obtain the coefficients of the wavelet function just according to the > statement aboveYes. Oh, and if you're confused why db4 has more than 4 coefficients (as Daubechies 4 should have) it is because in matlab the number after db refers to the number of vanishing moments of the wavelet.> Once we have obtained a plot of DWT of a certain signal, where actually do > we see this time-frequency representation? I hope you don't get bored with > my questions.I assume you use dyadic-decomposition. Something like this (use monospace font): x---->|S|--a0------>|S|---a1------>|S|-->a2 | | | -->|W|->b0 -->|W|-->b1 -->|W|-->b2 That should look like three level dyadic decomposition. S is scaling filter and W is wavelet filter. For clarity I didn't include downsampling to the ASCII art. Output from the filter bank are the signals b0,b1,b2 and a2. If x is N samples long, then there is N/2 b0 samples, N/4 b1 samples, B/8 b2 and a2 samples (in total N samples of output). Samples of b0 are in time order (not in any mixed order). So are b1, b2 and a2. So clearly you have time representation. Further every time the signal goes through S it gets lowpass filtered. That is, it loses high-frequencies. This means that a2 contains low-frequency information from x. Another way of saying is that a2 is an approximation of x. b2 contains high frequency info of a1 which in turn is low frequency info of x. So b2 is a frequency band and so is b1. b0 has gone trough only highpass filter so it's the high-frequency info of x. One can think bs as the details of the x, which are added in synthesis to the approximation aN. Note that because of downsampling the bandwidth of b0 is fs/4 (where fs is sampling rate of x) and b1 is fs/8, b2 and a2 are fs/16. This is why this tree is also called octave decomposition (or something close to that, i'm not sure about the exact english term). So in this case when you plot the signals you should have plot like this: ------------------------- b0: | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | ------------------------- b1: | | | | | | | | | | | | | | ------------------------- b2: | | | | ------------------------- a2: | | | | ------------------------- The | marks are borders of one sample and - are borders of one "band". Now the time goes from left to right and frequency from top to bottom. Now its also easy to verify that low-frequency components (b2,a2) have lower time resolution than b1 or b2 but higher frequency resolution and vice-versa.> Is it possible for me > to come to Tampere university to learn more about the wavelets?Yes, if you are a citizen of Finland. Perhaps through some exchange program otherwise (or if you are planning on moving to Finland). Check from here to be sure: http://www.tut.fi/public/index.cfm?MainSel=1601&Sel=1601&Show=1497&Siteid=32 -- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 23, 2005

>http://www.tut.fi/public/index.cfm?Thanks for the link , i'll try to get more information about it! I have found some good books about the wavelets, maybe they are already familiar to you : Wavelet applications in Engineering Electromagnetics by Tapan K. Sarkar... Wavelet and filters banks by Gilbert Strang /Truong Nguyen I will now keep reading...

Posted by ●November 27, 2005

>Umutesi Faith wrote: > >> As in matlab we have more than four coefficients , how then do weextend>> this rule to more than four coefficients. > >By using the four equations I gave you. Generally H0(z) is > >H0(z) = h0 + h1*z^-1 + h2*z^-2 + ... + h{N-1}*z^-(N-1) >- - - - - -- - - - - - - - -- - - - - - - - - - - - - - - - Ok ,i'll use this formula and see what results i get for more than 4 coefficients, and apply it to the matlab (db4) coefficients! Thanks a lot for your time!!

Posted by ●November 27, 2005

Umutesi Faith wrote:> As in matlab we have more than four coefficients , how then do we extend > this rule to more than four coefficients.By using the four equations I gave you. Generally H0(z) is H0(z) = h0 + h1*z^-1 + h2*z^-2 + ... + h{N-1}*z^-(N-1)> > I have tried to find out the relationship between the coefficients of the > synthesis filters, I got the following results(?): > synthesis scaling filter ( h3,h2,h1,h0) > synthesis wavelet filter (-h0,h1,-h2,h3)Looks ok to me.> > By the way, if I recall correctly, from the z transform we get the > frequency response by just replacing z = e^jw where w is the angular > frequency. Do we have to worry about the cutoff frequency of these > filters?No. Only thing one has to be concerned of is perfect reconstruction and this is preseverved if the scaling filter (lowpass filter) has certain properties and the other three filters are calculated from the scaling filters trought the simple equations I gave you. When matlab or some other source gives you the coefficients for the scaling function you don't have to worry a thing.> > Accocrding the the decomposition rule these filters work as halfband > filter which means that after the first level I get a lowfrequency part (0 > to1khz) and highfrequency part(1kHz to 2kHz) for a 2kHz signal for > instance. It could be quite diff�cult to desing own quadraturemirror> filters, so i hope if i use these matlab coefficients it could do the > same job!Actually designing new wavelets is quite easy, but finding wavelets that perform well for a given task is hard. If you are really intrested in wavelets I propose you read papers by Wim Sweldens (http://cm.bell-labs.com/who/wim/papers/papers.html) and especially this: http://cm.bell-labs.com/who/wim/papers/athome.pdf. Lifting is a way to implement dwt efficiently and losslessly (even in the precense of quantization errors) and Sweldens papers gets you started. -- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 26, 2005

H0(z), analysis scaling filter (ie. S) H1(z) = (-z)^-N*H0(-z^-1), analysis wavelet filter (ie. W) F0(z) = H1(-z), synthesis scaling filter F1(z) = -H0(-z), synthesis wavelet filter Now lets say that the scaling coefficients are h0,h1,h2, and h3. Then the z-transform of the filter is H0(z) = h0 + h1*z^-1 + h2*z^-2 + h3*z^-3. Yes. Oh, and if you're confused why db4 has more than 4 coefficients (as Daubechies 4 should have) it is because in matlab the number after db refers to the number of vanishing moments of the wavelet. ----- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - As in matlab we have more than four coefficients , how then do we extend this rule to more than four coefficients. S = (h0,h1,h2,h3) W = (h3,-h2,h1,-h0) I have tried to find out the relationship between the coefficients of the synthesis filters, I got the following results(?): synthesis scaling filter ( h3,h2,h1,h0) synthesis wavelet filter (-h0,h1,-h2,h3) By the way, if I recall correctly, from the z transform we get the frequency response by just replacing z = e^jw where w is the angular frequency. Do we have to worry about the cutoff frequency of these filters? Accocrding the the decomposition rule these filters work as halfband filter which means that after the first level I get a lowfrequency part (0 to1khz) and highfrequency part(1kHz to 2kHz) for a 2kHz signal for instance. It could be quite diff�cult to desing own quadrature mirror filters, so i hope if i use these matlab coefficients it could do the same job! Thanks!

Posted by ●November 23, 2005

>http://www.tut.fi/public/index.cfm?Thanks for the link , i'll try to get more information about it! I have found some good books about the wavelets, maybe they are already familiar to you : Wavelet applications in Engineering Electromagnetics by Tapan K. Sarkar... Wavelet and filters banks by Gilbert Strang /Truong Nguyen I will now keep reading...

Posted by ●November 22, 2005

Umutesi Faith wrote:>>More info: http://www.math.hmc.edu/faculty/ward/wavelets/pdfs/m185l2.pdf > > I have been at this link, well i guess the calculations of these > coefficients are not as easy as i thought. It shows quite simple way of > getting the scaling and wavelet coefficients instead > > S = (h0,h1,h2,h3) > W = (h3,-h2,h1,-h0) >I assumed that you were familiar with z-transform. Sorry about that. I'll try to explain what I meant. Here are the equations again (without the typo in H1): H0(z), analysis scaling filter (ie. S) H1(z) = (-z)^-N*H0(-z^-1), analysis wavelet filter (ie. W) F0(z) = H1(-z), synthesis scaling filter F1(z) = -H0(-z), synthesis wavelet filter Now lets say that the scaling coefficients are h0,h1,h2, and h3. Then the z-transform of the filter is H0(z) = h0 + h1*z^-1 + h2*z^-2 + h3*z^-3. OK, not let's use the equations above to derive the coefficients for other filters. First analysis wavelet: In our case N is 3 so, H1(z) = (-z)^-3 * H0(-z^-1) = (-z)^-3 * (h0 - h1*z^1 + h2*z^2 - h3*z^3) = -h0*z^-3 + h1*z^-2 - h2*z^-1 + h3. If you wonder why the signs change like that, it is simply because for example (-z)^-2 = z^-2 and (-z)^-1 = -z^-1. The exponent of the z tells us the place of the coefficient (0 corresponds to first coefficient and -1 to the second). => H1 = (h3, -h2, h1, -h0). Which is exactly the same that you said.> So, does it mean that when i perform load dbN in matlab command window , > what i get are the coefficients of the scaling function and from this i > can obtain the coefficients of the wavelet function just according to the > statement aboveYes. Oh, and if you're confused why db4 has more than 4 coefficients (as Daubechies 4 should have) it is because in matlab the number after db refers to the number of vanishing moments of the wavelet.> Once we have obtained a plot of DWT of a certain signal, where actually do > we see this time-frequency representation? I hope you don't get bored with > my questions.I assume you use dyadic-decomposition. Something like this (use monospace font): x---->|S|--a0------>|S|---a1------>|S|-->a2 | | | -->|W|->b0 -->|W|-->b1 -->|W|-->b2 That should look like three level dyadic decomposition. S is scaling filter and W is wavelet filter. For clarity I didn't include downsampling to the ASCII art. Output from the filter bank are the signals b0,b1,b2 and a2. If x is N samples long, then there is N/2 b0 samples, N/4 b1 samples, B/8 b2 and a2 samples (in total N samples of output). Samples of b0 are in time order (not in any mixed order). So are b1, b2 and a2. So clearly you have time representation. Further every time the signal goes through S it gets lowpass filtered. That is, it loses high-frequencies. This means that a2 contains low-frequency information from x. Another way of saying is that a2 is an approximation of x. b2 contains high frequency info of a1 which in turn is low frequency info of x. So b2 is a frequency band and so is b1. b0 has gone trough only highpass filter so it's the high-frequency info of x. One can think bs as the details of the x, which are added in synthesis to the approximation aN. Note that because of downsampling the bandwidth of b0 is fs/4 (where fs is sampling rate of x) and b1 is fs/8, b2 and a2 are fs/16. This is why this tree is also called octave decomposition (or something close to that, i'm not sure about the exact english term). So in this case when you plot the signals you should have plot like this: ------------------------- b0: | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | ------------------------- b1: | | | | | | | | | | | | | | ------------------------- b2: | | | | ------------------------- a2: | | | | ------------------------- The | marks are borders of one sample and - are borders of one "band". Now the time goes from left to right and frequency from top to bottom. Now its also easy to verify that low-frequency components (b2,a2) have lower time resolution than b1 or b2 but higher frequency resolution and vice-versa.> Is it possible for me > to come to Tampere university to learn more about the wavelets?Yes, if you are a citizen of Finland. Perhaps through some exchange program otherwise (or if you are planning on moving to Finland). Check from here to be sure: http://www.tut.fi/public/index.cfm?MainSel=1601&Sel=1601&Show=1497&Siteid=32 -- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 22, 2005

>More info: http://www.math.hmc.edu/faculty/ward/wavelets/pdfs/m185l2.pdfI have been at this link, well i guess the calculations of these coefficients are not as easy as i thought. It shows quite simple way of getting the scaling and wavelet coefficients instead S = (h0,h1,h2,h3) W = (h3,-h2,h1,-h0) So, does it mean that when i perform load dbN in matlab command window , what i get are the coefficients of the scaling function and from this i can obtain the coefficients of the wavelet function just according to the statement above.i have checked the freq(db4) it is a lowpass filter as you said! Once we have obtained a plot of DWT of a certain signal, where actually do we see this time-frequency representation? I hope you don't get bored with my questions. you really have a good knowledge about the wavelets! Is it possible for me to come to Tampere university to learn more about the wavelets? kiitos again!

Posted by ●November 20, 2005

Umutesi Faith wrote:> > In my previous post scale meant "how much more you see" >>when you downsample. So in this sence scale doubles when you downsample > > - - - - - - - - - - - - - - - - - - - - - - - - - - - - > Exactly, this was the point i wanted to understand,and i think you > explained it well but it is really hard to see it when having 1D signal > for instance of 64 samples and the downsampling gives 32 samples, this > means then that we see " twice" as long signal, How??You see more in sence that one sample of the downsampled signal corresponds to longer part of the original signal. So when you have limited window to the signal (ie. in the case of FIR filtering) then after downsampling you "see" longer part of the signal through the same window. Of course if the window is already of the same length than the original signal, you can't "see" more.> > > The downsampling by two in DWT doesn't require that the incoming >>coefficients (or the signal at scale j) are bandlimited to fs/4, where > fs >>is the sampling rate before downsampling. > > - - - - - - - - - - - - - - - - - - - - - > As far as i know, while doing 1D DWT ,the original signal has to be > sampled according the Nyquist theorem ( fs = 2fmax) >Actually this is somewhat false (altough true in practice ;)). DWT assumes that the orignal signal is "sampled" with the scaling function and the sampled signal is just scaling coefficients (ie. level 0 decomposition of the original continuous signal). Nyquist theorem assumes that the signal is "sampled" with the sinc function (or equivalently bandlimited to fs/2 and sampled with dirac delta). By saying that signal is "sampled" with function I mean that the original signal is described as a linear combination of time sifted functions and the samples are the weigths for the functions. If signal is sampled according to Nyquist theorem then the original signal can be reconstructed perfectly from it's samples by expressing it as a linear combination of sifted sinc-functions weighted with the samples. As far as I know, this does not necessarily hold for scaling-functions. But as I said, what you say is true in practice. We don't have ADCs that sample with scaling functions. So we often just treat the samples as level 0 scaling-coefficients (which is a "wavelet-crime" according to Strang&Nguyen). But what I _originally_ meant was simply that you don't have to design wavelet or scaling filters to bandlimit the signal before downsampling.> - - - - -- - - - -- - - - - - - -- - - - - > By the way, i would like to load the daubechies decomposition filters > coefficients, i have perfomed the following command > load db4 ( in matlab command windonw) > > db4 = > > Columns 1 through 7 > > 0.1629 0.5055 0.4461 -0.0198 -0.1323 0.0218 0.0233 > > Column 8 > > -0.0075 > > I got the result above. Are these coefficients for lowpass or highpass > filters and how to find out what are the decomposition and synthesis > coefficients.Well, you can always check whether it is lowpass by freqz(db4). (It's lowpass). Let H0(z) be the z-transform of the db4. Then the highpass filter (wavelet) H1(z) = z^-N*H0(z^-1). If H0 and H1 are analysis filters then F0 and F1 are synthesis filters given by: F0(z) = H1(-z) F1(z) = -H0(-z)> Is it correct to say the the lowpass filter is actually the > scaling function and the highpass filter the wavelet function?? >Scaling and wavelet functions are continuous-time altough they may not be continuous in the sence that it may not be differentiable at _any point_. But coefficients of the lowpass filter are the coefficient which define the scaling function and coefficients of the highpass filter are the coefficient which define the wavelet function. Scaling function is defined through so called dilation equation and wavelet function is defined through wavelet equation. More info: http://www.math.hmc.edu/faculty/ward/wavelets/pdfs/m185l2.pdf> Kiitos!Ole hyv�! :) -- ---- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 20, 2005

In my previous post scale meant "how much more you see">when you downsample. So in this sence scale doubles when you downsample- - - - - - - - - - - - - - - - - - - - - - - - - - - - Exactly, this was the point i wanted to understand,and i think you explained it well but it is really hard to see it when having 1D signal for instance of 64 samples and the downsampling gives 32 samples, this means then that we see " twice" as long signal, How?? The downsampling by two in DWT doesn't require that the incoming>coefficients (or the signal at scale j) are bandlimited to fs/4, wherefs>is the sampling rate before downsampling.- - - - - - - - - - - - - - - - - - - - - As far as i know, while doing 1D DWT ,the original signal has to be sampled according the Nyquist theorem ( fs = 2fmax) - - - - -- - - - -- - - - - - - -- - - - - By the way, i would like to load the daubechies decomposition filters coefficients, i have perfomed the following command load db4 ( in matlab command windonw) db4 = Columns 1 through 7 0.1629 0.5055 0.4461 -0.0198 -0.1323 0.0218 0.0233 Column 8 -0.0075 I got the result above. Are these coefficients for lowpass or highpass filters and how to find out what are the decomposition and synthesis coefficients. Is it correct to say the the lowpass filter is actually the scaling function and the highpass filter the wavelet function?? Kiitos!

Posted by ●November 17, 2005

Sorry if this post comes twice. I sent the same post yesterday but it never appeared on the comp.dsp group, so perhaps I have better luck this time. Umutesi Faith wrote:> Thanks for the reply! > So do u mean that the further we do the decomposition the higher gets the > scale! The downsampling changes the scale,right! So what is affecting the > resolution then, in the DWT?Downsampling changes the scale, right. I'm not exactly sure what you mean by the term scale. In my previous post scale meant "how much more you see" when you downsample. So in this sence scale doubles when you downsample by two. You "see" twice as long signal. But in wavelet literature the term scale often refers to the level of decomposition and there is different ways of numbering the different scales. One may start with scale 0 as the original signal and scale 1 as the first decomposition level. Or scale K as original and K-1 as first level and 0 as the final level of decomposition. Resolution clearly gets lower each time we do the downsampling. This is because we loose the details in favor of seeing more of the signal. From dyadic decomposition (the basic DWT), we get "details" from every scale (level) and if the decomposition is carried infinitely far we could reconstruct the original signal from the details.> Actually this Nyquist rule refers especially to the discrete signal that > we are using and i think it makes sense, since we are sampling it > according the Nyquist theorem!! >The downsampling by two in DWT doesn't require that the incoming coefficients (or the signal at scale j) are bandlimited to fs/4, where fs is the sampling rate before downsampling. Take for example the simplest possible wavelet transform, lazy wavelet (use monospace font): x --->|D2|--w1-->|U2|--- | | |z^-1| |z^-1| | | -->|D2|--w2-->|U2|-> + -----> x D2 is downsampling by two (ie. discard every other sample) and U2 is upsampling by two. The diagram shows wavelet transform (analysis and synthesis) with only one level. Filters are only unit-delays so it's clear that frequency content stays intact. Signals w1 and w2 are highly aliased due to the downsampling. But at the summation alias components cancel out, resulting in perfect reconstruction altough the signals were not bandlimited. -- Jani Huhtanen Tampere University of Technology, Pori

Posted by ●November 16, 2005

Umutesi Faith wrote: