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From: robert bristow-johnson on 30 Jul 2010 01:37 On Jul 29, 8:29 pm, Vladimir Vassilevsky <nos...(a)nowhere.com> wrote: > > No, you didn't get the idea. > > If the signal can be approximated as polynomial x = P(t), and > nonlinearity is also polynomial F(x) static, then you can find Nyquist > bandlimited Q(t) = F(P(t)) as the closed form solution. No filtering, no > upsampling, only algebra. > i now get what you're saying. let's assume that this is real-time and you have the current sample and the L-1 previous samples. to approximate a the recent segment of signal as a polynomial (of order L-1) is a mess if you want to get the traditional coefficients. if you leave it in the Lagrange factored form, it's not so messy, but what can you do with it? running it (in either form) into a static polynomial F(x) is gonna be a mess and will increase the order and the implied bandlimit. just evaluating it for the specific sample times is like sampling that high order polynomial at sample times. but that compound polynomial will have frequency components much greater than you Nyquist frequency unless you do this at the upsampled rate. so the images get folded over into your baseband and you can't do anything about them. so besides messy, i think there is an aliasing problem with the idea. r b-j
From: Vladimir Vassilevsky on 30 Jul 2010 10:39 robert bristow-johnson wrote: > On Jul 29, 8:29 pm, Vladimir Vassilevsky <nos...(a)nowhere.com> wrote: > >>No, you didn't get the idea. >> >>If the signal can be approximated as polynomial x = P(t), and >>nonlinearity is also polynomial F(x) static, then you can find Nyquist >>bandlimited Q(t) = F(P(t)) as the closed form solution. No filtering, no >>upsampling, only algebra. >> > i now get what you're saying. let's assume that this is real-time and > you have the current sample and the L-1 previous samples. to > approximate a the recent segment of signal as a polynomial (of order > L-1) is a mess if you want to get the traditional coefficients. if > you leave it in the Lagrange factored form, it's not so messy, but > what can you do with it? running it (in either form) into a static > polynomial F(x) is gonna be a mess and will increase the order and the > implied bandlimit. just evaluating it for the specific sample times > is like sampling that high order polynomial at sample times. but that > compound polynomial will have frequency components much greater than > you Nyquist frequency unless you do this at the upsampled rate. so > the images get folded over into your baseband and you can't do > anything about them. > > so besides messy, i think there is an aliasing problem with the idea. No there is no problem with aliasing. That's the whole point! As F(P(t)) is polynomial, you have Fourier(F(P(t)) as *closed* form solution. So you can drop the frequencies higher then Nyquist, and then do inverse Fourier. Inverse Fourier also has the closed form solution in this case. So the whole algorithm is pure algebra. Yes this is messy, but it is doable. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
From: rickman on 30 Jul 2010 11:52 On Jul 30, 10:39 am, Vladimir Vassilevsky <nos...(a)nowhere.com> wrote: > robert bristow-johnson wrote: > > On Jul 29, 8:29 pm, Vladimir Vassilevsky <nos...(a)nowhere.com> wrote: > > >>No, you didn't get the idea. > > >>If the signal can be approximated as polynomial x = P(t), and > >>nonlinearity is also polynomial F(x) static, then you can find Nyquist > >>bandlimited Q(t) = F(P(t)) as the closed form solution. No filtering, no > >>upsampling, only algebra. > > > i now get what you're saying. let's assume that this is real-time and > > you have the current sample and the L-1 previous samples. to > > approximate a the recent segment of signal as a polynomial (of order > > L-1) is a mess if you want to get the traditional coefficients. if > > you leave it in the Lagrange factored form, it's not so messy, but > > what can you do with it? running it (in either form) into a static > > polynomial F(x) is gonna be a mess and will increase the order and the > > implied bandlimit. just evaluating it for the specific sample times > > is like sampling that high order polynomial at sample times. but that > > compound polynomial will have frequency components much greater than > > you Nyquist frequency unless you do this at the upsampled rate. so > > the images get folded over into your baseband and you can't do > > anything about them. > > > so besides messy, i think there is an aliasing problem with the idea. > > No there is no problem with aliasing. That's the whole point! As F(P(t)) > is polynomial, you have Fourier(F(P(t)) as *closed* form solution. So > you can drop the frequencies higher then Nyquist, and then do inverse > Fourier. Inverse Fourier also has the closed form solution in this case. > So the whole algorithm is pure algebra. > Yes this is messy, but it is doable. > > Vladimir Vassilevsky > DSP and Mixed Signal Design Consultanthttp://www.abvolt.com Yes, Vlad is right. Once you have the polynomials and you multiply them, the resulting equation has terms representing signal within the Nyquist band and others representing signal outside the Nyquist band. As he said, you just drop the terms representing signal outside the Nyquist band... That reminds me of a joke a friend told me. An engineer, a chemist and an economist are shipwrecked on a desert island. They have no food except for a case of beans that washed up from the shipwreck. Wanting to open the cans, the engineer says they can build a fire to heat the cans and let the pressure build until the can bursts, but that would spew most of the contents into the fire and dirt. The chemist suggests that they let the cans soak in the salt water corroding them until the metal is weak enough to open them by hand, but that would likely create holes that let in the sea water spoiling the food. The economists says, "Assume we have a can opener"... My friend was an economist and had just explained how nearly all of economic theory is based on assumptions like that everyone has perfect knowledge and will make the decisions that maximize their personal utility. Now do you get it? "Assume we have polynomials..." Rick
From: rickman on 30 Jul 2010 12:03 On Jul 29, 10:22 pm, Fred Marshall <fmarshall_xremove_the...(a)xacm.org> wrote: > Dirk Bruere at NeoPax wrote: > > > > > So interpolating to 96kHz will not preserve any extra information > > during subsequent arithmetic? > > It might. Let's see: > > Assume that the sampling is done such that the bandwidth is pretty much > used up. i.e. there's no guard band of little energy around fs/2. > > Now, what happens when you interpolate? > You stuff zeros in time to double the sample rate. > You lowpass filter at double the sample rate to remove the components > from fs/2 to 3fs/2. Because of the initial assumption about lack of > guard band, you may lose some of the high frequencies in doing this. > Depends on the filter. Either that or you may retain some of the "new" > stuff above fs/2 and below 3fs/2. > > Assuming that this can be done perfectly though, then you have more > guard band to play with. The question is, can you take advantage of it > somehow? > > Let's take a wild leap and say that you can. This suggests that there > is energy above the original fs/2. Maybe it's in quantization noise > coming out of computations and subsequent rounding. And, maybe you can > filter that out..... But you already have 24 bits so I can't begin to > predict. > > Then, presumably you want to get back down to 48kHz, right? > This means you have to lowpass again to fs/4 or the original fs/2. > It might be useful (as above) but it's too complicated to comment on ... > for me! > > Anyone done this to advantage? > > Fred I don't see an advantage to doing any of this. In fact, there are processing disadvantage to upsampling. For one, a low pass digital filter requires more coefficients to get the same transition if the sample rate is higher, not to mention that you have to process more samples, unless you are downsampling at the same time. Other processing will take longer just because of the higher sample rate. If any of your processing is non-linear any components above the audible range can alias into the audible range. One of the advantages of digital sampling is that once it is in a digital signal it can be processed perfectly (other than the introduction of quantizing and round off noise). You don't need to have a higher sample rate for the processing. You can however get a better signal out of an ADC at a higher sample rate, but it depends on what is meant by "better". The same part run at a higher rate often has a poorer SNR. There must be some reason why higher sample rates are better. Many digital audio products can run at 96 kHz or even 192 kHz. Rick
From: robert bristow-johnson on 30 Jul 2010 12:23
On Jul 30, 11:52 am, rickman <gnu...(a)gmail.com> wrote: > On Jul 30, 10:39 am, Vladimir Vassilevsky <nos...(a)nowhere.com> wrote: > > > > > robert bristow-johnson wrote: > > > On Jul 29, 8:29 pm, Vladimir Vassilevsky <nos...(a)nowhere.com> wrote: > > > >>No, you didn't get the idea. > > > >>If the signal can be approximated as polynomial x = P(t), and > > >>nonlinearity is also polynomial F(x) static, then you can find Nyquist > > >>bandlimited Q(t) = F(P(t)) as the closed form solution. No filtering, no > > >>upsampling, only algebra. > > > > i now get what you're saying. let's assume that this is real-time and > > > you have the current sample and the L-1 previous samples. to > > > approximate a the recent segment of signal as a polynomial (of order > > > L-1) is a mess if you want to get the traditional coefficients. if > > > you leave it in the Lagrange factored form, it's not so messy, but > > > what can you do with it? running it (in either form) into a static > > > polynomial F(x) is gonna be a mess and will increase the order and the > > > implied bandlimit. just evaluating it for the specific sample times > > > is like sampling that high order polynomial at sample times. but that > > > compound polynomial will have frequency components much greater than > > > you Nyquist frequency unless you do this at the upsampled rate. so > > > the images get folded over into your baseband and you can't do > > > anything about them. > > > > so besides messy, i think there is an aliasing problem with the idea. > > > No there is no problem with aliasing. That's the whole point! well, the whole point is in question. there *is* a problem with aliasing (unless you sufficiently upsample before applying the F(x) polynomial). > > As F(P(t)) is polynomial, of what order? > > you have Fourier(F(P(t)) as *closed* form solution. no dispute here. but what *is* that polynomial and where (or how often) is it sampled? > > So you can drop the frequencies higher then Nyquist, how do you do that without upsampling? without upsampling there *are* no frequencies higher than Nyquist. > > and then do inverse Fourier. > > Inverse Fourier also has the closed form solution in this case. > > So the whole algorithm is pure algebra. > > Yes this is messy, but it is doable. let's set aside the messy, but get very specific about the doable thing that is done. > Yes, Vlad is right. Once you have the polynomials and you multiply > them, the resulting equation has terms representing signal within the > Nyquist band and others representing signal outside the Nyquist band. > As he said, you just drop the terms representing signal outside the > Nyquist band... > .... > Now do you get it? "Assume we have polynomials..." to "get it", let's pretend we have a painless, roundoff error free, "polyfit()" function where we can get the coefficients of the polynomial that passes through whatever set of uniform samples we give it. assume no mess, no muss, no fuss. we begin with the L most current samples, x[n] to x[n-L+1]. let's define the sample interval, T, to be 1. so, what is the algorithm? do we begin with an (L-1)th-order polynomial that hits all of those L samples? is that correct? if so, let's call that polynomial L-1 p_n(t) = SUM{ a_(k,n) * t^k } k=0 and that p(i) = x[n+i] for integers -L < i <= 0 . and getting the coefficients a_(k,n) is no problem. now what do we do with it? what is the next step? r b-j |