Hi, I'd like an explanation of the PSD of a sequence of independent random PAM symbols or amplitude +1 or -1. Now any standard textbook tells us that the PSD is the Fourier Transform of the autocorrelation of the signal, in my case the symbols are INDEPENDENT and => the autocorrelation is non-zero only over the symbol duration 'T', beyond this it goes to zero. Going through the maths I come out with and answer PSD = sinc^2(wT/2), it is basically just the square of the ESD of a rectangular pulse. Again any standard textbook confirms that this is the 'correct' answer. Here is my problem though; sinc^2(wT/2) has a maximum at w = 0, ie it is saying that the most powerful component of the signal is at DC. But if I simply take the mean of my sequence I get zero (the symbols are RANDOM +/-1 and => the average is zero). So, my signal has mean zero (=> zero DC component) yet the accepted 'correct' PSD for this signal implies that its most powerful component is at D.C. Can anyone explain how this can be correct ? Thanks, Usjes. --------------------------------------- Posted through http://www.DSPRelated.com
PSD of a sequence of independent random PAM symbols ?
Started by ●May 24, 2015
Reply by ●May 24, 20152015-05-24
On 24.05.15 23.54, Usjes wrote:> I'd like an explanation of the PSD of a sequence of independent random PAM > symbols or amplitude +1 or -1. Now any standard textbook tells us that the > PSD is the Fourier Transform of the autocorrelation of the signal, in my > case the symbols are INDEPENDENT and => the autocorrelation is non-zero > only over the symbol duration 'T', beyond this it goes to zero. Going > through the maths I come out with and answer PSD = sinc^2(wT/2), it is > basically just the square of the ESD of a rectangular pulse. Again any > standard textbook confirms that this is the 'correct' answer. Here is my > problem though; sinc^2(wT/2) has a maximum at w = 0, ie it is saying that > the most powerful component of the signal is at DC. But if I simply take > the mean of my sequence I get zero (the symbols are RANDOM +/-1 and => the > average is zero). So, my signal has mean zero (=> zero DC component) yet > the accepted 'correct' PSD for this signal implies that its most powerful > component is at D.C. Can anyone explain how this can be correct ?Probably they took the symbols 1 and 0 rather than -1. Marcel
Reply by ●May 24, 20152015-05-24
On Sunday, May 24, 2015 at 2:54:32 PM UTC-7, Usjes wrote:> Hi, > > I'd like an explanation of the PSD of a sequence of independent random PAM > symbols or amplitude +1 or -1. Now any standard textbook tells us that the > PSD is the Fourier Transform of the autocorrelation of the signal, in my > case the symbols are INDEPENDENT and => the autocorrelation is non-zero > only over the symbol duration 'T', beyond this it goes to zero.... So, my signal has mean zero (=> zero DC component) yet> the accepted 'correct' PSD for this signal implies that its most powerful > component is at D.C. Can anyone explain how this can be correct ? > > Thanks, > > Usjes. > --------------------------------------- > Posted through http://www.DSPRelated.comYou say the PSD is the FT of the autocorrelation. The autocorrelation is positive whether the signal is +1 or -1. Dale B. Dalrymple
Reply by ●May 25, 20152015-05-25
On Sun, 24 May 2015 16:54:30 -0500, Usjes wrote:> Hi, > > I'd like an explanation of the PSD of a sequence of independent random > PAM symbols or amplitude +1 or -1. Now any standard textbook tells us > that the PSD is the Fourier Transform of the autocorrelation of the > signal, in my case the symbols are INDEPENDENT and => the > autocorrelation is non-zero only over the symbol duration 'T', beyond > this it goes to zero. Going through the maths I come out with and answer > PSD = sinc^2(wT/2), it is basically just the square of the ESD of a > rectangular pulse. Again any standard textbook confirms that this is the > 'correct' answer. Here is my problem though; sinc^2(wT/2) has a maximum > at w = 0, ie it is saying that the most powerful component of the signal > is at DC. But if I simply take the mean of my sequence I get zero (the > symbols are RANDOM +/-1 and => the average is zero). So, my signal has > mean zero (=> zero DC component) yet the accepted 'correct' PSD for this > signal implies that its most powerful component is at D.C. Can anyone > explain how this can be correct ?The power spectral _density_ is maximum at DC, but finite. The spectral width of _absolutely pure_ DC is zero. (Some finite number) * 0 = 0. -- www.wescottdesign.com
Reply by ●May 25, 20152015-05-25
> >Probably they took the symbols 1 and 0 rather than -1. > > >MarcelNo, the sequence using symbols +/-1 has PSD max at frequency 0. A sequence using symbols 0,1 would also have a PSD max at frequency zero, but it is not so strange in that case. --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●May 25, 20152015-05-25
>On Sunday, May 24, 2015 at 2:54:32 PM UTC-7, Usjes wrote: >> Hi, >> >> I'd like an explanation of the PSD of a sequence of independent random >PAM >> symbols or amplitude +1 or -1. Now any standard textbook tells us that >the >> PSD is the Fourier Transform of the autocorrelation of the signal, in >my >> case the symbols are INDEPENDENT and => the autocorrelation is >non-zero >> only over the symbol duration 'T', beyond this it goes to zero. >... > So, my signal has mean zero (=> zero DC component) yet >> the accepted 'correct' PSD for this signal implies that its most >powerful >> component is at D.C. Can anyone explain how this can be correct ? >> >> Thanks, >> >> Usjes. >> --------------------------------------- >> Posted through http://www.DSPRelated.com > >You say the PSD is the FT of the autocorrelation. The autocorrelation is >positive whether the signal is +1 or -1. > >Dale B. DalrympleIndeed, the mathematics is correct, I know 'cos I did it myself :) This doesn't answer my question, PSD is usually interpreted as giving the components of the signal at different frequencies. In this case it seems to be implying that the most powerful component of the signal is at DC yet taking the average shows that the component at that frequency is zero. Hence my confusion. --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●May 25, 20152015-05-25
>On Sun, 24 May 2015 16:54:30 -0500, Usjes wrote: >> >The power spectral _density_ is maximum at DC, but finite. > >The spectral width of _absolutely pure_ DC is zero. > >(Some finite number) * 0 = 0. > >-- >www.wescottdesign.comHi Tim, Yes, this was the only explanation I could think of myself but I'm not entirely happy with it as it essentially seems to be saying the signal has no component at any frequency. For example, supposing I pass it through a filter with an infinitely narrow notch at DC, will the output of the filter be the same as its input given that I have only removed the content where there was none (signal mean = 0)? Now what if I repeat this with a cascade of an infinite number of such filters each at different frequencies, at the very end of this cascade I will have removed the content at all frequencies but the signal will remain as it was at the input of the first stage ? Very odd. I suppose it might be interesting to generate a sinc PSD, delete the component at DC and then take the IFFT and see what kind of pulse shape I get in the time domain. Anyway, it is useful to know that this is the generally accepted interpretation of the PSD even if I do still have some misgivings about it. Thanks, Usjes. --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●May 25, 20152015-05-25
On Mon, 25 May 2015 17:20:48 -0500, Usjes wrote:>>On Sun, 24 May 2015 16:54:30 -0500, Usjes wrote: >> >> > >>The power spectral _density_ is maximum at DC, but finite. >> >>The spectral width of _absolutely pure_ DC is zero. >> >>(Some finite number) * 0 = 0. >> >>-- >>www.wescottdesign.com > > Hi Tim, > > Yes, this was the only explanation I could think of myself but I'm not > entirely happy with it as it essentially seems to be saying the signal > has no component at any frequency.This is one of those problems you run into when you're dealing with infinities or infinitesimals.> For example, supposing I pass it > through a filter with an infinitely narrow notch at DC, will the output > of the filter be the same as its input given that I have only removed > the content where there was none (signal mean = 0)?Heh heh. Maybe. I hope that you have found this answer helpful in your day-to-day work. What? It isn't? Actually, if you pass it through a filter with an infinitely narrow notch at DC, then a whole bunch of things could happen, all depending on what other assumptions the various puckish graybeards in the group may choose to assume. If it's a zero-delay DC filter with linear phase, and the filter started with the correct output value, then the DC error will be infinitesimal -- but not only will the filter need to have been in existence for an infinitely long time, it will have to have internal "knowledge" of the pulse train states into the infinite future. If it's a _causal_ zero-delay DC filter with linear phase then you'll need to wait around a bit for an output, because the delay will be infinite. I could dream up more, but basically the bottom line is that an infinitely narrow filter is impossible here in the boring old real world.> Now what if I repeat > this with a cascade of an infinite number of such filters each at > different frequencies, at the very end of this cascade I will have > removed the content at all frequencies but the signal will remain as it > was at the input of the first stage ? Very odd.Yup. For any one notch filter at a real frequency the answer is kind of the same as above, except any filter offset will be a tone instead of a bias. I'm not sure what happens with an infinite number of infinitesimally wide filters -- I'm pretty sure that the answer is undefined, because you're multiplying 0 * infinity.> I suppose it might be > interesting to generate a sinc PSD, delete the component at DC and then > take the IFFT and see what kind of pulse shape I get in the time domain.You're thinking in Matlab. The bins of an FFT are not infinitesimally wide. You'll get a result that's the same as just taking your (necessarily finite-length) sample and subtracting its average.> Anyway, it is useful to know that this is the generally accepted > interpretation of the PSD even if I do still have some misgivings about > it.It can be hard to wrap your head around some of these notions, but it's a good idea to do so. Experimenting in Matlab may gain you some intuition, but you still need to be guided by the math to know if you're developing intuition about a real phenomenon or just an artifact of the experimental method. -- www.wescottdesign.com
Reply by ●May 26, 20152015-05-26
>Hi, > >I'd like an explanation of the PSD of a sequence of independent random >PAM >symbols or amplitude +1 or -1. Now any standard textbook tells us that >the >PSD is the Fourier Transform of the autocorrelation of the signal, in my >case the symbols are INDEPENDENT and => the autocorrelation is non-zero >only over the symbol duration 'T', beyond this it goes to zero. Going >through the maths I come out with and answer PSD = sinc^2(wT/2), it is >basically just the square of the ESD of a rectangular pulse. Again any >standard textbook confirms that this is the 'correct' answer. Here is my >problem though; sinc^2(wT/2) has a maximum at w = 0, ie it is saying >that >the most powerful component of the signal is at DC. But if I simply take >the mean of my sequence I get zero (the symbols are RANDOM +/-1 and => >the >average is zero). So, my signal has mean zero (=> zero DC component) yet >the accepted 'correct' PSD for this signal implies that its most >powerful >component is at D.C. Can anyone explain how this can be correct ? > >Thanks, > >Usjes. >--------------------------------------- >Posted through http://www.DSPRelated.comHere's and intuitive argument (maybe) ... If you take any finite length bipolar-NRZ sequence where the integral over the whole sequence is equal to zero, then the Fourier transform of that sequence will always be zero at DC. So taking the FT over a Manchester symbol gives 0 at DC as well as any balanced line code symbol (8B/10B for example). A single rectangular pulse obviously has a strong DC component so the argument boils down to the statistics of the underlying sequence. Sy(f) = |P(f)|^2 * Sx(f) Suppose I transmit 4 symbols. This gives 16 possible transmit sequences. 6 of these sequences are balanced and will have a 0 DC value if the FT is taken across the 4 symbols. The other 10 do not have balance and will have non-zero DC components. In computing the PSD you segment the sequence, take the FT over each segment and then average. Certainly, if each and every sequence was perfectly balanced, then the PSD would have a 0 at DC, but this will never be the case for random data, if it were, you would have the same thing as a 64B/68B line code or something like that. The DC component get lower and lower in magnitude the longer the segment becomes, but so does the magnitude at every other frequency. -Doug --------------------------------------- Posted through http://www.DSPRelated.com
