Rune Allnor wrote:> On 6 Jun, 19:47, julius <juli...@gmail.com> wrote: >> On Jun 6, 9:06 am, Jerry Avins <j...@ieee.org> wrote: >> >> >> >>> Do you divide by zero at some point? Deconvolution is tricky. >>> Jerry >> Yep, there are lots and lots of tricks and techniques that has to be >> learned. One "easy" modification is to do the pointwise division only >> if the denominator for that point is above a certain threshold. >> Otherwise set estimate of that point to zero. > > Doesn't work. If the denominator has a vanishing magnitude > for some w, so does the numerator (why?). You are effectively > messing around with a 0/0 type of expression.Suppose the denominator is a signal passed through a notch filter and the numerator is the original. Do you then claim 0/0? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
dividing two signals using fft
Started by ●June 6, 2007
Reply by ●June 7, 20072007-06-07
Reply by ●June 8, 20072007-06-08
On 7 Jun, 14:51, Jerry Avins <j...@ieee.org> wrote:> Rune Allnor wrote: > > On 6 Jun, 19:47, julius <juli...@gmail.com> wrote: > >> On Jun 6, 9:06 am, Jerry Avins <j...@ieee.org> wrote: > > >>> Do you divide by zero at some point? Deconvolution is tricky. > >>> Jerry > >> Yep, there are lots and lots of tricks and techniques that has to be > >> learned. One "easy" modification is to do the pointwise division only > >> if the denominator for that point is above a certain threshold. > >> Otherwise set estimate of that point to zero. > > > Doesn't work. If the denominator has a vanishing magnitude > > for some w, so does the numerator (why?). You are effectively > > messing around with a 0/0 type of expression. > > Suppose the denominator is a signal passed through a notch filter and > the numerator is the original. Do you then claim 0/0?Whenever the denominator vanishes so does the numerator. The converse it not necessarily true, but that situation poses no problems with respect to spectrum division. Rune
Reply by ●June 8, 20072007-06-08
On 7 Jun., 14:51, Jerry Avins <j...@ieee.org> wrote:> Rune Allnor wrote: > > On 6 Jun, 19:47, julius <juli...@gmail.com> wrote: > >> On Jun 6, 9:06 am, Jerry Avins <j...@ieee.org> wrote: > > >>> Do you divide by zero at some point? Deconvolution is tricky. > >>> Jerry > >> Yep, there are lots and lots of tricks and techniques that has to be > >> learned. One "easy" modification is to do the pointwise division only > >> if the denominator for that point is above a certain threshold. > >> Otherwise set estimate of that point to zero. > > > Doesn't work. If the denominator has a vanishing magnitude > > for some w, so does the numerator (why?). You are effectively > > messing around with a 0/0 type of expression. > > Suppose the denominator is a signal passed through a notch filter and > the numerator is the original. Do you then claim 0/0?I think Rune is actually correct. The problem is that you want to estimate the spectrum of a signal that passed through a bandstop filter, given just the output of the filter. Both the output and the filter have DFT bins close or equal to zero in that region. Guessing that the input signal's spectrum in the bandstop region is zero, as Julius suggested, is completely arbitrary. For this specific case, I would suggest estimating the spectrum in the bandstop region from the neighbouring region (for example the average of the neighbour bins). Regards, Andor
Reply by ●June 8, 20072007-06-08
Rune Allnor wrote: ...> Whenever the denominator vanishes so does the numerator. > The converse it not necessarily true, but that situation > poses no problems with respect to spectrum division.The spectrum of an impulse is the numerator. The spectrum of that impulse passed through a notch filter is the denominator. What makes the numerator vanish at the notch frequency? Sofi (and maybe you too) should read http://www.dspguide.com/ch17/2.htm The first illustrations are irrelevant. They are from a previous section. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●June 8, 20072007-06-08
On 8 Jun, 15:23, Jerry Avins <j...@ieee.org> wrote:> Rune Allnor wrote: > > ... > > > Whenever the denominator vanishes so does the numerator. > > The converse it not necessarily true, but that situation > > poses no problems with respect to spectrum division. > > The spectrum of an impulse is the numerator. The spectrum of that > impulse passed through a notch filter is the denominator. What makes the > numerator vanish at the notch frequency?First, if the input signal really is an impulse, one measures the impulse response directly and there is no need to perform any sort of spctrum division to find the transfer function. So x[n] is some signal, the only assumption about it is that it has been measured. Hence, it can perfectly well contain a notch in the spectrum somewhere. Hence: y[n] = x[n] (*) h[n] Y(w) = X(w)H(w) H(w) = Y(w)/X(w) = numerator / denominator (just to cover the possibility that I have mixed up the two terms.) If X(w') = 0 for some w', then Y(w') = H(w')*0 = 0 no matter what the value of H(w') is at the same w'. Hence, the computed H(w') becomes H(w') = Y(w')/X(w') = 0/0 which is bad, no matter how one treats the X(w')=0 situation. If X(w") = x" =/= 0, H(w") = 0 then Y(w") = x"*0 = 0 and the estimate becomes H(w") = 0/x" = 0 which is no problem. Rune
Reply by ●June 8, 20072007-06-08
Rune Allnor wrote:> On 8 Jun, 15:23, Jerry Avins <j...@ieee.org> wrote: >> Rune Allnor wrote: >> >> ... >> >>> Whenever the denominator vanishes so does the numerator. >>> The converse it not necessarily true, but that situation >>> poses no problems with respect to spectrum division. >> The spectrum of an impulse is the numerator. The spectrum of that >> impulse passed through a notch filter is the denominator. What makes the >> numerator vanish at the notch frequency? > > First, if the input signal really is an impulse, one measures > the impulse response directly and there is no need to perform > any sort of spctrum division to find the transfer function. > > So x[n] is some signal, the only assumption about it is that it > has been measured. Hence, it can perfectly well contain a > notch in the spectrum somewhere. > > Hence: > > y[n] = x[n] (*) h[n] > Y(w) = X(w)H(w) > > H(w) = Y(w)/X(w) = numerator / denominator (just to > cover the possibility that I have mixed up the two terms.) > > If X(w') = 0 for some w', then Y(w') = H(w')*0 = 0 > no matter what the value of H(w') is at the same w'. > Hence, the computed H(w') becomes > > H(w') = Y(w')/X(w') = 0/0 > > which is bad, no matter how one treats the X(w')=0 > situation. > > If X(w") = x" =/= 0, H(w") = 0 then > > Y(w") = x"*0 = 0 > > and the estimate becomes > > H(w") = 0/x" = 0 > > which is no problem.If impulse bothers you, use any signal with a flat spectrum. For that matter, consider any signal with energy at every frequency in consideration whether flat or not. Pass the signal through a black box. Form the fraction (input signal)/(output signal). Why must (input signal) become zero whenever (output signal) does? Didn't you claim that it does? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●June 8, 20072007-06-08
On 8 Jun, 18:39, Jerry Avins <j...@ieee.org> wrote:> Rune Allnor wrote: > > On 8 Jun, 15:23, Jerry Avins <j...@ieee.org> wrote: > >> Rune Allnor wrote: > > >> ... > > >>> Whenever the denominator vanishes so does the numerator. > >>> The converse it not necessarily true, but that situation > >>> poses no problems with respect to spectrum division. > >> The spectrum of an impulse is the numerator. The spectrum of that > >> impulse passed through a notch filter is the denominator. What makes the > >> numerator vanish at the notch frequency? > > > First, if the input signal really is an impulse, one measures > > the impulse response directly and there is no need to perform > > any sort of spctrum division to find the transfer function. > > > So x[n] is some signal, the only assumption about it is that it > > has been measured. Hence, it can perfectly well contain a > > notch in the spectrum somewhere. > > > Hence: > > > y[n] = x[n] (*) h[n] > > Y(w) = X(w)H(w) > > > H(w) = Y(w)/X(w) = numerator / denominator (just to > > cover the possibility that I have mixed up the two terms.) > > > If X(w') = 0 for some w', then Y(w') = H(w')*0 = 0 > > no matter what the value of H(w') is at the same w'. > > Hence, the computed H(w') becomes > > > H(w') = Y(w')/X(w') = 0/0 > > > which is bad, no matter how one treats the X(w')=0 > > situation. > > > If X(w") = x" =/= 0, H(w") = 0 then > > > Y(w") = x"*0 = 0 > > > and the estimate becomes > > > H(w") = 0/x" = 0 > > > which is no problem. > > If impulse bothers you, use any signal with a flat spectrum.If one has sufficient (i.e. complete) contol of the source, this would be a very wise thing to do. There are, however, situations where one has to settle for only being able to *measure* the input (i.e. not controlling what signal excites the system), in which case zeros in the input spectrum becomes an issue.> For that > matter, consider any signal with energy at every frequency in > consideration whether flat or not.Variations over the same theme; assuming sufficient control over the source.> Pass the signal through a black box. > Form the fraction (input signal)/(output signal). Why must (input > signal) become zero whenever (output signal) does? Didn't you claim that > it does?I hope not; if I wrote anything to that effect, it is was because of my imperfect mastery of the English language. The output must be zero when the input is zero. Rune
Reply by ●June 8, 20072007-06-08
Rune Allnor wrote: ...> I hope not; if I wrote anything to that effect, it is was > because of my imperfect mastery of the English language. > The output must be zero when the input is zero.aHH. i'M RELIEVED. oUR MISUNDERSTANDING MAY WELL HAVE ARISEN FROM MY OBSCURE WAY OF STATING WHAT i MEANT. (Caps lock on. Too lazy to retype.) The problem arises when someone naively tries to make an equalizer. (input signal) The fraction -------------- (output signal) is the response of an ideal equalizer for a signal passing through a frequency-selective channel. The ideal isn't always obtainable. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
