Hi all, I am new here and would like to ask questions about discrete integrator.>>>How can I characterise a discrete integrator??for example I can measure the time constant of a continous integrator, can i do so with discete integrator?? Is there any other characteristics?? like gain??>>>If I build a discrete integrator how can i measure the characteristicyou have mentioned?? For example input a step input and measure its declay time, etc. Thanks for your help Vincent
Simple Discrete integrator question
Started by ●December 7, 2006
Reply by ●December 7, 20062006-12-07
vinching wrote:> I am new here and would like to ask questions about discrete > integrator.I assume you mean the recurrence y[n] = x[n] + a y[n-1] with transfer function H(z) = 1/(1 - a/z), a in (0,1).>>>>How can I characterise a discrete integrator?? > for example I can measure the time constant of a continous > integrator, can i do so with discete integrator??It's easy to see that the impulse response of H(z) above is h[n] = a^n, n >= 0. Now a^n = e^(n ln(a)), so you could call tau = 1/ln(a) the time constant in samples and tau' = tau/f_s the time constant in seconds, where fs is your sampling frequency.> Is there any other characteristics?? like gain??The integrator's gain of course depends on frequency. Frequency wraps around the unit circle in the z-domain, so the gain (magnitude spectrum) is M(w) = |H(e^(j w))| for angular frequency w. If you meant the maximum gain which occurs at w = 0, that tells us it's M_max = M(0) = |H(1)| = 1/(1 - a).>>>>If I build a discrete integrator how can i measure the >>>>characteristic you have mentioned??Because the maximum gain occurs at zero frequency, the step response approaches M_max as time grows. You can measure your implementation's behavior by feeding it a step and generating output until it stops changing, then compare the final value to the ideal M_max above. You can measure the effective time constant similarly over some "large" number of samples N. You'd apply the filter to the impulse x = (1,0,0,0, ...) to get y[N]. This is approximately h[N], so tau_eff = N / ln(y[N]). "Measuring" the whole spectrum could mean many different things, but the results of http://groups.google.com/groups?q=group:comp.dsp+measure+filter+response seem relevant. Martin -- Better to remain silent and be thought a fool, than to speak out and remove all doubt. --Abraham Lincoln
Reply by ●December 7, 20062006-12-07
vinching wrote:> I am new here and would like to ask questions about discrete > integrator.I assume you mean the recurrence y[n] = x[n] + a y[n-1] with transfer function H(z) = 1/(1 - a/z), a in (0,1).>>>>How can I characterise a discrete integrator?? > for example I can measure the time constant of a continous > integrator, can i do so with discete integrator??It's easy to see that the impulse response of H(z) above is h[n] = a^n, n >= 0. Now a^n = e^(n ln(a)), so you could call tau = 1/ln(a) the time constant in samples and tau' = tau/f_s the time constant in seconds, where fs is your sampling frequency.> Is there any other characteristics?? like gain??The integrator's gain of course depends on frequency. Frequency wraps around the unit circle in the z-domain, so the gain (magnitude spectrum) is M(w) = |H(e^(j w))| for angular frequency w. If you meant the maximum gain which occurs at w = 0, that tells us it's M_max = M(0) = |H(1)| = 1/(1 - a).>>>>If I build a discrete integrator how can i measure the >>>>characteristic you have mentioned??Because the maximum gain occurs at zero frequency, the step response approaches M_max as time grows. You can measure your implementation's behavior by feeding it a step and generating output until it stops changing, then compare the final value to the ideal M_max above. You can measure the effective time constant similarly over some "large" number of samples N. You'd apply the filter to the impulse x = (1,0,0,0, ...) to get y[N]. This is approximately h[N], so tau_eff = N / ln(y[N]). "Measuring" the whole spectrum could mean many different things, but the results of http://groups.google.com/groups?q=group:comp.dsp+measure+filter+response seem relevant. Martin -- Better to remain silent and be thought a fool, than to speak out and remove all doubt. --Abraham Lincoln
Reply by ●December 8, 20062006-12-08
"Martin Eisenberg" <martin.eisenberg@udo.edu> wrote in message news:1165518969.711899@localhost...> vinching wrote: > > > I am new here and would like to ask questions about discrete > > integrator. > > I assume you mean the recurrence y[n] = x[n] + a y[n-1] > with transfer function H(z) = 1/(1 - a/z), a in (0,1). >That's not a pure integrator - only when a=1. Tam -- Posted via a free Usenet account from http://www.teranews.com
Reply by ●December 8, 20062006-12-08
"Martin Eisenberg" <martin.eisenberg@udo.edu> wrote in message news:1165520435.913210@localhost...> vinching wrote: > > > I am new here and would like to ask questions about discrete > > integrator. > > I assume you mean the recurrence y[n] = x[n] + a y[n-1] > with transfer function H(z) = 1/(1 - a/z), a in (0,1). > >That's a low-pass filter. Some uniformed people call it a leaky integrator of course. Tam -- Posted via a free Usenet account from http://www.teranews.com
Reply by ●December 8, 20062006-12-08
Heid the baw - goal!! wrote:> "Martin Eisenberg" <martin.eisenberg@udo.edu> wrote in message > news:1165520435.913210@localhost... >> vinching wrote: >> >> > I am new here and would like to ask questions about discrete >> > integrator. >> >> I assume you mean the recurrence y[n] = x[n] + a y[n-1] >> with transfer function H(z) = 1/(1 - a/z), a in (0,1). > > That's a low-pass filter. Some uniformed people call it a leaky > integrator of course.The reason for calling it that may be any mixture of ignorance and convenience. Vincent can clarify if I missed his understanding. Martin -- Sphinx of black quartz, judge my vow! --David Lemon
Reply by ●December 8, 20062006-12-08
Martin Eisenberg wrote:> Heid the baw - goal!! wrote: >> "Martin Eisenberg" <martin.eisenberg@udo.edu> wrote ...It is usually unproductive to reply to trolls. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●December 8, 20062006-12-08
On 8 Dez., 16:57, Jerry Avins <j...@ieee.org> wrote:> Martin Eisenberg wrote: > > Heid the baw - goal!! wrote: > >> "Martin Eisenberg" <martin.eisenb...@udo.edu> wrote ... >It is usually unproductive to reply to trolls.Why troll? Tam's objection is solid: a stable lowpass filter is not an integrator. Regards, Andor
Reply by ●December 8, 20062006-12-08
Andor wrote:> > On 8 Dez., 16:57, Jerry Avins <j...@ieee.org> wrote: >> Martin Eisenberg wrote: >>> Heid the baw - goal!! wrote: >>>> "Martin Eisenberg" <martin.eisenb...@udo.edu> wrote ... >> It is usually unproductive to reply to trolls. > > Why troll? Tam's objection is solid: a stable lowpass filter is not an > integrator.Read all his messages. I have yet to see a productive answer. A short quibble seems to be his recurring style. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●December 8, 20062006-12-08
"Andor" <andor.bariska@gmail.com> wrote in news:1165593781.238754.117100@ 16g2000cwy.googlegroups.com:> > > On 8 Dez., 16:57, Jerry Avins <j...@ieee.org> wrote: >> Martin Eisenberg wrote: >> > Heid the baw - goal!! wrote: >> >> "Martin Eisenberg" <martin.eisenb...@udo.edu> wrote ... >>It is usually unproductive to reply to trolls. > > Why troll? Tam's objection is solid: a stable lowpass filter is not an > integrator. > > Regards, > Andor >A low pass filter with an infinite time constant is the mathematical equivalent of an integrator. One could thus think of a low pass filter with a smaller time constant as a leaky integrator. If you want to The term became popular because often we want to integrate, but we really want the integration to leak a bit to avoid saturation, and in the analog world, the integrator is only as good as the "real" capacitors used to implement them. Think about a perfect integrator, and subtract a low-pass "leak" term, and what you'll get back is a low pass filter. -- Scott Reverse name to reply
