sync filter: Can you help me to set 3 parameter?
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From: gpezzella on 14 Jul 2010 07:14 Hi Jerry Every time I take new sample I calculate the actual value of Y[i] and save it in a Y[i] array. Since I have few memory, which is the minumum number of Y[i] that I must save? Thanks P.S. Remember that Ft = 50 Hz Sample Rate = 2.48KHz >On 7/9/2010 5:42 AM, gpezzella wrote: >> Ok I will use this notation from now >> >> I previous post (7/1/2010 3:21 AM) you wrote me the follow equation: >> <a> "y[1] = k*x[i] + (1-k)*y[i-1] that behaves like a single RC". >> >> In the day 7/8/2010 I discovery another thread that use: >> <b> "Y[i] = K * X[i] + [1-K] * Y[i-1]" >> which is coincident to yours: >> <c> "Y[i] = [ X[i] - Y[i-1] ] * K + Y[i-1]" >> >> Today instead you write that: >> <d> "Y[i] = k*Y[i-1] + (k-1)*X[i]" is the correct equation >> >> Question 1: Which is the good one between a,b,c,d? > >They are all good. Where I initially have the coefficients as k and 1-k, >your other gave them as 1-k and k. Because k is an arbitrary constant, >the two notations are equivalent. After your response, I switched to >your notation. > >> Question 2: After choose the correct equation, is the follow method correct >> for calculate K coefficient? >> -------------------------------------------------------------------------------- >> ts >> -(--) Where: ts = sample period (seconds) >> T T = time constant (seconds) >> (1) K = 1 - e K = change in o/p after 1 ts >> >> >> The value for K is then used as the coefficient in the RC LPF code. >> >> If the above equation is correct I should use for my case: >> 1)Ft = 50 Hz >> 2)Sample Rate = 2.48KHz >> >> >> 1 >> T = ------- = 0.00318 >> 2*Pi*50 >> >> ts = 1/2.48 Hz = 0.0004 >> >> >> 0.0004 >> -(--------) >> 0.00318 >> K = 1 - e = 1 - 0.88 = 0.12 >> ------------------------------------------------------------------------------ > >Yes. (I would have written the last line either as e^(-.0004/.00318) or >exp(-.0004/.00318).) > >Jerry >-- >Engineering is the art of making what you want from things you can get. >����������������������������������������������������������������������� >
From: gpezzella on 14 Jul 2010 08:33 Hi Jerry Every time I take new sample I calculate the actual value of Y[i] and save it in a Y[i] array. Since I have few memory, which is the minumum number of Y[i] that I must save? Thanks P.S. Remember that Ft = 50 Hz Sample Rate = 2.48KHz >On 7/9/2010 5:42 AM, gpezzella wrote: >> Ok I will use this notation from now >> >> I previous post (7/1/2010 3:21 AM) you wrote me the follow equation: >> <a> "y[1] = k*x[i] + (1-k)*y[i-1] that behaves like a single RC". >> >> In the day 7/8/2010 I discovery another thread that use: >> <b> "Y[i] = K * X[i] + [1-K] * Y[i-1]" >> which is coincident to yours: >> <c> "Y[i] = [ X[i] - Y[i-1] ] * K + Y[i-1]" >> >> Today instead you write that: >> <d> "Y[i] = k*Y[i-1] + (k-1)*X[i]" is the correct equation >> >> Question 1: Which is the good one between a,b,c,d? > >They are all good. Where I initially have the coefficients as k and 1-k, >your other gave them as 1-k and k. Because k is an arbitrary constant, >the two notations are equivalent. After your response, I switched to >your notation. > >> Question 2: After choose the correct equation, is the follow method correct >> for calculate K coefficient? >> -------------------------------------------------------------------------------- >> ts >> -(--) Where: ts = sample period (seconds) >> T T = time constant (seconds) >> (1) K = 1 - e K = change in o/p after 1 ts >> >> >> The value for K is then used as the coefficient in the RC LPF code. >> >> If the above equation is correct I should use for my case: >> 1)Ft = 50 Hz >> 2)Sample Rate = 2.48KHz >> >> >> 1 >> T = ------- = 0.00318 >> 2*Pi*50 >> >> ts = 1/2.48 Hz = 0.0004 >> >> >> 0.0004 >> -(--------) >> 0.00318 >> K = 1 - e = 1 - 0.88 = 0.12 >> ------------------------------------------------------------------------------ > >Yes. (I would have written the last line either as e^(-.0004/.00318) or >exp(-.0004/.00318).) > >Jerry >-- >Engineering is the art of making what you want from things you can get. >����������������������������������������������������������������������� >
From: Jerry Avins on 14 Jul 2010 15:43 On 7/14/2010 7:14 AM, gpezzella wrote: > Hi Jerry > > Every time I take new sample I calculate the actual value of Y[i] and save > it in a Y[i] array. No. A variable, If you have one to dedicate, a register. > Since I have few memory, which is the minumum number of Y[i] that I must > save? You need only the last Y you calculated. The new one you calculate replaces it. Remember: this is a one-pole filter. You may need something with sharper cut-off. The simplest is a cascade of two or more of these, but there are computationally more efficient ones. http://www-users.cs.york.ac.uk/~fisher/mkfilter/trad.html is an on-line application to design them. Try a two- or three-pole filter with a 3 dB cutoff point at 60 Hz. Jerry -- Engineering is the art of making what you want from things you can get.
From: gpezzella on 15 Jul 2010 11:48 Dear Jerry, GREAT LINK!! Thanks If for example I choose: ----------------------------------- filtertype = Chebyshev passtype = Lowpass ripple = -1 order = 5 samplerate = 2480 corner1 = 50 corner2 = adzero = logmin = Recurrence relation: y[n] = ( 1 * x[n- 5]) + ( 5 * x[n- 4]) + ( 10 * x[n- 3]) + ( 10 * x[n- 2]) + ( 5 * x[n- 1]) + ( 1 * x[n- 0]) + ( 0.8881071970 * y[n- 5]) + ( -4.5278119979 * y[n- 4]) + ( 9.2530651832 * y[n- 3]) + ( -9.4749879058 * y[n- 2]) + ( 4.8616237413 * y[n- 1]) ---------------------------------- 1) Always I must save Y[n] into a variable? 2) If yes, for check if there are been frequency under 50Hz I must only compare Y[n] with a threshold, correct? Thanks >On 7/14/2010 7:14 AM, gpezzella wrote: >> Hi Jerry >> >> Every time I take new sample I calculate the actual value of Y[i] and save >> it in a Y[i] array. > >No. A variable, If you have one to dedicate, a register. > >> Since I have few memory, which is the minumum number of Y[i] that I must >> save? > >You need only the last Y you calculated. The new one you calculate >replaces it. Remember: this is a one-pole filter. You may need something >with sharper cut-off. The simplest is a cascade of two or more of these, >but there are computationally more efficient ones. >http://www-users.cs.york.ac.uk/~fisher/mkfilter/trad.html is an on-line >application to design them. Try a two- or three-pole filter with a 3 dB >cutoff point at 60 Hz. > >Jerry >-- >Engineering is the art of making what you want from things you can get. >
From: Jerry Avins on 15 Jul 2010 12:10
On 7/15/2010 11:48 AM, gpezzella wrote: > Dear Jerry, > > GREAT LINK!! Thanks > > If for example I choose: > ----------------------------------- > filtertype = Chebyshev > passtype = Lowpass > ripple = -1 > order = 5 > samplerate = 2480 > corner1 = 50 > corner2 = > adzero = > logmin = > > Recurrence relation: > y[n] = ( 1 * x[n- 5]) > + ( 5 * x[n- 4]) > + ( 10 * x[n- 3]) > + ( 10 * x[n- 2]) > + ( 5 * x[n- 1]) > + ( 1 * x[n- 0]) > > + ( 0.8881071970 * y[n- 5]) > + ( -4.5278119979 * y[n- 4]) > + ( 9.2530651832 * y[n- 3]) > + ( -9.4749879058 * y[n- 2]) > + ( 4.8616237413 * y[n- 1]) > ---------------------------------- Why do you need so sharp a cutoff? There will be attenuation at the corner frequency, so you should set it higher than what you want to pass unattenuated. > 1) Always I must save Y[n] into a variable? Your recursion relation needs 5 past x[n] and 5 past y[n]. You need to save them all, then update them for the next iteration. > 2) If yes, for check if there are been frequency under 50Hz I must only > compare Y[n] with a threshold, correct? That should work. Jerry -- Engineering is the art of making what you want from things you can get. |