sync filter: Can you help me to set 3 parameter?
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From: Jerry Avins on 8 Jul 2010 10:57 On 7/8/2010 5:14 AM, gpezzella wrote: > Dear Jerry > > just yesterday I have discovery another thread > (http://www.dsprelated.com/showmessage/13897/1.php) in this forum that talk > about the equation that you wrote to me: > > -------------------------------- > Y(i) = K * X(i) + (1-K) * Y(i-1) > -------------------------------- > > The difference is only apparent: > > ----------------------------------- > Y(i) = [X(i) - Y(i-1)] * K + Y(i-1) > ----------------------------------- > > > Now in that thread the author say this: > > Where the desired time constant is known, use the following method. > Assuming an input step from 0.0 to +1.0, an RC LPF will change its > output voltage towards the new input by an amount equal to: > > 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 > > > > Is this my equation? > ---------------------------------------- > Y(i) = [X(i) - Y(i-1)] * 0.12 + Y(i-1) > ---------------------------------------- First, notation. Subscripts are generally unavailable in plain text, so we substitute parentheses or brackets. Typical DSP practice makes a virtue of necessity. Parentheses are used for continuous variables such as time, while square brackets are are used for enumerations. Y(i) = [X(i) - Y(i-1)] * 0.12 + Y(i-1) is incorrect. Y[i] needs to be some (large) fraction of Y[i-1], augmented by a part of X[i]. To achieve proper low-frequency performance, the fraction and the part must sum to 1. So Y[i] = k*Y[i-1] + (k-1)*X[i]. You calculated k=.88. On a fixed-point processor, I would change that yo ..875, so that multiplications by k and by k-1 can be simply done with shifts. That is, k*Y[i-1] = Y[i-1]-Y[i-1]>>3, and (k-1)*X[i]=X[i]>>3. Note that the lower bits if X[i] are discarded whather you shift or multiply. You need to determine if that's a problem. There are ways to deal with it, but that's another topic. Jerry -- Engineering is the art of making what you want from things you can get.
From: Tim Wescott on 8 Jul 2010 11:48 On 07/08/2010 07:57 AM, Jerry Avins wrote: > On 7/8/2010 5:14 AM, gpezzella wrote: >> Dear Jerry >> >> just yesterday I have discovery another thread >> (http://www.dsprelated.com/showmessage/13897/1.php) in this forum that >> talk >> about the equation that you wrote to me: >> >> -------------------------------- >> Y(i) = K * X(i) + (1-K) * Y(i-1) >> -------------------------------- >> >> The difference is only apparent: >> >> ----------------------------------- >> Y(i) = [X(i) - Y(i-1)] * K + Y(i-1) >> ----------------------------------- >> >> >> Now in that thread the author say this: >> >> Where the desired time constant is known, use the following method. >> Assuming an input step from 0.0 to +1.0, an RC LPF will change its >> output voltage towards the new input by an amount equal to: >> >> 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 >> >> >> >> Is this my equation? >> ---------------------------------------- >> Y(i) = [X(i) - Y(i-1)] * 0.12 + Y(i-1) >> ---------------------------------------- > > First, notation. Subscripts are generally unavailable in plain text, so > we substitute parentheses or brackets. Typical DSP practice makes a > virtue of necessity. Parentheses are used for continuous variables such > as time, while square brackets are are used for enumerations. > > Y(i) = [X(i) - Y(i-1)] * 0.12 + Y(i-1) is incorrect. Y[i] needs to be > some (large) fraction of Y[i-1], augmented by a part of X[i]. To achieve > proper low-frequency performance, the fraction and the part must sum to > 1. So Y[i] = k*Y[i-1] + (k-1)*X[i]. > > You calculated k=.88. On a fixed-point processor, I would change that yo > .875, so that multiplications by k and by k-1 can be simply done with > shifts. That is, k*Y[i-1] = Y[i-1]-Y[i-1]>>3, and (k-1)*X[i]=X[i]>>3. > Note that the lower bits if X[i] are discarded whather you shift or > multiply. You need to determine if that's a problem. There are ways to > deal with it, but that's another topic. Actually, if it's a DSP chip (and on some newer 'regular' chips) there's no extra cost for a multiply, and particularly on a DSP chip a shift may be slower than a multiply (because most DSP chips give you single-cycle multiply-and-accumulate, but few give you single-cycle shift-and-accumulate). -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
From: Jerry Avins on 8 Jul 2010 11:57 On 7/8/2010 11:48 AM, Tim Wescott wrote: > On 07/08/2010 07:57 AM, Jerry Avins wrote: ... >> You calculated k=.88. On a fixed-point processor, I would change that yo >> .875, so that multiplications by k and by k-1 can be simply done with >> shifts. That is, k*Y[i-1] = Y[i-1]-Y[i-1]>>3, and (k-1)*X[i]=X[i]>>3. >> Note that the lower bits if X[i] are discarded whather you shift or >> multiply. You need to determine if that's a problem. There are ways to >> deal with it, but that's another topic. > > Actually, if it's a DSP chip (and on some newer 'regular' chips) there's > no extra cost for a multiply, and particularly on a DSP chip a shift may > be slower than a multiply (because most DSP chips give you single-cycle > multiply-and-accumulate, but few give you single-cycle > shift-and-accumulate). I should have pointed that out. I remember the OP writing that his resources are very limited, and I switched into Z-80 mode. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
From: gpezzella on 9 Jul 2010 05:42 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? Question 2: After chhose 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 ------------------------------------------------------------------------------
From: Jerry Avins on 9 Jul 2010 09:48
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. ����������������������������������������������������������������������� |