Dear all, Currently, I am working on implementing recursive low pass filter with design equation extracted from manual y(n) = 2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6) The frequency response of the filter is good. However, the manual also mention, in actual implementation, there is a normalization factor of 1/8 based upon the gain of the filter. That is, the equation become: y(n) = 1/8 * (2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6)) But the frequency response of this filter is bad. Is it some common practice to design a filter with desired good frequency response but later add some normalization factor to normalize the signal? What is the advantage to do so? Thank you for your help. Henry 

problem in normalization factor in low pass filter implementation
Started by ●July 14, 2002
Reply by ●July 15, 200220020715
Henry Are you sure the gain factor (1/8) is not supposed to be applied to the x[n] components only? What happens if you try that? Typically the gain/normalization factor is intended to insure unity gain (or desired gain) in the passband, but the frequency response of the filter (shape) should remain unchanged. Jeff Brower DSP sw/hw engineer Signalogic i_am_henry wrote: > > Dear all, > > Currently, I am working on implementing recursive low pass filter > with design equation extracted from manual > y(n) = 2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6) > > The frequency response of the filter is good. > > However, the manual also mention, in actual implementation, there > is a normalization factor of 1/8 based upon the gain of the filter. > That is, the equation become: > y(n) = 1/8 * (2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6)) > > But the frequency response of this filter is bad. > > Is it some common practice to design a filter with desired good > frequency response but later add some normalization factor to > normalize the signal? What is the advantage to do so? > > Thank you for your help. > > Henry 
Reply by ●July 16, 200220020716
henry, the correct equation after scaling should be: y(n) = 2*y(n1)  y(n2) + 1/8 *( x(n)  2*x(n3) + x(n6) ). This scaling is required to keep the signal level same after filtering. It is required mostly when we are working with finite resolution systems with a fixed dynamic range, as the signal may saturate/overflow otherwise or we'll loose precision. priyank  i_am_henry <> wrote: > Dear all, > > Currently, I am working on implementing recursive low pass filter > with design equation extracted from manual > y(n) = 2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6) > > The frequency response of the filter is good. > > However, the manual also mention, in actual implementation, there > is a normalization factor of 1/8 based upon the gain of the filter. > That is, the equation become: > y(n) = 1/8 * (2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6)) > > But the frequency response of this filter is bad. > > Is it some common practice to design a filter with desired good > frequency response but later add some normalization factor to > normalize the signal? What is the advantage to do so? > > Thank you for your help. > > Henry > __________________________________________________ 
Reply by ●July 16, 200220020716
Henry You're not reading carefully. I said all of the "x[n]" components, like this: y(n) = 2*y(n1)  y(n2) + x(n)/8  x(n3)/4 + x(n6)/8 Jeff Brower DSP sw/hw engineer Signalogic i_am_henry wrote: > > Hi Jeff, > > Thank you for you help first. > > Yes, I have applied the gain/normalization factor (1/8) to all the > input components according to the equation. > > However, for > y(n) = 2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6) > the filter response tranfer function is > Yn 1  2*z^3 + z^6 >  =  .............. (1) > Xn 1  2*z^1 + z^2 > For > y(n) = 1/8 * (2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6)) > the filter response tranfer function is > Yn 1  2*z^3 + z^6 >  =  .............. (2) > Xn 8  2*z^1 + z^2 > > The filter frequency respose of Equ. (2) is different from (1). > > Regards > Henry Chang > Assistant Computer Officer > Chinese Univerisity of HK > >  In matlab@y..., Jeff Brower <jbrower@s...> wrote: > > Henry > > > > Are you sure the gain factor (1/8) is not supposed to be applied to > the x[n] > > components only? What happens if you try that? > > > > Typically the gain/normalization factor is intended to insure unity > gain (or desired > > gain) in the passband, but the frequency response of the filter > (shape) should remain > > unchanged. > > > > Jeff Brower > > DSP sw/hw engineer > > Signalogic > > > > > > i_am_henry wrote: > > > > > > Dear all, > > > > > > Currently, I am working on implementing recursive low pass > filter > > > with design equation extracted from manual > > > y(n) = 2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6) > > > > > > The frequency response of the filter is good. > > > > > > However, the manual also mention, in actual implementation, > there > > > is a normalization factor of 1/8 based upon the gain of the > filter. > > > That is, the equation become: > > > y(n) = 1/8 * (2*y(n1)  y(n2) + x(n)  2*x(n3) + x(n6)) > > > > > > But the frequency response of this filter is bad. > > > > > > Is it some common practice to design a filter with desired good > > > frequency response but later add some normalization factor to > > > normalize the signal? What is the advantage to do so? > > > > > > Thank you for your help. > > > > > > Henry 