"Gert Baars" <g.baars13@chello.nl> wrote in message news:d65c1$43280129$3ec23590$23949@news.chello.nl...> The easiest way is to compare only the center tap which origins > from sinc(0). Here hn = 2*Fc/Fs and the Hamming window = 1. > > This equals the centre tap of the ScopeFir design as long as > Fc >= 1 KHz with Fs = 12KHz with LPF. > The difference also gets very small if #TAPS gets very large > no matter what Fc is.I asked what the "by hand" design algorithm was ...... not the comparison method or results. Fred
coefficients compared
Started by ●September 13, 2005
Reply by ●September 14, 20052005-09-14
Reply by ●September 15, 20052005-09-15
I get good results with correction. The problem is that any window is much too small if Fc/Fs gets very small because sinc(x) will be 1 everywhere in the window so the actual h[n] will become shaped exactly like the (hamming)window. Concerning your question; Because H(W) of a HPF equals 1-H(W) of a LPF? Jerry Avins wrote:> Gert Baars wrote: > >> Jerry Avins wrote: >> >>> Gert Baars wrote: >>> >>> ... >>> >>>> Now I have to find out how I differ from ScopeFIR's method >>>> (I selected the Windowed Sinc type of filter). >>> >>> >>> >>> >>> ScopeFIR uses a different method. It involves no window. There is no >>> reason to expect the results to be identical. In fact, for a given >>> number of taps, you can expect the Parks/McClellan design to match >>> the specs more closely. >>> >>> Jerry >> >> >> >> >> What I think ScopeFIR does is correct afterwards. I tried adding up all >> coefficients with my method (including the Hammingwindow) and then >> afterwards correct all coefficients so they DO add up to 1. The results >> sound pretty good. > > > Windowed sinc is good; numerical approximation methods are a bit better. > Coefficients summing to one gives unity gain at DC. For a high-pass > filter, they will sum to zero. In fact, with an odd number of symmetric > taps, subtracting the sum of a low-pass's taps from the middle one makes > an inverting high-pass. Do you see why? > > Jerry
Reply by ●September 15, 20052005-09-15
Gert Baars wrote:> I get good results with correction. The problem is that any window > is much too small if Fc/Fs gets very small because sinc(x) will be > 1 everywhere in the window so the actual h[n] will become shaped > exactly like the (hamming)window.It seems to me that when the sinc is so heavily truncated that it's 1 everywhere, you have at nest a moving average filter, and a rectangular window may be as good as any.> Concerning your question; Because H(W) of a HPF equals 1-H(W) of a LPF?Yes ... Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
