for comparison, here is an 80-tap FIR that was numerically optimized for -40 dBc peak -in-band error and some reasonable stopband rejection. One should have a look at the requirements for both, then reconsider the length. frequency response (blue), error vector magnitude (red) http://www.dsprelated.com/blogimages/MarkusNentwig/comp.dsp/111220_RRC1.png impulse response: http://www.dsprelated.com/blogimages/MarkusNentwig/comp.dsp/111220_RRC1.png http://www.dsprelated.com/blogimages/MarkusNentwig/comp.dsp/111220_ir.dat

# why does my FIR output contain discontinuities?

Started by ●December 17, 2011

Reply by ●December 20, 20112011-12-20

Reply by ●December 20, 20112011-12-20

typo in the frequency response URL: frequency response (blue), error vector magnitude (red) http://www.dsprelated.com/blogimages/MarkusNentwig/comp.dsp/111220_RRC2.png

Reply by ●December 20, 20112011-12-20

Here is example of using firls for even filter with two equal peaks (even symmetry).set dc gain to 1, transition band as per suitable formula, stop band at zero. B = .15; %rolloff Rs = .5; %symbols upsampled by 2 T = 1/Rs; Fn = Rs/2; %transition band f = linspace((1-B)/4,(1+B)/4,256); for i = 1:256 A(i) = sqrt(.5 + .5*sin(pi/(2*Fn)*(Fn - f(i))/B)); end %stopband A(224:end) = 0; h = firls(79,[0 f*2 1],[1 A 0],[1 ones(1,256/2-1)*1 1]); kadhiem

Reply by ●December 20, 20112011-12-20

an interesting comparison. Both are (or "should be") optimal, as they are the result of a LMS algorithm. The difference is in the weighting of the error that is minimized, how to trade off between different frequencies. The "firls" design minimizes the error over the whole bandwidth. An in-band vector error due to ripple is considered as "important" as out-of-band leakage, and the sum of all errors over the whole bandwidth is minimized. I'm putting a bit more emphasis on stopband rejection, but the numbers are quite arbitrary. The radio standard will demand a "spectral emission mask", and most likely the point where the stopband start (let's say, bin 8000 in the following pictures) will become the critical requirement that determines your filter size, when a given error vector magnitude may not be exceeded. stopband: blue = firls; black = "my" weighted LS design http://www.dsprelated.com/blogimages/MarkusNentwig/comp.dsp/111220_out1.png passband ripple http://www.dsprelated.com/blogimages/MarkusNentwig/comp.dsp/111220_out2.png

Reply by ●December 20, 20112011-12-20

argh... I meant the other way round. It's late... blue = "my" weighted LS design Black, "firls" from Kadhiem's script

Reply by ●December 21, 20112011-12-21

Hi all, Thank you all for your suggestions and contributions to this thread. Regarding the 'original' question about discontinuities in the filter time series output, the cause turned out to be as a result of truncation of the impulse response. As the impulse response length is decreased (by reducing N), errors appear in the filter output every T samples N = 256; T = 64; alpha = 0.5; hd = fdesign.pulseshaping(T,'Square Root Raised Cosine','N,Beta',N, alpha); h = design(hd); figure(1); subplot(2,1,1) plot(-N/2:N/2,h.Numerator) subplot(2,1,2) y = filter(h.Numerator,1,upsample((2 * randi([0 1],1,2000))-1,T)); plot(y(N/2:100*T))

Reply by ●December 21, 20112011-12-21

Hi all, Thank you all for your suggestions and contributions to this thread. Regarding the 'original' question about discontinuities in the filter time series output, the cause turned out to be as a result of truncation of the impulse response. As the impulse response length is decreased (by reducing N), errors appear in the filter output every T samples N = 256; T = 64; alpha = 0.5; hd = fdesign.pulseshaping(T,'Square Root Raised Cosine','N,Beta',N, alpha); h = design(hd); figure(1); subplot(2,1,1) plot(-N/2:N/2,h.Numerator) subplot(2,1,2) y = filter(h.Numerator,1,upsample((2 * randi([0 1],1,2000))-1,T)); plot(y(N/2:100*T))