Hi, This is my first post to comp.dsp. I found the filter below but I haven't been able to identify the type of filter it is (so I can understand and modify its parameters, properties, look for the theory, etc). It is used to separate U and V color components of chroma prior to demodulation by rotation according to the phase of the carrier. Using a sine as input, the output of y is 23 degrees behind the input (phase I mean), and z is 67 degrees ahead, meaning y and z are 90 degs apart. y = cy[0] * (x[ 0] - x[1]); y += cy[1] * (x[-1] - x[2]); y += cy[2] * (x[-2] - x[3]); y += cy[3] * (x[-3] - x[4]); y += cy[4] * (x[-4] - x[5]); y += cy[5] * (x[-5] - x[6]); y += cy[6] * (x[-6] - x[7]); y += cy[7] * (x[-7] - x[8]); y >>= 4; z = cz[0] * (x[ 0] + x[1]); z += cz[1] * (x[-1] + x[2]); z += cz[2] * (x[-2] + x[3]); z += cz[3] * (x[-3] + x[4]); z += cz[4] * (x[-4] + x[5]); z += cz[5] * (x[-5] + x[6]); z += cz[6] * (x[-6] + x[7]); z += cz[7] * (x[-7] + x[8]); z >>= 4; Thanks, and I didn't post the coeffs because I'm not sure if they're patented or something) Zero

# What kind of filter is this?

Started by ●November 3, 2003

Reply by ●November 3, 20032003-11-03

"Zerofall" <zerofall@hotmail.com> wrote in message news:628e317a.0311031518.18002a3d@posting.google.com...> Hi, > > This is my first post to comp.dsp. I found the filter below but I haven'tbeen> able to identify the type of filter it is (so I can understand and modifyits> parameters, properties, look for the theory, etc). It is used to separateU and V> color components of chroma prior to demodulation by rotation according tothe> phase of the carrier. > > Using a sine as input, the output of y is 23 degrees behind the input(phase> I mean), and z is 67 degrees ahead, meaning y and z are 90 degs apart. > > > y = cy[0] * (x[ 0] - x[1]); > y += cy[1] * (x[-1] - x[2]); > y += cy[2] * (x[-2] - x[3]); > y += cy[3] * (x[-3] - x[4]); > y += cy[4] * (x[-4] - x[5]); > y += cy[5] * (x[-5] - x[6]); > y += cy[6] * (x[-6] - x[7]); > y += cy[7] * (x[-7] - x[8]); > y >>= 4; > > z = cz[0] * (x[ 0] + x[1]); > z += cz[1] * (x[-1] + x[2]); > z += cz[2] * (x[-2] + x[3]); > z += cz[3] * (x[-3] + x[4]); > z += cz[4] * (x[-4] + x[5]); > z += cz[5] * (x[-5] + x[6]); > z += cz[6] * (x[-6] + x[7]); > z += cz[7] * (x[-7] + x[8]); > z >>= 4; > > Thanks, and I didn't post the coeffs because I'm not sure if they'repatented or> something)I'm just learning all this kind of stuff at Uni myself... however, it looks like the algorithm is a pair of eight tap FIR filters (correction if I'm wrong please). In that case, the only really inmportant aspect is the actual value of the coefficients. They determine the response of the filter. If all it does is shift the phase of the input signal, then I would say it's a phase shifting filter (all-pass??) I'm sure you'll get better info from the others in the group... Bevan Weiss

Reply by ●November 3, 20032003-11-03

Zerofall wrote:> > Hi, > > This is my first post to comp.dsp. I found the filter below but I haven't been > able to identify the type of filter it is (so I can understand and modify its > parameters, properties, look for the theory, etc). It is used to separate U and V > color components of chroma prior to demodulation by rotation according to the > phase of the carrier. > > Using a sine as input, the output of y is 23 degrees behind the input (phase > I mean), and z is 67 degrees ahead, meaning y and z are 90 degs apart. > > y = cy[0] * (x[ 0] - x[1]); > y += cy[1] * (x[-1] - x[2]); > y += cy[2] * (x[-2] - x[3]); > y += cy[3] * (x[-3] - x[4]); > y += cy[4] * (x[-4] - x[5]); > y += cy[5] * (x[-5] - x[6]); > y += cy[6] * (x[-6] - x[7]); > y += cy[7] * (x[-7] - x[8]); > y >>= 4; > > z = cz[0] * (x[ 0] + x[1]); > z += cz[1] * (x[-1] + x[2]); > z += cz[2] * (x[-2] + x[3]); > z += cz[3] * (x[-3] + x[4]); > z += cz[4] * (x[-4] + x[5]); > z += cz[5] * (x[-5] + x[6]); > z += cz[6] * (x[-6] + x[7]); > z += cz[7] * (x[-7] + x[8]); > z >>= 4; > > Thanks, and I didn't post the coeffs because I'm not sure if they're patented or > something)They are FIR filters. The first is anti-symmetric the second symmetric. The frequency response of the first is pure real, and the second is pure imaginary. The lag and lead you noted is dependent on frequency of the sine wave you input. If you want to make a gain vs frequency plot use these formulas: Y = sum(i:0,7) cy[i]*sin(2*i*pi*f) Z = sum(i:0,7) cz[i]*cos(2*i*pi*f) -jim -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Reply by ●November 3, 20032003-11-03

It's a pair of 16-tap, linear phase FIR filters. The code is optimized to reduce the number of multiplications to 1 for each pair of equal magnitude coefficients. The y filter is even and the z filter odd. Given that, what you said about your test, and because there are two of them, the filters are probably a Hilbert transform pair.

Reply by ●November 3, 20032003-11-03

jim wrote:> They are FIR filters. The first is anti-symmetric the second symmetric. The > frequency response of the first is pure real, and the second is pure imaginary.Sorry, that's backwards - A symmetric filter has a real frequency response. An anti-symmetric has the imaginary frequency response. -jim -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Reply by ●November 3, 20032003-11-03

Bevan Weiss wrote:> "Zerofall" <zerofall@hotmail.com> wrote in message > news:628e317a.0311031518.18002a3d@posting.google.com... > >>Hi, >> >>This is my first post to comp.dsp. I found the filter below but I haven't > > been > >>able to identify the type of filter it is (so I can understand and modify > > its > >>parameters, properties, look for the theory, etc). It is used to separate > > U and V > >> color components of chroma prior to demodulation by rotation according to > > the > >>phase of the carrier. >> >>Using a sine as input, the output of y is 23 degrees behind the input > > (phase > >> I mean), and z is 67 degrees ahead, meaning y and z are 90 degs apart. >> >> >>y = cy[0] * (x[ 0] - x[1]); >>y += cy[1] * (x[-1] - x[2]); >>y += cy[2] * (x[-2] - x[3]); >>y += cy[3] * (x[-3] - x[4]); >>y += cy[4] * (x[-4] - x[5]); >>y += cy[5] * (x[-5] - x[6]); >>y += cy[6] * (x[-6] - x[7]); >>y += cy[7] * (x[-7] - x[8]); >>y >>= 4; >> >>z = cz[0] * (x[ 0] + x[1]); >>z += cz[1] * (x[-1] + x[2]); >>z += cz[2] * (x[-2] + x[3]); >>z += cz[3] * (x[-3] + x[4]); >>z += cz[4] * (x[-4] + x[5]); >>z += cz[5] * (x[-5] + x[6]); >>z += cz[6] * (x[-6] + x[7]); >>z += cz[7] * (x[-7] + x[8]); >>z >>= 4; >> >>Thanks, and I didn't post the coeffs because I'm not sure if they're > > patented or > >>something) > > > I'm just learning all this kind of stuff at Uni myself... > however, it looks like the algorithm is a pair of eight tap FIR filters > (correction if I'm wrong please). > In that case, the only really inmportant aspect is the actual value of the > coefficients. They determine the response of the filter. If all it does is > shift the phase of the input signal, then I would say it's a phase shifting > filter (all-pass??) > > I'm sure you'll get better info from the others in the group... > > Bevan Weiss > >It's not really possible to characterize a filter by probing it at a single frequency. I don't think you can patent coefficients. You can copyright a road map, but you can't patent a route. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●November 4, 20032003-11-04

Hi, Wow, so many replies, thank you, I'm posting the coefficients and the rotation by the carrier's phase (cordic style), maybe it will clarify things. y = 704 * (x[ 0] - x[1]); y += 1559 * (x[-1] - x[2]); y += 1078 * (x[-2] - x[3]); y += -457 * (x[-3] - x[4]); y += -1020 * (x[-4] - x[5]); y += -282 * (x[-5] - x[6]); y += 199 * (x[-6] - x[7]); y += 141 * (x[-7] - x[8]); y >>= 4; z = 1700 * (x[ 0] + x[1]); z += 364 * (x[-1] + x[2]); z += -1525 * (x[-2] + x[3]); z += -2182 * (x[-3] + x[4]); z += 821 * (x[-4] + x[5]); z += 962 * (x[-5] + x[6]); z += 199 * (x[-6] + x[7]); z += -340 * (x[-7] + x[8]); z >>= 4; fs = cos1[(Phase - 90deg)]; fc = cos1[(Phase )]; u = (fs*y + fc*z) >> 8; v = (fc*y - fs*z) >> 8; What confuses me is that it uses "(x[0]-x[1]" and (x[-1]-x[2]) and so on, cause the coefficients are multiplying at differentials between samples that get farther and farther from each other (0 1,-1 2, -2 3, etc), or is it as Matt said that the actual 16-tap filter would be: y = 0; y -= 141 * x[8]; y -= 199 * x[7]; y -= -282 * x[6]; y -= -1020 * x[5]; y -= -457 * x[4]; y -= 1078 * x[3]; y -= 1559 * x[2]; y -= 704 * x[1]; y += 704 * x[0]; y += 1559 * x[-1]; y += 1078 * x[-2]; y += -457 * x[-3]; y += -1020 * x[-4]; y += -282 * x[-5]; y += 199 * x[-6]; y += 141 * x[-7]; also meaning that (as Jim said) it is anti-symmetric (because the coeffs would be: ......,-1559,-704,704,1559,...... ), representing the "real" freq resp, that I think would represent the "cosine samples" for the UV extraction, which would be according to the fact that in this case the "y" value would be rotated (using cordic) to become V, and "z" would be U, components of chroma (Chroma = U*sin(wt)+V*cos(wt)). Ouch, I hope I got everything right, of course a newbee like me will most likely get really messed up. Can anyone point out a website (preferable, because in my country bookstores and libraries don't have many DSP books) where I could find some theory/lectures about this type of filters, because I'm looking forward to design filters like these to perform I-Q extraction (I'm not sure if it would be enough to just add 33degrees to the Cordic rotation, starting from the same filters). Also, how do I change the email address so I don't get spam, Google wants to check the new *fony* address I want to put in. Thanks Zero

Reply by ●November 6, 20032003-11-06