Hi all, I need the above, but I can't find it in any of the books I have handy and every google search I try turns up nothing but references to Matlab :-(. The Chebyshev I analogue prototype is: H(s) = 1 / (s^2 + s / Q + 1) and I was thinking of subtracting the above from 1 but I'm not sure that's correct. Anyone have a book handy? TIA, Erik -- +-----------------------------------------------------------+ Erik de Castro Lopo nospam@mega-nerd.com (Yes it's valid) +-----------------------------------------------------------+ "Even Napoleon had his Watergate" -- Michael Spautz

# 2nd order Chebyshev II analogue prototype

Started by ●October 2, 2004

Reply by ●October 4, 20042004-10-04

"Erik de Castro Lopo" <nospam@mega-nerd.com> wrote in message news:415E2BFA.6470C4F2@mega-nerd.com...> Hi all, > > I need the above, but I can't find it in any of the books I have > handy and every google search I try turns up nothing but references > to Matlab :-(. > > The Chebyshev I analogue prototype is: > > H(s) = 1 / (s^2 + s / Q + 1) > > and I was thinking of subtracting the above from 1 but I'm not sure > that's correct. > > Anyone have a book handy? > > TIA, > Erik > -- > +-----------------------------------------------------------+ > Erik de Castro Lopo nospam@mega-nerd.com (Yes it's valid) > +-----------------------------------------------------------+ > "Even Napoleon had his Watergate" -- Michael SpautzHi Eric: A 2nd order Chebychev type II is the same as a Butterworth, or 1/(s^2 + 1.4142s + 1). For a 2nd order, the stopband notches don't happen. From Proakis and Manolakis: "the family of type II Chebyshev filters contains both poles and zeros and exhibits a monotonic behavior in the passband and an equiripple behavior in the stopband." Regards Ian

Reply by ●October 4, 20042004-10-04

Ian Buckner wrote:> > Hi Erik: > > A 2nd order Chebychev type II is the same as a Butterworth, > or 1/(s^2 + 1.4142s + 1). For a 2nd order, the stopband notches > don't happen. > > From Proakis and Manolakis: "the family of type II Chebyshev > filters contains both poles and zeros and exhibits a monotonic > behavior in the passband and an equiripple behavior in the > stopband."Hi Ian, I've been looking at this a little further since I posted the question, but I wasn't quite ready to repond. Anyway, it seems that: X s^2 + 1 H(s) = ------------- s^2 + Y s + 1 has a frequency response that has a magnitude of 1 at a s == 0, a magnitude of X at s == infinity, a notch in roughly the right position and if X and Y are chosen correctly, has a maximally flat passband. I'm using GNU Octave and I'm doing: w = linspace (0, 3, 1000) ; plot (w, abs (freqs ([0.3 0 1], [1 1.18 1], w))) and that plot looks pretty close to what I'm after. Is there any reason why this is wrong? Erik -- +-----------------------------------------------------------+ Erik de Castro Lopo nospam@mega-nerd.com (Yes it's valid) +-----------------------------------------------------------+ "Python is the most efficient language I've ever used. It's 10 times better than any of the other tools I have used. It's free, it's object-oriented, it adapts to everything, it runs on everything. There is almost an indescribable 'quality without a name' attraction on my part." --Bruce Eckel, Author of Thinking in Java

Reply by ●October 5, 20042004-10-05

"Erik de Castro Lopo" <nospam@mega-nerd.com> wrote in message news:41613A24.DB838F82@mega-nerd.com...> Ian Buckner wrote: > > > > Hi Erik: > > > > A 2nd order Chebychev type II is the same as a Butterworth, > > or 1/(s^2 + 1.4142s + 1). For a 2nd order, the stopband notches > > don't happen. > > > > From Proakis and Manolakis: "the family of type II Chebyshev > > filters contains both poles and zeros and exhibits a monotonic > > behavior in the passband and an equiripple behavior in the > > stopband." > > Hi Ian, > > I've been looking at this a little further since I posted the > question, but I wasn't quite ready to repond. > > Anyway, it seems that: > > X s^2 + 1 > H(s) = ------------- > s^2 + Y s + 1 > > has a frequency response that has a magnitude of 1 at a s == 0, > a magnitude of X at s == infinity, a notch in roughly the right > position and if X and Y are chosen correctly, has a maximally > flat passband. > > I'm using GNU Octave and I'm doing: > > w = linspace (0, 3, 1000) ; > plot (w, abs (freqs ([0.3 0 1], [1 1.18 1], w))) > > and that plot looks pretty close to what I'm after. > > Is there any reason why this is wrong? > > Erik > --Hi Erik: The filter design package I use always has one less notch than is topologically possible for even order Chebychev type II low pass filters. From the help file, the package is using that degree of freedom to make the passband not just monotonic but also maximally flat. Matlab includes the extra notch for even order filters. For a -20dB stopband, and stopband frequency of 1 rad/s, Matlab returns: 1+ 0.5s^2 ------------------ 1+ 3s + 5s^2 I don't know which is "correct", or if you can pick X and Y to achieve both maximally flat passband and equiripple stop (might be achievable for 2nd order, the stopband is always equiripple). Regards Ian

Reply by ●October 5, 20042004-10-05

"Erik de Castro Lopo" <nospam@mega-nerd.com> wrote in message news:41613A24.DB838F82@mega-nerd.com...> Ian Buckner wrote: > > > > Hi Erik: > > > > A 2nd order Chebychev type II is the same as a Butterworth, > > or 1/(s^2 + 1.4142s + 1). For a 2nd order, the stopband notches > > don't happen. > > > > From Proakis and Manolakis: "the family of type II Chebyshev > > filters contains both poles and zeros and exhibits a monotonic > > behavior in the passband and an equiripple behavior in the > > stopband." > > Hi Ian, > > I've been looking at this a little further since I posted the > question, but I wasn't quite ready to repond. > > Anyway, it seems that: > > X s^2 + 1 > H(s) = ------------- > s^2 + Y s + 1 > > has a frequency response that has a magnitude of 1 at a s == 0, > a magnitude of X at s == infinity, a notch in roughly the right > position and if X and Y are chosen correctly, has a maximally > flat passband. > > I'm using GNU Octave and I'm doing: > > w = linspace (0, 3, 1000) ; > plot (w, abs (freqs ([0.3 0 1], [1 1.18 1], w))) > > and that plot looks pretty close to what I'm after. > > Is there any reason why this is wrong? > > ErikHi Erik: The filter design package I use always has one less notch than is topologically possible for even order Chebychev type II low pass filters. From the help file, the package is using that degree of freedom to make the passband not just monotonic but also maximally flat. Matlab includes the extra notch for even order filters. For a -20dB stopband, and stopband frequency of 1 rad/s, Matlab returns: 1+ 0.5s^2 ------------------ 1+ 3s + 5s^2 The way Matlab sets the parameters makes me think they focus on making the stopband equiripple, and don't also try and achieve maximally flat passband. I don't know which is "correct", or if you can pick X and Y to achieve both maximally flat passband and equiripple stop (might be achievable for 2nd order, the stopband is always equiripple). Regards Ian

Reply by ●October 5, 20042004-10-05

Hi Chebyshev II filter as ALLA analog filters have equal number of poles and zeros if you count zeros at infinity. Second-order Chevyshev II and CCauer filters of even order have two finite zeros. However, the magnitude response must tend to zero as w tend to infinity, i.e., there must be at least on zero at infinity, in order (transfer function) to be realized by lumped element LC filters. No problem in the digital domain, however. There exists modification to even order Chebyshev II and cauer filters so that they can be realized by LC filters. Regards, Lasse -- Lars Wanhammar