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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  n...@mega-nerd.com (Yes it's valid)
+-----------------------------------------------------------+
"Even Napoleon had his Watergate" -- Michael Spautz

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"Erik de Castro Lopo" <n...@mega-nerd.com> wrote in message
news:4...@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  n...@mega-nerd.com (Yes it's valid)
> +-----------------------------------------------------------+
> "Even Napoleon had his Watergate" -- Michael Spautz

Hi 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

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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  n...@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

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"Erik de Castro Lopo" <n...@mega-nerd.com> wrote in message
news:4...@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

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"Erik de Castro Lopo" <n...@mega-nerd.com> wrote in message
news:4...@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

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

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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

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