Reply by robert bristow-johnson●March 13, 20062006-03-13

laura_pretty05@yahoo.com.hk wrote:

> in fact, what is the meaning of prewarping?

here's another one. i once posted this to the music-dsp list to answer
precisely the same question. someone liked it enough that the put it
in their code archive as an "explanation". it really is nothing other
than the textbook stuff.
____
prewarping is simply recognizing the warping that the BLT introduces.
to determine frequency response, we evaluate the digital H(z) at
z=exp(j*w*T) and we evaluate the analog Ha(s) at s=j*W . the following
will confirm the jw to unit circle mapping and will show exactly what
the
mapping is (this is the same stuff in the textbooks):
the BLT says: s = (2/T) * (z-1)/(z+1)
substituting: s = j*W = (2/T) * (exp(j*w*T) - 1) / (exp(j*w*T) + 1)
(exp(j*w*T/2) - exp(-j*w*T/2))
j*W = (2/T) * ---------------------------------------------
(exp(j*w*T/2) + exp(-j*w*T/2))
= (2/T) * (j*2*sin(w*T/2)) / (2*cos(w*T/2))
= j * (2/T) * tan(w*T/2)
or
analog W = (2/T) * tan(w*T/2)
so when the real input frequency is w, the digital filter will behave
with
the same amplitude gain and phase shift as the analog filter will have
at a
hypothetical frequency of W. as w*T approaches pi (Nyquist) the digital
filter behaves as the analog filter does as W -> inf. for each degree
of
freedom that you have in your design equations, you can adjust the
analog
design frequency to be just right so that when the deterministic BLT
warping does its thing, the resultant warped frequency comes out just
right. for a simple LPF, you have only one degree of freedom, the
cutoff
frequency. you can precompensate it so that the true cutoff comes out
right but that is it, above the cutoff, you will see that the LPF dives
down to -inf dB faster than an equivalent analog at the same
frequencies.
r b-j

Reply by Jerry Avins●March 13, 20062006-03-13

laura_pretty05@yahoo.com.hk wrote:

> Noway2,
> in fact, what is the meaning of prewarping?
> Laura

Laura,
http://www.bores.com/courses/intro/iir/5_warp.htm has a nice
explanation. http://www.bores.com/courses/intro/iir/index.htm backs up
one step. http://www.bores.com/courses/intro/index.htm comes before that.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Reply by ●March 13, 20062006-03-13

Noway2,
in fact, what is the meaning of prewarping?
Laura
robert bristow-johnson =E5=AF=AB=E9=81=93=EF=BC=9A

> Noway2 wrote:
> > The impulse invariance IS the method to convert H(s) to H(z)
>
> it isn't the only way do convert H(s) to H(z) and seldom is, from what
> i can tell, the preferred method. Impulse invariant essentially causes
> aliasing in the frequency response. sometimes that's okay, sometimes
> not.
>
> > it does require taking the inverse laplace and then taking the Z
> > transform of the terms.
>
> you left out one important step: after inverse L.T., you *sample* the
> impulse response (getting a discrete-time impulse response), and
> Z-transform that sucker. hence, at least at the sampling instances, it
> is the same impulse response (but it's not in-between sampling
> instances due to bandlimited reconstruction of the sampled impulse
> response). this sampling of the analog impulse response is what causes
> aliasing or folding of the frequency response about the Nyquist
> frequency.
>
> > If you are having trouble with this you may want to check out the
> > Bilinear transform which lets you plug in a (z) substitution for the s.
> > While it doesn't require taking the inverse laplace, partial
> > fractions, and directly taking the z transform, the algebra can be very
> > messy.
>
> you can do bilinear transform by simply applying the BLT to the poles
> and zeros. no more messy than that. (well, i guess with compensation
> for BLT frequency warping, called "prewarping" sometimes, there is a
> little more mess to it.)
>=20
> r b-j

Reply by robert bristow-johnson●March 13, 20062006-03-13

Noway2 wrote:

> The impulse invariance IS the method to convert H(s) to H(z)

it isn't the only way do convert H(s) to H(z) and seldom is, from what
i can tell, the preferred method. Impulse invariant essentially causes
aliasing in the frequency response. sometimes that's okay, sometimes
not.

> it does require taking the inverse laplace and then taking the Z
> transform of the terms.

you left out one important step: after inverse L.T., you *sample* the
impulse response (getting a discrete-time impulse response), and
Z-transform that sucker. hence, at least at the sampling instances, it
is the same impulse response (but it's not in-between sampling
instances due to bandlimited reconstruction of the sampled impulse
response). this sampling of the analog impulse response is what causes
aliasing or folding of the frequency response about the Nyquist
frequency.

> If you are having trouble with this you may want to check out the
> Bilinear transform which lets you plug in a (z) substitution for the s.
> While it doesn't require taking the inverse laplace, partial
> fractions, and directly taking the z transform, the algebra can be very
> messy.

you can do bilinear transform by simply applying the BLT to the poles
and zeros. no more messy than that. (well, i guess with compensation
for BLT frequency warping, called "prewarping" sometimes, there is a
little more mess to it.)
r b-j

Reply by ●March 13, 20062006-03-13

Tam,
When expanding the H(s) in partial fractions, it is so long and very
time consuming. Can use it in matlab?? but i am not familiar with
matlab.
Laura
HelpmaBoab =E5=AF=AB=E9=81=93=EF=BC=9A

> <laura_pretty05@yahoo.com.hk> wrote in message
> news:1142180317.710183.32790@z34g2000cwc.googlegroups.com...
> > Now, I use the impulse invariance to design a lowpass filter. I need to
> > find the H(s) from pole pairs. I don't know how toconvert H(s) to
> > H(z). Can anyone help me? Thanks!!
> >
> > Laura
> >
>=20
> Expand H(s) in partial fractions and use the tables.
>=20
>=20
> Tam

Reply by Noway2●March 13, 20062006-03-13

The impulse invariance IS the method to conververt H(s) to H(z) though
it does require taking the inverse laplace and then taking the Z
transform of the terms.
If you are having trouble with this you may want to check out the
Bilinear transform which lets you plug in a (z) substitution for the s.
While it doesn't require taking the inverse laplace, partial
fractions, and directly taking the z transform, the algebra can be very
messy.

Reply by HelpmaBoab●March 13, 20062006-03-13

<laura_pretty05@yahoo.com.hk> wrote in message
news:1142180317.710183.32790@z34g2000cwc.googlegroups.com...

> Now, I use the impulse invariance to design a lowpass filter. I need to
> find the H(s) from pole pairs. I don't know how toconvert H(s) to
> H(z). Can anyone help me? Thanks!!
>
> Laura
>

Expand H(s) in partial fractions and use the tables.
Tam

Reply by ●March 12, 20062006-03-12

Now, I use the impulse invariance to design a lowpass filter. I need to
find the H(s) from pole pairs. I don't know how toconvert H(s) to
H(z). Can anyone help me? Thanks!!
Laura