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From: laura_pretty05 on 12 Mar 2006 11:18
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

From: HelpmaBoab on 13 Mar 2006 00:29

<laura_pretty05(a)yahoo.com.hk> wrote in message
news:1142180317.710183.32790(a)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


From: Noway2 on 13 Mar 2006 09:05
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.

From: laura_pretty05 on 13 Mar 2006 12:30
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 寫道：

> <laura_pretty05(a)yahoo.com.hk> wrote in message
> news:1142180317.710183.32790(a)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

From: robert bristow-johnson on 13 Mar 2006 13:03

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

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