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

# DSP-impulse invariance method

Started by ●March 12, 2006

Reply by ●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 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 ●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 ●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

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 ●March 13, 20062006-03-13

laura_pretty05@yahoo.com.hk wrote:> Noway2, > in fact, what is the meaning of prewarping? > LauraLaura, 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

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