I am designing low pass IIR filter using Impulse Invariance method. I've got transfer function H(z) which depends on impulse sample period T. I was asked to choose an appropriate impulse sample period T for H(z) such that the input signal x(t) = 5 cos(2pi(8000)t ? pi/3) ? 4 cos(2pi(40000)t + pi/2) sampled at a rate of Fs = 100 kHz has only the 8000 Hz sinusoid in the passband. I don't understand what is relation between impulse sample period T and what sinusoid will be in the passband of the filter. My understanding is that impulse sample period T determines where aliasing starts. Whether 40000Hz simusoid wil or will not be in passband depends on cutoff frequency. I am confused, can anyone to clarify the situation. Thanks in advance.

# Question about IIR filter design using Impulse Invariance method

Started by ●November 13, 2004

Reply by ●November 13, 20042004-11-13

Bobby wrote:> I am designing low pass IIR filter using Impulse Invariance method. > I've got transfer function H(z) which depends on impulse sample period T. > > I was asked to choose an appropriate impulse sample period T for H(z) > such that the input signal > > x(t) = 5 cos(2pi(8000)t ? pi/3) ? 4 cos(2pi(40000)t + pi/2) > > sampled at a rate of Fs = 100 kHz has only the 8000 Hz sinusoid in > the passband. > > I don't understand what is relation between impulse sample period T > and what sinusoid will be in the passband of the filter. > > My understanding is that impulse sample period T determines where > aliasing starts. Whether 40000Hz simusoid wil or will not be in passband > depends on cutoff frequency. I am confused, can anyone to clarify the > situation. > Thanks in advance.The cutoff frequency of a filter is a fraction of the sample frequency -- certainly not more than half -- that depends on its coefficients. Given the sampling period, you can place the cutoff where you want by choosing appropriate coefficients. With any given set of coefficients, you can choose a sample period to place the cutoff where you want provided that the period remains short enough to avoid aliasing. The question marks in the line x(t) = 5 cos(2pi(8000)t ? pi/3) ? 4 cos(2pi(40000)t + pi/2) probably represent a mathematical symbol that my program doesn't reproduce. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●November 13, 20042004-11-13

Yes , the signal should look as follows: x(t) = 5 cos(2pi(8000)t - pi/3) - 4 cos(2pi(40000)t + pi/2) "Jerry Avins" <jya@ieee.org> wrote in message news:2vnfr2F2id4vaU1@uni-berlin.de...> Bobby wrote: > > > I am designing low pass IIR filter using Impulse Invariance method. > > I've got transfer function H(z) which depends on impulse sample periodT.> > > > I was asked to choose an appropriate impulse sample period T for H(z) > > such that the input signal > > > > x(t) = 5 cos(2pi(8000)t ? pi/3) ? 4 cos(2pi(40000)t + pi/2) > > > > sampled at a rate of Fs = 100 kHz has only the 8000 Hz sinusoid in > > the passband. > > > > I don't understand what is relation between impulse sample period T > > and what sinusoid will be in the passband of the filter. > > > > My understanding is that impulse sample period T determines where > > aliasing starts. Whether 40000Hz simusoid wil or will not be inpassband> > depends on cutoff frequency. I am confused, can anyone to clarify the > > situation. > > Thanks in advance. > > The cutoff frequency of a filter is a fraction of the sample frequency > -- certainly not more than half -- that depends on its coefficients. > Given the sampling period, you can place the cutoff where you want by > choosing appropriate coefficients. With any given set of coefficients, > you can choose a sample period to place the cutoff where you want > provided that the period remains short enough to avoid aliasing. > > The question marks in the line > > x(t) = 5 cos(2pi(8000)t ? pi/3) ? 4 cos(2pi(40000)t + pi/2) > > probably represent a mathematical symbol that my program doesn'treproduce.> > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������

Reply by ●November 13, 20042004-11-13

Bobby wrote:> Yes , the signal should look as follows: > x(t) = 5 cos(2pi(8000)t - pi/3) - 4 cos(2pi(40000)t + pi/2)Then clearly, you must locate the cutoff somewhere between 8 and 40 KHz. Given that the sample rate is 100 KHz, the cutoff must be between .08 and .4 relative to the sample rate. Does that help? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●November 14, 20042004-11-14

"Jerry Avins" <jya@ieee.org> wrote in message news:2vnvntF2lldunU1@uni-berlin.de...> Bobby wrote: > > > Yes , the signal should look as follows: > > x(t) = 5 cos(2pi(8000)t - pi/3) - 4 cos(2pi(40000)t + pi/2) > > Then clearly, you must locate the cutoff somewhere between 8 and 40 KHz. > Given that the sample rate is 100 KHz, the cutoff must be between .08 > and .4 relative to the sample rate. Does that help? > > Jerry > -- > Engineering is the art of making what you want from things you can get. >����������������������������������������������������������������������� Hi Bobby : is there a z where H(z) = 0 and where z<>0 ? best of luck - Mike

Reply by ●November 14, 20042004-11-14

"Mike Yarwood" <mpyarwood@btopenworld.com> wrote in message news:cn7ti5$3d7$1@hercules.btinternet.com...><snip>> Hi Bobby : is there a z where H(z) = 0 and where z<>0 ? > best of luck - Mike >Should have added : and H(2z) <> 0 . Best of luck = Mike

Reply by ●November 15, 20042004-11-15

Thank you Jerry for your response. "Jerry Avins" <jya@ieee.org> wrote in message news:2vnvntF2lldunU1@uni-berlin.de...> Bobby wrote: > > > Yes , the signal should look as follows: > > x(t) = 5 cos(2pi(8000)t - pi/3) - 4 cos(2pi(40000)t + pi/2) > > Then clearly, you must locate the cutoff somewhere between 8 and 40 KHz. > Given that the sample rate is 100 KHz, the cutoff must be between .08 > and .4 relative to the sample rate. Does that help? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������

Reply by ●November 15, 20042004-11-15

Thanks Mike, I think I've solved this problem. "Mike Yarwood" <mpyarwood@btopenworld.com> wrote in message news:cn8cs4$q5c$1@hercules.btinternet.com...> > "Mike Yarwood" <mpyarwood@btopenworld.com> wrote in message > news:cn7ti5$3d7$1@hercules.btinternet.com... > > > <snip> > > > Hi Bobby : is there a z where H(z) = 0 and where z<>0 ? > > best of luck - Mike > > > Should have added : and H(2z) <> 0 . > > Best of luck = Mike > >

Reply by ●November 15, 20042004-11-15

"Bobby" <waystark@yahoo.com> wrote in message news:<Xtsld.83987$vO1.486568@nnrp1.uunet.ca>...> I am designing low pass IIR filter using Impulse Invariance method. > I've got transfer function H(z) which depends on impulse sample period T. > > I was asked to choose an appropriate impulse sample period T for H(z) > such that the input signal > > x(t) = 5 cos(2pi(8000)t ? pi/3) ? 4 cos(2pi(40000)t + pi/2) > > sampled at a rate of Fs = 100 kHz has only the 8000 Hz sinusoid in > the passband.There is a problem with terminology here. The standard use of the terms "sample period T" and "sampling frequency Fs" is T = 1/Fs so that given Fs= 100 kHz I can't see that you have any choise other than setting T = 1e-5 s. The problem you describe seems to be a contradicion in terms, you can't choose a T while keeping a given Fs, if you use the standard terminology. Some crucial piece of information is missing here.> I don't understand what is relation between impulse sample period T > and what sinusoid will be in the passband of the filter.That depends on the filter coefficients as well.> My understanding is that impulse sample period T determines where > aliasing starts.Correct.> Whether 40000Hz simusoid wil or will not be in passband > depends on cutoff frequency.That's right. You have two ways of adjusting the _physical_ response of a filter: Either keep the sampling parameters fixed, and play with the coefficients, or keep the coefficients fixed and play with the sampling parameters.> I am confused, can anyone to clarify the > situation.You need to sort out the terminology and the problem. Do you have a set of given filter coefficients? Would these be designed from a Fs = 100 kHz spec? Is your task to use this spec and adjust the sampling frequency so that these filter coefficients match the physical filtering task? Do you work with an application where it is easier to adjust the sampling frequency, than computing a new set of filter coefficients? Could you be so kind to tell us more about what application you work with?> Thanks in advance.Rune