# Question about IIR filter design using Impulse Invariance method

Started by November 13, 2004
```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.

```
```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.

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.
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
```
```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 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.
>
> 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.
> &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;

```
```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.
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
```
```"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.
>
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Hi Bobby : is there a z where H(z) = 0  and where z<>0 ?
best of luck - Mike

```
```"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

```
```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.
> &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;

```
```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
>
>

```
```"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