Reply by Peter Nachtwey May 8, 20062006-05-08
Jerry Avins wrote:
> Peter Nachtwey wrote: > > > "Jerry Avins" <jya@ieee.org> wrote in message > > news:voWdnfojucgzr8PZRVn-qw@rcn.net... > > | Peter Nachtwey wrote: > > | > > | ... > > | > > | > Look at the link It works through your problem step by step and it > > works, > > | > but I started in the s domain. > > | > ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch.pdf > > | > > | ... > > | > > | How does the graph "Notch Filter Response" on page 5 differ from that of > > | a short to ground? > > | > > | Jerry > > | -- > > | Engineering is the art of making what you want from things you can get. > > | > > I am just showing the filter works well for an input frequency of 50 Hz. If > > the input frequency > > is changed then the filter starts to pass the input signal. The Q on my > > filter is large so the notch isn't > > very narrow. I changed only the input frequency to 49 hz to show the > > difference.. > > ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch49Hz.pdf > > You have Mathcad. Try it. > > We're at cross purposes here. The graph on page 5 of > ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch.pdf is a > straight line. Something in the worksheet is broken. Look again.
There are two .pdf files now ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch.pdf ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch49Hz.pdf Notice that the second one has a 49Hz input that does get passed passed through a little bit.
> I just went through the equations. It's not a frequency response plot, > but a time-domain plot (I assumed the axis to have been mislabeled; > should have known better) with input at 50 Hz.
I like to test my work to see if I get the desired results. I don't like to screw up in front of the whole world. Peter Nachtwey
Reply by Jerry Avins May 7, 20062006-05-07
Peter Nachtwey wrote:

> "Jerry Avins" <jya@ieee.org> wrote in message > news:voWdnfojucgzr8PZRVn-qw@rcn.net... > | Peter Nachtwey wrote: > | > | ... > | > | > Look at the link It works through your problem step by step and it > works, > | > but I started in the s domain. > | > ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch.pdf > | > | ... > | > | How does the graph "Notch Filter Response" on page 5 differ from that of > | a short to ground? > | > | Jerry > | -- > | Engineering is the art of making what you want from things you can get. > | > I am just showing the filter works well for an input frequency of 50 Hz. If > the input frequency > is changed then the filter starts to pass the input signal. The Q on my > filter is large so the notch isn't > very narrow. I changed only the input frequency to 49 hz to show the > difference.. > ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch49Hz.pdf > You have Mathcad. Try it.
We're at cross purposes here. The graph on page 5 of ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch.pdf is a straight line. Something in the worksheet is broken. Look again. I just went through the equations. It's not a frequency response plot, but a time-domain plot (I assumed the axis to have been mislabeled; should have known better) with input at 50 Hz. Sorry: Question withdrawn. 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;
Reply by Peter Nachtwey May 7, 20062006-05-07
"Jerry Avins" <jya@ieee.org> wrote in message 
news:voWdnfojucgzr8PZRVn-qw@rcn.net...
| Peter Nachtwey wrote:
|
|  ...
|
| > Look at the link   It works through your problem step by step and it 
works,
| > but I started in the s domain.
| > ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch.pdf
|
|  ...
|
| How does the graph "Notch Filter Response" on page 5 differ from that of
| a short to ground?
|
| Jerry
| -- 
| Engineering is the art of making what you want from things you can get.
|
I am just showing the filter works well for an input frequency of 50 Hz.  If 
the input frequency
is changed then the filter starts to pass the input signal.   The Q on my 
filter is large so the notch isn't
very narrow.   I changed only the input frequency to 49 hz to show the 
difference..
ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch49Hz.pdf
You have Mathcad.  Try it.

Peter Nachtwey


 &#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; 


Reply by Jerry Avins May 7, 20062006-05-07
Peter Nachtwey wrote:

  ...

> Look at the link It works through your problem step by step and it works, > but I started in the s domain. > ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch.pdf
... How does the graph "Notch Filter Response" on page 5 differ from that of a short to ground? 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;
Reply by Peter Nachtwey May 7, 20062006-05-07
"terrp" <terrp@hotmail.com> wrote in message 
news:_t6dnWlhYrBe38bZRVn-tg@giganews.com...
| Hi all,
|       I was given a matlab project to use a unit circle with 2poles and
| 2zeroes to create a notch filter.

Does it have to be in the z domain?   It is easier to start with the s 
domain.

| The component to be removed is 50 Hz,
| sampling frequency is 200 Hz.

The notch and sample frequency are selected to make placing the zeros easy

| I tried putting 2 zeroes at z=1, angle= pie/2;

Every one got that part.   I assume you really mean +/- PI/2

| and 2 poles at z=0.9, angle = pie/2.

Now how did you arrive at this?
Did you remember to prewarp the frequency?
No one has mentioned that yet.  This is important when the sample frequency
is not MUCH higher than the notch frequency.  I think this is part of the 
point
of the exercise.

| However, when i look at the spectra
| using SPTOOL MATLAB, nothing seems to be removed from the spectra. I tried
| playing with the values at gain, but it seems the frequency components at
| 20 Hz and above is removed.

I didn't verify that but your poles don't match mine.
Netiher did any of the other guesses.

| Any advice on how to create a notch filter using poles and zeroes to
| remove 50 Hz components from signal. (MATLAB SPTOOL). Can't figure it
| out....

Look at the link   It works through your problem step by step and it works,
but I started in the s domain.
ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20Notch.pdf
So what does the pre-warping do to the poles?   Mine don't look right.
I too had to think about why.
How does the sample frequency affect the poles in the z domain?

Peter Nachtwey


Reply by Jerry Avins May 6, 20062006-05-06
Mad Prof wrote:
> "Jerry Avins" <jya@ieee.org> wrote in message > news:MKGdnYh7J52eucHZnZ2dnUVZ_v-dnZ2d@rcn.net... > >>Mad Prof wrote: >> >>>"Jerry Avins" <jya@ieee.org> wrote in message >>>news:-dCdnVcixZYYaMbZRVn-iA@rcn.net... >>> >>>>Mad Prof wrote: >>>> >>>> ... >>>> >>>> >>>>>The easy answer is to consider 50Hz as a fraction of 200Hz ie 50/200 = >>> >>>1/4. >>> >>>>>So you need complex zeros at 1/4 sampling freq which is pi/2 and -pi/2 >>>>>(since 2pi is fs). >>>>>So we have zeros ta z=0+j1 and z=-0-j1. The filter is therefore >>>>> >>>>>(z-j1)(z+j1)=z^2+1. >>>> >>>>What about the poles to sharpen the notch? >>>> >>>>Jerry >>>>-- >>>>Engineering is the art of making what you want from things you can get. >>>>&#4294967295;&#4294967295;&#4294967295; >>> >>>I think for a college exercise that maybe this is all he wants. After > > all we > >>>should really design Butterworth filters etc but this is just an > > exercise on > >>>the unit circle. >> >>If you read the original question, you will see that he was asked to >>place two zeros and two poles. >> >>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; > > > They can be at z=0 ie 1+z^-2=0
They can, but there are better locations. 0.99 @ +/-pi/2 seem promising. 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;
Reply by Mad Prof May 6, 20062006-05-06
"Jerry Avins" <jya@ieee.org> wrote in message
news:MKGdnYh7J52eucHZnZ2dnUVZ_v-dnZ2d@rcn.net...
> Mad Prof wrote: > > "Jerry Avins" <jya@ieee.org> wrote in message > > news:-dCdnVcixZYYaMbZRVn-iA@rcn.net... > >> Mad Prof wrote: > >> > >> ... > >> > >>> The easy answer is to consider 50Hz as a fraction of 200Hz ie 50/200 = > > 1/4. > >>> So you need complex zeros at 1/4 sampling freq which is pi/2 and -pi/2 > >>> (since 2pi is fs). > >>> So we have zeros ta z=0+j1 and z=-0-j1. The filter is therefore > >>> > >>> (z-j1)(z+j1)=z^2+1. > >> What about the poles to sharpen the notch? > >> > >> Jerry > >> -- > >> Engineering is the art of making what you want from things you can get. > >> &#4294967295;&#4294967295;&#4294967295; > > I think for a college exercise that maybe this is all he wants. After
all we
> > should really design Butterworth filters etc but this is just an
exercise on
> > the unit circle. > > If you read the original question, you will see that he was asked to > place two zeros and two poles. > > 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;
They can be at z=0 ie 1+z^-2=0 M.P
Reply by Jerry Avins May 6, 20062006-05-06
Mad Prof wrote:
> "Jerry Avins" <jya@ieee.org> wrote in message > news:-dCdnVcixZYYaMbZRVn-iA@rcn.net... >> Mad Prof wrote: >> >> ... >> >>> The easy answer is to consider 50Hz as a fraction of 200Hz ie 50/200 = > 1/4. >>> So you need complex zeros at 1/4 sampling freq which is pi/2 and -pi/2 >>> (since 2pi is fs). >>> So we have zeros ta z=0+j1 and z=-0-j1. The filter is therefore >>> >>> (z-j1)(z+j1)=z^2+1. >> What about the poles to sharpen the notch? >> >> Jerry >> -- >> Engineering is the art of making what you want from things you can get. >> &#4294967295;&#4294967295;&#4294967295; > I think for a college exercise that maybe this is all he wants. After all we > should really design Butterworth filters etc but this is just an exercise on > the unit circle.
If you read the original question, you will see that he was asked to place two zeros and two poles. 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;
Reply by Mad Prof May 5, 20062006-05-05
"Jerry Avins" <jya@ieee.org> wrote in message
news:-dCdnVcixZYYaMbZRVn-iA@rcn.net...
> Mad Prof wrote: > > ... > > > The easy answer is to consider 50Hz as a fraction of 200Hz ie 50/200 =
1/4.
> > So you need complex zeros at 1/4 sampling freq which is pi/2 and -pi/2 > > (since 2pi is fs). > > So we have zeros ta z=0+j1 and z=-0-j1. The filter is therefore > > > > (z-j1)(z+j1)=z^2+1. > > What about the poles to sharpen the notch? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > &#4294967295;&#4294967295;&#4294967295;
I think for a college exercise that maybe this is all he wants. After all we should really design Butterworth filters etc but this is just an exercise on the unit circle. M.P
Reply by Jerry Avins May 5, 20062006-05-05
Mad Prof wrote:

   ...

> The easy answer is to consider 50Hz as a fraction of 200Hz ie 50/200 = 1/4. > So you need complex zeros at 1/4 sampling freq which is pi/2 and -pi/2 > (since 2pi is fs). > So we have zeros ta z=0+j1 and z=-0-j1. The filter is therefore > > (z-j1)(z+j1)=z^2+1.
What about the poles to sharpen the notch? 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;