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Notch filter design using poles and zeroes(2nd order) Newbie question

Started by terrp May 5, 2006
Hi all,
       I was given a matlab project to use a unit circle with 2poles and
2zeroes to create a notch filter. The component to be removed is 50 Hz,
sampling frequency is 200 Hz.
I tried putting 2 zeroes at z=1, angle= pie/2; 
and 2 poles at z=0.9, angle = pie/2. 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.

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....


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....


In order to learn how they work, I suggest starting with only the
zeroes.... maybe only one zero... then add the other zero.. by the way
are they two independent zeros or a complex zero pair?

Then add the poles and same question for them.

Mark

On Fri, 05 May 2006 07:19:15 -0500, "terrp" <terrp@hotmail.com> wrote:

>Hi all, > I was given a matlab project to use a unit circle with 2poles and >2zeroes to create a notch filter. The component to be removed is 50 Hz, >sampling frequency is 200 Hz. >I tried putting 2 zeroes at z=1, angle= pie/2; >and 2 poles at z=0.9, angle = pie/2. 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. > >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....
Hi, put one zero at z = 0 + j1 (|z|=1 at angle pie/2), and put the other zero at z = 0 - j1 (|z|=1 at angle -pie/2). You'll have a real-valued-coefficients filter with notches at +50 Hz and -50 Hz. If you want to make those two notches very narrow, then put one pole at z = 0 + jM (|z|=M at angle pie/2), and put the other pole at z = 0 - jM (|z|=M at angle -pie/2). Then experiment by letting M be in the range of, say, 0.6 -to- 0.99 to see how M affects the width of the notches. [-Rick-]
Check the post from Rune here, it's a very good one:


<http://groups.google.com/group/comp.soft-sys.matlab/browse_frm/thread/bad839e1a15920bb/5248378bbcea10f8?q=notch&rnum=2#5248378bbcea10f8>

"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. The component to be removed is 50 Hz, > sampling frequency is 200 Hz. > I tried putting 2 zeroes at z=1, angle= pie/2; > and 2 poles at z=0.9, angle = pie/2. 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. > > 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.... > >
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. M.P
Rick Lyons wrote:
> On Fri, 05 May 2006 07:19:15 -0500, "terrp" <terrp@hotmail.com> wrote: > > >Hi all, > > I was given a matlab project to use a unit circle with 2poles and > >2zeroes to create a notch filter. The component to be removed is 50 Hz, > >sampling frequency is 200 Hz. > >I tried putting 2 zeroes at z=1, angle= pie/2; > >and 2 poles at z=0.9, angle = pie/2. 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. > > > >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.... > > Hi, > > put one zero at z = 0 + j1 (|z|=1 at > angle pie/2), and put the other zero at > z = 0 - j1 (|z|=1 at angle -pie/2). > You'll have a real-valued-coefficients filter > with notches at +50 Hz and -50 Hz. > > If you want to make those two notches very > narrow, then put one pole at z = 0 + jM (|z|=M at > angle pie/2), and put the other pole at > z = 0 - jM (|z|=M at angle -pie/2). > Then experiment by letting M be in the range > of, say, 0.6 -to- 0.99 to see how M affects > the width of the notches.
All this talk about pies is making me hungry! ;-)
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. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
"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. > &#2013266095;&#2013266095;&#2013266095;
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
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. >> &#2013266095;&#2013266095;&#2013266095; > 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. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
"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. > >> &#2013266095;&#2013266095;&#2013266095; > > 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. > &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
They can be at z=0 ie 1+z^-2=0 M.P