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

Started by 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:

```
```"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...
>
>    ...
>
> > 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...
>>
>>    ...
>>
>>> 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...
> > "Jerry Avins" <jya@ieee.org> wrote in message
> > news:-dCdnVcixZYYaMbZRVn-iA@rcn.net...
> >>
> >>    ...
> >>
> >>> 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

```