# Anti-Aliasing filter

Started by August 23, 2005
```Ok I am sampling with an ordinary A/D (not sigma-delta) at 11025kHz
which mean I need any noise to be attenuated at half sampling which is
at 5512.5Hz. I have read that the nosie level needs to be less that the
R.M.S Quantisation Level of my A/D. Now I swing +or - 10 volts with 16
bits. So my Quantisation level is

delta = dynamic range/2^16=20/65536=0.000305176 volts.

(or 0.305mV)

Now my Quantisaton RMS voltage is delta/sqrt(12)

=88.096 micro volts.

So my noise needs to be lower than this at Fs/2 and then it will be at
least as low as the quantistaion noise.

Suppose I use a Butterworth filter (3dB passband freq)with passband
freq of 4kHz. To go from full swing of 10 volts to 88.096 micro volts
requires an attenuation in the stopband of 100dB approx - with a
stopband freq of Fs/2=5512Hz.

The order of my filter will then be

n=log(10^0.1*100-1)^0.5/log(1.378)

which is approx 36th order!!

What am I doing wrong...is my 4kHz passband assuming too much - do I
have to sacrifice this?

```
```OOps- I forgot that should be a full swing in rms ie 10/Sqrt(2) and not
10 when calculating the attenuation but it doesnt help much.

```
```naebad wrote:

> Ok I am sampling with an ordinary A/D (not sigma-delta) at 11025kHz
> which mean I need any noise to be attenuated at half sampling which is
> at 5512.5Hz. I have read that the nosie level needs to be less that the
> R.M.S Quantisation Level of my A/D. Now I swing +or - 10 volts with 16
> bits. So my Quantisation level is
>
> delta = dynamic range/2^16=20/65536=0.000305176 volts.
>
> (or 0.305mV)
>
> Now my Quantisaton RMS voltage is delta/sqrt(12)
>
> =88.096 micro volts.
>
> So my noise needs to be lower than this at Fs/2 and then it will be at
> least as low as the quantistaion noise.
>
> Suppose I use a Butterworth filter (3dB passband freq)with passband
> freq of 4kHz. To go from full swing of 10 volts to 88.096 micro volts
> requires an attenuation in the stopband of 100dB approx - with a
> stopband freq of Fs/2=5512Hz.
>
> The order of my filter will then be
>
> n=log(10^0.1*100-1)^0.5/log(1.378)
>
> which is approx 36th order!!
>
> What am I doing wrong...is my 4kHz passband assuming too much - do I
> have to sacrifice this?
>
>
This is why most systems engineers want to oversample and do some
filtering in the digital domain -- it's much easier to start with a
reasonable order for your analog filter and work from there.

You only have to have that severe of attenuation if you want to slam
your system with a full-range signal at 5.5kHz.  If you know for sure
that your actual spectral content will be lower you can work with that.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```You will probably need to relax your filter requirements a bit to come up with a
reasonable order filter.  Either start rolling off sooner, or accept more noise.
Granted it is nice to get your aliasing components down below the ADC noise
floor for the worst case input (full scale at 1/2FS), but maybe you can live
with less?  Or maybe you know your input signal will never be that worst case?
Your other option is to sample at a higher frequency and filter digitally.  Best
wishes!

--
Jon Harris
SPAM blocker in place:
Remove 99 (but leave 7) to reply

> Ok I am sampling with an ordinary A/D (not sigma-delta) at 11025kHz
> which mean I need any noise to be attenuated at half sampling which is
> at 5512.5Hz. I have read that the nosie level needs to be less that the
> R.M.S Quantisation Level of my A/D. Now I swing +or - 10 volts with 16
> bits. So my Quantisation level is
>
> delta = dynamic range/2^16=20/65536=0.000305176 volts.
>
> (or 0.305mV)
>
> Now my Quantisaton RMS voltage is delta/sqrt(12)
>
> =88.096 micro volts.
>
> So my noise needs to be lower than this at Fs/2 and then it will be at
> least as low as the quantistaion noise.
>
> Suppose I use a Butterworth filter (3dB passband freq)with passband
> freq of 4kHz. To go from full swing of 10 volts to 88.096 micro volts
> requires an attenuation in the stopband of 100dB approx - with a
> stopband freq of Fs/2=5512Hz.
>
> The order of my filter will then be
>
> n=log(10^0.1*100-1)^0.5/log(1.378)
>
> which is approx 36th order!!
>
> What am I doing wrong...is my 4kHz passband assuming too much - do I
> have to sacrifice this?
>
>

```
```"naebad" <minnaebad@yahoo.co.uk> wrote in message

[snip]

> What am I doing wrong...is my 4kHz passband assuming too much - do I
> have to sacrifice this?
>

It depends on what you are trying to do with the signal. You may be able to
live with aliasing of stuff just above fs/2. This will cut your filter poles
in half. e.g. fs=11kHz and signal BW of 4kHz; interference between 5.5kHz
and 7kHz folds back into the band between 5.5k and 4k (which you are
ignoring anyway). So, your AA filter needs to be down -96dB (assuming full
scale interference is possible) at 7kHz, not 5.5kHz. That's an 18 pole

Others have suggested relaxing the stop band attenuation. A related trick is
to use a Cauer filter alignment. IIRC you specify passband ripple and stop
band attenuation for a Cauer. The stop band doesn't drop forever (like a
Butterworth or Bessel) - it has little bumps in the response. What you lose
in stop band, you gain in faster transition band. The net result of Cauer
and 7kHz+ stop band will be many fewer poles (my WAG would be about 9).

Bob

```
```Bob wrote:
>
> [snip]
>
>
>>What am I doing wrong...is my 4kHz passband assuming too much - do I
>>have to sacrifice this?
>>
>
>
> It depends on what you are trying to do with the signal. You may be able to
> live with aliasing of stuff just above fs/2. This will cut your filter poles
> in half. e.g. fs=11kHz and signal BW of 4kHz; interference between 5.5kHz
> and 7kHz folds back into the band between 5.5k and 4k (which you are
> ignoring anyway). So, your AA filter needs to be down -96dB (assuming full
> scale interference is possible) at 7kHz, not 5.5kHz. That's an 18 pole
>
> Others have suggested relaxing the stop band attenuation. A related trick is
> to use a Cauer filter alignment. IIRC you specify passband ripple and stop
> band attenuation for a Cauer. The stop band doesn't drop forever (like a
> Butterworth or Bessel) - it has little bumps in the response. What you lose
> in stop band, you gain in faster transition band. The net result of Cauer
> and 7kHz+ stop band will be many fewer poles (my WAG would be about 9).

Tim's suggestion, oversampling, is at the heart of most practical
approaches.

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:o8OdnYAImb2oPJHeRVn-qw@rcn.net...
> Bob wrote:
> >

> Tim's suggestion, oversampling, is at the heart of most practical
> approaches.
>
> Jerry

True. My suggestions are useful if oversample isn't an option e.g. because
the ADC to oversample doesn't exist or is too expensive. 11kHz at 16bits is
easily oversampled.

Bob

```
```"Bob" <SkiBoyBob@excite.com> wrote in
news:3n3p8dF19jontU1@individual.net:

>
> "Jerry Avins" <jya@ieee.org> wrote in message
> news:o8OdnYAImb2oPJHeRVn-qw@rcn.net...
>> Bob wrote:
>> >
>
>> Tim's suggestion, oversampling, is at the heart of most practical
>> approaches.
>>
>> Jerry
>
> True. My suggestions are useful if oversample isn't an option e.g.
> because the ADC to oversample doesn't exist or is too expensive. 11kHz
> at 16bits is easily oversampled.
>
> Bob
>
>
>

I have a question that no one has asked.

Do you need to use a SAR converter? The antialiasing filter problem
basically goes away with a sigma delta converter since the actual
sampling rate is very much higher than the effective sample rate.

The penalty for using a sigma delta converter is group delay. This is
usually the only reason a SAR converter is used instead of a sigma delta
at these frequencies.

You might be able to use a "trick". If you have sufficient signal
processing power, you can oversample and use digital filters. In the
sigma delta case, you can decimate with IIR filters to sample at a lower
sampling rate with a relatively small group delay. Analog antialiasing
filters are going to have about the same delay as digital IIR filters

--
Al Clark
Danville Signal Processing, Inc.
--------------------------------------------------------------------
Purveyors of Fine DSP Hardware and other Cool Stuff
Available at http://www.danvillesignal.com
```
```Jerry Avins wrote:

> Tim's suggestion, oversampling, is at the heart of most practical
> approaches.

There are systems where oversampling is impossible. For example - hard
disk drives, where the sampling rate for the VCM control is closely
related to the spindle RPM, which is very difficult to increase.

--JS
```
```Al Clark wrote:
> "Bob" <SkiBoyBob@excite.com> wrote in
> news:3n3p8dF19jontU1@individual.net:
>
>
>>"Jerry Avins" <jya@ieee.org> wrote in message
>>news:o8OdnYAImb2oPJHeRVn-qw@rcn.net...
>>
>>>Bob wrote:
>>>
>>>>
>>
>>>Tim's suggestion, oversampling, is at the heart of most practical
>>>approaches.
>>>
>>>Jerry
>>
>>True. My suggestions are useful if oversample isn't an option e.g.
>>because the ADC to oversample doesn't exist or is too expensive. 11kHz
>>at 16bits is easily oversampled.
>>
>>Bob
>>
>>
>>
>
>
> I have a question that no one has asked.
>
> Do you need to use a SAR converter? The antialiasing filter problem
> basically goes away with a sigma delta converter since the actual
> sampling rate is very much higher than the effective sample rate.

It does not. The internal oversampling allows averaging conversions at
low precision (typically, one bit) to achieve a single sample at high
precision PROVIDED that anti-alias filtering for the final sample rate
has been applied.

> The penalty for using a sigma delta converter is group delay. This is
> usually the only reason a SAR converter is used instead of a sigma delta
> at these frequencies.
>
> You might be able to use a "trick". If you have sufficient signal
> processing power, you can oversample and use digital filters. In the
> sigma delta case, you can decimate with IIR filters to sample at a lower
> sampling rate with a relatively small group delay. Analog antialiasing
> filters are going to have about the same delay as digital IIR filters

It comes down again to oversampling. The type of converter makes no
difference.

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;
```