Calculate cut-off frequency of digital IIR lowpass-filter

Started by November 12, 2013
```How to calculate the -3d cut-off frequency of a given digital lowpass-filter?
Given is a IIR filter in the form of H(z) = a / (1 - (1-a) z^-1).

How to calculate the cut-off frequency? I thought about evaluating the magnitude of the DTFT (z-transform on unit-circle), for values where the DTFT exists.

Is this correct? Or are there other possibilities for doing that?
```
```On Tuesday, November 12, 2013 2:11:37 PM UTC-5, Marc Jacob wrote:
> How to calculate the -3d cut-off frequency of a given digital lowpass-filter?
>
> Given is a IIR filter in the form of H(z) = a / (1 - (1-a) z^-1).
>
>
>
> How to calculate the cut-off frequency? I thought about evaluating the magnitude of the DTFT (z-transform on unit-circle), for values where the DTFT exists.
>
>
>
> Is this correct? Or are there other possibilities for doing that?

Marc,

Yes, letting z^-1 = e^-jwt and finding where the magnitude = 1/2 is probably the sanest way if you wish to work from first principles.

Another way is to look at a table of z-transform pairs and figure it that way.

You should find this entry or one like it in the table

1/ (1-b*z^-1)    <->  (b^n)*u[n]    for |z| < |b|

IHTH,
Clay

```
```Hi Clay,

> Yes, letting z^-1 = e^-jwt and finding where the magnitude = 1/2 is probably the sanest way if you wish to work from first principles.

You mean 1/sqrt(2), do you?

> Another way is to look at a table of z-transform pairs and figure it that way.
> You should find this entry or one like it in the table
> 1/ (1-b*z^-1)    <->  (b^n)*u[n]    for |z| < |b|

I know these tables, but how to derive the cut-off frequency omega from that?

Cheers Marc

```
```On Wednesday, November 13, 2013 8:11:37 AM UTC+13, Marc Jacob wrote:
> How to calculate the -3d cut-off frequency of a given digital lowpass-filter?
>
> Given is a IIR filter in the form of H(z) = a / (1 - (1-a) z^-1).
>
>
>
> How to calculate the cut-off frequency? I thought about evaluating the magnitude of the DTFT (z-transform on unit-circle), for values where the DTFT exists.
>
>
>
> Is this correct? Or are there other possibilities for doing that?

Unfortunately there is no equivalent easy way (unlike in analogue) of the simple time-constant where you just say 1/T as the cut-off 3dB frequency. You need to do it by computation. You could derive a formula of course but it wouldn't be pretty.
```
```On 11/12/13 12:23 PM, Marc Jacob wrote:
> Hi Clay,
>
>
>> Yes, letting z^-1 = e^-jwt and finding where the magnitude = 1/2 is probably the sanest way if you wish to work from first principles.
>
> You mean 1/sqrt(2), do you?
>
>> Another way is to look at a table of z-transform pairs and figure it that way.
>> You should find this entry or one like it in the table
>> 1/ (1-b*z^-1)<->   (b^n)*u[n]    for |z|<  |b|
>
> I know these tables, but how to derive the cut-off frequency omega from that?
>

plug in

z  =  e^(j*omega)  =  cos(omega)  +  j*sin(omega)

into the expression from the table.  then plug-and-chug.

--

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

```
```>     z  =  e^(j*omega)  =  cos(omega)  +  j*sin(omega)
> into the expression from the table.  then plug-and-chug.

Just wanted to confirm that I already managed it.

Cheers Marc

```
```On 11/12/2013 1:11 PM, Marc Jacob wrote:

> How to calculate the -3d cut-off frequency of a given digital
> lowpass-filter? Given is a IIR filter in the form of H(z) = a / (1 -
> (1-a) z^-1).

Fc = - Fs ln a / 2 Pi

> How to calculate the cut-off frequency? I thought about evaluating
> the magnitude of the DTFT (z-transform on unit-circle), for values
> where the DTFT exists.

Wow

> Is this correct? Or are there other possibilities for doing that?

Wow

```
```On Wednesday, November 13, 2013 1:13:43 PM UTC+13, Vladimir Vassilevsky wrote:
> On 11/12/2013 1:11 PM, Marc Jacob wrote:
>
>
>
> > How to calculate the -3d cut-off frequency of a given digital
>
> > lowpass-filter? Given is a IIR filter in the form of H(z) = a / (1 -
>
> > (1-a) z^-1).
>
>
>
> Fc = - Fs ln a / 2 Pi
>
>
>
> > How to calculate the cut-off frequency? I thought about evaluating
>
> > the magnitude of the DTFT (z-transform on unit-circle), for values
>
> > where the DTFT exists.
>
>
>
> Wow
>
>
>
> > Is this correct? Or are there other possibilities for doing that?
>
>
>
> Wow

Well I don't know if it's the same but I get

Fc = (1/2Pi) Fs arccos(1-x)

where x=a^2/(2(1-a))

```
```> Well I don't know if it's the same but I get
> Fc = (1/2Pi) Fs arccos(1-x)
> where x=a^2/(2(1-a))

I cannot agree with your solution. I'm getting

fc = (1/2Pi)  arccos( (1-2a^2)(2x) + x/2)

where x = (1-a)

I compared these results with Matlab, and they match.

Cheers Marc

```
```On Thursday, November 14, 2013 10:11:23 AM UTC+13, Marc Jacob wrote:
> > Well I don't know if it's the same but I get
>
> > Fc = (1/2Pi) Fs arccos(1-x)
>
> > where x=a^2/(2(1-a))
>
>
>
> I cannot agree with your solution. I'm getting
>
>
>
> fc = (1/2Pi)  arccos( (1-2a^2)(2x) + x/2)
>
>
>
> where x = (1-a)
>
>
>
> I compared these results with Matlab, and they match.
>
>
>
> Cheers Marc

I'm not going to argue, I did my calculation fairly quickly, it looks like Vlad Vampyre is wrong though too.
```