Calculate cut-off frequency of digital IIR lowpass-filter

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

You missed Fs

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?

From the plug and chug approach I find

omega = acos((a^2+2a-2)/(2(a-1)))

which is the cutoff frequency (mag^2 = 1/2) in radians per sample.

Just multiply omega by Fs/(2pi) to convert to cycles per time.

Clay

On Friday, November 15, 2013 5:39:36 AM UTC+13, cl...@claysturner.com wrote:
> 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?
> 
> 
> 
> From the plug and chug approach I find
> 
> 
> 
> omega = acos((a^2+2a-2)/(2(a-1)))
> 
> 
> 
> which is the cutoff frequency (mag^2 = 1/2) in radians per sample.
> 
> 
> 
> Just multiply omega by Fs/(2pi) to convert to cycles per time.
> 
> 
> 
> Clay

Same as my result above, 

(a^2+2a-2)/(2(a-1))=  (2(a-1) +a^2) / 2(a-1) = 1- (a^2/2(1-a))

so why did the other guy say he checked with Matlab and gets a different answer?

On Thursday, November 14, 2013 7:36:52 PM UTC-5, gyans...@gmail.com wrote:
> On Friday, November 15, 2013 5:39:36 AM UTC+13, cl...@claysturner.com wrote:
> 
> > 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?
> 
> > 
> 
> > 
> 
> > 
> 
> > From the plug and chug approach I find
> 
> > 
> 
> > 
> 
> > 
> 
> > omega = acos((a^2+2a-2)/(2(a-1)))
> 
> > 
> 
> > 
> 
> > 
> 
> > which is the cutoff frequency (mag^2 = 1/2) in radians per sample.
> 
> > 
> 
> > 
> 
> > 
> 
> > Just multiply omega by Fs/(2pi) to convert to cycles per time.
> 
> > 
> 
> > 
> 
> > 
> 
> > Clay
> 
> 
> 
> Same as my result above, 
> 
> 
> 
> (a^2+2a-2)/(2(a-1))=  (2(a-1) +a^2) / 2(a-1) = 1- (a^2/2(1-a))
> 
> 
> 
> so why did the other guy say he checked with Matlab and gets a different answer?

Good question!

An alternate formulation is omega = 2 asin*( |a| / (2*sqrt(1-a)) )

Clay