1
$\begingroup$

I have been trying to figure out how it is I have an IIR filter with large negative group delay. It's my understanding that if a filter is causal and stable, the group delay can't be negative at all frequencies. Is this correct?

The essence of my problem can be seen in the following MATLAB code. This first filter gives four poles inside the unit circle and four zeros at zero. The group delay is both positive and negative with a mean of zero. I believe the contribution of each pole and zero adds and negates one sample of average group delay, respectively.

  • (1a) grpdelay(1,[1 0 0 0 .1],100000,'whole')

Now, if I create a semi-stable filter, the poles on the unit circle contribute 1/2 sample, so the average group delay is -2. Actually it's -2 everywhere. If the poles are outside the unit circle, there's no contribution; so, the average is -4 for the four zeros at zero.

  • (1b) grpdelay(1,[1 0 0 0 1],100000,'whole')

  • (1c) grpdelay(1,[1 0 0 0 2],100000,'whole')

OK so far. Now if I just change the last coefficient to be complex valued, I believe this just rotates the poles, but now there is no contribution for poles inside the circle, poles on the circle subtract 1/2 sample instead of adding, and poles outside the circle subtract 1 sample. And thus, you get negative group delay everywhere, even when the poles are inside the circle.

  • (2a) grpdelay(1,[1 0 0 0 .1j],100000,'whole')

  • (2b) grpdelay(1,[1 0 0 0 1j],100000,'whole')

  • (2c) grpdelay(1,[1 0 0 0 2j],100000,'whole')

I briefly glanced at the code for grpdelay but did not make a serious attempt to understand it. Also, if you flip this around and use an FIR filter, changing the last coefficient to complex does not change the average group delay contribution by each zero.

  • Example (1a):

    enter image description here

  • Example (2a):

    enter image description here

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Let me summarize what I understand from your question: for real-valued coefficients, the group delay is positive at some frequencies and negative at others for the causal and stable filter (poles inside the unit circle); for poles on and outside the circle, the group delay is negative everywhere. However, if the last coefficients is imaginary, you get negative group delay everywhere, even in the case of a causal and stable filter. Is that what you're saying? Can you add a link to your group delay plots? $\endgroup$ – Matt L. Sep 18 '16 at 19:17
  • $\begingroup$ I don't have the reputation points to post more than two links. Also, the coefficient only needs to be any non-real value. The average group delay will be the same. $\endgroup$ – Todd Sep 18 '16 at 21:26
  • 1
    $\begingroup$ Mathworks has confirmed this is a bug. $\endgroup$ – Todd Sep 20 '16 at 21:49
  • $\begingroup$ OK, that's interesting to know! $\endgroup$ – Matt L. Sep 21 '16 at 7:07
1
$\begingroup$

This is a bug, either in your code, or in your version of the Matlab function grpdelay. As you correctly pointed out, making the last denominator coefficient imaginary just rotates the pole-zero diagram. This means that for the given example, the zeros stay where they were (in the origin), and the pole angles are rotated by $\pi/8$; or course, the pole radii remain unchanged.

For the group delay function this means that it only gets shifted (along the frequency axis), but its maximum and minimum values (and its overall shape) must remain the same.

Using Octave I get the expected result:

g1 = grpdelay(1,[1 0 0 0 .1],1000,'whole');
g2 = grpdelay(1,[1 0 0 0 .1j],1000,'whole');
plot([g1(:),g2(:)]), axis([0,1000,-.4,.5]), grid on

enter image description here

Improve this answer
$\endgroup$
3
  • $\begingroup$ I didn't write any code. I guess it's a problem with the grpdelay function then. Thanks for confirming. $\endgroup$ – Todd Sep 19 '16 at 14:14
  • $\begingroup$ @Todd: Welcome! If it's really a bug in grpdelay you should let the mathworks people know. You can link to this post. $\endgroup$ – Matt L. Sep 19 '16 at 14:18
  • $\begingroup$ OK will do. I should mention this is version 2016a. $\endgroup$ – Todd Sep 19 '16 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.