0
$\begingroup$

Given the FIR transfer function:

h(z) = $ .36 + .384z^{-1} + .1608z^{-2} +.9712z^{-3} + .352z^{-4} + .18z^{-5} - .2z^{-6} $

How do you determine if this transfer function has a linear - phase response.

$\endgroup$
  • 1
    $\begingroup$ Wikipedia: "A discrete-time filter with linear phase may be achieved by an FIR filter which is either symmetric or anti-symmetric", so you have to check whether your coefficients fulfill this prerequisites $\endgroup$ – Irreducible Nov 12 '19 at 10:46
1
$\begingroup$

Real-valued and causal FIR filters have a (generalized) linear phase response if and only if their coefficients satisfy either

$$h[n]=h[N-1-n]\tag{1}$$

or

$$h[n]=-h[N-1-n]\tag{2}$$

where $N$ is the filter length (number of taps).

There are four types of linear phase FIR filters, depending on the type of symmetry (even as in $(1)$ or odd as in $(2)$), and the filter length $N$ (even or odd). They have different properties, which is explained in this more detailed answer.

$\endgroup$
  • $\begingroup$ Therefore in this case we say that the filter length is equal to the number of taps? If so... Then the approach for this problem would be to take the inverse z transform , find h[n] and plug in the filter length (N) and evaluate whether they are equal or not. To see if transform satisfies the coefficient condition? $\endgroup$ – dspLearner Nov 13 '19 at 1:15
  • $\begingroup$ @dspLearner: Yes, the number of taps is the same as the filter length. Your approach is correct, but it's actually very easy. The inverse Z transform is trivial in the case of FIR filters, you can read the filter coefficients directly from the transfer function: $H(z)=h[0]+h[1]z^{-1}+h[2]z^{-2}+\ldots$. $\endgroup$ – Matt L. Nov 13 '19 at 6:26
  • $\begingroup$ Wow..thank you...This makes the approach much easier than learned in class. @Matt L. This mean that when approaching any problem in this form...use this coefficient condition in the first part of the answer, and if the coefficients are equal there would be symmetry about one of the coefficients if plotted as unit impulses on a discrete number line. $\endgroup$ – dspLearner Nov 13 '19 at 8:36

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.