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The length should be found such that the group delay is minimum

It is given that the impulse response is real. One zero of the Transfer function is at $0.6e^{j\frac{\pi}{4}}$, and another one is at -2.

Since the impulse response is real (and the phase is linear), there must be three other zeros (from what I read): reciprocal, conjugate, and conjugate reciprocal of $0.6e^{j\frac{\pi}{4}}$

So I computed the transfer function using these five zeros, but the coefficients of powers of $z$ weren't symmetric at all (which is required for linear phase).

I must be missing something. Please help.

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  • $\begingroup$ Are you sure this is supposed to be a type 4 linear phase filter? I think there should be 6 zeros in total, and it's going to be a type 1 filter. $\endgroup$ – Matt L. Oct 9 at 12:23
  • $\begingroup$ If type 4 is required, you need to add another zero. $\endgroup$ – Matt L. Oct 9 at 12:25
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You are missing a couple of zeros.

  • First of all, you must also include the reciprocal of the one located at $z=-2$, that is one at $z=-0.5$
  • You should also have one more zero coming from the fact that types 3 and 4 are anti-symmetric. This zero must be located at $z=1$, and is responsible for the minus sign in the anti-mirror image polynomial equation:

    $H(z)=-z^{-L} H(\frac{1}{z})$

Given that you now have an odd number of zeros, you don't need to add any other one. If it were not the case, if you had an even number of zeros (type 3 filter), then you should add one in $z=-1$.

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  • $\begingroup$ So the answer is 8, right? ( As the highest power of z will be 7) $\endgroup$ – Ryder Rude Oct 9 at 12:49

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