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I have a filter with the transfer function

$$H(z) = 1 - 2z^{-2} + z^{-4}.$$ The task is to find the phase function $\theta (\omega).$


My attempt is to start by expressing the frequency response \begin{align} H(\omega) &=H(z)\big\vert_{z=e^{j\omega}}\\ &= 1 - 2e^{-2j\omega} + e^{-4j\omega}\\ &= e^{-2j\omega}( e^{2j\omega} - 2 + e^{-2j\omega})\\ &= e^{-2j\omega}(2\cos(2\omega)-2) \end{align}

I think there is a relation of the sort $$H(\omega) = e^{j\theta (\omega)}|H(\omega)|$$

which in my case would give me $$\theta (\omega) = -2\omega.$$

The correct answer is $$\theta (\omega) = -2\omega + \pi.$$ How do I get the extra term?

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You're on the right track, but your error comes when you map:

$$ H(\omega) = e^{-j2\omega}\left(2\cos(2\omega)-2\right) $$

onto the split magnitude/phase representation:

$$ H(\omega) = e^{j\theta(\omega)}|H(\omega)| $$

Remember a requirement of $|H(\omega)|$; since it is a magnitude, it must always be nonnegative. You implicitly made the assumption that:

$$ |H(\omega)| = 2\cos(2\omega) - 2 $$

The problem is that $2\cos(2\omega) - 2$ is never positive. It covers the range $[-4,0]$. So, we must break the expression down slightly differently:

$$ \begin{aligned} H(\omega) &= -e^{-j2\omega}\left(2-2\cos(2\omega)\right) \\ &= e^{-j2\omega}e^{j\pi}\left(2-2\cos(2\omega)\right) \\ &= e^{j(-2\omega+\pi)}\left(2-2\cos(2\omega)\right) \end{aligned} $$

Above, we took advantage of the fact that $e^{j\pi} = -1$. That is, if you apply a phase shift of $\pm \pi$, it's equivalent to multiplying by $-1$. This expression does fit the split magnitude/phase form that you want, because $2 - 2\cos(2\omega)$ is nonnegative; it fits the requirements to represent a magnitude. Therefore, it follows that:

$$ \angle H(\omega) = -2\omega + \pi $$

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  • $\begingroup$ That's quite explicit for a homework type question, isn't it? :) $\endgroup$ – Matt L. Apr 11 '18 at 11:30
  • $\begingroup$ I'm feeling generous today. :) $\endgroup$ – Jason R Apr 11 '18 at 11:33
  • $\begingroup$ This does not change the rest of your answer, but the range should be $[-4,0]$. $\endgroup$ – fibonatic Apr 11 '18 at 12:24
  • $\begingroup$ @fibonatic: Indeed it should be. Fixed. $\endgroup$ – Jason R Apr 11 '18 at 12:43
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HINT:

Answer these two basic questions, and you should be able to solve your problem:

  1. What's the sign of the term $2\cos(2\omega)-2$? Can it be a magnitude?
  2. What does a term of $\pm \pi$ in the phase do to the sign of an expression?
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