I would appreciate it very much if someone would be able to provide some clarity on plotting phase responses.

For instance, given that the frequency response of a filter can be written as $$H(\exp(j*\Omega))=\cos(\Omega)$$ How would I go about determining the phase response?

The answer key demonstrates a piecewise solution:

enter image description here

This is what I understand so far:

  • $\cos(\Omega)$ is strictly real, so the phase carries values of $0$, or $\pm\pi$

  • Between the values of $-\pi/2$ and $\pi/2$, $\cos(\Omega)$ is positive, so the phase is $0$

I understand since the function $\cos(\Omega)$ is strictly real, phases where the function is not positive should either be $\pi$ or $-\pi$ (as opposed to $-\pi/2$ or $\pi/2$, for when it's imaginary), but how do you know for which range it's $\pi$, and for which range it's $-\pi$?

This might be a stupid question, but I've spent a lot of time thinking it through, so I'm thinking I'm missing something, and would appreciate any clarity.

Thank you so much!


1 Answer 1


Your understanding is correct. There is no difference between a phase of $\pi$ and a phase of $-\pi$. You can always add or subtract integer multiples of $2\pi$ to the phase without changing anything because $e^{j2\pi n}=1$ for $n\in\mathbb{Z}$. Clearly you have $e^{j\pi}=e^{-j\pi}=-1$.

The probable reason why they used opposite signs in the phase plot for positive and negative frequencies is because then you can directly see from the figure that the phase (of a real-valued signal) is an odd function, but that's a purely cosmetic matter.


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