I am approaching to the study of FIR systems. In particular, I was analyzing the graphs of amplitude and phase of the transfer function, when I had some trouble understanding how the phase behaves. Let me explain this with an example:

Suppose that our transfer function is $\displaystyle{H(z)=\frac{1-z^{-5}}{1-z ^ {-1}}}$. Then the DTFT is $\displaystyle{H(e^{i\omega}) = e^ {-2i\omega } \frac{\sin(\frac{5}{2}\omega)}{{\sin({\frac{1}{2}\omega }})}}$. If we set as $x$ coordinate the variable $\omega$ and as $y$ coordinate the phase, we observe that the graph is piecewise linear, with slope -2, and with jumps at the points where transfer function vanishes (in this case, $\omega = \pm\frac{2\pi} 5, \pm \frac{4\pi} 5$, taking as domain the interval $[-\pi , \pi ]$).

In the same way, if we consider $\displaystyle{H_1(z) = 1- z ^{-2}}$, therefore $\displaystyle{H_ 1 (e ^{i\omega}) = 2 ie ^ {-i\omega}\sin(\omega) }$ and we get a phase function piecewise linear, with slope $-1$ and jump for $\omega = 0$.

My question is: why are there jumps at these points? I see that the slope is given by the $\arctan\left(\frac{Im(H)}{Re(H)}\right)$, but I don't see why at these points the function jumps (and in particular, the correlation with the fact that it jumps in $\omega $ such that $H(e^{i\omega })= 0$ ).

If someone can give me an explanation, it would be really appreciated.


1 Answer 1


It depends on how you define the phase. For a linear phase FIR filter you could write the frequency response as


where $A(\omega)$ is a real-valued (or purely imaginary) function that can become positive and negative such that the phase $\phi(\omega)$ is a linear function without jumps. This is the case for the first frequency response in your question. Note that the term $\sin(5\omega /2)\,/\sin(\omega/2)$ takes both signs.

Another way to write the frequency response is


Now the phase $\varphi(\omega)$ jumps by $\pi$ at all frequencies where the corresponding amplitude function $A(\omega)$ changes sign. Since $|H(e^{j\omega})|$ can't change sign, the phase has to compensate for that and you get a piecewise linear phase with jumps at the zeros (of odd multiplicity) of $H(e^{j\omega})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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