# Phase of DTFT transform of impulse response

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.

$$H(e^{j\omega})=A(\omega)e^{j\phi(\omega)}\tag{1}$$
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.
$$H(e^{j\omega})=\big|H(e^{j\omega})\big|e^{j\varphi(\omega)}\tag{2}$$
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 $$H(e^{j\omega})$$.