Plotting angle/phase of DTFT

Question

I can plot the magnitude, but don't have the faintest idea how to plot the phase. I solved for a DTFT $X(e^{jw}) = e^{-j\omega3/2}\frac{sin(2\omega)}{sin(\omega/2)}$ If someone could show me the steps I need to take to graph $\measuredangle X(e^{jw})$ between $\omega=\pm\pi$ I'd really appreciate it. I actually am not even sure of the meaning/significance of a phase of a DTFT, so it'd be extremely helpful if someone would start there.

Answer 1

For a better understanding, fix $\omega=\omega_0$. You can look at $X(e^{j\omega_0})$ as a complex number.

A complex number $X=a+jb$ can be represented by $|X|e^{j\measuredangle X}$ in polar coordinate, where $$|X|=\sqrt{a^2+b^2}$$ is the amplitude and $$\measuredangle X=\tan^{-1}\frac{b}{a}$$ is the phase.

In your question, $X(e^{jw}) = e^{-j\omega3/2}\frac{\sin(2\omega)}{\sin(\omega/2)}$ is already in polar form. Therefore, $$|X(e^{jw})|=|\frac{\sin(2\omega)}{\sin(\omega/2)}|$$ and

$$\measuredangle X(e^{jw})=\begin{cases}-\frac{3}{2}\omega+\pi,& \text{if }\frac{\sin(2\omega)}{\sin(\omega/2)}<0\\-\frac{3}{2}\omega, &\text{otherwise}\end{cases}$$

since when $\frac{\sin(2\omega)}{\sin(\omega/2)}$ becomes negative, it produces a phase shift equal to $\pi$ and when it is positive it's impact on the phase is zero. The reason is $-1=e^{j\pi}$.