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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.

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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 $-1=e^{j\pi}$.

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  • $\begingroup$ my dsp professor said something about in order to graph it, I need to graph the angle of the exponential, then graph the angle of the periodic sinc, then graph the sum of the two. would you mind showing me how this is done? His graph of the angle of the periodic sinc part looked like an odd function with values of 0 from +- pi/2, pi from -pi to pi/2 and -pi from pi/2 to pi. $\endgroup$
    – Austin
    Commented Sep 14, 2016 at 15:28
  • $\begingroup$ Perhaps, s/he wants you to do it in the Cartesian coordinate by writing $e^{-j\omega3/2}=\cos(3\omega/2)+j\sin(3\omega/2)$ and $\frac{sin(2\omega)}{sin(\omega/2)}+j0$. The angle of the first term is $\tan^{-1}(\frac{\sin(3\omega/2)}{\cos(3\omega/2)})=\tan^{-1}(\tan(3\omega/2))=3\omega/2$ and the angle of the second term is $\tan^{-1}(\frac{0}{\frac{sin(2\omega)}{sin(\omega/2)}})=0$, then add the two. Ugly way to do it! $\endgroup$
    – msm
    Commented Sep 14, 2016 at 21:25
  • $\begingroup$ @Jake In general you should consider all cases i.e. when $\frac{\sin(2\omega)}{\sin(\omega/2)}$ becomes negative. $\endgroup$
    – msm
    Commented Sep 20, 2016 at 4:37

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