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If I need to graph the magnitude and angle of a discrete Fourier transform which happens to be $X(e^{j\omega}) = 4\cos(4\omega)$, I know how to graph the magnitude, but how do you graph the angle?

I would make an attempt, but I don't even know how to start. I'm studying for a final tomorrow and got stuck on this. I have the solution sheet right in front of me and don't understand it. The graph of the angle of $X(e^{j\omega})$ on the solution sheet is drawn as a pulse wave with period $\pi/2$ shifted $\pi/4$ from the origin.

Austin
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1 Answers1

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Since $X(e^{j\omega}) = 4\cos(4\omega)$ is purely real, we have $\angle X(e^{j\omega}) = 0$ if $X(e^{j\omega}) = 4\cos(4\omega) \ge 0$ and $\angle X(e^{j\omega}) = \pi$ if $X(e^{j\omega}) = 4\cos(4\omega) < 0$. When graphed, this makes a square wave.

JimmyK4542
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  • why does it stay high in between? I was thinking it would just be impulses of $\pi$ at the zero crossings of the cosine. Ohhh, nvm I was looking at the magnitude graph for >= 0, not the original graph. Thanks a lot ;) – Austin Dec 15 '15 at 03:39