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From a window function $x(t)=u(t+2)-u(t-2)$, we can get the Fourier Transform $X(j\omega)=\frac{2\sin(2\omega)}{\omega}$.

Then, I want to calculate its magnitude spectrum and phase spectrum.

The magnitude spectrum is the magnitude distribution at every $\omega$, so it's simply absolute value of $X(j\omega)$, is this correct?

And how to calculate the phase spectrum?

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HINT:

Magnitude $M(\omega)\ge 0$ and phase $\phi(\omega)$ are defined by

$$X(\omega)=M(\omega)e^{j\phi(\omega)}\tag{1}$$

Note that $(1)$ is generally complex-valued. In your example, the Fourier transform $X(\omega)$ is clearly real-valued. This restricts the possible values of the phase $\phi(\omega)$. What are those two values of $\phi(\omega)$ for which $X(\omega)$ in $(1)$ is real-valued? How do they relate to the sign of $X(\omega)$? I'm sure you can take it from here.

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  • $\begingroup$ Thanks for you help. Since it's real-valued, the phase should be 0 for positive and $+\pi$ or $-\pi$ for negative. But, $X(j\omega)$ is smooth, should the phase be $\pi$ at only the peak of each negative loop, like shifted $\delta$ function? Or should it last for the whole negative loop? $\endgroup$ – keanehui Apr 2 '20 at 12:18
  • $\begingroup$ I tried to express $X(j\omega)$ in the complex form, I got $\frac{j}{\omega}(e^{-j2\omega}-e^{j2\omega})$ $\endgroup$ – keanehui Apr 2 '20 at 12:20
  • $\begingroup$ @keanehui: The phase must equal $\pi$ for all frequencies for which $X(\omega)$ is negative, otherwise the phase equals zero. $\endgroup$ – Matt L. Apr 2 '20 at 12:26

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