Say I have the function $$x(t)=j \operatorname{rect}(t)$$
Is the phase spectrum even or odd?
I am confused whether the phase spectrum is an odd/even function of $\omega$ (angular frequency, Fourier transform variable).
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3 Answers
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This is a homework type question, so this answer will only provide a hint. Let
$$x(t)=jx_R(t)\tag{1}$$
where $x_R(t)$ is a real-valued function. Consequently, $x(t)$ is purely imaginary. From the properties of the Fourier transform it follows that the Fourier transform of $x(t)$ is simply
$$X(f)=jX_R(f)\tag{2}$$
where $X_R(f)$ is the (not necessarily real-valued) Fourier transform of $x_R(t)$. Now you probably know that $X_R(f)$ satisfies
$$X_R(f)=X_R^*(-f)\tag{3}$$
Consequently, $X(f)$ must satisfy
$$X(f)=-X^*(-f)\tag{4}$$
Can you draw your own conclusion concerning the magnitude and phase of $X(f)$?
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$j$ can be written as $e^{j\frac{\pi}{2}}$. And if we have any real valued function $x(t)$, by multiplying this $e^{j\frac{\pi}{2}}$ with $x(t)$, you will just shift the phase-response of $x(t)$ up by $\frac{\pi}{2}$. Mathematically, this can be seen that if $X(f)$ is the Fourier representation of $x(t)$, then: $$ X(f) = |X(f)|\cdot e^{j\phi(f)} $$ Where $|X(f)|$ is the magnitude response and $\phi(f)$ is the phase response. So, the Fourier representation of $j\cdot x(t)$ will just be : $$ |X(f)|\cdot e^{j\phi(f)}\cdot e^{j\frac{\pi}{2}} = |X(f)|\cdot e^{j\left(\phi(f)+\frac{\pi}{2}\right)} $$ We cannot say any thing on whether phase response is even or odd from this.
Observing the fact that $x(t)$ was a real-valued function to start with, we know that $X^*(f) = X(-f)$, meaning the Fourier representation of real valued functions are conjugate-symmetric. Write the magnitude and phase response of both $X^*(f)$ and $X(-f)$ and equate them to check for even or oddness.
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This might be helpful: https://users.aber.ac.uk/ruw/teach/340/ft_symmetry.php
Your example is straight forward. You are just multiplying with a constant factor $j$ and that factor carries through the Fourier Transform, i.e. it's simply
$$\mathcal F\big\{j \cdot x(t)\big\} = j \cdot \mathcal F\big\{x(t)\big\}$$
