# How do the magnitude and phase spectrum of an imaginary function look like?

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).

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 $$X(f)$$?

$$j$$ 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.

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
$$\mathcal F\big\{j \cdot x(t)\big\} = j \cdot \mathcal F\big\{x(t)\big\}$$