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I am learning about complex sampling.
I am confused why $~e^{ j 2\pi f~ n}~$ has only a real spectrum. I would have thought the $j ~\sin(2 \pi f n)$ would produce a single spike in imaginary spectrum just like there is a single spike in real axis from $\cos(2 \pi f n)$.
I understand that the spectrum is one sided because the negative complex exponentials cancel out, but why is there not a one sided real and imaginary spectrum?
Many thanks
The imaginary part of the spectrum corresponds to the odd part of the time-domain sequence. Since the given signal is even, i.e.,
$$x[n]=e^{jn\omega_0}=x^*[-n]=e^{-j(-n)\omega_0}=e^{jn\omega_0}$$
the imaginary part of the spectrum is zero, i.e., the spectrum is purely real-valued.
Note that for complex-valued signals, even in this context means $x[n]=x^*[-n]$, i.e., its real-part is even and its imaginary part is odd.
In sum, if a sequence $x[n]$ satisfies
$$x[n]=x^*[-n]$$
(i.e., the sequence is even and, consequently, its odd part is zero), then its DTFT is purely real-valued.
Your last paragraph seems to have the concept exactly backwards.
A strictly real cos(w) or a purely imaginary i*sin(w) in the time domain is two sided in the frequency domain, where the two sides cancel out the other component.
Otherwise exp(i*w)m alone is a spiral in complex space with its excursions into both real and imaginary space being non-zero (e.g. see one of Euler's identities). You need a mirror image spiral to cancel out the twist of one spiral with its opposite rotation.
A complex signal can have either a strictly real or strictly imaginary (or any ratio of the two) spectrum (depending on phase), as long as it does not have a equal magnitude mirror image, complex conjugated, to completely cancel out the imaginary component.
