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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

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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.

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  • $\begingroup$ Hi @MattL. I do not doubt you are right, my script shows its imag spectrum is zero. You say the entire signal is even, but it has an odd component from the sine and I thought the sum of an even function and an odd function is neither odd nor even. $\endgroup$ – Natalie Johnson Jan 13 at 21:04
  • $\begingroup$ @NatalieJohnson: This is about complex-valued sequences, for which the odd part is defined by $\frac12\big(x[n]-x^*[-n]\big)$. So in this sense, your sequence is even (because its odd part is zero). $\endgroup$ – Matt L. Jan 14 at 9:09
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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.

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  • $\begingroup$ The spectrum of $~e^{ j 2\pi f~ n}~$ is one sided with a single the positive frequency that is not mirrored in the negative frequency... I see this in my python script. "The negative exponetials cancel out" - what I meant was when you write cos + isin in the complex exponential form, the negative frequency exponentials cancel out leaving just positive frequency so the spectrum of this complex is one sided (only positive frequency). Maybe we mean the same thing and I was not clear... $\endgroup$ – Natalie Johnson Jan 13 at 20:57

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