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This seems to be quite a basic question but is there some detailed proof why interference power (i.e., power spectral density) coming from the positive and negative frequencies are independent? I am considering two real baseband OFDM signals which are interfering with each other and where $2N-f=-f$ is relation between positive and negative frequencies while $2N$ is total number of frequencies (tones). Assume that two signals are not sync and that is the source of interference leakage coming from the negative spectrum. To simplify, I want to prove why is the received PSD equal to : $PSD_{rec.,k}=A^2_{pos.,k}+A^2_{neg.,k}$ instead of $PSD_{rec.,k}=A^2_{pos.,k}+2A_{pos.,k}A_{neg.,k}+A^2_{neg.,k}$ where $A^2$ is squared amplitude (i.e.,DFT coefficient) of tone $k$. Could anyone recommend me some literature or give me a short answer on this question?

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  • $\begingroup$ Can you clarify what do you mean by 'interference power coming from pos. and neg. frequencies'? What is the source of the interference? Are you working with passband or baseband signals? $\endgroup$ – MBaz May 23 '16 at 13:28
  • $\begingroup$ I am considering baseband transmission. I am having OFDM system which is not sync with another OFDM system and therefore receives influence also from negative side of the spectrum. I am just looking for the reasoning why the PSD contributions from the pos. and neg. frequencies on some specific frequency are independent. $\endgroup$ – Cali May 23 '16 at 13:56
  • $\begingroup$ Actually now I have already answered my question. I know they are independent but I am trying to show it analyticalally. Looking for someone how can explain why or recommend some good literature. $\endgroup$ – Cali May 23 '16 at 14:41
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    $\begingroup$ Well, if you can answer your question, do so! You're encouraged to post answers to your own questions. $\endgroup$ – Marcus Müller May 23 '16 at 15:03
  • $\begingroup$ But my problem is that I do not know how to prove it. Simply by reading some research articles I saw that they are independent but I didn't find detailed proof for that. I what to know what is the reason behind it, fact it self doesn't help me so much. $\endgroup$ – Cali May 23 '16 at 15:15
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This is a bit waffly, but gives the general idea.

The usual reason for this is because of expectations used when we talk about Power Spectral Densities: $$ S_{xx}(\omega) = \int_{-\infty}^{+\infty} \gamma(\tau) e^{-i\omega \tau} d\tau = \int_{-\infty}^{+\infty} E\left[ x(t) x(t+\tau) \right] e^{-i\omega \tau} d\tau $$

So what you have to look at is how the $E[\cdot]$ works on the cross-products between the positive and negative frequencies. Assuming $x$ is just: $$ x(t) = \sum_{k=1}^{N_+} a_k e^{j\omega_{+,k} t} + \sum_{k=1}^{N_-} b_k e^{j\omega_{-,k}t} $$ then each of the cross terms will look like: $$ \lim_{T\rightarrow \infty} \frac{1}{T} \int_{-T}^{+T}e^{j\omega_+t}e^{-j\omega_-t}dt = \delta(\omega_+ - \omega_-) $$ which will be zero if $\omega_+ \not= \omega_-$. That makes all the cross-terms zero, so you're just left with the $x_+(t)x_+(t+\tau)$ and $x_-(t)x_-(t+\tau)$ terms, i.e. your $A^2_{pos,k}$ and $A^2_{neg,k}$ terms.

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  • $\begingroup$ thank you for your answer. So summarizing everything into one sentence I could say that expectation operator eliminates cross-product terms between positive and negative frequencies? $\endgroup$ – Cali May 25 '16 at 15:49
  • $\begingroup$ @Cali : Yes, that's not a bad way of phrasing it. $\endgroup$ – Peter K. May 25 '16 at 16:25
  • $\begingroup$ Could I just get one additional clarification: I understand that PSD is actually Fourier transform of auto-correlation function which is exactly what is written in the first equation. What I do not understand completely is how this relates with the second equation that you have written, i.e., Dirac function. How do I combine this two things together? Maybe if you could edit your answer with one or two intermediate steps. Sorry if this is a stupid question, I am still learning a lot of stuff regarding the spectral analysis. $\endgroup$ – Cali May 25 '16 at 17:47
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    $\begingroup$ @Cali : Added an intermediate step: I'm assuming that $x$ is a sum of sinusoids, some of which are positive frequency and some of which are negative. $\endgroup$ – Peter K. May 25 '16 at 17:57
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    $\begingroup$ @Fat32 : Thanks for the correction! $\endgroup$ – Peter K. May 25 '16 at 18:40

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