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Consider a signal $x(t)$, which is input to a pulse shaping filter with transfer function $g_t(t)$:

$$x(t) = \sum_n d_n \delta(t-nT_s)$$

with $n$ an index from negative to positive infinity and $d_n$ random, equiprobable binary symbols such as 1 and -1 (antipodal) or 0 and 1 (on-off keying).

Well, the power spectral density of the filtered output is given by: $$S_{ss} = \frac{1}{T_s} E\left[|d_n|^2 \right] |G_t(f)|^2 $$ where $E[\cdot]$ denotes the expected value.

I'm confused by this formula since up to now I thought the power spectral density is given by the signal mutliplied with the absolute value squared of the transfer function. But why do I have to consider the expectation value in this case and why am I scaling by the sampling time?

Thx for any help!

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  • $\begingroup$ you have very weird and inconsistent notational convention. can we fix that? $\endgroup$ – robert bristow-johnson Oct 18 '17 at 14:53
  • $\begingroup$ sure. I'm not sure what you mean, but you're welcome to change whatever you feel necessary. $\endgroup$ – user25356 Oct 18 '17 at 14:54
  • $\begingroup$ The reason has to do with $d_n$ being random. See also dsp.stackexchange.com/questions/42381/… $\endgroup$ – MBaz Oct 18 '17 at 15:05
  • $\begingroup$ d_n is a stream of 1 and 0 which is suppossed to be the information I want to transmit. Not sure if this is random. Bit errors occur random. Hmm.. $\endgroup$ – user25356 Oct 18 '17 at 15:39
  • $\begingroup$ The information is modeled as random. Of course, you can also find the PSD for a specific, deterministic sequence, but then the result would only be valid for that one case. $\endgroup$ – MBaz Oct 18 '17 at 15:47

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