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Let $$n(t) = Re\{z(t)e^{j2\pi f_ct}\},\ $$ where $z(t)$ is the low pass equivalent of white noise $n(t)$. We know the autocorrelation of $n(t)$ is $$R_n(\tau) = \frac{N_0}{2} \delta(\tau),\ $$

$\delta(\tau)$ being the Dirac delta function and the PSD is $S_n(f) = N_0/2$.

From Proakis' Digital Communication book, I learned that the autocorrelation of $z(t)$, $R_z(\tau) = N_0\delta(\tau)$, and PSD is naturally, $N_0$.

Now consider $$z(t) = x(t) + jy(t)$$ I also know that $$R_z(\tau) = R_x(\tau) + jR_y(\tau)$$ So what are $R_x(\tau)$ and $R_y(\tau)$? Proakis' book mentions $R_z(\tau) = R_x(\tau) = R_y(\tau)$. But that seems a bit off to me.

Since, $R_z(\tau)$ is real then shouldn't $R_y(\tau)$ be zero being as it is the imaginary term in a real quantity?

Reference: Proakis, Digital Communication, Third edition, Chapter 4: Characterization of communication signals and systems, section 4-1-4, Representation of Bandpass Stationary Stochastic Processes.

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  • $\begingroup$ Can you please scan and include the pertinent page? $\endgroup$ – P2000 Jul 16 at 3:29
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Note that for the complex noise envelope $z(t)=x(t)+jy(t)$, the autocorrelation $R_z(\tau)$ is defined by (cf. Eq. $(4.1.47)$ in Proakis)

$$R_z(\tau)=\frac12E\big\{z^*(t)z(t+\tau))\big\}=\frac12\big[R_x(\tau)+R_y(\tau)\big]+j\frac12\big[R_{xy}(\tau)-R_{yx}(\tau)\big]\tag{1}$$

As shown in the chapter you refer to, for the real-valued bandpass noise $n(t)$ to be stationary, the following must be true:

$$\begin{align}R_x(\tau)&=R_y(\tau)\tag{2}\\R_{xy}(\tau)&=-R_{yx}(\tau)\tag{3}\end{align}$$

Plugging $(2)$ and $(3)$ into $(1)$ we obtain

$$R_z(\tau)=R_x(\tau)+jR_{xy}(\tau)\tag{4}$$

Note that this expression is different from the one you suggested in your question.

If the power spectrum of $z(t)$ is even, as is the case for bandpass white noise, the autocorrelation function $R_z(\tau)$ must be real-valued, and, consequently, $R_{xy}(\tau)=0$. But note that we don't require $R_y(\tau)=0$.

As a final note, defining a complex envelope is only useful for bandpass processes, such as bandpass white noise, but not for ordinary white noise, because the latter is no bandpass process.

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  • $\begingroup$ Thanks for the answer. I guess it all boils down to me misreading something. Thanks for pointing it out in (4)! $\endgroup$ – Kartik Jul 17 at 5:00

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