# Low pass equivalent of bandpass white noise

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.

• Can you please scan and include the pertinent page? – P2000 Jul 16 at 3:29

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.

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