Complex Equivalent Representation of Passband Stochastic process

In the previous section, we considered the baseband equivalents of deterministic passband signals. In this section, we will derive the baseband equivalents of passband stochastic processes.

A random process $X(t)$ will be called a passband process if the power spectral density (PSD) of the process is zero outside of a range of frequencies centered around $f_c$ (check this defintion, I defined the processes as passband or baseband regardless of whether the process is stationary or not and whether it is zero-mean or not. Proakis defines the process as being passband only if the mean is zero and it is WSS). Similarly, it will be called a baseband stochastic (or, random) process if its PSD is zero outside of a range of frequencies centered around $0$ Hz. If the process is wide-sense stationary (WSS), then the PSD is the Fourier transform of the autocorrelation function and, hence, this is equivalent to saying that the autocorrelation $R_{XX}(\tau)$ is a deterministic passband or baseband signal.

Then, almost every realization of the passband random process and, hence, the random process process can be expressed in terms of a baseband equivalent random process in a manner similar to what we did for deterministic signals. The passband process $X(t)$ can be expressed as

\begin{eqnarray} X(t)&=& X_i(t)\cos(\omega _c t) - X_q(t)\sin(\omega _c t)\\ &=& a(t)\cos(\omega _c t + \theta(t))\\ &=& \Re \{ X_l(t)e^{j 2 \pi f_c t}\} \end{eqnarray}

Here $X_l(t) = X_i(t) + j X_q(t)$ is the complex baseband equivalent of $X(t)$ and $X_i(t)$ and $X_q(t)$ are called the in-phase and quadrature components of $X(t)$. Futher, $X_i(t),X_q(t)$ and $a(t)$ are real baseband random processes.

  • We will first show that $X_i(t)$ and $X_q(t)$ are jointly WSS processes and if $X(t)$ is a zero-mean process, then $X_i(t)$ and $X_q(t)$ are also zero-mean processes.

The Hilbert transform is very useful in showing this result. Recall that the Hilbert transform of the signal $X(t)$ is the result of passing it through an LTI system with an impulse response $h(t)$ and frequency response $H(f) = -j sgn(f)$.


Just like how the in-phase and quadrature components of a deterministic passband signal can be written in terms of the signal and its Hilbert transform, we can write $X_i(t)$ and $X_q(t)$ in terms of $X(t)$ and its Hilbert transform, namely $\hat{X}(t)$ as follows.

\begin{eqnarray} X_i(t) & = & X(t) \cos (\omega_c t) + \hat{X}(t) \sin (\omega_c t) \\ X_q(t) & = & \hat{X}(t) \cos (\omega_c t) - X(t) \sin (\omega_c t) \end{eqnarray}

Since $\hat{X}(t)$ is the result of passing $X(t)$ through an LTI system with impulse reponse $h(t)$, i.e., a Hilbert transformer, first of all $X(t)$ and $\hat{X}(t)$ are jointly WSS. Further, the autocorrelation and crosscorrelation between them can be expressed as

\begin{eqnarray} R_{\hat{X} X}(\tau) & = & R_{XX}(\tau) \star h(\tau) = \hat{R}_{XX}(\tau) \\ R_{X \hat{X}}(\tau) & = & R_{XX}(\tau) \star h^*(-\tau) = R_{XX}(\tau) \star -h(\tau) = -\hat{R}_{XX}(\tau) \\ S_{\hat{X} \hat{X}}(f) & = & S_{XX}(f) |H(f)|^2 = S_{XX}(f) \\ R_{\hat{X} \hat{X}}(\tau) & = & R_{XX}(\tau) \end{eqnarray}

Now, the plan is to expand the autocorrelation function of $X_i(t)$ and the cross correlation between $X_i(t)$ and $X_q(t)$ and use the above results to show that the autocorrelation function is only a function of $\tau$. This is done in 2.9-4 in Proakis 5th Ed. This essentially shows that $X_i(t)$ and $X_q(t)$ are jointly WSS processes. It is easy to see from 2 that if $X(t)$ is a zero-mean process, then $\hat{X}(t)$, $X_i(t)$ and $X_q(t)$ are all zero-mean processes.

Relationship between passband correlations and correlation functions of I and Q processes

  • In the remaining part of this section, we will explore the relationship between the autocorrelation function $R_{XX}(\tau)$ and the autocorrelation functions $R_{X_i X_i}(\tau )$,$R_{X_q X_q}(\tau )$, $R_{X_i X_q}(\tau )$ and $R_{X_q X_i}(\tau )$. We will begin with the definition of
\begin{eqnarray} R_{XX}(\tau)&=& E[X(t) X(t+\tau)]\\ &=& E\left[ \{ X_i(t)\cos(\omega _c t) - X_q(t)\sin(\omega _c t)\} \{(X_i(t+\tau)\cos(\omega _c(t+\tau)) - X_q(t+\tau)\sin(\omega _c (t+\tau))\} \right]\\ &=& R_{X_i X_i}(\tau)\cos(\omega _c t) \cos(\omega _c (t+\tau)) -R_{X_q X_i}(\tau)\cos(\omega_c t)\sin(\omega _c(t+\tau))+R_{X_q X_q}(\tau)\sin(\omega _c t) \sin(\omega_c (t+\tau)) -R_{X_i X_q}(\tau)\sin(\omega _c t)\cos(\omega_c (t+\tau)) \end{eqnarray}

using trig identities,

\begin{align} R_{XX}(\tau)= \frac{1}{2}\left[ \{ R_{X_i X_i}(\tau)+R_{X_q X_q}(\tau)\} \cos(\omega _c t)- \{R_{X_i X_i}(\tau)-R_{X_q X_q}(\tau)\} \cos(2 \omega _c (t+\tau)) + \{ R_{X_qX_i}(\tau)-R_{X_i X_q}(\tau) \} \sin(\omega_c t) - \{ R_{X_q X_i}(\tau)+R_{X_i X_q}(\tau) \} \sin(2\omega _c (t+\tau)) \right] \end{align}

since $X(t)$ is wide-sense stationary $R_{XX}(\tau )$ must only be a function of $\tau$. Therefore, $- \{R_{X_i X_i}(\tau)-R_{X_q X_q}(\tau)\} \cos(2 \omega _c (t+\tau)) + \{ R_{X_qX_i}(\tau)-R_{X_i X_q}(\tau) \} \sin(\omega_c t)$ must be zero for all $t$. This is possible only if

\begin{eqnarray} R_{X_i X_i}(\tau)&=&R_{X_q X_q}(\tau)\\ R_{X_q X_i}(\tau)&=&-R_{X_i X_q}(\tau). \end{eqnarray}

Since $X_i(t)$ and $X_q(t)$ are two jointly WSS processes, their cross correlations satisfy
$R_{X_q X_i}(\tau)=R_{X_i X_q}(-\tau)$. Therefore, we get,

\begin{align} R_{X_i X_q}(\tau)=-R_{X_i X_q}(-\tau) \end{align}

Using this in Eqn.5, $R_{XX}(\tau)$ simplifies to

\begin{eqnarray} R_{XX}(\tau)&=&R_{X_i X_i}(\tau)\cos(\omega _c \tau) - R_{X_q X_i}(\tau)\sin(\omega _c \tau)\\ &=& \Re \left[ \{R_{X_i X_i}(\tau) + j R_{X_q X_i}(\tau)\} e^{j \omega _c \tau} \right] \end{eqnarray}

This shows that $R_{X_i X_i}(\tau) + j R_{X_q X_i}(\tau)$ is the complex baseband equivalent of the passband signal $R_{XX}(\tau)$. Hence,

\begin{eqnarray} R_{X_i X_i}(\tau) = R_{X_q X_q}(\tau) & = & R_{XX}(\tau) \cos(\omega_c \tau) + \hat{R}_{XX}(\tau) \sin(\omega_c \tau) \\ R_{X_i X_q}(\tau) = -R_{X_q X_i}(\tau) & = & R_{XX}(\tau) \sin(\omega_c \tau) - \hat{R}_{XX}(\tau) \cos(\omega_c \tau) \end{eqnarray}

Relationship between passband correlations and complex-baseband correlations

Often, it is useful to express the autocorrelation function of $x(t)$ interms of the autocorrelation function of the complex baseband equivalent random process $X_l(t)$. In this section, we will relate these two by relating $X_l(t)$ to the autocorrelations and crosscorrelations between $X_i(t)$ and $X_q(t)$.

Recall that the complex envelope of $X(t)$ is $X(t) = X_i(t)+j X_q(t)$. Let us consider the complex-valued autocorrelation function of $R_{X_l X_l}(\tau)$ given by

\begin{eqnarray} R_{X_l X_l}(\tau) & = & E[X_l(t) X_l^*(t-\tau)]\\ &=& \left[ R_{X_i X_i}(\tau) + R_{X_q X_q}(\tau)\right] + j \left[ R_{X_q X_i}(\tau) - R_{X_i X_q}(\tau) \right] \\ &=& 2R_{X_i X_i}(\tau) + 2 j R_{X_q X_i}(\tau) \end{eqnarray}


\begin{eqnarray} R_{XX}(\tau)&=&\frac{1}{2} \Re \{R_{X_l X_l}(\tau)e^{j \omega _c \tau}\}\\ S_{XX}(f)&=&\frac{1}{4}[S_{X_l X_l}(f-f_c) + S_{X_l X_l}^*(-f-f_c)]. \end{eqnarray}

Compare the above relationship to similar relationships that we obtained in the deterministic case where(12)
\begin{align} S(f)=\frac{1}{2} \left[ S_l(f-f_c)+S_l^*(-f-f_c) \right] \end{align}

Comparing Eqn.11 with Eqn.12, we can get equivalent relationships between $R_{X_l X_l}(f)$ and $R_{XX}(f)$ using the relation,

\begin{eqnarray} S_{X_l X_l}(f)&=& 4S(f-f_c)U(f)\\ &=& 4S(f+f_c) \hspace{20mm}|f|<f_o \end{eqnarray}

Relationship between PSDs

We will now try to write the PSD of the I and Q components, namely $S_{X_i X_i}(f)$, $S_{X_q X_q}(f)$ and the cross power spectral density $S_{X_q X_i}(f)$ in terms of the PSD of the passband random process $S_{XX}(f)$. Notice that,

\begin{align} R_{X_l X_l}(\tau)=2 R_{X_i X_i}(\tau) + j 2 R_{X_q X_i}(\tau). \end{align}

From this, we can see that

\begin{eqnarray} R_{X_i X_i}(\tau)&=&\frac{1}{2}Re[R_{X_l X_l}(\tau)]\\ R_{X_q X_i}(\tau)&=&\frac{1}{2}Im[R_{X_l X_l}(\tau)] \end{eqnarray}

Recall that the Fourier transform of the real (imaginary) part of a signal is the symmetric (skew-symmetric) part of Fourier transform of the original signal and, therefore,
$S_{X_i X_i}(f)$ is half the symmetric part of $S_{X_l X_l}(f)$ which is given by

\begin{eqnarray} S_{X_i X_i}(f)&=&S_{XX}(f+f_c) + S_{XX}(f-f_c) \hspace{5mm} |f|\leq f_0 \\ S_{X_i X_q}(f)&=& -j \left[ S_{XX}(f+f_c) - S_{XX}(f-f_c) \right] \hspace{5mm} |f|\leq f_0 \end{eqnarray}

Notice that $S_{XX}(f-f_c)$ is actually a flipped version of $S_{XX}(f+f_c)$ because $S_{XX}(f)$ is symmetric around zero to begin with.

Gaussian Processes

Suppose $X(t)$ is a Gaussian passband process (in addition to being zero-mean and WSS), then $X_i(t)$, $X_q(t)$ are jointly Gaussian processes. Further, $X_l(t)$ will be a proper Gaussian process. It will also be a circular Gaussian process. In this case, if $S_{XX}(f+f_c) = S_{XX}(f-f_c)$ for $|f| < f_c$, then $X_i(t)$ and $X_q(t)$ will be independent as well.

Example: (Example 2.9-1 from Proakis Fifth Edition, Page 81) One of the most often used passband random processes in digital communications is a bandlimited white Gaussian noise process. Such a process $N(t)$ is defined as a zero-mean Gaussian process whose PSD is given by

\begin{eqnarray} S_{XX}(f) = \left{ \begin{array}{ll} \frac{N_0}{2} & |f-f_c| < W\\ 0 & \mbox{otherwise} \end{array} \right. \end{eqnarray}

The PSD is also shown in the figure below. What are the correlation functions and power spectral densities of the baseband equivalent random process?


Using 16, we can see that

\begin{eqnarray} S_{X_i X_i}(f) = S_{X_q X_q}(f) & = & N_0, \ \ \ |f| < W \\ S_{X_i X_q}(f) = S_{X_q X_i}(f) & = & 0 \\ S_{X_l X_l}(f) & = & 2 N_0, \ \ \ |f| < W \end{eqnarray}

This implies that

\begin{eqnarray} R_{X_i X_i}(\tau) = R_{X_q X_q}(\tau) & = & N_0 W sinc(2W \tau) \\ R_{X_i X_q}(\tau) = R_{X_q X_i}(\tau) & = & 0 \end{eqnarray}

Since $X_i(t)$ and $X_q(t)$ are Gaussian process and they are uncorrelated, they are also independent.

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