BANDPASS PROCESSES

Must work about these for multiplying many signals using same channel (like radio)
Now need to worry about decomposition
Any X(t) can be decomposed into its sin portion and its cosine portion. What we want to do is represent X(t) as a modulated signal

$$X(t)=X_c(t)\cos 2\pi f_0 t- X_s(t) \sin 2\pi f_0 t$$

$$\hat{X}(t)=X_c(t)\sin 2\pi f_0 t+ X_s(t)\cos 2\pi f_0 t$$

The X_c(t) and X_s(t) are "baseband" processes
Hilbert tells us

$$\hat{X}^(t)=Hil(X(t))$$

$$X_c(t)=X(t)\cos 2\pi f_0 t+ \hat{X}(t)\sin 2\pi f_0 t$$

$$X_s(t)=X(t)\cos 2\pi f_0 t- \hat{X}(t)\sin 2\pi f_0 t$$

FALLOUT
Thm.
X(t) is a stationary BP process (BW=2W)
$\Rightarrow X_s(t)$ and X_c(t) are jointly stationary lowpass (baseband) processes each having the power of X(t)

$$R_{X_c}(\tau)=R_{X_s}(\tau)=R_X(\tau)\cos 2\pi f_0 \tau+ \hat{R}_X(\tau)\sin 2\pi f_0 \tau$$

$$R_{X_cX_s}(\tau)=R_X(\tau)\sin 2\pi f_0 \tau - \hat{R}_X(\tau)\cos 2\pi f_0 \tau$$

$$\hat{R}_X(\tau)=Hil(R_X(\tau))$$

BP

$$S_{X_c}(f)=S_{X_s}(f)= \left\{\begin{array}{ll} \displaysty... ... \displaystyle {0} & {\qquad \mbox{o.w.}} \end{array}\right.$$

Since S_X(f) limited to |f-f₀|<W then S_Xc/s(f) limited to |f|<W

$$S_{X_c X_s}(f)=\left\{\begin{array}{ll} \displaystyle {j(S_... ...f_0} \\ \displaystyle {0} & {\qquad o.w.} \end{array}\right.$$

same idea (notice S_{Xs Xc}(0)= always)

Important $\Rightarrow$ if X(t) is such that

$$S_X(f+f_0)=S_X(f-f_0) \qquad \qquad (\vert f\vert<f_0)$$

then cross corr is zero!
i.e. $R_{X_c X_s}(\tau)\equiv 0$ everywhere!
So, say X(t) is also zero mean $\rightarrow X_c(t)$ and X_s(t) also zero mean (easy to see)
Therefore, X_c(t) and X_s(t) uncorrelated
Say X(t) is also gaussian $\rightarrow X_c$ and X_s jointly gaussian
Therefore, X_c and X_s INDEPT

FINALLY, P_X=R_X(0)=R_Xc(0)=R_Xs(0)
Powers in X_c(t) nd X_s(t) same as in X(t)

END