Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2000

# Part 2

$\mathbf{X}\left(t\right)$ is a WSS process with its psd zero outside the interval $\left[-\omega_{max},\ \omega_{max}\right]$ . If $R\left(\tau\right)$ is the autocorrelation function of $\mathbf{X}\left(t\right)$ , prove the following: $R\left(0\right)-R\left(\tau\right)\leq\frac{1}{2}\omega_{max}^{2}\tau^{2}R\left(0\right).$ (Hint: $\left|\sin\theta\right|\leq\left|\theta\right|$ ).

# Solution 1 (retrived from here)

ref. pds means the power spectral density (More information on the Power Spectrum).

If $\mathbf{X}\left(t\right)$ is real, then $R_{\mathbf{X}}\left(\tau\right)$ is real and even function.

$S_{\mathbf{X}}\left(\omega\right)=\int_{-\infty}^{\infty}R_{\mathbf{X}}\left(\tau\right)e^{-i\omega\tau}d\tau=\int_{-\infty}^{\infty}\left(R_{\mathbf{X}}\left(\tau\right)\cos\left(\omega\tau\right)-R_{\mathbf{X}}\left(\tau\right)i\sin\left(\omega\tau\right)\right)d\tau $$=2\int_{0}^{\infty}R_{\mathbf{X}}\left(\tau\right)\cos\left(\omega\tau\right)d\tau\Longrightarrow\;\therefore S_{\mathbf{X}}\left(\omega\right)\text{ is real and even function.} R_{\mathbf{X}}\left(\tau\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_{\mathbf{X}}\left(\omega\right)e^{i\omega\tau}d\omega=\frac{1}{\pi}\int_{0}^{\omega_{max}}S_{\mathbf{X}}\left(\omega\right)\cos\left(\omega\tau\right)d\omega. R_{\mathbf{X}}\left(0\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_{\mathbf{X}}\left(\omega\right)e^{i\omega0}d\omega=\frac{1}{\pi}\int_{0}^{\omega_{max}}S_{\mathbf{X}}\left(\omega\right)d\omega. R_{\mathbf{X}}\left(0\right)-R_{\mathbf{X}}\left(\tau\right)=\frac{1}{\pi}\int_{0}^{\omega_{max}}S_{\mathbf{X}}\left(\omega\right)\left(1-\cos\left(\omega\tau\right)\right)d\omega=\frac{1}{\pi}\int_{0}^{\omega_{max}}S_{\mathbf{X}}\left(\omega\right)\left(2\sin^{2}\left(\frac{\omega\tau}{2}\right)\right)d\omega$$ \leq\frac{2}{\pi}\left|\int_{0}^{\omega_{max}}S_{\mathbf{X}}\left(\omega\right)\sin^{2}\left(\frac{\omega\tau}{2}\right)d\omega\right|\leq\frac{2}{\pi}\int_{0}^{\omega_{max}}\left|S_{\mathbf{X}}\left(\omega\right)\right|\left|\sin\left(\frac{\omega\tau}{2}\right)\right|^{2}d\omega $$\leq\frac{2}{\pi}\int_{0}^{\omega_{max}}\left|S_{\mathbf{X}}\left(\omega\right)\right|\left(\frac{\omega^{2}\tau^{2}}{4}\right)d\omega\leq\frac{\omega_{max}^{2}\tau^{2}}{2\pi}\int_{0}^{\omega_{max}}\left|S_{\mathbf{X}}\left(\omega\right)\right|d\omega$$ \leq\frac{\omega_{max}^{2}\tau^{2}}{2\pi}\pi R_{\mathbf{X}}\left(0\right)=\frac{\omega_{max}^{2}\tau^{2}}{2}R_{\mathbf{X}}\left(0\right).$

$\therefore R_{\mathbf{X}}\left(0\right)-R_{\mathbf{X}}\left(\tau\right)\leq\frac{1}{2}\omega_{max}^{2}\tau^{2}R_{\mathbf{X}}\left(0\right).$

$\because\cos\left(\omega\tau\right)=\cos^{2}\left(\frac{\omega\tau}{2}\right)-\sin^{2}\left(\frac{\omega\tau}{2}\right)=1-2\sin^{2}\left(\frac{\omega\tau}{2}\right).$

Write it here.