Latest revision as of 10:34, 13 September 2013

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| $ ).

Share and discuss your solutions below.

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). $

Solution 2

Write it here.

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