Revision as of 11:36, 23 November 2010

7.1 QE 2000 August

1.

(a)

The Laplacian density function is given by $ f\left(x\right)=\frac{A}{2}e^{-A\left|x\right|}\text{ where }A>0. $ Determine its characteristic function.

$ \Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\int_{-\infty}^{\infty}\frac{A}{2}e^{-A\left|x\right|}\cdot e^{i\omega x}dx=\frac{A}{2}\left[\int_{-\infty}^{0}e^{x\left(A+i\omega\right)}dx+\int_{0}^{\infty}e^{x\left(-A+i\omega\right)}dx\right] $$ =\frac{A}{2}\left[\frac{e^{x\left(A+i\omega\right)}}{A+i\omega}\biggl|_{-\infty}^{0}+\frac{e^{x\left(-A+i\omega\right)}}{-A+i\omega}\biggl|_{0}^{\infty}\right]=\frac{A}{2}\left[\frac{1}{A+i\omega}-\frac{1}{-A+i\omega}\right] $$ =\frac{A}{2}\cdot\frac{A-i\omega+A+i\omega}{A^{2}+\omega^{2}}=\frac{A^{2}}{A^{2}+\omega^{2}}. $

(b)

Determine a bound on the probability that a RV is within two standard deviations of its mean.

$ P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|\leq2\sigma\right\} \right)=1-P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|>2\sigma\right\} \right). $ By Chebyshev Inequality, $ P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|>2\sigma\right\} \right)\leq\frac{\sigma^{2}}{\left(2\sigma\right)^{2}}=\frac{1}{4} $ .

$ P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|\leq2\sigma\right\} \right)\geq\frac{3}{4}. $

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| $ ).

ref. pds means the power spectral density 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). $

3.

Inquiries arrive at a recorded message device according to a Poisson process of rate 15 inquiries per minute. Find the probability that in a 1-minute period, 3 inquiries arrive during the first 10 seconds and 2 inquiries arrive during the last 15 seconds.

$ \lambda=\frac{15}{60\text{ sec}}=\frac{1}{4}\text{ sec}^{-1}. $

$ P\left(\left\{ N\left(t_{1},t_{2}\right)=k\right\} \right)=\frac{\left(\left(\lambda\left(t_{2}-t_{1}\right)\right)^{k}e^{-\lambda\left(t_{2}-t_{1}\right)}\right)}{k!}. $

$ P\left(\left\{ N\left(0,10\right)=3\right\} \cap\left\{ N\left(45,60\right)=2\right\} \right)=P\left(\left\{ N\left(0,10\right)=3\right\} \right)P\left(\left\{ N\left(45,60\right)=2\right\} \right) $$ =\frac{\left(\frac{1}{4}\times10\right)^{3}e^{-\frac{1}{4}\times10}}{3!}\times\frac{\left(\frac{1}{4}\times15\right)^{2}e^{-\frac{1}{4}\times15}}{2!} $$ =\frac{1}{12}\cdot\left(\frac{5}{2}\right)^{3}\left(\frac{15}{4}\right)^{2}e^{-\frac{25}{4}}. $

4.

A RV is given by $ \mathbf{Z}=\sum_{n=0}^{8}\mathbf{X}_{n} $ where $ \mathbf{X}_{n} $ 's are i.i.d. RVs with characteristic function given by $ \Phi_{\mathbf{X}}\left(\omega\right)=\frac{1}{1-j\omega/2}. $

(a)

Determine the characteristic function of $ \mathbf{Z} $ .

$ \Phi_{\mathbf{Z}}\left(\omega\right)=E\left[e^{i\omega\mathbf{Z}}\right]=E\left[e^{i\omega\sum_{n=0}^{8}\mathbf{X}_{n}}\right]=E\left[\prod_{n=0}^{8}e^{i\omega\mathbf{X}_{n}}\right]=\prod_{n=0}^{8}E\left[e^{i\omega\mathbf{X}_{n}}\right]=\left(\frac{1}{1-j\omega/2}\right)^{9}. $

(b)

Determine the pdf of $ \mathbf{Z} $ . You can leave your answer in integral form.

$ f_{\mathbf{Z}}\left(z\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\Phi_{\mathbf{Z}}\left(\omega\right)e^{-i\omega z}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\frac{1}{1-j\omega/2}\right)^{9}e^{-i\omega z}d\omega. $

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