(New page: ==Question from ECE QE CS Q1 August 2000== ---- ==Share and discuss your solutions below.== ---- =Solution 1 (retrived from [[ECE600_QE_2000_August|her...) |
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| − | + | [[Category:ECE]] | |
| + | [[Category:QE]] | ||
| + | [[Category:CNSIP]] | ||
| + | [[Category:problem solving]] | ||
| + | [[Category:random variables]] | ||
| + | [[Category:probability]] | ||
| + | |||
| + | <center> | ||
| + | <font size= 4> | ||
| + | [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] | ||
| + | </font size> | ||
| + | |||
| + | <font size= 4> | ||
| + | Communication, Networking, Signal and Image Processing (CS) | ||
| + | |||
| + | Question 1: Probability and Random Processes | ||
| + | </font size> | ||
| + | |||
| + | August 2000 | ||
| + | </center> | ||
| + | ---- | ||
| + | ---- | ||
| + | =Part 4= | ||
| + | A RV is given by <math class="inline">\mathbf{Z}=\sum_{n=0}^{8}\mathbf{X}_{n}</math> where <math class="inline">\mathbf{X}_{n}</math> 's are i.i.d. RVs with characteristic function given by <math class="inline">\Phi_{\mathbf{X}}\left(\omega\right)=\frac{1}{1-j\omega/2}.</math> | ||
| + | |||
| + | '''(a)''' Determine the characteristic function of <math class="inline">\mathbf{Z}</math> . | ||
| + | |||
| + | '''(b)''' Determine the pdf of <math class="inline">\mathbf{Z}</math> . You can leave your answer in integral form. | ||
---- | ---- | ||
==Share and discuss your solutions below.== | ==Share and discuss your solutions below.== | ||
---- | ---- | ||
=Solution 1 (retrived from [[ECE600_QE_2000_August|here]])= | =Solution 1 (retrived from [[ECE600_QE_2000_August|here]])= | ||
| + | '''(a)''' | ||
| + | |||
| + | <math class="inline">\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}.</math> | ||
| + | |||
| + | '''(b)''' | ||
| + | <math class="inline">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.</math> | ||
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Latest revision as of 10:34, 13 September 2013
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2000
Contents
Part 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} $ .
(b) Determine the pdf of $ \mathbf{Z} $ . You can leave your answer in integral form.
Solution 1 (retrived from here)
(a)
$ \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)
$ 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. $
Solution 2
Write it here.
