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=Solution 1 (retrived from [[ECE600_QE_2000_August|here]])=
 
=Solution 1 (retrived from [[ECE600_QE_2000_August|here]])=
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'''(a)'''
  
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<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>
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'''(b)'''
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<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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Revision as of 08:56, 27 June 2012

Question from ECE QE CS Q1 August 2000

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.

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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

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