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===6 Finals===
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= 6 Finals =
  
'''Example (Important because it was dealt in the class)'''
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'''Example (Important because it was dealt in the class)'''  
  
Let <math>\mathbf{X}\left(t\right)</math> be a WSS random process with PSD <math>S_{\mathbf{XX}}\left(\omega\right)</math> and let <math>\mathbf{Y}\left(t\right)</math> be the “smoothed” random process given by <math>\mathbf{Y}\left(t\right)=\frac{1}{2T}\int_{t-T}^{t+T}\mathbf{X}\left(\alpha\right)d\alpha.</math>
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Let <math>\mathbf{X}\left(t\right)</math> be a WSS random process with PSD <math>S_{\mathbf{XX}}\left(\omega\right)</math> and let <math>\mathbf{Y}\left(t\right)</math> be the “smoothed” random process given by <math>\mathbf{Y}\left(t\right)=\frac{1}{2T}\int_{t-T}^{t+T}\mathbf{X}\left(\alpha\right)d\alpha.</math>  
  
<br>
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<br>This can be represented by a LTI system  
This can be represented by a LTI system  
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  [[Image:pasted22.png]]
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  [[Image:Pasted22.png]]
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with impulse response <math>h\left(t\right)=\frac{1}{2T}\mathbf{1}_{\left[-T,T\right]}\left(t\right)</math>. What is the PSD <math>S_{\mathbf{YY}}\left(\omega\right)</math> of <math>\mathbf{Y}\left(t\right)</math> ?
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with impulse response <math>h\left(t\right)=\frac{1}{2T}\mathbf{1}_{\left[-T,T\right]}\left(t\right)</math>. What is the PSD <math>S_{\mathbf{YY}}\left(\omega\right)</math> of <math>\mathbf{Y}\left(t\right)</math>&nbsp;?  
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'''Solution'''
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<math>H\left(\omega\right)=\int_{-\infty}^{\infty}h\left(t\right)e^{-i\omega t}dt=\int_{-\infty}^{\infty}\frac{1}{2T}\mathbf{1}_{\left[-T,T\right]}\left(t\right)e^{-i\omega t}dt=\frac{1}{2T}\int_{-T}^{T}e^{-i\omega t}dt</math><math>=\left.\frac{1}{2T}\frac{e^{-i\omega t}}{-i\omega}\right|_{-T}^{T}=\frac{1}{2T}\frac{e^{-i\omega T}-e^{i\omega T}}{-i\omega}=\frac{1}{2T}\frac{\left(\cos\omega T-i\sin\omega T\right)-\left(\cos\omega T+i\sin\omega T\right)}{-i\omega}</math><span class="texhtml"> </span>
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'''Solution'''
 
  
<math>H\left(\omega\right)=\int_{-\infty}^{\infty}h\left(t\right)e^{-i\omega t}dt=\int_{-\infty}^{\infty}\frac{1}{2T}\mathbf{1}_{\left[-T,T\right]}\left(t\right)e^{-i\omega t}dt=\frac{1}{2T}\int_{-T}^{T}e^{-i\omega t}dt</math><math>=\left.\frac{1}{2T}\frac{e^{-i\omega t}}{-i\omega}\right|_{-T}^{T}=\frac{1}{2T}\frac{e^{-i\omega T}-e^{i\omega T}}{-i\omega}=\frac{1}{2T}\frac{\left(\cos\omega T-i\sin\omega T\right)-\left(\cos\omega T+i\sin\omega T\right)}{-i\omega}</math><math>=\frac{1}{2T}\frac{2\sin\omega T}{\omega}=\frac{\sin\omega T}{\omega T}.</math>
 
  
 
<math>S_{\mathbf{YY}}\left(\omega\right)=S_{\mathbf{XX}}\left(\omega\right)\left|H\left(\omega\right)\right|^{2}=S_{\mathbf{XX}}\left(\omega\right)\left|\frac{\sin\omega T}{\omega T}\right|^{2}.</math>  
 
<math>S_{\mathbf{YY}}\left(\omega\right)=S_{\mathbf{XX}}\left(\omega\right)\left|H\left(\omega\right)\right|^{2}=S_{\mathbf{XX}}\left(\omega\right)\left|\frac{\sin\omega T}{\omega T}\right|^{2}.</math>  
  
Note that <math>h\left(t\right)</math> acts as a crude low-pass filter that attenuates high-frequency power.
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Note that <math>h\left(t\right)</math> acts as a crude low-pass filter that attenuates high-frequency power.  
  
[[Image:pasted23.png]]
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[[Image:Pasted23.png]]  
  
Example (True or False)
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Example (True or False)  
  
Let <math>\mathbf{X}\left(t\right)</math> and <math>\mathbf{Y}\left(t\right)</math> be two zero-mean statistically independent, jointly wide-sense stationary random processes. Then the cross-correlation function <math>R_{\mathbf{XY}}\left(\tau\right)=0</math> .
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Let <math>\mathbf{X}\left(t\right)</math> and <math>\mathbf{Y}\left(t\right)</math> be two zero-mean statistically independent, jointly wide-sense stationary random processes. Then the cross-correlation function <math>R_{\mathbf{XY}}\left(\tau\right)=0</math> .  
  
'''Solution'''
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'''Solution'''  
  
''True.''
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''True.''  
  
 
<math>R_{\mathbf{XY}}\left(t_{1},t_{2}\right)=E\left[\mathbf{X}\left(t_{1}\right)\mathbf{Y}^{*}\left(t_{2}\right)\right]=E\left[\mathbf{X}\left(t_{1}\right)\right]E\left[\mathbf{Y}^{*}\left(t_{2}\right)\right]=0\cdot0=0.</math>  
 
<math>R_{\mathbf{XY}}\left(t_{1},t_{2}\right)=E\left[\mathbf{X}\left(t_{1}\right)\mathbf{Y}^{*}\left(t_{2}\right)\right]=E\left[\mathbf{X}\left(t_{1}\right)\right]E\left[\mathbf{Y}^{*}\left(t_{2}\right)\right]=0\cdot0=0.</math>  
  
'''Example (True or False)'''
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'''Example (True or False)'''  
  
The cross-correlation function <math>R_{\mathbf{XY}}\left(\tau\right)</math> of two real, jointly wide-sense stationary random process <math>\mathbf{X}\left(t\right)</math> and <math>\mathbf{Y}\left(t\right)</math> is an even function of <math>\tau</math> .
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The cross-correlation function <math>R_{\mathbf{XY}}\left(\tau\right)</math> of two real, jointly wide-sense stationary random process <math>\mathbf{X}\left(t\right)</math> and <math>\mathbf{Y}\left(t\right)</math> is an even function of <span class="texhtml">τ</span> .  
  
'''Solution'''
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'''Solution'''  
  
''False.''
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''False.''  
  
*[[ECE 600 Finals MRB 1992 Final|1992 Final]]
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*[[ECE 600 Finals MRB 1992 Final|1992 Final]]  
*[[ECE 600 Finals MRB 1994 Final|1994 Final]]
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*[[ECE 600 Finals MRB 1994 Final|1994 Final]]  
 
*[[ECE 600 Finals MRB 2004 Final|2004 Spring Final]]
 
*[[ECE 600 Finals MRB 2004 Final|2004 Spring Final]]

Revision as of 12:27, 19 November 2010

6 Finals

Example (Important because it was dealt in the class)

Let $ \mathbf{X}\left(t\right) $ be a WSS random process with PSD $ S_{\mathbf{XX}}\left(\omega\right) $ and let $ \mathbf{Y}\left(t\right) $ be the “smoothed” random process given by $ \mathbf{Y}\left(t\right)=\frac{1}{2T}\int_{t-T}^{t+T}\mathbf{X}\left(\alpha\right)d\alpha. $


This can be represented by a LTI system

Pasted22.png

with impulse response $ h\left(t\right)=\frac{1}{2T}\mathbf{1}_{\left[-T,T\right]}\left(t\right) $. What is the PSD $ S_{\mathbf{YY}}\left(\omega\right) $ of $ \mathbf{Y}\left(t\right) $ ?

Solution

$ H\left(\omega\right)=\int_{-\infty}^{\infty}h\left(t\right)e^{-i\omega t}dt=\int_{-\infty}^{\infty}\frac{1}{2T}\mathbf{1}_{\left[-T,T\right]}\left(t\right)e^{-i\omega t}dt=\frac{1}{2T}\int_{-T}^{T}e^{-i\omega t}dt $$ =\left.\frac{1}{2T}\frac{e^{-i\omega t}}{-i\omega}\right|_{-T}^{T}=\frac{1}{2T}\frac{e^{-i\omega T}-e^{i\omega T}}{-i\omega}=\frac{1}{2T}\frac{\left(\cos\omega T-i\sin\omega T\right)-\left(\cos\omega T+i\sin\omega T\right)}{-i\omega} $

1 2sinωT sinωT
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=
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2T ω ωT


$ S_{\mathbf{YY}}\left(\omega\right)=S_{\mathbf{XX}}\left(\omega\right)\left|H\left(\omega\right)\right|^{2}=S_{\mathbf{XX}}\left(\omega\right)\left|\frac{\sin\omega T}{\omega T}\right|^{2}. $

Note that $ h\left(t\right) $ acts as a crude low-pass filter that attenuates high-frequency power.

Pasted23.png

Example (True or False)

Let $ \mathbf{X}\left(t\right) $ and $ \mathbf{Y}\left(t\right) $ be two zero-mean statistically independent, jointly wide-sense stationary random processes. Then the cross-correlation function $ R_{\mathbf{XY}}\left(\tau\right)=0 $ .

Solution

True.

$ R_{\mathbf{XY}}\left(t_{1},t_{2}\right)=E\left[\mathbf{X}\left(t_{1}\right)\mathbf{Y}^{*}\left(t_{2}\right)\right]=E\left[\mathbf{X}\left(t_{1}\right)\right]E\left[\mathbf{Y}^{*}\left(t_{2}\right)\right]=0\cdot0=0. $

Example (True or False)

The cross-correlation function $ R_{\mathbf{XY}}\left(\tau\right) $ of two real, jointly wide-sense stationary random process $ \mathbf{X}\left(t\right) $ and $ \mathbf{Y}\left(t\right) $ is an even function of τ .

Solution

False.

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn