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===== <math>\color{blue}\text{Solution 1:}</math> ===== | ===== <math>\color{blue}\text{Solution 1:}</math> ===== | ||
| − | <math> \mathbf{X}(t) \text{ is SSS if } F_{(t_1+\tau)...(t_n+\tau)}(x_1,...,x_n) \text{ does not depend on } \tau. \text{ To show that, we can show that } \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) \text{ does not depend on } \tau: | + | <font face="Times New Roman" font size="5"><math> |
| − | </math> | + | \mathbf{X}(t) \text{ is SSS if } F_{(t_1+\tau)...(t_n+\tau)}(x_1,...,x_n) \text{ does not depend on } \tau. \text{ To show that, we can show that } \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) \text{ does not depend on } \tau: |
| + | </math></font> | ||
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\Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{Y(t_j+\tau)} \right ] = \Phi_{(t_1+\tau)...(t_n+\tau)}(1) | \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{Y(t_j+\tau)} \right ] = \Phi_{(t_1+\tau)...(t_n+\tau)}(1) | ||
</math> | </math> | ||
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=\sum_{i,j=1}^{n}{\omega_j^2 cov(t_j,t_j)} + \sum_{i,j=1}^{n}{\omega_i \omega_j cov(t_j,t_j)} | =\sum_{i,j=1}^{n}{\omega_j^2 cov(t_j,t_j)} + \sum_{i,j=1}^{n}{\omega_i \omega_j cov(t_j,t_j)} | ||
</math> | </math> | ||
| + | |||
| + | |||
<font face="Times New Roman" font size="5"><math> | <font face="Times New Roman" font size="5"><math> | ||
Revision as of 13:41, 29 July 2012
Contents
ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS)
Question 1, August 2011, Part 2
- Part 1,2]
$ \color{blue}\text{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.} $
$ \color{blue}\text{Solution 1:} $
$ \mathbf{X}(t) \text{ is SSS if } F_{(t_1+\tau)...(t_n+\tau)}(x_1,...,x_n) \text{ does not depend on } \tau. \text{ To show that, we can show that } \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) \text{ does not depend on } \tau: $
$ \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{i\sum_{j=1}^{n}{\omega_jX(t_j+\tau)}} \right ] $
$ \text{Define } Y(t_j+\tau) = \sum_{j=1}^{n}{\omega_jX(t_j+\tau)} \text{, so} $
$ \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{Y(t_j+\tau)} \right ] = \Phi_{(t_1+\tau)...(t_n+\tau)}(1) $
$ \text{Since } Y(t) \text{ is Gaussian, it is characterized just by its mean and variance. So, we just need to show that mean and variance of } Y(t) \text{do not depend on } \tau. \text{Since } Y(t) \text{ is WSS, its mean is constant and does not depend on . For variance} $
$ var(Y(t_j+\tau)) = E \left [(\sum_{j=1}^{n}{w_j(X(t_j+\tau)-\mu)^2} \right ] $
$ =\sum_{j=1}^{n}{\omega_j^2E \left [ (X(t_j+\tau)-\mu)^2 \right ]} + \sum_{i,j=1}^{n}{\omega_i \omega_j E \left[ (X(t_i+\tau)-\mu)(X(t_j+\tau)-\mu) \right]} $
$ =\sum_{i,j=1}^{n}{\omega_j^2 cov(t_j,t_j)} + \sum_{i,j=1}^{n}{\omega_i \omega_j cov(t_j,t_j)} $
$ \text{Which does not depend on } \tau. $
$ \color{blue}\text{Solution 2:} $
$ \mathbf{X}(t) \text{ is SSS if } F_{(t_1+\tau)...(t_n+\tau)}(x_1,...,x_n) \text{ does not depend on } \tau. \text{ To show that, we can show that } \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) \text{ does not depend on } \tau: $
"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011
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- Part 1: solutions and discussions
- Part 2: solutions and discussions
