(Created page with "Category:ECE Category:QE Category:CNSIP Category:problem solving Category:image processing = ECE Ph.D. Qualifying Exam in Com...") |
|||
| Line 12: | Line 12: | ||
== Solution: == | == Solution: == | ||
| − | a) <math> | + | a) |
| + | |||
| + | <math> | ||
\text{Since}\ Y_{x}\ \text{is a Poisson random variable,} \\ | \text{Since}\ Y_{x}\ \text{is a Poisson random variable,} \\ | ||
\Rightarrow | \Rightarrow | ||
E[Y_{x}]=\lambda_{x}\\ | E[Y_{x}]=\lambda_{x}\\ | ||
</math> | </math> | ||
| − | (b)<math> | + | |
| + | (b) | ||
| + | |||
| + | <math> | ||
For Poisson r.v., <math>E[Y_{x}]=Var[Y_{x}]\\ | For Poisson r.v., <math>E[Y_{x}]=Var[Y_{x}]\\ | ||
\Rightarrow | \Rightarrow | ||
| − | Var[Y_{x}]=\lambda_{x}</math> | + | Var[Y_{x}]=\lambda_{x} |
| + | </math> | ||
| + | |||
(c) | (c) | ||
The attenuation of photons obeys: | The attenuation of photons obeys: | ||
<math>\frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x}</math> | <math>\frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x}</math> | ||
| + | |||
(d) | (d) | ||
The solution is: | The solution is: | ||
Revision as of 19:23, 2 December 2015
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)
Question 5, August 2012, Part 2
- Part 1 , 2
Solution:
a)
$ \text{Since}\ Y_{x}\ \text{is a Poisson random variable,} \\ \Rightarrow E[Y_{x}]=\lambda_{x}\\ $
(b)
$ For Poisson r.v., <math>E[Y_{x}]=Var[Y_{x}]\\ \Rightarrow Var[Y_{x}]=\lambda_{x} $
(c) The attenuation of photons obeys:
$ \frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x} $
(d) The solution is:
$ \lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t} $ (e) Based on the result of (d)
$ \lambda_{T}=\lambda_{0}e^{-\int_0^T \mu(t)\partial t}\\ \Rightarrow \frac{\lambda_{T}}{\lambda_{0}}=e^{-\int_0^T \mu(t)\partial t}\\ \Rightarrow \int_0^T \mu(t)\partial t=-ln{\frac{\lambda_{T}}{\lambda_{0}}}=ln{\frac{\lambda_{0}}{\lambda_{T}}} $
