| Line 20: | Line 20: | ||
</math> | </math> | ||
| − | (b) | + | (b) 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} | Var[Y_{x}]=\lambda_{x} | ||
</math> | </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: | + | |
<math>\lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t}</math> | <math>\lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t}</math> | ||
| − | (e) | + | (e) Based on the result of (d) |
| − | Based on the result of (d) | + | |
<math> | <math> | ||
Revision as of 19:24, 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., $ 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}}} $
