| Line 6: | Line 6: | ||
= [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] in Communication Networks Signal and Image processing (CS) = | = [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] in Communication Networks Signal and Image processing (CS) = | ||
| − | = [[ | + | = [[ECE-QE_CS5-2015|August 2015]], Part 2= |
:[[ CS5_2015_Aug_prob1_solution | Part 1 ]],[[ CS5_2015_Aug_prob2_solution | 2 ]] | :[[ CS5_2015_Aug_prob1_solution | Part 1 ]],[[ CS5_2015_Aug_prob2_solution | 2 ]] | ||
Revision as of 19:35, 2 December 2015
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)
August 2015, Part 2
- Part 1 , 2
Solution:
a) Since $ Y_{x} $ is Poisson random variable, $ 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}}} $
