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= [[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| Part 1 ]],[[CS5_2015_Aug_prob2_solution| 2 ]] |
---- | ---- | ||
== Solution: == | == Solution: == | ||
− | a) | + | a) Since <math> Y_{x} </math> is Poisson random variable, <math> E[Y_{x}]=\lambda_{x}</math>. |
− | + | (b) For Poisson r.v., <math>E[Y_{x}]=Var[Y_{x}]\\ | |
− | + | ||
− | + | ||
− | + | ||
− | + | ||
− | + | ||
− | (b) | + | |
− | + | ||
− | + | ||
− | 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> |
Latest revision as of 21:46, 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}}} $