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:[[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 ]] | ||
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| − | == Solution: == | + | == Solution 1: == |
| + | |||
| + | a) <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}]\\ | ||
| + | \Rightarrow | ||
| + | Var[Y_{x}]=\lambda_{x} | ||
| + | </math> | ||
| + | |||
| + | (c) The attenuation of photons obeys: | ||
| + | |||
| + | <math>\frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x}</math> | ||
| + | |||
| + | (d) The solution is: | ||
| + | |||
| + | <math>\lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t}</math> | ||
| + | (e) Based on the result of (d) | ||
| + | |||
| + | <math> | ||
| + | \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}}} | ||
| + | </math> | ||
| + | == Solution 2: == | ||
a) Since <math> Y_{x} </math> is Poisson random variable, <math> E[Y_{x}]=\lambda_{x}</math>. | a) Since <math> Y_{x} </math> is Poisson random variable, <math> E[Y_{x}]=\lambda_{x}</math>. | ||
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<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) Based on the result of (d) | (e) Based on the result of (d) | ||
Revision as of 22:40, 3 December 2015
Contents
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)
August 2015, Part 2
Solution 1:
a) $ 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}}} $
Solution 2:
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}}} $
