ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)

August 2015, Part 2

Part 1 , 2

Solution 1:

a) $ \lambda_{x} $ (as Y is is Poisson r.v).

(b) For Poisson r.v., $ E[Y_{x}]=Var[Y_{x}]=\lambda_{x} $

(c) The attenuation of photons obeys:

$ \frac{\partial \lambda_{x}}{\partial x}=-\mu(x)\lambda_{x} $

(d) The solution to c) gives $ \lambda_{x}=\lambda_{0}e^{-\int_0^x \mu(t)\partial t} $

(e) Given (d), and substitute x with t we can get that $ \int_0^x \mu(T)\partial t=-log{\frac{\lambda_{T}}{\lambda_{0}}}=log{\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}}} $


Back to ECE QE page:

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett