| (11 intermediate revisions by the same user not shown) | |||
| Line 8: | Line 8: | ||
= [[ECE-QE_CS5-2014|August 2014]], Problem 1 = | = [[ECE-QE_CS5-2014|August 2014]], Problem 1 = | ||
| − | :[[CS5_2014_Aug_prob1| Problem 1 ]],[[ | + | :[[CS5_2014_Aug_prob1| Problem 1 ]],[[CS5_2014_Aug_prob2| 2 ]] |
---- | ---- | ||
===Solution 1=== | ===Solution 1=== | ||
a) <math>E\left[Y_x\right] = \lambda_x</math> | a) <math>E\left[Y_x\right] = \lambda_x</math> | ||
| + | |||
| + | <span style="color:green"> Lack of proof. Should mention the property of Poisson distribution to show the equivalence. See the proof in solution 2.</span> | ||
b) Because the rate of absorption is proportional to the number of photons and the density of the material, so the attenuation of photons obeys the following equation | b) Because the rate of absorption is proportional to the number of photons and the density of the material, so the attenuation of photons obeys the following equation | ||
| Line 36: | Line 38: | ||
== Solution 2: == | == Solution 2: == | ||
| − | a) <math> | + | a). As we know <math>P\left\{Y_x=k\right\} = \frac{e^{-\lambda_x}\lambda_x^k}{k!}</math> is a Potion distribution, it is known that the expectation of a Poisson RV is <math>\lambda_x</math>. |
| − | \ | + | |
| − | + | Proof: | |
| − | \ | + | |
| + | <math> | ||
| + | \begin{split} | ||
| + | E[Y_x] &= \sum^{+ \infty}_{k > 0} k \frac{e^{-\lambda_x}\lambda_x^k}{k!}\\ | ||
| + | &= \sum^{+ \infty}_{k = 1} \frac{e^{-\lambda_x}\lambda_x^k}{(k-1)!}\\ | ||
| + | &= \sum^{+ \infty}_{k = 1} \frac{e^{-\lambda_x}\lambda_x^{k-1}}{(k-1)!}\lambda_x\\ | ||
| + | &= \lambda_xe^{-\lambda_x}\sum^{+ \infty}_{k = 0} \frac{\lambda_x^k}{k!}\\ | ||
| + | &= \lambda_xe^{-\lambda_x}e^{\lambda_x}\\ | ||
| + | &= \lambda_x\\ | ||
| + | \end{split} | ||
</math> | </math> | ||
| − | + | So <math>E[Y_x] = \lambda_x</math> | |
| − | + | <span style="color:red"> Here, it used <math>\sum^{+ \infty}_{k = 0} \frac{\lambda_x^k}{k!} = e^{\lambda_x}</math> to derive the final conclusion.</span> | |
| + | b). Because the number of photons will decrease when increasing the depth, | ||
<math> | <math> | ||
| − | + | d\lambda_x = -\lambda_x\mu(x)dx | |
| − | + | </math> | |
| − | + | ||
| − | d | + | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | = | + | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | and | |
| − | + | ||
| − | + | ||
| − | + | <math> | |
| − | + | \frac{d\lambda_x}{dx} = -\lambda_x\mu(x) | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | d | + | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
</math> | </math> | ||
| − | c) | + | c). The final differential equation in b). is an ordinary differential equation. We can get the expression as |
| + | |||
| + | <math>\lambda_x = \lambda_0e^{-\int^x_0\mu(t)dt}</math> | ||
| + | |||
| + | where <math>\lambda_0</math> is the initial number of photons. | ||
| + | |||
| + | d). From part c). <math>\frac{\lambda_x}{\lambda_0} = e^{-\int^x_0\mu(t)dt}</math>, so we have | ||
<math> | <math> | ||
| − | + | \int^x_0\mu(t)dt = -\log\left(\frac{\lambda_T}{\lambda_0}\right) | |
| − | + | </math> | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | = | + | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | e). Because from a). and c)., we can get | |
| − | + | ||
| − | , | + | <math> |
| − | + | \int^x_0\mu(t)dt = -\log\left(\frac{Y_T}{Y_0}\right) | |
</math> | </math> | ||
| − | + | <span style="color:red"> This photon attenuation question is very similar to other questions: for example 2017S-ECE637-Exam1, Problem 3. Related topics are projection problems(e.g.: 2013S-ECE637-Exam1, Problem 2; 2012S-ECE637-Exam1, Problem 3) and scan problems(e.g.: 2016QE-CS5, Problem 1). </span> | |
| − | + | ||
| − | |||
---- | ---- | ||
| − | [[ECE-QE_CS5- | + | [[ECE-QE_CS5-2014|Back to QE CS question 1, August 2014]] |
[[ECE_PhD_Qualifying_Exams|Back to ECE QE page]]: | [[ECE_PhD_Qualifying_Exams|Back to ECE QE page]]: | ||
Latest revision as of 15:42, 18 May 2017
Contents
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)
August 2014, Problem 1
- Problem 1 , 2
Solution 1
a) $ E\left[Y_x\right] = \lambda_x $
Lack of proof. Should mention the property of Poisson distribution to show the equivalence. See the proof in solution 2.
b) Because the rate of absorption is proportional to the number of photons and the density of the material, so the attenuation of photons obeys the following equation
$ \frac{d\lambda_x}{dx} = -\mu(x)\lambda_x $
c) Solve the differential equation in b), we have
$ \lambda_x = \lambda_0e^{-\int^x_0\mu(t)dt} $
d) So the integral of the density, $ \int^T_0\mu(x)dx $ can be written as $ \int^T_0\mu(x)dx = -\log\left(\frac{\lambda_T}{\lambda_0}\right) $
e) $ \int^T_ \mu(x)dx \simeq -\log \left( \frac{Y_T}{Y_0} \right) $
Solution 2:
a). As we know $ P\left\{Y_x=k\right\} = \frac{e^{-\lambda_x}\lambda_x^k}{k!} $ is a Potion distribution, it is known that the expectation of a Poisson RV is $ \lambda_x $.
Proof:
$ \begin{split} E[Y_x] &= \sum^{+ \infty}_{k > 0} k \frac{e^{-\lambda_x}\lambda_x^k}{k!}\\ &= \sum^{+ \infty}_{k = 1} \frac{e^{-\lambda_x}\lambda_x^k}{(k-1)!}\\ &= \sum^{+ \infty}_{k = 1} \frac{e^{-\lambda_x}\lambda_x^{k-1}}{(k-1)!}\lambda_x\\ &= \lambda_xe^{-\lambda_x}\sum^{+ \infty}_{k = 0} \frac{\lambda_x^k}{k!}\\ &= \lambda_xe^{-\lambda_x}e^{\lambda_x}\\ &= \lambda_x\\ \end{split} $
So $ E[Y_x] = \lambda_x $
Here, it used $ \sum^{+ \infty}_{k = 0} \frac{\lambda_x^k}{k!} = e^{\lambda_x} $ to derive the final conclusion.
b). Because the number of photons will decrease when increasing the depth, $ d\lambda_x = -\lambda_x\mu(x)dx $
and
$ \frac{d\lambda_x}{dx} = -\lambda_x\mu(x) $
c). The final differential equation in b). is an ordinary differential equation. We can get the expression as
$ \lambda_x = \lambda_0e^{-\int^x_0\mu(t)dt} $
where $ \lambda_0 $ is the initial number of photons.
d). From part c). $ \frac{\lambda_x}{\lambda_0} = e^{-\int^x_0\mu(t)dt} $, so we have
$ \int^x_0\mu(t)dt = -\log\left(\frac{\lambda_T}{\lambda_0}\right) $
e). Because from a). and c)., we can get
$ \int^x_0\mu(t)dt = -\log\left(\frac{Y_T}{Y_0}\right) $
This photon attenuation question is very similar to other questions: for example 2017S-ECE637-Exam1, Problem 3. Related topics are projection problems(e.g.: 2013S-ECE637-Exam1, Problem 2; 2012S-ECE637-Exam1, Problem 3) and scan problems(e.g.: 2016QE-CS5, Problem 1).
