(6 intermediate revisions by 3 users not shown)
Line 4: Line 4:
 
[[Category:problem solving]]
 
[[Category:problem solving]]
  
−
= [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]: Automatic Control (AC)- Question 1, August 2007=
+
<center>
−
X and Y are iid random variable with
+
<font size= 4>
 +
[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
 +
</font size>
  
−
<math> P(X=i) = P(Y=i) = \frac {1}{2^i}\ ,i = 1,2,3,... </math>
+
<font size= 4>
 +
Automatic Control (AC)
  
−
a) Find <math> P(min(X,Y)=k)\ </math>.
+
Question 1: Feedback Control Systems
 +
</font size>
  
−
b) Find <math> P(X=Y)\ </math>.
+
August 2007
−
 
+
</center>
−
c) Find <math> P(Y>X)\ </math>.
+
−
 
+
−
d) Find <math> P(Y=kX)\ </math>.
+
−
----
+
−
=Solution 1 (retrived from [[Automatic_Controls:Linear_Systems_(HKNQE_August_2007)_Problem_1|here]])=
+
−
 
+
−
*To find <math> P(min(X,Y)=k)\ </math>, let  <math> Z = min(X,Y)\ </math>. Then finding the pmf of Z uses the fact that X and Y are iid
+
−
        <math> P(Z=k) = P(X \ge k,Y \ge k) = P(X \ge k)P(Y \ge k) = P(X \ge k)^2 </math>
+
−
 
+
−
        <math> P(Z=k) = \left ( \sum_{i=k}^N \frac {1}{2^i} \right )^2 = \left ( \frac {1}{2^k} \right )^2 = \frac {1}{4^k} </math>
+
−
 
+
−
*To find <math> P(X=Y)\ </math>, note that  X and Y are iid and summing across all possible i,
+
−
        <math> P(X=Y) = \sum_{i=1}^\infty P(X=i, Y=i) = \sum_{i=1}^\infty P(X=i)P(Y=i) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
+
−
 
+
−
*To find <math> P(Y>X)\ </math>, again note that X and Y are iid and summing across all possible i,
+
−
        <math> P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i)</math>
+
−
 
+
−
:Next, find <math> P(Y<i)\ </math>
+
−
        <math> P(Y>i) = 1 - P(Y \le i) </math>
+
−
 
+
−
        <math> P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i} </math>
+
−
 
+
−
      <math> \therefore P(Y>i) = \frac {1}{2^i} </math>
+
−
 
+
−
:Plugging this result back into the original expression yields
+
−
        <math> P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
+
−
 
+
−
 
+
−
*To find <math> P(Y=kX)\ </math>, note that X and Y are iid and summing over all possible combinations one arrives at
+
−
        <math> P(Y=kX) = \sum_{i=1}^\infty i = 1^\infty P(Y=ki, X=i) = \sum_{i=1}^\infty P(Y=ki)P(X=i) </math>
+
−
:Thus,
+
−
        <math> P(Y=kX) = \sum_{i=1}^\infty \frac {1}{2^{ki}} \frac {1}{2^i} = \sum_{i=1}^\infty \frac {1}{2^{(k+1)i}} = \frac {1}{2^{(k+1)}-1} </math>
+
 
----
 
----
−
==Solution 2==
 
−
Write it here.
 
 
----
 
----
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Latest revision as of 21:59, 6 August 2018


ECE Ph.D. Qualifying Exam

Automatic Control (AC)

Question 1: Feedback Control Systems

August 2007

Back to ECE Qualifying Exams (QE) page

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=ECE_PhD_QE_Automatic_Control_2007_Problem1&oldid=75033"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang