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\frac{d\lambda_x}{dx} = -\mu(x)\lambda_x
 
\frac{d\lambda_x}{dx} = -\mu(x)\lambda_x
 
</math>
 
</math>
 +
 
c) Solve the differential equation in b), we have
 
c) Solve the differential equation in b), we have
  

Revision as of 14:52, 18 May 2017


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 $

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) $ \frac{R}{255}^\alpha=r_{linear}\\ \Rightarrow \gamma=log_{\frac{R}{255}}{(R^{\alpha})}=\frac{ln{(R^{\alpha})}}{ln{\frac{R}{255}}}=\frac{\alpha{ln{R}}}{ln{R}-ln{255}} $

$ \gamma $ should be 1.

b)

$ P_r= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right) = \left( \begin{array}{ccc} a \\ d \\ g \end{array} \right) \\ \Rightarrow x_r=\frac{a}{a+d+g} , y_r=\frac{d}{a+d+g} \\ P_g= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right) = \left( \begin{array}{ccc} b \\ e \\ h \end{array} \right) \\ \Rightarrow x_g=\frac{b}{b+e+h} , y_g=\frac{e}{b+e+h} \\ P_b= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{ccc} 0 \\ 0 \\ 1\end{array} \right) = \left( \begin{array}{ccc} c \\ f \\ i \end{array} \right) \\ \Rightarrow x_g=\frac{c}{c+f+i} , y_g=\frac{f}{c+f+i} $

c)

$ W= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{ccc} 1 \\ 1 \\ 1\end{array} \right) = \left( \begin{array}{ccc} a+b+c \\ d+e+f \\ g+h+i \end{array} \right) \\ \Rightarrow x_g=\frac{a+b+c}{a+b+c+d+e+f+g+h+i} , y_g=\frac{d+e+f}{a+b+c+d+e+f+g+h+i} $

d) Pro1 2015 Aug.PNG

e) Gamma correction a quantization will create an effect of dynamic range compression for pixels with small values. This will create dark block of shadings in a gradient region instead of a smooth transition.


Back to QE CS question 1, August 2013

Back to ECE QE page:

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