Difference between revisions of "QE637 sol2012 Q2" - Rhea

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Problem 1,2

Solution1:

a)

$ \begin{align} & {{h}_{1}}(m,n)=\sum\limits_{j=-N}^{N}{{a}_{j}\delta (m,n-j)=}{a}_{n}\delta (m) \\ & \delta (m,n)=\left\{ \begin{matrix} 1\ m=n=0 \\ 0\qquad O.W \\ \end{matrix}, \right. \delta (m,n-j)=\left\{ \begin{matrix} 1\qquad n=j \\ 0\qquad O.W \\ \end{matrix} \right. \\ \end{align} $

b)

$ \begin{align} & {{h}_{2}}(m,n)=\sum\limits_{i=-N}^{N}{{b}_{i}\delta (m-i,n)=}{b}_{m}\delta (n) \\ & \delta (m,n)=\left\{ \begin{matrix} 1\ m=n=0 \\ 0\qquad O.W \\ \end{matrix}, \right. \delta (m-i,n)=\left\{ \begin{matrix} 1\ m=i;\ n=0 \\ 0\qquad O.W \\ \end{matrix} \right. \\ \end{align} $

c)

$ \begin{align} & h(m,n)=\sum\limits_{i=-N}^{N}{\sum\limits_{j=-N}^{N}{{{b}_{i}}\ {{a}_{j}}^{{}}\delta (m-i,n-j)}}={{b}_{m}}\ {{a}_{n}} \\ & z(m,n)=\sum\limits_{i=-N}^{N}{{{b}_{i}}\ y(m-i,n)=}\sum\limits_{i=-N}^{N}{{{b}_{i}}\ \left( \sum\limits_{j=-N}^{N}{{{a}_{j}}\ x(m-i,n-j)} \right)=\sum\limits_{i=-N}^{N}{\sum\limits_{j=-N}^{N}{{{b}_{i}}\ {{a}_{j}}\ x(m-i,n-j)}}} \\ & \delta (m-i,n-j)=\left\{ \begin{matrix} 1\ m=i;\ n=j \\ 0\qquad O.W \\ \end{matrix} \right. \\ \end{align} $

d)

Number of multiplies per output point to implement each individual system = 2N+1. So, the number of multiplies per output point to implement each of the two individual systems is 2(2N+1) = 4N+2.

Number of multiplies per output point to implement the complete system with a single convolution is $ \left( 2N+1 \right)\left( 2N+1 \right)\text{ }=4{{N}^{2}}+4N+1 $

e)

Implementing the two systems in sequence requires less computation, but it is more complex and more sensitive to noise. Implementing the two systems in a single complete system requires more computation, but it is simpler, less sensitive to noise, and more stable.

Solution 2:

a)

$ \begin{align} & {{h}_{1}}(m,n)=\sum\limits_{j=-N}^{N}{{a}_{j}\delta (m,n-j)=}\sum\limits_{j=-N}^{N}{{a}_{j}\delta (m)\ \delta(n-j)=} {a}_{n}\ \delta (m) \\ \end{align} $

The answer is correct, but it's not obvious how the solution is derived. (See Solution 1)

b)

$ \begin{align} & {{h}_{2}}(m,n)=\sum\limits_{i=-N}^{N}{{b}_{i}\delta (m-i,n)=}\sum\limits_{i=-N}^{N}{{b}_{i}\delta (m-i)\ \delta(n)=}{b}_{m}\ \delta (n) \\ \end{align} $

The answer is correct, but it's not obvious how the solution is derived. (See Solution 1)

c)

$ \begin{align} & h(m,n)=\sum\limits_{i=-N}^{N}{\sum\limits_{j=-N}^{N}{{{b}_{i}}\ {{a}_{j}}\ \delta (m-i,n-j)}}=\sum\limits_{i=-N}^{N}{\sum\limits_{j=-N}^{N}{{{b}_{i}}\ {{a}_{j}}\ \delta (m-i)\ \delta (n-j)}}={{b}_{m}}\ {{a}_{n}} \\ \end{align} $

The answer is correct, but it's not obvious how the solution is derived. (See Solution 1)

d)

Individually: $ 2(2N+1)=4N+2 $
Complete system: $ {( 2N+1)}^{2}=4{{N}^{2}}+4N+1 $

e)

Fewer multipliers are required when implementing individually, but the system is more complicated.

More complete for the complete system.

The Student didn't mention advantage/disadvantage of the complete system, it is just mentioned that "More complete for the complete system". (See Solution 1)

Related Problem

Consider the following 2-D LSI systems. The first system (S1) has input x(m, n) and output y(m, n), and the second system (S2) has input y(m, n) and output z(m, n).

The third system (S3) is formed by the composition of (S1) and (S2) with input x(m, n) and output z(m,n) and impulse response $ {{h}_{3}}(m,n) $.

a) Calculate the 2-D impulse response, $ {{h}_{1}}(m,n) $, of the first system (S1).

b) Calculate the 2-D impulse response, $ {{h}_{2}}(m,n) $, of the second system (S2).

c) Calculate the 2-D impulse response, $ {{h}_{3}}(m,n) $, of the complete system (S3).

d) Calculate the 2-D transfer function, $ {{H}_{1}}({z}_{1},{z}_{2}) $, of the first system (S1).

e) Calculate the 2-D transfer function, $ {{H}_{3}}({z}_{1},{z}_{2}) $, of the first system (S3).

Refer to ECE637 Spring 2014 Exam1 Problem1.

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