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[[Category:2010 Fall ECE 438 Boutin]]
 
[[Category:2010 Fall ECE 438 Boutin]]
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== Solution of [[Hw8ECE438F10|HW8]] ==
 
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== Solution to Question 1 of HW5 ==
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== Question 1 ==
 
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----
  
 +
According to the definition
 +
 +
<math>
 +
\begin{align}
 +
X_2[k]&=\sum_{n=0}^{N-1}x_2[n]e^{-\frac{j2\pi nk}{N}} \\
 +
X_1[z]&=\sum_{n=0}^{N-1}x_1[n]z^{-n}
 +
\end{align}
 +
</math>
 +
 +
Pull in <math>z=\frac{1}{2}e^{-j\frac{2\pi}{N}k}</math> in <math>X_1(z)</math>
 +
 +
<math>
 +
\begin{align}
 +
X_2[k]&=X_1(z)|_{z=\frac{1}{2}e^{-j\frac{2\pi}{N}k}} \\
 +
&=\sum_{n=0}^{N-1}x_1 [n](\frac{1}{2}e^{-j\frac{2\pi}{N}k})^{-n} \\
 +
&=\sum_{n=0}^{N-1}x_1 [n](\frac{1}{2})^{-n}e^{j\frac{2\pi n}{N}k} \\
 +
&\text{Change variable m=-n } \\
 +
&=\sum_{m=0}^{-(N-1)}x_1 [-m](\frac{1}{2})^{m}e^{-j\frac{2\pi m}{N}k} \\
 +
&\text{Change variable l=m+N } \\
 +
&=\sum_{l=N}^{1}x_1 [N-l](\frac{1}{2})^{l-N}e^{-j\frac{2\pi (l-N)}{N}k} \\
 +
&=\sum_{l=1}^{N}x_1 [N-l](2)^{N-l}e^{-j\frac{2\pi l}{N}k}
 +
\end{align}
 +
</math>
  
 +
Therefore,
  
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<math>x_2 [n]=x_1 [(-n)\text{ mod N }]2^{((-n)\text{ mod N })}</math>, where n=0,1,...,N-1
 
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== Solution to Question 2 of HW5 ==
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== Question 2 ==
 
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== Solution to Question 3 of HW5 ==
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== Question 3 ==
 
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----
  
 +
a)
  
 +
<math>
 +
\begin{align}
 +
X(\omega)&=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\
 +
&=\sum_{n=-\infty}^{\infty}(2\delta [n]+6\delta [n-1]-\delta [n-2])e^{-j\omega n} \\
 +
&=2+6e^{-j\omega}-e^{-j2\omega}
 +
\end{align}
 +
</math>
  
 +
<math>
 +
\begin{align}
 +
Y(\omega)&=\sum_{n=-\infty}^{\infty}y[n]e^{-j\omega n} \\
 +
&=\sum_{n=-\infty}^{\infty}x[-n]e^{-j\omega n} \\
 +
&=\sum_{n=-\infty}^{\infty}(2\delta [-n]+6\delta [-n-1]-\delta [-n-2])e^{-j\omega n} \\
 +
&=2+6e^{j\omega}-e^{j2\omega}
 +
\end{align}
 +
</math>
 +
 +
b)
 +
 +
Denote v[n]=x[n]*y[n]
 +
 +
Then the DTFT of v[n] is given by
 +
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<math>V(\omega)=X(\omega)Y(\omega)</math>
 +
 +
By using answer of part a,
 +
 +
<math>
 +
\begin{align}
 +
V(\omega)&= (2+6e^{-j\omega}-e^{-j2\omega})(2+6e^{j\omega}-e^{j2\omega}) \\
 +
&=41+6e^{-j\omega}+6e^{j\omega}-2e^{j2\omega}-2e^{-j2\omega}
 +
\end{align}
 +
</math>
 +
 +
Recall the definition of DTFT,
 +
 +
<math>V(\omega)=\sum_{n=-\infty}^{\infty}v[n]e^{-j\omega n}</math>
 +
 +
Therefore,
 +
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<math>\sum_{n=-\infty}^{\infty}v[n]e^{-j\omega n}=41+6e^{-j\omega}+6e^{j\omega}-2e^{j2\omega}-2e^{-j2\omega}</math>
 +
 +
By comparison the coefficients on both sides of the equation, we get
 +
 +
<math>v[n]=-2\delta [n+2]+6\delta [n+1]+41\delta [n]+6\delta [n-1]-2\delta [n-2]</math>
 +
 +
c)
 +
 +
<math>z[0]=x[0 \text{ mod 4}]=x[0]=2</math>
 +
 +
<math>z[1]=x[-1 \text{ mod 4}]=x[3]=0</math>
 +
 +
<math>z[2]=x[-2 \text{ mod 4}]=x[2]=-1</math>
 +
 +
<math>z[3]=x[-3 \text{ mod 4}]=x[1]=6</math>
 +
 +
<math>z[n]=2\delta [n]-\delta [n-2]+6\delta [n-3]</math>
 +
 +
In order to compute the 4-pt circular convolution of x[n] and z[n]
 +
 +
We first compute 4-pt DFT of x[n] and z[n].
 +
 +
Denote <math>t[n]=x[n]\circledast_4 z[n]</math>
 +
 +
T[k] is the 4-pt DFT of t[n], where k=0,1,2,3
 +
 +
<math>
 +
\begin{align}
 +
X(k)&=\sum_{n=0}^{3}x[n]e^{-\frac{j2\pi nk}{4}} \\
 +
&=2+6e^{-\frac{j2\pi k}{4}}-e^{-\frac{j4\pi k}{4}} \\
 +
&=2+6e^{-\frac{j\pi k}{2}}-e^{-j\pi k}
 +
\end{align}
 +
</math>
 +
 +
<math>
 +
\begin{align}
 +
Z(k)&=\sum_{n=0}^{3}z[n]e^{-\frac{j2\pi nk}{4}} \\
 +
&=2-e^{-\frac{j4\pi k}{4}}+6e^{-\frac{j6\pi k}{4}} \\
 +
&=2-e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}}
 +
\end{align}
 +
</math>
 +
 +
<math>
 +
\begin{align}
 +
T(k)=X(k)Z(k)&=(2+6e^{-\frac{j\pi k}{2}}-e^{-j\pi k})(2-e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}}) \\
 +
&=41+12e^{-\frac{j\pi k}{2}}-4e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}}-6e^{-\frac{j5\pi k}{2}} \\
 +
&=41+6e^{-\frac{j\pi k}{2}}-4e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}}
 +
\end{align}
 +
</math>
 +
 +
Then the circular convolution can be computed by doing IDFT to T(k)
 +
 +
<math>
 +
\begin{align}
 +
T[k]&=\sum_{n=0}^{3}t[n]e^{-\frac{j2\pi nk}{4}} \\
 +
&=41+6e^{-\frac{j\pi k}{2}}-4e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}}
 +
\end{align}
 +
</math>
 +
 +
By comparing the coefficients, we can get
 +
 +
<math>x[n]\circledast_4 z[n]=t[n]=41\delta [n]+6\delta [n-1]-4\delta [n-2]+6\delta [n-3]</math>
 +
 +
d)
 +
 +
In order to avoid aliasing in the circular convolution, we must guarantee that
 +
 +
<math>N\ge length(x)+length(z)-1=3+4-1=6</math>
 
----
 
----
 +
 +
I don't understand why n = 3 would be included when calculating z[n]. Isn't the range of n [0,3)?
 +
 +
----
 +
 
Back to [[Hw8ECE438F10|HW8]]
 
Back to [[Hw8ECE438F10|HW8]]
  
 
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]]
 
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]]

Latest revision as of 15:16, 2 December 2010

Solution of HW8

Question 1

According to the definition

$ \begin{align} X_2[k]&=\sum_{n=0}^{N-1}x_2[n]e^{-\frac{j2\pi nk}{N}} \\ X_1[z]&=\sum_{n=0}^{N-1}x_1[n]z^{-n} \end{align} $

Pull in $ z=\frac{1}{2}e^{-j\frac{2\pi}{N}k} $ in $ X_1(z) $

$ \begin{align} X_2[k]&=X_1(z)|_{z=\frac{1}{2}e^{-j\frac{2\pi}{N}k}} \\ &=\sum_{n=0}^{N-1}x_1 [n](\frac{1}{2}e^{-j\frac{2\pi}{N}k})^{-n} \\ &=\sum_{n=0}^{N-1}x_1 [n](\frac{1}{2})^{-n}e^{j\frac{2\pi n}{N}k} \\ &\text{Change variable m=-n } \\ &=\sum_{m=0}^{-(N-1)}x_1 [-m](\frac{1}{2})^{m}e^{-j\frac{2\pi m}{N}k} \\ &\text{Change variable l=m+N } \\ &=\sum_{l=N}^{1}x_1 [N-l](\frac{1}{2})^{l-N}e^{-j\frac{2\pi (l-N)}{N}k} \\ &=\sum_{l=1}^{N}x_1 [N-l](2)^{N-l}e^{-j\frac{2\pi l}{N}k} \end{align} $

Therefore,

$ x_2 [n]=x_1 [(-n)\text{ mod N }]2^{((-n)\text{ mod N })} $, where n=0,1,...,N-1

Question 2

a)

$ x[n]=6\delta[n]+5 \delta[n-1]+4 \delta[n-2]+3 \delta[n-3]+2 \delta[n-4]+\delta[n-5]\,\! $

Using the 6-point DFT formula,

$ \begin{align} X[k] &=\sum_{n=0}^{5}\left(6\delta[n]+5 \delta[n-1]+4 \delta[n-2]+3 \delta[n-3]+2 \delta[n-4]+\delta[n-5]\right)e^{-j\frac{2\pi}{6}kn} \\ &= 6 + 5e^{-j\frac{2\pi}{6}k} + 4e^{-j\frac{2\pi}{6}2k} + 3e^{-j\frac{2\pi}{6}3k} + 2e^{-j\frac{2\pi}{6}4k} + e^{-j\frac{2\pi}{6}5k} \\ \end{align} $


b)

We use the 6-point inverse-DFT formula to obtain $ y_6[n] $

$ y_6[n]=\frac{1}{6}\sum_{k=0}^{5} W_6^{-2k} X[k] e^{j\frac{2\pi}{6}nk} = \frac{1}{6}\sum_{k=0}^{5} X[k] e^{j\frac{2\pi}{6}(n+2)k} \quad \text{where} \;\; W_N=e^{-j\frac{2\pi}{N}} $

If you compare this with the 6-point inverse-DFT of $ X[k] $

$ x_6[n]=\frac{1}{6}\sum_{k=0}^{5}X[k] e^{j\frac{2\pi}{6}nk} $

then, you will notice that $ y_6[n]=x_6[(n+2)\text{mod}6] $. Thus, it becomes

$ y_6[n]=4\delta[n]+3 \delta[n-1]+2\delta[n-2]+\delta[n-3]+6 \delta[n-4]+5\delta[n-5]\,\! $

(Producting $ W^{-2k} $ to $ X[k] $ yields circular-shifting to the left by 2 in the periodic discrete-time signal)



c)

$ h[n]=\delta[n]+\delta[n-1]+\delta[n-2]\,\! $

computing the circular convolution with $ x[n] $ and $ h[n] $,

$ \begin{align} y[n] =& x[n]\circledast_6 h[n] \\ =& \quad \{\quad 6,\quad 5,\quad 4,\quad 3,\quad 2,\quad 1\} \\ & +\! \{\quad 1,\quad 6,\quad 5,\quad 4,\quad 3,\quad 2\} \\ & +\! \{\quad 2,\quad 1,\quad 6,\quad 5,\quad 4,\quad 3\} \\ =& \quad \{\quad 9,\;\;12,\;\;\!15,\;\;12,\quad 9,\quad 6\} \\ =& 9\delta[n]+12\delta[n-1]+15\delta[n-2]+12\delta[n-3]+9\delta[n-4]+6\delta[n-5] \\ \end{align} $



d)

In order for the periodic repetition (with period N) of the usual convolution between x[n] and h[n] to be the same with the N-point circular convolution,

$ N \geq L+M-1 $ where L is the length of x[n] and M is the length of h[n].

Therefore, $ N\geq8 $.


Question 3

a)

$ \begin{align} X(\omega)&=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\ &=\sum_{n=-\infty}^{\infty}(2\delta [n]+6\delta [n-1]-\delta [n-2])e^{-j\omega n} \\ &=2+6e^{-j\omega}-e^{-j2\omega} \end{align} $

$ \begin{align} Y(\omega)&=\sum_{n=-\infty}^{\infty}y[n]e^{-j\omega n} \\ &=\sum_{n=-\infty}^{\infty}x[-n]e^{-j\omega n} \\ &=\sum_{n=-\infty}^{\infty}(2\delta [-n]+6\delta [-n-1]-\delta [-n-2])e^{-j\omega n} \\ &=2+6e^{j\omega}-e^{j2\omega} \end{align} $

b)

Denote v[n]=x[n]*y[n]

Then the DTFT of v[n] is given by

$ V(\omega)=X(\omega)Y(\omega) $

By using answer of part a,

$ \begin{align} V(\omega)&= (2+6e^{-j\omega}-e^{-j2\omega})(2+6e^{j\omega}-e^{j2\omega}) \\ &=41+6e^{-j\omega}+6e^{j\omega}-2e^{j2\omega}-2e^{-j2\omega} \end{align} $

Recall the definition of DTFT,

$ V(\omega)=\sum_{n=-\infty}^{\infty}v[n]e^{-j\omega n} $

Therefore,

$ \sum_{n=-\infty}^{\infty}v[n]e^{-j\omega n}=41+6e^{-j\omega}+6e^{j\omega}-2e^{j2\omega}-2e^{-j2\omega} $

By comparison the coefficients on both sides of the equation, we get

$ v[n]=-2\delta [n+2]+6\delta [n+1]+41\delta [n]+6\delta [n-1]-2\delta [n-2] $

c)

$ z[0]=x[0 \text{ mod 4}]=x[0]=2 $

$ z[1]=x[-1 \text{ mod 4}]=x[3]=0 $

$ z[2]=x[-2 \text{ mod 4}]=x[2]=-1 $

$ z[3]=x[-3 \text{ mod 4}]=x[1]=6 $

$ z[n]=2\delta [n]-\delta [n-2]+6\delta [n-3] $

In order to compute the 4-pt circular convolution of x[n] and z[n]

We first compute 4-pt DFT of x[n] and z[n].

Denote $ t[n]=x[n]\circledast_4 z[n] $

T[k] is the 4-pt DFT of t[n], where k=0,1,2,3

$ \begin{align} X(k)&=\sum_{n=0}^{3}x[n]e^{-\frac{j2\pi nk}{4}} \\ &=2+6e^{-\frac{j2\pi k}{4}}-e^{-\frac{j4\pi k}{4}} \\ &=2+6e^{-\frac{j\pi k}{2}}-e^{-j\pi k} \end{align} $

$ \begin{align} Z(k)&=\sum_{n=0}^{3}z[n]e^{-\frac{j2\pi nk}{4}} \\ &=2-e^{-\frac{j4\pi k}{4}}+6e^{-\frac{j6\pi k}{4}} \\ &=2-e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}} \end{align} $

$ \begin{align} T(k)=X(k)Z(k)&=(2+6e^{-\frac{j\pi k}{2}}-e^{-j\pi k})(2-e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}}) \\ &=41+12e^{-\frac{j\pi k}{2}}-4e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}}-6e^{-\frac{j5\pi k}{2}} \\ &=41+6e^{-\frac{j\pi k}{2}}-4e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}} \end{align} $

Then the circular convolution can be computed by doing IDFT to T(k)

$ \begin{align} T[k]&=\sum_{n=0}^{3}t[n]e^{-\frac{j2\pi nk}{4}} \\ &=41+6e^{-\frac{j\pi k}{2}}-4e^{-j\pi k}+6e^{-\frac{j3\pi k}{2}} \end{align} $

By comparing the coefficients, we can get

$ x[n]\circledast_4 z[n]=t[n]=41\delta [n]+6\delta [n-1]-4\delta [n-2]+6\delta [n-3] $

d)

In order to avoid aliasing in the circular convolution, we must guarantee that

$ N\ge length(x)+length(z)-1=3+4-1=6 $

I don't understand why n = 3 would be included when calculating z[n]. Isn't the range of n [0,3)?

Back to HW8

Back to ECE 438 Fall 2010

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