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a)
 
a)
  
−
<math>y[n]= \frac{x[n]+x[n-1]}{2} </math><br>  
+
<math>y[n]= \frac{x[n]+x[n-2]}{2} </math>
  
−
Applying Z-transform on both sides and grouping terms, we can obtain the transfer function<br/>
+
Applying Z-transform on both sides and grouping terms, we can obtain the transfer function
  
 
<math>\begin{align}
 
<math>\begin{align}
−
Y[z]&= \frac{X[z]+X[z].z^{-1}}{2} \\
+
Y[z]&= \frac{X[z]+X[z]z^{-2}}{2} \\
−
\frac{Y[z]}{X[z]}&= \frac{1+z^{-1}}{2} \\
+
\frac{Y[z]}{X[z]}&= \frac{1+z^{-2}}{2} \\
−
H[z] &= \frac{1+z^{-1}}{2} \\
+
H[z] &= \frac{1+z^{-2}}{2} \\
 
\end{align}</math>  
 
\end{align}</math>  
  
 
Frequency Response <math>H(\omega)</math>  
 
Frequency Response <math>H(\omega)</math>  
 +
 
<math>\begin{align}
 
<math>\begin{align}
−
H[e^{j\omega }] &= \frac{1+e^{-j\omega }}{2} \\
+
H[e^{j\omega }] &= \frac{1+e^{-2j\omega }}{2} \\
−
&= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}+e^{-j\frac{\omega }{2}}}{2} \right) \\
+
&= e^{-j\omega } \left( \frac{e^{j\omega}+e^{-j\omega}}{2} \right) \\
−
&= e^{-j\frac{\omega }{2}} cos \left( \frac{\omega }{2} \right) \\
+
&= e^{-j\omega } cos \left( \omega \right) \\
 
\end{align}</math>  
 
\end{align}</math>  
 +
 +
Zero-pole plot
  
 
[[Image:HW6Q1fig1.jpg]]
 
[[Image:HW6Q1fig1.jpg]]
 +
 +
Frequency response
 +
 +
[[Image:HW6Q1fig2.jpg]]
  
 
b)
 
b)
Line 29: Line 36:
 
<math>y[n]= \frac{x[n]-x[n-1]}{2}</math>  
 
<math>y[n]= \frac{x[n]-x[n-1]}{2}</math>  
  
−
Applying Z-transform on both sides and grouping terms, we can obtain the transfer function<br>
+
Applying Z-transform on both sides and grouping terms, we can obtain the transfer function
  
 
<math>\begin{align}
 
<math>\begin{align}
−
Y[z]&= \frac{X[z]-X[z].z^{-1}}{2} \\
+
Y[z]&= \frac{X[z]-X[z]z^{-1}}{2} \\
 
\frac{Y_2[z]}{X[z]}&= \frac{1-z^{-1}}{2} \\
 
\frac{Y_2[z]}{X[z]}&= \frac{1-z^{-1}}{2} \\
 
H[z] &= \frac{1-z^{-1}}{2} \\
 
H[z] &= \frac{1-z^{-1}}{2} \\
Line 38: Line 45:
  
 
Frequency Response <math>H(\omega)</math>
 
Frequency Response <math>H(\omega)</math>
 +
 
<math>\begin{align}
 
<math>\begin{align}
 
H[e^{j\omega }] &= \frac{1-e^{-j\omega }}{2} \\
 
H[e^{j\omega }] &= \frac{1-e^{-j\omega }}{2} \\
Line 45: Line 53:
 
\end{align}</math>  
 
\end{align}</math>  
  
−
[[Image:HW6Q1fig2.jpg]]
+
Zero-pole plot
 +
 
 +
[[Image:HW6Q1fig3.jpg]]
 +
 
 +
Frequency response
 +
 
 +
[[Image:HW6Q1fig4.jpg]]
 +
 
 
----
 
----
 
==Question 2==
 
==Question 2==
  
 +
a. Substituting values directly would yield the following - <br>
  
−
[[Image:HW6Q2fig1.jpg]]
+
<math>\begin{align}
 +
n &= ...\text{ -3, -2, -1, 0, 1, 2, 3, 4, 5} ...\\
 +
y\left[n\right] &= ...\text{  0,  -2,  -4, -1,  2, -1, -4, -2, 0} ...\\
 +
\end{align}</math>
  
 +
b. h[n] =&nbsp;? <br> Substitute x[n] = <span class="texhtml">δ[''n'']</span> to obtain y[n] = h[n], <br> h[n] = <span class="texhtml">δ[''n'']</span> + 2<span class="texhtml">δ[''n'' − 1]</span> + <span class="texhtml">δ[''n'' − 2]</span> <br>
 +
 +
Now x[n] = -2<span class="texhtml">δ[''n'' + 2]</span> + <span class="texhtml">δ[''n'']</span> - 2<span class="texhtml">δ[''n'' − 2]</span> <br>
 +
 +
<math>\begin{align}
 +
y[n] &= x[n] * h[n] \\
 +
&= (-2\delta[n+2] + \delta[n] - 2\delta[n-2]) * (\delta[n] + 2\delta[n-1] + \delta[n-2]) \\
 +
&= -2\delta[n+2] - 4\delta[n+1] -2\delta[n] + \delta[n] + 2\delta[n-1] + \delta[n-2] - 2\delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\
 +
&= -2\delta[n+2] - 4\delta[n+1] - \delta[n] + 2\delta[n-1] - \delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\
 +
\end{align}</math>
 +
 +
c. x[n] = <span class="texhtml">''e''<sup>''j''ω''n''</sup></span> <br> (i) <br> <math>\begin{align}
 +
y[n] &=  e^{j\omega n} + 2 e^{j\omega (n-1)} + e^{j\omega (n-2)} \\
 +
&= e^{j\omega n}(1 + 2 e^{-j\omega } + e^{-2j\omega }) \\
 +
H(e^{j\omega}) &= 1 + 2 e^{-j\omega } + e^{-2j\omega } \\
 +
\end{align}</math>
 +
 +
(ii) <br> h[n] = <span class="texhtml">δ[''n'']</span> + 2<span class="texhtml">δ[''n'' − 1]</span> + <span class="texhtml">δ[''n'' − 2]</span> <br> H(<span class="texhtml">''e''<sup>''j''ω</sup></span>) = 1 + 2 <span class="texhtml">''e''<sup> − ''j''ω</sup></span> + <span class="texhtml">''e''<sup> − 2''j''ω</sup></span> <br>
 +
 +
(i) and (ii) are the same.<br>
 +
 +
d. x[n] = -2<span class="texhtml">δ[''n'' + 2]</span> + <span class="texhtml">δ[''n'']</span> - 2<span class="texhtml">δ[''n'' − 2]</span> <br> X(<span class="texhtml">''e''<sup>''j''ω</sup></span>) = -2 <span class="texhtml">''e''<sup>2''j''ω</sup></span> + 1 - 2<span class="texhtml">''e''<sup> − 2''j''ω</sup></span> <br>
 +
 +
<math>\begin{align}
 +
Y(e^{j\omega}) &= X(e^{j\omega})H(e^{j\omega}) \\
 +
&= (-2e^{2j\omega } + 1 - 2e^{-2j\omega } ).(1 + 2e^{-j\omega } + e^{-2j\omega }) \\
 +
&= -2e^{2j\omega } - 4e^{j\omega } - 2 + 1 + 2e^{-j\omega } + 2e^{-2j\omega } - 2e^{-2j\omega } - 4e^{-3j\omega } - 2e^{-4j\omega } \\
 +
&= -2e^{2j\omega } - 4e^{j\omega } - 1 - 2e^{-j\omega } - e^{-2j\omega } - 4e^{-3j\omega } - 2e^{-4j\omega } \\
 +
\\
 +
\text{Using Inverse DTFT,} \\
 +
y[n] &= -2\delta[n+2] - 4\delta[n+1] - \delta[n] + 2\delta[n-1] - \delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\
 +
\end{align}</math>
 +
 +
All 3 approaches lead to the same y[n].
  
 
----
 
----
 
==Question 3==
 
==Question 3==
  
 +
a. Sketch the locations of the poles and zeros.
  
 +
<math>\begin{align}
 +
H(z) &= \frac{1-\frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}} \\
 +
H(z) &= \frac{(z+\frac{1}{\sqrt{2}})(z-\frac{1}{\sqrt{2}})}
 +
{  (z-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(z-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \\
 +
\end{align}</math>
 +
 +
Zeros: <br> <math>z_1 = \frac{1}{\sqrt{2}}, z_2 = -\frac{1}{\sqrt{2}}</math><br> Poles: <br> <math>p_1 = \frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}, p_2 = \frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}}</math>
 +
 +
[[Image:Zp1.jpg]]
 +
 +
:b. Determine the magnitude and phase of the frequency response <span class="texhtml">''H''(ω)</span>, for
 +
 +
<math>\omega = 0</math> <br/>
 +
[[Image:Zp2.jpg]]
 +
<math>
 +
\left| H(e^{j\omega}) \right| = \left| H(e^{j0}) \right| = \left| H(z=1) \right|</math><br> <math> = \left| \frac{(1+\frac{1}{\sqrt{2}})(1-\frac{1}{\sqrt{2}})} {  (1-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(1-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 0.921</math><br> <math>\angle H(e^{j0}) = \angle c + \angle d - \angle a - \angle b = 0</math>
 +
 +
<br> <math>\omega =\frac{\pi}{4}</math><br>
 +
[[Image:Zp3.JPG]]
 +
<math>
 +
\left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{\pi}{4}}) \right| = \left| H(z=\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}) \right|</math><br>
 +
<math> = \left| \frac{(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})} {  (\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 2
 +
</math>
 +
<br>
 +
<math>\angle H(e^{j\frac{\pi}{4}}) = \angle c + \angle d - \angle a - \angle b = \frac{\pi}{2} + arctan^{-1} \left( \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} \right) - \frac{\pi}{4} - arctan^{-1} \left( \frac{\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}}{1-\sqrt{2}-\frac{1}{2\sqrt{2}}} \right) = 0</math>
 +
 +
<math>\omega =\frac{\pi}{2}</math><br> [[Image:Zp4.jpg]] <math>\left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{\pi}{2}}) \right| = \left| H(z=j) \right|</math><br>
 +
 +
<math> = \left| \frac{(j+\frac{1}{\sqrt{2}})(j-\frac{1}{\sqrt{2}})} {  (j-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(j-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 1.455</math><br>
 +
 +
<br>
 +
<math>\angle H(e^{j\frac{\pi}{2}}) = (\angle c + \angle d) - \angle a - \angle b = (\pi) - (arctan^{-1}\left( \frac{1-\frac{1}{2\sqrt{2}}}{\frac{-1}{\sqrt{2}}} \right) + \pi) - (arctan^{-1}\left( \frac{1+\frac{1}{2\sqrt{2}}}{1-\frac{1}{\sqrt{2}}} \right) + \pi) = -0.7563
 +
</math>
 +
 +
<math>\omega =\frac{3\pi}{4}</math><br/>
 +
[[Image:Zp5.JPG]]
 +
<math>\left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{3\pi}{4}}) \right| = \left| H(z=\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}) \right|</math><br>
 +
 +
<math> = \left| \frac{(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})} {  (\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = \frac{2}{3}
 +
</math>
 +
 +
<math>\angle H(e^{j\frac{3\pi}{4}}) = \angle c + \angle d - \angle a - \angle b = (arctan^{-1}\left( \frac{\frac{1}{\sqrt{2}}}{\frac{-2}{\sqrt{2}}} \right) + \pi) + \frac{\pi}{2} - (arctan^{-1}\left( \frac{\frac{1}{\sqrt{2}} - \frac{1}{2\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{2\sqrt{2}}} \right) + \pi) + \frac{3\pi}{4} = -0.9273
 +
</math>
 +
 +
 +
<math>\omega =\pi</math><br/>
 +
[[Image:Zp6.JPG]]
 +
<math>
 +
\left| H(e^{j\omega}) \right| = \left| H(e^{j\pi}) \right| = \left| H(z=-1) \right|</math><br>
 +
<math> = \left| \frac{(-1+\frac{1}{\sqrt{2}})(-1-\frac{1}{\sqrt{2}})} {  (-1-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(-1-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 0.255
 +
</math><br>
 +
 +
<math>\angle H(e^{j\pi}) = (\angle c + \angle d) - \angle a - \angle b = 2\pi - 0 - 0 = 2\pi</math>
 +
 +
c. Is the system stable? Explain why or why not? <br/>
 +
The system is causal and the ROC extends outwards from the outermost pole since |<math>p_1</math>| = |<math>p_2</math>| < 1 and this ROC contains the unit circle. Therefore the system is stable. <br/>
 +
 +
d. Find the difference equation for y[n] in terms of x[n], corresponding to this transfer function H(z). <br/>
 +
 +
<math>H(z) = \frac{Y(z)}{X(z)} = \frac{1-\frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}}</math><br/>
 +
<math>Y(z)(1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}) = X(z)(1-\frac{1}{2}z^{-2})</math><br/>
 +
 +
Taking inverse,<br/>
 +
<math>y[n]-\frac{1}{\sqrt{2}}y[n-1] + \frac{1}{4}y[n-2] = x[n] - \frac{1}{2}x[n-2]</math><br/>
 +
<math>y[n] = x[n] - \frac{1}{2}x[n-2] +\frac{1}{\sqrt{2}}y[n-1] - \frac{1}{4}y[n-2]</math><br/>
 +
<br>
  
 
----
 
----
 
==Question 4==
 
==Question 4==
  
 +
<br/>
 +
<math>
 +
y[n]=\frac{1}{8} \left( x[n]+x[n-1]+x[n-2]+x[n-3]+x[n-4]+x[n-5]+x[n-6]+x[n-7]\right)
 +
</math>
  
 +
a.<br/>
 +
<math>
 +
h[n]=\frac{1}{8} \left( \delta[n]+\delta[n-1]+\delta[n-2]+\delta[n-3]+\delta[n-4]+\delta[n-5]+\delta[n-6]+\delta[n-7] \right)
 +
</math> <br/>
 +
This is a finite duration response.<br/>
  
−
[[Image:HW6Q4fig2.jpg]]
+
b.  
 +
<math>
 +
H[z]=\frac{1}{8} \left( 1+z^{-1}+z^{-2}+z^{-3}+z^{-4}+z^{-5}+z^{-6}+z^{-7} \right)
 +
</math> <br/>
 +
<math>
 +
H[z]=\frac{1}{8} \left( \frac{1-z^{-8}}{1-z^{-1}} \right)
 +
</math> <br/>
  
 +
c.
 +
 +
<math>
 +
H[z]=\frac{1}{8} \left( \frac{z^{8}-1}{z^{7}(z-1)} \right)
 +
</math> <br/>
 +
[[Image:Zp7.jpg]]
 +
<br/>
 +
Poles: <br/>
 +
<math>p_1 = 1</math><br/> 
 +
<math>p_2 = p_3 = ... = p_8 = 0</math><br/>
 +
 +
Zeros: <br/>
 +
<math>z^{8} - 1 = 0 </math><br/>
 +
<math>z^{8} = e^{j2\pi } </math><br/>
 +
 +
Generalizing,<br/>
 +
<math>z_k = e^{j2\pi k/8 }</math> for k = 0,1,2,...,7<br/>
  
 
----
 
----
Line 70: Line 222:
 
==Question 5==
 
==Question 5==
  
 +
a. <br/>
 +
<math>
 +
y[n]= \frac{1}{8} \left( x[n]-x[n-8]+y[n-1] \right)
 +
</math>
 +
 +
Using z-transform,<br/>
 +
<math>
 +
Y(z) = \frac{1}{8} \left( X(z)-X(z)z^{-8} + Y(z)z^{-1} \right)
 +
</math><br/>
 +
<math>
 +
Y(z) (1 - \frac{1}{8}z^{-1}) = X(z) \frac{1}{8}(1 - z^{-8})
 +
</math><br/>
 +
<math>
 +
H(z) = \frac{Y(z)}{X(z)}  =  \frac{1}{8} \left( \frac{ 1 - z^{-8} } {1 - \frac{1}{8}z^{-1}} \right)
 +
</math><br/>
 +
b.
 +
 +
[[Image:HW6Q5fig1.jpg]]
 +
 +
c.
 +
H(z) can be re-written as <br/>
 +
<math>
 +
H[z]=\frac{1}{8} \left( \frac{1}{1-\frac{1}{8}z^{-1}} - \frac{z^{-8}}{1-\frac{1}{8}z^{-1}} \right)
 +
</math> <br/>
 +
Assuming the ROC: <math>|z|>\frac{1}{8}</math>Taking inverse Z-transform of H(z) - <br/>
 +
<math>
 +
h[n]=\frac{1}{8} \left( (\frac{1}{8})^n u[n] - (\frac{1}{8})^{n-8} u[n-8] \right)
 +
</math> <br/>
 +
 +
This is a infinite duration response.<br/>
  
 
----
 
----
  
−
Back to [[Hw5_ECE438F11|Homework 6]]
+
Back to [[Hw6_ECE438F11|Homework 6]]
  
 
Back to [[2011_Fall_ECE_438_Boutin|ECE 438 Fall 2011]]
 
Back to [[2011_Fall_ECE_438_Boutin|ECE 438 Fall 2011]]

Latest revision as of 10:15, 30 October 2011

Homework 6, ECE438, Fall 2011, Prof. Boutin

Question 1

a)

$ y[n]= \frac{x[n]+x[n-2]}{2} $

Applying Z-transform on both sides and grouping terms, we can obtain the transfer function

$ \begin{align} Y[z]&= \frac{X[z]+X[z]z^{-2}}{2} \\ \frac{Y[z]}{X[z]}&= \frac{1+z^{-2}}{2} \\ H[z] &= \frac{1+z^{-2}}{2} \\ \end{align} $

Frequency Response $ H(\omega) $

$ \begin{align} H[e^{j\omega }] &= \frac{1+e^{-2j\omega }}{2} \\ &= e^{-j\omega } \left( \frac{e^{j\omega}+e^{-j\omega}}{2} \right) \\ &= e^{-j\omega } cos \left( \omega \right) \\ \end{align} $

Zero-pole plot

HW6Q1fig1.jpg

Frequency response

HW6Q1fig2.jpg

b)

$ y[n]= \frac{x[n]-x[n-1]}{2} $

Applying Z-transform on both sides and grouping terms, we can obtain the transfer function

$ \begin{align} Y[z]&= \frac{X[z]-X[z]z^{-1}}{2} \\ \frac{Y_2[z]}{X[z]}&= \frac{1-z^{-1}}{2} \\ H[z] &= \frac{1-z^{-1}}{2} \\ \end{align} $

Frequency Response $ H(\omega) $

$ \begin{align} H[e^{j\omega }] &= \frac{1-e^{-j\omega }}{2} \\ &= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2} \right) \\ &= je^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2j} \right) \\ &= je^{-j\frac{\omega }{2}} sin \left( \frac{\omega }{2} \right) \\ \end{align} $

Zero-pole plot

HW6Q1fig3.jpg

Frequency response

HW6Q1fig4.jpg

Question 2

a. Substituting values directly would yield the following -

$ \begin{align} n &= ...\text{ -3, -2, -1, 0, 1, 2, 3, 4, 5} ...\\ y\left[n\right] &= ...\text{ 0, -2, -4, -1, 2, -1, -4, -2, 0} ...\\ \end{align} $

b. h[n] = ?
Substitute x[n] = δ[n] to obtain y[n] = h[n],
h[n] = δ[n] + 2δ[n − 1] + δ[n − 2]

Now x[n] = -2δ[n + 2] + δ[n] - 2δ[n − 2]

$ \begin{align} y[n] &= x[n] * h[n] \\ &= (-2\delta[n+2] + \delta[n] - 2\delta[n-2]) * (\delta[n] + 2\delta[n-1] + \delta[n-2]) \\ &= -2\delta[n+2] - 4\delta[n+1] -2\delta[n] + \delta[n] + 2\delta[n-1] + \delta[n-2] - 2\delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\ &= -2\delta[n+2] - 4\delta[n+1] - \delta[n] + 2\delta[n-1] - \delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\ \end{align} $

c. x[n] = ejωn
(i)
$ \begin{align} y[n] &= e^{j\omega n} + 2 e^{j\omega (n-1)} + e^{j\omega (n-2)} \\ &= e^{j\omega n}(1 + 2 e^{-j\omega } + e^{-2j\omega }) \\ H(e^{j\omega}) &= 1 + 2 e^{-j\omega } + e^{-2j\omega } \\ \end{align} $

(ii)
h[n] = δ[n] + 2δ[n − 1] + δ[n − 2]
H(ejω) = 1 + 2 e − jω + e − 2jω

(i) and (ii) are the same.

d. x[n] = -2δ[n + 2] + δ[n] - 2δ[n − 2]
X(ejω) = -2 e2jω + 1 - 2e − 2jω

$ \begin{align} Y(e^{j\omega}) &= X(e^{j\omega})H(e^{j\omega}) \\ &= (-2e^{2j\omega } + 1 - 2e^{-2j\omega } ).(1 + 2e^{-j\omega } + e^{-2j\omega }) \\ &= -2e^{2j\omega } - 4e^{j\omega } - 2 + 1 + 2e^{-j\omega } + 2e^{-2j\omega } - 2e^{-2j\omega } - 4e^{-3j\omega } - 2e^{-4j\omega } \\ &= -2e^{2j\omega } - 4e^{j\omega } - 1 - 2e^{-j\omega } - e^{-2j\omega } - 4e^{-3j\omega } - 2e^{-4j\omega } \\ \\ \text{Using Inverse DTFT,} \\ y[n] &= -2\delta[n+2] - 4\delta[n+1] - \delta[n] + 2\delta[n-1] - \delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\ \end{align} $

All 3 approaches lead to the same y[n].

Question 3

a. Sketch the locations of the poles and zeros.

$ \begin{align} H(z) &= \frac{1-\frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}} \\ H(z) &= \frac{(z+\frac{1}{\sqrt{2}})(z-\frac{1}{\sqrt{2}})} { (z-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(z-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \\ \end{align} $

Zeros:
$ z_1 = \frac{1}{\sqrt{2}}, z_2 = -\frac{1}{\sqrt{2}} $
Poles:
$ p_1 = \frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}, p_2 = \frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}} $

Zp1.jpg

b. Determine the magnitude and phase of the frequency response H(ω), for

$ \omega = 0 $
Zp2.jpg $ \left| H(e^{j\omega}) \right| = \left| H(e^{j0}) \right| = \left| H(z=1) \right| $
$ = \left| \frac{(1+\frac{1}{\sqrt{2}})(1-\frac{1}{\sqrt{2}})} { (1-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(1-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 0.921 $
$ \angle H(e^{j0}) = \angle c + \angle d - \angle a - \angle b = 0 $


$ \omega =\frac{\pi}{4} $
Zp3.JPG $ \left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{\pi}{4}}) \right| = \left| H(z=\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}) \right| $
$ = \left| \frac{(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})} { (\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 2 $
$ \angle H(e^{j\frac{\pi}{4}}) = \angle c + \angle d - \angle a - \angle b = \frac{\pi}{2} + arctan^{-1} \left( \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} \right) - \frac{\pi}{4} - arctan^{-1} \left( \frac{\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}}{1-\sqrt{2}-\frac{1}{2\sqrt{2}}} \right) = 0 $

$ \omega =\frac{\pi}{2} $
Zp4.jpg $ \left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{\pi}{2}}) \right| = \left| H(z=j) \right| $

$ = \left| \frac{(j+\frac{1}{\sqrt{2}})(j-\frac{1}{\sqrt{2}})} { (j-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(j-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 1.455 $


$ \angle H(e^{j\frac{\pi}{2}}) = (\angle c + \angle d) - \angle a - \angle b = (\pi) - (arctan^{-1}\left( \frac{1-\frac{1}{2\sqrt{2}}}{\frac{-1}{\sqrt{2}}} \right) + \pi) - (arctan^{-1}\left( \frac{1+\frac{1}{2\sqrt{2}}}{1-\frac{1}{\sqrt{2}}} \right) + \pi) = -0.7563 $

$ \omega =\frac{3\pi}{4} $
Zp5.JPG $ \left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{3\pi}{4}}) \right| = \left| H(z=\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}) \right| $

$ = \left| \frac{(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})} { (\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = \frac{2}{3} $

$ \angle H(e^{j\frac{3\pi}{4}}) = \angle c + \angle d - \angle a - \angle b = (arctan^{-1}\left( \frac{\frac{1}{\sqrt{2}}}{\frac{-2}{\sqrt{2}}} \right) + \pi) + \frac{\pi}{2} - (arctan^{-1}\left( \frac{\frac{1}{\sqrt{2}} - \frac{1}{2\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{2\sqrt{2}}} \right) + \pi) + \frac{3\pi}{4} = -0.9273 $


$ \omega =\pi $
Zp6.JPG $ \left| H(e^{j\omega}) \right| = \left| H(e^{j\pi}) \right| = \left| H(z=-1) \right| $
$ = \left| \frac{(-1+\frac{1}{\sqrt{2}})(-1-\frac{1}{\sqrt{2}})} { (-1-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(-1-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 0.255 $

$ \angle H(e^{j\pi}) = (\angle c + \angle d) - \angle a - \angle b = 2\pi - 0 - 0 = 2\pi $

c. Is the system stable? Explain why or why not?
The system is causal and the ROC extends outwards from the outermost pole since |$ p_1 $| = |$ p_2 $| < 1 and this ROC contains the unit circle. Therefore the system is stable.

d. Find the difference equation for y[n] in terms of x[n], corresponding to this transfer function H(z).

$ H(z) = \frac{Y(z)}{X(z)} = \frac{1-\frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}} $
$ Y(z)(1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}) = X(z)(1-\frac{1}{2}z^{-2}) $

Taking inverse,
$ y[n]-\frac{1}{\sqrt{2}}y[n-1] + \frac{1}{4}y[n-2] = x[n] - \frac{1}{2}x[n-2] $
$ y[n] = x[n] - \frac{1}{2}x[n-2] +\frac{1}{\sqrt{2}}y[n-1] - \frac{1}{4}y[n-2] $

Question 4


$ y[n]=\frac{1}{8} \left( x[n]+x[n-1]+x[n-2]+x[n-3]+x[n-4]+x[n-5]+x[n-6]+x[n-7]\right) $

a.
$ h[n]=\frac{1}{8} \left( \delta[n]+\delta[n-1]+\delta[n-2]+\delta[n-3]+\delta[n-4]+\delta[n-5]+\delta[n-6]+\delta[n-7] \right) $
This is a finite duration response.

b. $ H[z]=\frac{1}{8} \left( 1+z^{-1}+z^{-2}+z^{-3}+z^{-4}+z^{-5}+z^{-6}+z^{-7} \right) $
$ H[z]=\frac{1}{8} \left( \frac{1-z^{-8}}{1-z^{-1}} \right) $

c.

$ H[z]=\frac{1}{8} \left( \frac{z^{8}-1}{z^{7}(z-1)} \right) $
Zp7.jpg
Poles:
$ p_1 = 1 $
$ p_2 = p_3 = ... = p_8 = 0 $

Zeros:
$ z^{8} - 1 = 0 $
$ z^{8} = e^{j2\pi } $

Generalizing,
$ z_k = e^{j2\pi k/8 } $ for k = 0,1,2,...,7

Question 5

a.
$ y[n]= \frac{1}{8} \left( x[n]-x[n-8]+y[n-1] \right) $

Using z-transform,
$ Y(z) = \frac{1}{8} \left( X(z)-X(z)z^{-8} + Y(z)z^{-1} \right) $
$ Y(z) (1 - \frac{1}{8}z^{-1}) = X(z) \frac{1}{8}(1 - z^{-8}) $
$ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{8} \left( \frac{ 1 - z^{-8} } {1 - \frac{1}{8}z^{-1}} \right) $
b.

HW6Q5fig1.jpg

c. H(z) can be re-written as
$ H[z]=\frac{1}{8} \left( \frac{1}{1-\frac{1}{8}z^{-1}} - \frac{z^{-8}}{1-\frac{1}{8}z^{-1}} \right) $
Assuming the ROC: $ |z|>\frac{1}{8} $Taking inverse Z-transform of H(z) -
$ h[n]=\frac{1}{8} \left( (\frac{1}{8})^n u[n] - (\frac{1}{8})^{n-8} u[n-8] \right) $

This is a infinite duration response.

Back to Homework 6

Back to ECE 438 Fall 2011

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood