Difference between revisions of "ECE438 Week5 Quiz Q4sol" - Rhea

Latest revision as of 15:42, 19 September 2010

Solution to Q4 of Week 5 Quiz Pool

From the definition, we know that

$H(e^{jw}) = \frac{e^{jw}-j}{e^{jw}-2} \,\!$

$\text{For } w_1, \; |H(e^{j w_1})| = \bigg|\frac{e^{j\frac{\pi}{2}}-j}{e^{j\frac{\pi}{2}}-2}\bigg| = 0, \;\; \text{ since } e^{j\frac{\pi}{2}}=j. \,\!$

$\text{For } w_2, \; \text{ since } e^{-j\frac{\pi}{2}}=-j, \; H(e^{j w_2}) = \frac{e^{-j\frac{\pi}{2}}-j}{e^{-j\frac{\pi}{2}}-2} = \frac{-j-j}{-j-2} = \frac{2j}{2+j}. \,\!$

Therefore,

$|H(e^{j w_2})| = \bigg|\frac{2j}{2+j}\bigg| = \frac{\sqrt{0^2+2^2}}{\sqrt{2^2+1^2}} = \frac{2}{\sqrt{5}}. \,\!$

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@@ Line 1: / Line 1: @@
 [[Category:2010 Fall ECE 438 Boutin]]
+----
 == Solution to Q4 of Week 5 Quiz Pool ==
 ----
-From the first question, we knew that
+From the definition, we know that
+<math> H(e^{jw}) = \frac{e^{jw}-j}{e^{jw}-2} \,\!</math>
+<math> \text{For } w_1, \; |H(e^{j w_1})| = \bigg|\frac{e^{j\frac{\pi}{2}}-j}{e^{j\frac{\pi}{2}}-2}\bigg| = 0, \;\; \text{ since } e^{j\frac{\pi}{2}}=j. \,\!</math>
-<math> -a^{n}u[-n-1] = \mathcal{Z}^{-1}\bigg\{\frac{1}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \,\!</math>
+<math> \text{For } w_2, \; \text{ since } e^{-j\frac{\pi}{2}}=-j, \; H(e^{j w_2}) = \frac{e^{-j\frac{\pi}{2}}-j}{e^{-j\frac{\pi}{2}}-2} = \frac{-j-j}{-j-2} = \frac{2j}{2+j}.  \,\!</math>
-And the time-shifting property of Z-transform is defined as
+Therefore,
-<math> x[n-k] = \mathcal{Z}^{-1}\bigg\{z^{-k}X(z)\bigg\} \text{   when   } x[n] = \mathcal{Z}^{-1}\bigg\{X(z)\bigg\}\,\!</math>
+<math> |H(e^{j w_2})| = \bigg|\frac{2j}{2+j}\bigg| = \frac{\sqrt{0^2+2^2}}{\sqrt{2^2+1^2}} = \frac{2}{\sqrt{5}}. \,\!</math>
-Therefore, if we use the time-shifting property of Z-transform, then
-<math> -a^{n-3}u[-(n-3)-1] = \mathcal{Z}^{-1}\bigg\{\frac{z^{-3}}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \,\!</math>
-Combined with the result from the linearity of Z-transform, then
-<math>
-\begin{align}
-\mathcal{Z}^{-1}\bigg\{\frac{2z^{-3}}{1-az^{-1}}\bigg\}  \text{ for } |z|<|a| &= -2a^{n-3}u[-(n-3)-1], \\
-&= -2a^{n-3}u[-n+2]
-\end{align}
-\,\!</math>
 ----
 Back to [[ECE438_Week5_Quiz|Lab Week 5 Quiz Pool]]

Difference between revisions of "ECE438 Week5 Quiz Q4sol" - Rhea

Latest revision as of 15:42, 19 September 2010

Solution to Q4 of Week 5 Quiz Pool

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