(New page: Category:2010 Fall ECE 438 Boutin == Solution to Q4 of Week 5 Quiz Pool == ---- From the first question, we knew that <math> -a^{n}u[-n-1] = \mathcal{Z}^{-1}\bigg\{\frac{1}{1-az^{-...) |
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| − | From the | + | From the definition, we know that |
| − | <math> | + | <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> | + | <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> |
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| + | Therefore, | ||
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| + | <math> |H(e^{j w_2})| = \bigg|\frac{2j}{2+j}\bigg| = \frac{2}{\sqrt{5}}. \,\!</math> | ||
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Revision as of 15:13, 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{2}{\sqrt{5}}. \,\! $
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