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| − | <math> |H(e^{j w_2})| = \bigg|\frac{2j}{2+j}\bigg| = \frac{2}{\sqrt{5}}. \,\!</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> |
Revision as of 14:28, 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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