hwset3solution - 2-2 Solution a The-3 dB bandwidth is found...

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2-2 Solution a. The -3 dB bandwidth is found using the roots of | F ( ω ) | = | F ( ω o ) | 2 = 1 2 . Using the identity | 1 1 ± j | = 1 2 it is easy to see that the -3 dB frequencies are the roots of Q ( ω ω o ω o ω ) = ± 1 . Or ω 2 ± ω o Q ω ω 2 o = 0 . The roots are: ω o 2 Q ± ω o 2 Q 1 + 4 Q 2 Only two of these 4 roots are positive numbers. The -3 dB bandwidth is the magnitude of the di ff erence between the two positive (or negative) roots. The two positive roots are ω o 2 Q + ω o 2 Q 1 + 4 Q 2 and + ω o 2 Q + ω o 2 Q 1 + 4 Q 2 The di ff erence between these two roots is ω o Q . Hence the -3 dB bandwidth is ω o Q . b. T ( ω ) = G 1 AG + jQ ( ω ω o ω o ω ) . Using the same reasoning that was described in part a., we can immediately see that the -3 dB frequencies are the roots of Q ( ω ω o ω o ω ) = ± (1 AG ) . Or ω 2 ± ω o Q (1 AG ) ω ω 2 o = 0 . The two positive roots are ω o 2 Q (1 AG ) + ω o 2 Q (1 AG ) 2 + 4 Q 2 and + ω o 2 Q (1 AG ) + ω o 2 Q (1 AG ) 2 + 4 Q 2 The di ff erence between these two roots is ω o Q (1 AG ) . Hence the -3 dB bandwidth is ω o Q (1 AG ) . The bandwidth goes to zero as AG 1 .
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