Hence the 3 db bandwidth is q b t g 1 ag jq

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Unformatted text preview: ￿ + 1 + 4Q2 2Q 2Q The difference between these two roots is ωo . Hence the -3 dB bandwidth is Q + b. T (ω ) = G ω 1 − AG + jQ( ωo − ωo Q. ωo . ω) Using the same reasoning that was described in part a., we can immediately see that the -3 dB frequencies are the roots of ω ωo Q( − ) = ±(1 − AG). ωo ω Or ωo 2 ω2 ± (1 − AG)ω − ωo = 0. Q The two positive roots are ωo ωo ￿ − (1 − AG) + (1 − AG)2 + 4Q2 2Q 2Q and ωo ωo ￿ + (1 − AG) + (1 − AG)2 + 4Q2 2Q 2Q The difference between these two roots is ωo (1 − AG). Hence the -3 dB bandwidth is ωo (1 − AG). The Q Q bandwid...
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