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The Butterworth filter of order n is described by the magnitude squared of its frequency response given by $${\left| {{H_n}\left( {j\Omega } \right)} \right|^2} = \frac{1}{{\left[ {1 + {{\left( {\frac{\Omega }{{{\Omega _{\,C}}}}} \right)}^{2n}}} \right]}}.$$      The value of $$20\log \left| {{H_n}\left( {j\Omega } \right)} \right|$$   at $$\Omega = {\Omega _{\,C}}$$  is

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Correct Answer: Option B
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