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
A. -2 dB
B. -3.01 dB
C. -3 dB
D. -3.5 dB
Answer: Option B
This Question Belongs to Electronics And Communications Engineering >> Signal Processing
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