The highest modulating signal frequency that can be demodulated by a peak detector without attenuation is given by

Question

TuteeHUB · Accepted Answer

${f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {{m^2} - 1} }}{{2{\rm{\pi RC}}}}$

TuteeHUB · Answer

${f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {\left( {\frac{1}{{{m^2}}}} \right) - 1} }}{{2\pi RC}}$

TuteeHUB · Answer

${f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {1 - \frac{1}{{{m^2}}}} }}{{2\pi \;RC}}$

TuteeHUB · Answer

${f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {\left( {\frac{1}{{{m^2}}}} \right) + 1} }}{{2\pi RC}}$

1.	The highest modulating signal frequency that can be demodulated by a peak detector without attenuation is given by
A.	\({f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {\left( {\frac{1}{{{m^2}}}} \right) - 1} }}{{2\pi RC}}\)
B.	\({f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {{m^2} - 1} }}{{2{\rm{\pi RC}}}}\)
C.	\({f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {1 - \frac{1}{{{m^2}}}} }}{{2\pi \;RC}}\)
D.	\({f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {\left( {\frac{1}{{{m^2}}}} \right) + 1} }}{{2\pi RC}}\)
Answer» B. \({f_m}\left( {{\rm{max}}} \right) = \frac{{\sqrt {{m^2} - 1} }}{{2{\rm{\pi RC}}}}\)

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