A) 10.62 MHz
B) 10.61 kHz
C) 5.31 MHz
D) 5.31 kHz
Correct Answer: B
Solution :
Given: Resistance \[\operatorname{R}= 100 kilo ohm =\,\,100 \times 1{{0}^{3}}\Omega \] \[\operatorname{Capacitance}\,\,C=250picofarad=250\times {{10}^{-\,12}}F\] \[\tau = RC = 100 \times 1{{0}^{3}}\times 250 \times 1{{0}^{-}}^{12}\,sec\] \[=\,\,\,2.5\times 1{{0}^{7}}\times 1{{0}^{-}}^{12}\,sec\] \[=\,\,\,2.5\times {{10}^{-\,5}}\,sec\] The higher frequency which can be detected with tolerable distortion is \[f=\frac{1}{2\pi {{m}_{a}}RC}=\frac{1}{2\pi \times 0.6\times 2.5\times {{10}^{-5}}}Hz\] \[=\,\,\frac{100\times {{10}^{4}}}{25\times 1.2\pi }Hz\] \[=\,\,\frac{4}{1.2\pi }\times {{10}^{4}}\,Hz\] \[=\,\,\,1\,0.61 kHz\] This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less then the rate of decay modulated singnal voltage for proper detection of mdoulated signal.
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