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| 1. |
An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3KHz. Could this wave be demodulated by a diode detector which has the values of R and C as |
| Answer» <html><body><p>`R=1kOmega,<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>=0.01muE`<br/>`R=10kOmega,C=0.01muF`<br/>`R=10kOmega,C=0.1pF`<br/>None of these </p>Solution : `v_(c)=20MHz=20xx10^(6)Hz` <br/> `(1)/(v_(c))=(1)/(20)xx10^(-6)=0.05xx10^(-6)=0.5xx10^(-7)s` <br/> Band width =`20_(m)=3kHz=3xx10^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)Hz` <br/> `v_(m)=1.5xx10^(3)Hz` <br/> `(1)/(v_(m))=(1)/(1.<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)xx10^(-3)=0.7xx10^(-3)s`. <br/> Diode detector can demodulate when <br/> `(1)/(v_(c))ltltRCltlt(1)/(v_(m))` <br/> (i)RC=`1kOmegaxx0.01muF=10^(3)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(0.01xx10^(-6))=10^(-5)s`. it can be demodulated. <br/> (ii) RC=`10kOmegaxx1xx10^(-8)=10^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)s`It can be demodulated <br/> (iii) RC`=10kOmegaxx1pF=10^(4)xx10^(-12)=10^(-8)s` <br/> It cannot be demodulated.</body></html> | |