Calculate (i) the gain in dB at 10kHz and (ii) the cut-off frequency for
R = 100W and C = 1μF if the input voltage = 10V.


Any help is appreciated.

 

can you get the gain itself? because the gain in dB is simply 20log |G|, where |G| is the modulus of the gain, and that's log to the base 10.

 

But how would i work out the output voltage?

 

But how would i work out the output voltage?

what you have there is a potential divider. let X denote the impedence of the capacitor, so that X = 1/jwC. the output voltage would therefore be V_out = V_in[X / (X + R)]. from there, you can divide by V_in to get V_out/V_in, plug in your numbers, find the modulus and 20log it.

 

you're asking for help on your exam?

 

you're asking for help on your exam?

past papers are a great way to prepare for your REAL exams. in fact, it's the only way i prepare for them.

 

You're right! Thanks for the help.

 

you're asking for help on your exam?

If so, I think he already failed about 10 months ago (exam dated May/08).

 

Can you please check if this is correct (same question)

Vo = Vi/[jωRC + 1] (where Vi= 10 volts is given)

= 10/[ j(2∏ x E^4 x 100 x -j15.9) + 1 ]
= E^-7
So:

Gain = 20log(E^-7/10) = -160 dB

what goes inside the log is |Vo/Vi|, NOT Vo.

 

@ 10kHz ≈ -16dB (I believe you had a typo)
-3dB / 45° cutoff ≈ 1.6kHz
Filter is 2nd order low pass filter, attenuation -6dB/octave after cutoff point.

Always check other calculations to see if they "make sense", i.e. 10kHz is about 4x 1.6kHz, so either -dB per octave is way too low for attenuation, or the 10kHz calculation is incorrect.