Gain in dB

Discussion in 'Homework Help' started by Hello, Mar 30, 2009.

1. Hello Thread Starter Active Member

Dec 18, 2008
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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.

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Last edited: Mar 30, 2009
2. ABoul Member

Mar 30, 2009
15
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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.

3. Hello Thread Starter Active Member

Dec 18, 2008
82
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But how would i work out the output voltage?

4. ABoul Member

Mar 30, 2009
15
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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.

Mar 23, 2009
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6. ABoul Member

Mar 30, 2009
15
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past papers are a great way to prepare for your REAL exams. in fact, it's the only way i prepare for them.

7. Hello Thread Starter Active Member

Dec 18, 2008
82
0
You're right! Thanks for the help.

8. thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
If so, I think he already failed about 10 months ago (exam dated May/08).

9. ABoul Member

Mar 30, 2009
15
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what goes inside the log is |Vo/Vi|, NOT Vo.

10. thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
@ 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.