Hello vonsproken,
your answer depends on the precision wanted. At first, you can solve it in a very simplified manner, using a bode plot of the circuit. This circuit your professor proposed is a simple active low-pass filter. After the pole of the circuit, there'll be a slope of -6dB per octave, right? So, if there's a pole located one octave below the 9.6kHz frequency (that means, if there's a pole at 4.8kHz), the passband gain will be 6db higher than the gain at 9.6kHz. The big problem here is that these approximations aren't good enough because the frequencies involved aren't more than a decade apart, what makes the bode approximation a little bit lame. The engineer's solution: as we know that, near the pole, the gain decreases more quickly than the asymptote, if we place the pole a little bit after 4.8kHz, say, at 5.5kHz (keep in mind that this 5.5kHz is a wild guess =P), everything will be fine.
On to the calculations:
To achieve a passband gain of 6, the resistor in the feedback network should be 40.2kΩ.
The cut-off frequency is: fc = 1/(2*pi*R*C), and the values of R and fc are known (40.2k and 5.5k). Solving it for C, the formula yields: C≈720pF.
As everything until now were wild guesses, it is good to simulate the circuit just to see if all of this is not completely wrong. I used LTSpice to simulate the circuit, and got the following results:
- Gain at 1kHz = 5.9
- Gain at 9.6kHz = 2.97
Remember that all of this depended a LOT on my guess of the right location for the pole... And as the pole is very near 1kHz, the gain at this frequency will be certainly lower than 6. If we raised a little bit the value of the resistance, the values probably would be better. But again, this are all guesses, and in a real application you would have a bigger constraint: the commercial values.
But if you really want to calculate it, solve the equations. It isn't too difficult, but it demands more effort. Again, it really depends on what does your professor want. I had both kinds of professors at college: the practical ones, that would love a solution like this, and the math-lovers, that woud find it hideous.
Hope it helped.
Castilho.