Solving the Filtered Amplifier Puzzle

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Homework Statement



We have a filtered amplifier (diagram is attached). We do not know Cf or Rf.
We know that:
1 volt input at 5 Hz = 9.25 volt output
1 volt input at 200 Hz = 6.77 volt output
1 volt input at 500 Hz = 3.66 volt output

Homework Equations



The transfer function for the first amplifier is H(w) = -Zf/Zi = -10000/2000=-5
The transfer function for the second amplifier is H(w) = -Zf/Zi, where Zi = 5000 and
Zf = Rf/(Rf j w Cf + 1).


The Attempt at a Solution


If we multiply the two transfer functions above, we obtain
H(w) = V0/Vi = Rf/(1000(Rf j w Cf + 1))

I multiplied the frequencies by 2*pi and made appropriate substitutions in the above equation to get the following three expressions:
9.25 = Rf/(1000(31.4159 Rf j Cf + 1))
6.77 = Rf/(1000(1256.64 Rf j Cf + 1))
3.66 = Rf/(1000(3141.59 Rf j Cf + 1))

Three equations for only two unknowns?

I get that Cf = -5.3E-8 i and Rf = 9395.

What am I doing wrong?
 

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engineer23 said:
I multiplied the frequencies by 2*pi and made appropriate substitutions in the above equation to get the following three expressions:
9.25 = Rf/(1000(31.4159 Rf j Cf + 1))
6.77 = Rf/(1000(1256.64 Rf j Cf + 1))
3.66 = Rf/(1000(3141.59 Rf j Cf + 1))

Since you are only given the magnitudes (and not the phases) of the input and output signals you must work with the absolute value of the transfer function.