Solving the Filtered Amplifier Puzzle

[engineer23]

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?

 

[antonantal]

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.

 

[piercas]

It seems that you have made a mistake in your calculations. Multiplying the transfer functions will lead to a transfer function that is not consistent with the given output values. Instead, you should use the transfer function for the first amplifier (H(w) = -5) and the transfer function for the second amplifier (H(w) = -Zf/Zi), and equate them to the given output values. This will lead to three equations for three unknowns (Rf, Cf, and Zi), which can be solved simultaneously to find the values of Rf and Cf.