Simple transfer function - algebra giving me problems

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SUMMARY

The discussion focuses on deriving the transfer function for a non-inverting amplifier, specifically represented as \(\frac{V_o(s)}{V_i(s)} = \frac{Z_1(s) + Z_2(s)}{Z_1(s)}\). The impedances are defined as \(Z_1(s) = R_1 + \frac{1}{C_1s}\) and \(Z_2(s) = \frac{\frac{R_2}{C_2s}}{R_2 + \frac{1}{C_2s}}\). The final transfer function is expressed as \(G(s) = \frac{C_1C_2R_1R_2s^2 + (C_2R_2 + C_1R_2 + C_1R_1)s + 1}{C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1}\). The initial misinterpretation of the problem led to incorrect calculations, which were later corrected.

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trickae
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Homework Statement



Find a transfer function: \frac{V_o(s)}{V_i(s)} = \frac{Z_2 (s)}{-Z_1(s)}

Homework Equations



Z_1(s) = R_1 + \frac{1}{C_1s}
Z_2(s) = \frac {\frac{R_2}{C_2s}}{R_2 + \frac{1}{C_2s}}final solution should be:
G(s) = \frac{V_o(s)}{V_i(s)} = \frac{C_1C_2R_1R_2s^2 + (C_2R_2 + C_1R_2 + C_1R_1)s + 1}{C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1}

The Attempt at a Solution



- Give me a second I'm still typing up the latex commands

G(s) = \frac{V_o(s)}{V_i(s)}= \frac{-\frac {\frac{R_2}{C_2s}}{R_2 + \frac{1}{C_2s}}}{R_1 + \frac{1}{C_1s}}

= -\frac {\frac{R_2}{C_2s}}{(R_2 + \frac{1}{C_2s})(R_1 + \frac{1}{C_1s}) }

= \frac{-R_2}{(C_2s)(R_2 + \frac{1}{C_2s})(R_1 + \frac{1}{C_1s})}

= \frac{-R_2(C_1C_2s)}{(C_2s)(C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1)}

=\frac{-R_2(C_1)}{(C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1)}
which is no where near the solution.
 
Last edited:
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I misread the problem - apologies - the transfer function for a Non inverting amplifer is in the form

\frac{Z_1(s) + Z_2(s)}{Z_1(s)} - now i get the right answer
 
trickae said:
I misread the problem - apologies - the transfer function for a Non inverting amplifer is in the form

\frac{Z_1(s) + Z_2(s)}{Z_1(s)} - now i get the right answer

hehe good. Cause I quickly did it, and definitely did not get the "answer".

I feel bad for you typing all of that up in Latex. Probably took a few ;)
 

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