What is the Wien bridge frequency proof and how do you solve it?

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SUMMARY

The Wien bridge frequency proof demonstrates that when the bridge is balanced, the frequency can be determined using the equation w=2*pi*f. The general solution for AC bridges is expressed as Z4 = (Z3/Z1)*Z2, where specific impedance values are defined as Z1 = R1, Z2 = R2 - j/wC2, Z3 = R3, and Z4 = (R4/jwC4)/(R4 + jwC4). The solution requires equating the real and imaginary parts of Z4 to zero, leading to the simplified equation R2R4C2C4*w^2 + 1 = 0.

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[SOLVED] Wien Bridge Frequency Proof

Homework Statement



The ac bridge circuit shown is called a Wien bridge (also Wein bridge).
Wien Bridge.png


Show that when the bridge is balanced, frequency is found by the Eqn attached.
Eqn.png

Homework Equations



w=2*pi*f

Using the number from each branch as a guide, the general solution to AC bridges is Z4 = (Z3/Z1)*Z2

The Attempt at a Solution



I keep trying to solve for impedences to plug into the second equation provided, but my j values do not cancel and the equation left is rather messy (R2R4C2C4*w^2 +jw(RzR4C2 -R2C2 -R4C4) +1 = 0.

For my impedence values I have been using
Z1 = R1
Z2 = R2 -j/wC2
Z3 = R3
Z4 = (R4/jwC4)/(R4 +jwC4)

Any help with this would be greatly appreciated.
 
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I think Z_4 = \frac {-jR_4} {\omega C_4 (R_4 - \frac{j}{\omega C_4})}


Both the real and the imaginary parts of Z4 = (Z3/Z1)*Z2 must be 0
 
Actually, I think our Z4 values are the same, since -j = 1/j. But thank you! I managed to figure it out after you got me thinking about real and imaginary parts.

It's fairly simple, actually: I just had to equate the real parts from my equation so that R2R4C2C4*w^2 + 1 = 0 (and then I took absolute value).
 

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