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In an example about oscillator

In the circuit shown in Figure, \$R_1 = 10 k\$, \$R_2 = 1k\$, \$C = 0.002μF\$, and \$L = 1μH\$. If the circuit is oscillating, find the voltage gain \$A\$ of the forward amplifier and the frequency of oscillation enter image description here

and in the solution it is provided the feedback gain $$\beta=\frac{j\omega L}{R_1-\omega^2R_2CL+j\omega(L+R_1R_2C)} $$ But I can not figure out why \$\beta\$ is this because when I solved for it I found \$\beta =1\$ by considering unity gain as Buffer Amplifier.

I want to know how \$\beta\$ is provided so?

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The given expression for beta is correct - however, it applies for a circuit where the componenets C and L are grounded. Note that these two components must be grounded for a working oscillator circuit.

In this case, the expression for beta is simply calculated by muliplying the transfer functions of the R1-L highpass and the R2-C lowpass.

EDIT: For finding the required gain A you have to remember the oscillation condition (unity loop gain). In particular, the PHASE of the loop gain LG=A*beta must be zero.

Because A is real you have to find the frequency which makes beta real. Because the numerator of beta is imaginary, also the denominator must be imaginary. Therefore, set the real part of the denominator equal to zero - this gives you the oscillating frequency. Introduce this frequency into the beta expression and set A=1/beta.
This gives you the required gain A.

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