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The small signal equivalent circuit of the cross-coupled oscillator is drawn asThe resonance frequency of each of the two tank circuits is \omega=\omega_{0}=\frac{1}{\sqrt{L C}}The load of each of Q_{1} and Q_{2} reduces to a resistance R_{P}=\omega_{0} L Q, where Q is the quality factor of the inductance.considering the output resistance r_{0} of each of the Q_{1} and Q_{2}, we can write the gain of each of the two stages at \omega=\omega_{0} as A_{1}=A_{2}=-q_{m}\left(R_{p} \| \gamma_{0}\right)Thus each stage exhibits a 180^{\circ} phase shift, for a total phase shift around the loop of 360^{\circ}.Thus the circuit will provide sustained oscillations at w_{0}=\frac{1}{\sqrt{L C}}provided \left|A_{1} A_{2}\right|=\left|q_{m}\left(R_{p} \| \psi_{2}\right)\right|^{2}=1 \Rightarrow q_{m}\left(R_{p} \| b_{0}\right)=1Thus the minimum sequired value of q_{m}=\frac{1}{\left(R_{p} \| l_{0}\right)} at which each of Q_{1} and Q_{2} is operated for oscillations to be sustained ...