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I'm confused on where I went wrong finding \$ \zeta \$ (zeta) and \$ \omega_0\$. I think its where I originally did nodal at Avx. Any help would be appreciated.

-thanks

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  • \$\begingroup\$ \$ s^2+2.5s-1.5\times 10^5\$ cannot be correct because the negative term renders the TF unstable. All the coefficients of a 2nd order TF must be +ve for stability. \$\endgroup\$
    – Chu
    Commented Feb 18, 2019 at 8:27

1 Answer 1

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My simple analysis

Since V(L)=LdI/di = "Vx" is in series with A*Vx for A=4 , then R damping voltage sees (1+4=5) or 5*L as effective inductance.

Since Ic=CdV/dt = "Iy" boosted by B*Iy, for B= 3, then the circuit produces (3+1 = 4x) or 4 effective capacitance multiplier.

So the \$L'=5L,~ C'=4C,~~~~~ L'C'=10^{-4}= 5*2.5mH*4*2mF\$

I know that I can apply here \$\omega _0=\dfrac{1}{\sqrt{L'C'}}=100 ~rad/s \$

Z(L') @ ωo =2π100 rad/s 2.5mH = 1.571 Ω at resonance

R=50Ω is higher than Z(ωoL) in parallel with the current source so underdamped <<1

Damping Coefficient = ζ =0.0125=α/ωo=1.25/100 using your value for α

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