Electrical – Calculating the damping ratio, \$ \zeta \$, and the undamped natural frequency, \$ \omega_0\$
damping-factor
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
Best Answer
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 \$
I think you mainly need help understanding what it means for the system to be ‘second-order dominated on the verge of instability.’
It means that there is a pair of complex poles just barely to the left of the imaginary axis (or perhaps on the axis, which is marginally stable), with no other poles that are also close to the imaginary axis.
Best Answer
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