Damping: Part 1 ECE 3100 John Stahl Western Michigan University
advertisement
Damping: Part 1 ECE 3100 John Stahl Western Michigan University 1 2nd Order Circuit We are going to start by solving for the transfer function of the 2nd order circuit. There are many ways to solve for Vo(w). For this circuit we are going to use Thévenin's theorem to aid in circuit reduction. 2 Vth For this circuit we are going to start our analysis by Theveninizing the RC circuit. Solve for the open circuit voltage across the capacitor and impedance ππ‘β (π) = = ππ β π (π) ππ + π 1 1 ππ βπ ππ + π 1 1 = 1 πππΆ 1 + π 1 πππΆ = ππ = β π1 β 1 πππΆ XC is the reactance of the capacitor. πππΆ πππΆ 1 β π (π) 1 + πππΆπ 1 1 3 Zth The impedance seen across AB is found by shorting the AC source and reducing the circuit to the terminals (AB), which is R1 in parallel with XC. ππ‘β ππ β π 1 = ππ + π 1 ππ β π 1 = ππ + π 1 1 ππ = πππΆ 1 πππΆ β π 1 πππΆ = β 1 πππΆ + π 1 πππΆ ππ‘β = π 1 1 + πππΆπ 1 4 Solve for Vo(w) Redraw the circuit and replace V1, R1 and C with the Thévenin's circuit. We see théveninizing the circuit makes the output a voltage divider saving a lot of time and effort. ππ π = ππΏ βπ ππ‘β + π 2 + ππΏ π‘β ππ π = πππΏ βπ ππ‘β + π 2 + πππΏ π‘β ππ π = Replace XL with jwL. πππΏ 1 β π1 (π) π 1 1 + πππΆπ 1 1 + πππΆπ 1 + π 2 + πππΏ β 5 The equation is technically a transfer function H(w), but it needs some clean up to place it in a standard form. ππ π π» π = = π1 (π) = πππΏ π 1 + π 2 + πππΏ β 1 + πππΆπ 1 1 + πππΆπ 1 πππΏ π 1 + π 2 + πππΏ β 1 + πππΆπ 1 = ππ β ππ = −π2 πππΏ π 1 + π 2 + πππΏ + πππΆπ 1π 2 − π 2 πΆπΏπ 1 1 πππΏ πΆπΏπ 1 = β −π 2 πΆπΏπ 1 + ππ πΏ + πΆπ 1π 2 + π 1 + π 2 1 πΆπΏπ 1 1 ππ πΆπ 1 π» π = πΏ + πΆπ 1π 2 π 1 + π 2 −π 2 + ππ + πΆπΏπ 1 πΆπΏπ 1 6
