Second Order RLC Filters

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1 ECEN 60 Circuits/Electronics Spring P. Mathys Second Order RLC Filters RLC Lowpass Filter A passive RLC lowpass filter (LPF) circuit is shown in the following schematic. R L C v O (t) Using phasor analysis, v O (t) V O is computed as jωc R jωl jωc V S LC (jω) jω R L LC Setting ω 0 / LC and ζω 0 R/L, where ω 0 is the (undamped) natural frequency and ζ is the damping ratio, yields V S. ω 0 (jω) jω ζω 0 ω 0 V S v () O (t) ζω 0 v () O (t) ω 0 v O (t) ω 0. The blockdiagram that represents this differential equation is ω0 v () O (t) v() O (t) v O(t) ζω 0 ω 0 Unit Step Response. By definition, the unit step response g(t) of a circuit is the zero-state response (ZSR) to the input u(t). For the nd order LPF considered here the unit step response is of the form (if ζ ) g(t) K e s t K e s t, t 0 g () (t) s K e s t s K e s t, t 0,

2 with initial conditions g(0) 0 and g () (0) 0. The values of s and s are the solutions of the characteristic equation s ζω 0 s ω 0 0 s, ( ζ ± ζ ) ω 0, and the properties of g(t) change fundamentally depending on whether ζ >, ζ, or ζ <. Overdamped Case, ζ >. In this case the characteristic equation has two real solutions s α, α (ζ ζ ) ω 0, and s α, α (ζ ζ ) ω 0. Note that α < α. The unit step response is of the form g(t) K e α t K e α t, t 0 g () (t) α K e α t α K e α t, t 0. Using initial conditions g(0) 0 and g () (0) 0 yields [ ] [ K K, α K α K 0. α α K K ] [ ] 0. From this K and K are obtained as K α α α, K α α α, and thus the unit step response for a nd order overdamped LPF is Note that g(t) α e α t α e α t α α, t 0. g () (t) α α α α (e α t e α t ), t 0, and therefore g () (t) 0 requires that e α t e α t which can only happen at t 0 or t if α α. This implies that the extrema of g(t) occur at t 0 (where g(0) 0) and at t (where g( ) ) and thus g(t) has no overshoot. Critically Damped Case, ζ. In this case the characteristic equation has one real double solution s s α, α ω 0, and the unit step response is of the form g(t) K e αt K t e αt, t 0, g () (t) αk e αt K e αt αk t e αt, t 0. Using initial conditions g(0) 0 and g () (0) 0 yields K, K α,

3 and thus the unit step response for a nd order critically damped LPF is Note that g(t) ( αt) e αt, t 0. g () (t) α t e αt, t 0, and therefore g () (t) 0 requires either t 0 or t, which implies that the extrema of g(t) are 0 (at t 0) and (at t ) and thus g(t) has no overshoot. Underdamped Case, ζ <. In this case the characteristic equation has two complex solutions which are conjugates of each other s α jβ, and s s α jβ, where α ζω 0, β ζ ω 0. The unit step response is of the form g(t) K e s t K e s t, t 0 g () (t) s K e s t s K e s t, t 0. Substituting s, α ± jβ and using initial conditions g(0) 0 and g () (0) 0 yields [ ] [ ] [ ] K K, K. (αjβ)k (αjβ)k 0. αjβ αjβ 0 From this K and K K are obtained as K K β jα β ρ e jφ, K β jα β ρ e jφ, where ρ α β β ζ, and φ π α tan β π ζ tan. ζ Thus, the unit step response for a nd order underdamped LPF is g(t) ρ e αt ( e j(βtφ) e j(βtφ)) ρ e αt cos(βt φ), t 0, or, with ρ and φ substituted g(t) eαt cos ( βt tan ζ ), t 0. ζ ζ To obtain a formula for g () (t) easily, first note that s K (s K ) and s K (α jβ) β jα β α β jβ. 3

4 Then use g () (t) ( s K e jβt s K e jβt) e αt to obtain g () (t) α β β ( e e αt jβt e jβt ) α β e αt sin βt, t 0. j β To find the times where the extrema of g(t) occur, set g () (t) 0 and solve for t. The sine has zero crossings for βt kπ and, for k 0, g(t) g(0) 0 clearly has a minimum. The largest and most interesting maximum (due to underdamping) of g(t) happens when bt π t π/β. The value of the maximum is computed as g max g ( π β ) ρ e πα/β cos(π φ) eπζ/ ζ ζ cos ( tan ζ ζ ). Using the identity cos ( tan x x ) x, the final result is g max e πζ/ ζ Overshoot in %: 00 e πζ/ ζ. The following graph shows the unit step response g(t) for several values of ζ..6 Unit Step Responses of Second Order LPF, ω g(t) ζ0. 0. ζ0.5 ζ0.707 ζ ζ t in msec 4

5 Frequency Response. From the phasor analysis the system function of the LPF is obtained as H V O ω0 V S ω0 ω j ζω 0 ω. The magnitude and the phase of H are H Note that, at ω ω 0, ω 0 (ω 0 ω ) (ζω 0 ω), and H tan ζω 0 ω ω 0 ω. H ω0 ζ, and H ω 0 90, and thus ζ and ω 0 can be easily determined from the magnitude and phase of H. RLC Bandpass Filter A passive RLC bandpass filter (BPF) circuit is shown in the following schematic. R L C v O (t) Using phasor analysis, v O (t) V O is computed as R jωl (jω) LC jωl (jω) LC V S Setting ω 0 / LC and ζω 0 /(RC) yields jω ζω 0 (jω) jω ζω 0 ω 0 jω RC (jω) jω RC LC V S. V S v () O (t) ζω 0 v () O (t) ω 0 v O (t) ζω 0 v () S (t). The blockdiagram that represents this differential equation is (after integrating both sides so that the input is rather than v () S (t)) v O (t) ζω 0 v () O (t) v O(t) vo (τ)dτ ζω 0 ω 0 5

6 Frequency Response. From the phasor analysis the system function of the BPF is obtained as H V O j ζω 0 ω V S ω0 ω j ζω 0 ω. The magnitude and the phase of H are H ζω 0 ω (ω 0 ω ) (ζω 0 ω), and H π/ ζω tan 0 ω ω0 ω. Note that, at ω ω 0 (the center frequency of the BPF), H ω0, and H ω0 0. To obtain the lower half-power (or -3dB) frequency ω 3 of the BPF, set ω 0 ω 3 ζω 0ω 3 ω 3 ( ζ ζ ) ω 0. Similarly, the upper half-power (or -3dB) frequency ω 3 of the BPF is obtained from ω 3 ω 0 ζω 0 ω 3 ω 3 ( ζ ζ ) ω 0. Thus, the half-power (or -3dB) bandwidth of the BPF is equal to ζω 0, and ζ and ω 0 can therefore be determined from the magnitude and phase of H. c , P. Mathys. Last revised: --07, PM. 6