ECEN 2260 Circuits/Electronics 2 Spring 20072-10-07 P. Mathys

Second Order RLC Filters

1 RLC Lowpass Filter

A passive RLC lowpass filter (LPF) circuit is shown in the following schematic.

◦

◦ ◦

◦

•

•

R L

C

−

vS(t)

+

−

vO(t)

+

Using phasor analysis, vO(t)⇔ VO is computed as

VO =

1jωC

R + jωL + 1jωC

VS =1

LC

(jω)2 + jω RL

+ 1LC

VS .

Setting ω0 = 1/√

LC and 2ζω0 = R/L, where ω0 is the (undamped) natural frequency andζ is the damping ratio, yields

VO =ω2

0

(jω)2 + jω 2ζω0 + ω20

VS ⇐⇒ v(2)O (t) + 2ζω0 v

(1)O (t) + ω2

0 vO(t) = ω20 vS(t) .

The blockdiagram that represents this differential equation is

ω20 +

∫ ∫ •

ω20

•

2ζω0

+

+

−

++

vS(t) v(2)O (t) v

(1)O (t) vO(t)

Unit Step Response. By definition, the unit step response g(t) of a circuit is the zero-stateresponse (ZSR) to the input vS(t) = u(t). For the 2’nd order LPF considered here the unitstep response is of the form (if ζ 6= 1)

g(t) = K1 es1t + K2 es2t + 1 , t ≥ 0 =⇒ g(1)(t) = s1K1 es1t + s2K2 es2t , t ≥ 0 ,

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with initial conditions g(0) = 0 and g(1)(0) = 0. The values of s1 and s2 are the solutions ofthe characteristic equation

s2 + 2ζω0s + ω20 = 0 =⇒ s1,2 =

(− ζ ±

√ζ2 − 1

)ω0 ,

and the properties of g(t) change fundamentally depending on whether ζ > 1, ζ = 1, orζ < 1.

Overdamped Case, ζ > 1. In this case the characteristic equation has two real solutions

s1 = −α1 , α1 = (ζ −√

ζ2 − 1) ω0 , and s2 = −α2 , α2 = (ζ +√

ζ2 − 1) ω0 .

Note that α1 < α2. The unit step response is of the form

g(t) = K1 e−α1t +K2 e−α2t +1 , t ≥ 0 =⇒ g(1)(t) = −α1K1 e−α1t−α2K2 e−α2t , t ≥ 0 .

Using initial conditions g(0) = 0 and g(1)(0) = 0 yields

K1 + K2 = −1 ,

α1K1 + α2K2 = 0 .=⇒

[1 1

α1 α2

] [K1

K2

]=

[−1

0

].

From this K1 and K2 are obtained as

K1 =−α2

α2 − α1

, K2 =α1

α2 − α1

,

and thus the unit step response for a 2’nd order overdamped LPF is

g(t) = 1− α2 e−α1t − α1 e−α2t

α2 − α1

, t ≥ 0 .

Note thatg(1)(t) =

α1 α2

α2 − α1

(e−α1t − e−α2t) , t ≥ 0 ,

and therefore g(1)(t) = 0 requires that e−α1t = e−α2t which can only happen at t = 0 ort =∞ if α1 6= α2. This implies that the extrema of g(t) occur at t = 0 (where g(0) = 0) andat t =∞ (where g(∞) = 1) and thus g(t) has no overshoot.

Critically Damped Case, ζ = 1. In this case the characteristic equation has one realdouble solution

s1 = s2 = −α , α = ω0 ,

and the unit step response is of the form

g(t) = K1 e−αt + K2 t e−αt + 1 , t ≥ 0 ,

=⇒ g(1)(t) = −αK1 e−αt + K2 e−αt − αK2 t e−αt , t ≥ 0 .

Using initial conditions g(0) = 0 and g(1)(0) = 0 yields

K1 = −1 , K2 = −α ,

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and thus the unit step response for a 2’nd order critically damped LPF is

g(t) = 1− (1 + αt) e−αt , t ≥ 0 .

Note thatg(1)(t) = α2t e−αt , t ≥ 0 ,

and therefore g(1)(t) = 0 requires either t = 0 or t = ∞, which implies that the extrema ofg(t) are 0 (at t = 0) and 1 (at t =∞) and thus g(t) has no overshoot.

Underdamped Case, ζ < 1. In this case the characteristic equation has two complexsolutions which are conjugates of each other

s1 = −α + jβ , and s2 = s∗1 = −α− jβ , where α = ζω0 , β =√

1− ζ2 ω0 .

The unit step response is of the form

g(t) = K1 es1t + K2 es2t + 1 , t ≥ 0 =⇒ g(1)(t) = s1K1 es1t + s2K2 es2t , t ≥ 0 .

Substituting s1,2 = −α± jβ and using initial conditions g(0) = 0 and g(1)(0) = 0 yields

K1 + K2 = −1 ,

(−α+jβ)K1 − (α+jβ)K2 = 0 .=⇒

[1 1

−α+jβ −α−jβ

] [K1

K2

]=

[−1

0

].

From this K1 and K2 = K∗1 are obtained as

K1 =−β + jα

2β= ρ ejφ , K2 =

−β − jα

2β= ρ e−jφ ,

where

ρ =

√α2 + β2

2β=

1

2√

1− ζ2, and φ = π − tan−1 α

β= π − tan−1 ζ√

1− ζ2.

Thus, the unit step response for a 2’nd order underdamped LPF is

g(t) = 1 + ρ e−αt(ej(βt+φ) + e−j(βt+φ)

)= 1 + 2ρ e−αt cos(βt + φ) , t ≥ 0 ,

or, with ρ and φ substituted

g(t) = 1− e−αt√1− ζ2

cos(βt− tan−1 ζ√

1− ζ2

), t ≥ 0 .

To obtain a formula for g(1)(t) easily, first note that s2K2 = (s1K1)∗ and

s1 K1 = (−α + jβ)−β + jα

2β=

α2 + β2

2jβ.

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Then use g(1)(t) =(s1K1 ejβt + s2K2 e−jβt

)e−αt to obtain

g(1)(t) =α2 + β2

βe−αt

(ejβt − e−jβt

2j

)=

α2 + β2

βe−αt sin βt , t ≥ 0 .

To find the times where the extrema of g(t) occur, set g(1)(t) = 0 and solve for t. The sinehas 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 whenbt = π ⇒ t = π/β. The value of the maximum is computed as

gmax = g(π

β

)= 1 + 2ρ e−πα/β cos(π + φ) = 1 +

e−πζ/√

1−ζ2√1− ζ2

cos(tan−1 ζ√

1− ζ2

).

Using the identity

cos(tan−1 x√

1− x2

)=√

1− x2 ,

the final result is

gmax = 1 + e−πζ/√

1−ζ2=⇒ Overshoot in %: 100 e−πζ/

√1−ζ2

.

The following graph shows the unit step response g(t) for several values of ζ.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

t in msec

g(t)

Unit Step Responses of Second Order LPF, ω0=3141.5927

ζ=0.2ζ=0.5ζ=0.70711ζ=1ζ=2

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Frequency Response. From the phasor analysis the system function of the LPF is obtainedas

H =VO

VS

=ω2

0

ω20 − ω2 + j 2ζω0 ω

.

The magnitude and the phase of H are

|H| = ω20√

(ω20 − ω2)2 + (2ζω0 ω)2

, and ∠H = − tan−1 2ζω0 ω

ω20 − ω2

.

Note that, at ω = ω0,

|H|ω0 =1

2ζ, and ∠Hω0 = −90◦ ,

and thus ζ and ω0 can be easily determined from the magnitude and phase of H.

2 RLC Bandpass Filter

A passive RLC bandpass filter (BPF) circuit is shown in the following schematic.

◦

◦ ◦

◦

•

•

•

•

R

L C

−

vS(t)

+

−

vO(t)

+

Using phasor analysis, vO(t)⇔ VO is computed as

VO =

jωL(jω)2LC+1

R + jωL(jω)2LC+1

VS =jω 1

RC

(jω)2 + jω 1RC

+ 1LC

VS .

Setting ω0 = 1/√

LC and 2ζω0 = 1/(RC) yields

VO =jω 2ζω0

(jω)2 + jω 2ζω0 + ω20

VS ⇐⇒ v(2)O (t)+2ζω0 v

(1)O (t)+ω2

0 vO(t) = 2ζω0 v(1)S (t) .

The blockdiagram that represents this differential equation is (after integrating both sides

so that the input is vS(t) rather than v(1)S (t))

2ζω0 +∫ ∫

ω20

•

2ζω0

+

+

−

++

∫vO(τ)dτvS(t) v

(1)O (t) vO(t)

vO(t)

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Frequency Response. From the phasor analysis the system function of the BPF is obtainedas

H =VO

VS

=j 2ζω0ω

ω20 − ω2 + j 2ζω0ω

.

The magnitude and the phase of H are

|H| = 2ζω0 ω√(ω2

0 − ω2)2 + (2ζω0 ω)2, and ∠H = π/2− tan−1 2ζω0 ω

ω20 − ω2

.

Note that, at ω = ω0 (the center frequency of the BPF),

|H|ω0 = 1 , and ∠Hω0 = 0◦ .

To obtain the lower half-power (or -3dB) frequency ω3− of the BPF, set

ω20 − ω2

3− = 2ζω0ω3− =⇒ ω3− =(√

1 + ζ2 − ζ)ω0 .

Similarly, the upper half-power (or -3dB) frequency ω3+ of the BPF is obtained from

ω23+ − ω2

0 = 2ζω0ω3+ =⇒ ω3+ =(√

1 + ζ2 + ζ)ω0 .

Thus, the half-power (or -3dB) bandwidth of the BPF is equal to 2ζω0, and ζ and ω0 cantherefore be determined from the magnitude and phase of H.

c©2003–2007, P. Mathys. Last revised: 2-12-07, PM.

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