ECEN 2260
Circuits/Electronics 2
Spring 2007
2-10-07
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
◦ +
•
vS (t)
vO (t)
C
− ◦
•
◦ −
Using phasor analysis, vO (t) ⇔ VO is computed as
VO =
1
jωC
R + jωL +
1
jωC
VS =
1
LC
(jω)2 + jω
R
L
+
1
LC
VS .
√
Setting ω0 = 1/ LC and 2ζω0 = R/L, where ω0 is the (undamped) natural frequency and
ζ is the damping ratio, yields
VO =
ω02
VS
(jω)2 + jω 2ζω0 + ω02
(2)
(1)
vO (t) + 2ζω0 vO (t) + ω02 vO (t) = ω02 vS (t) .
⇐⇒
The blockdiagram that represents this differential equation is
vS (t)
ω02
+
(2)
+
−
vO (t)
(1)
R
•
vO (t)
2ζω0
+
+
+
R
•
vO (t)
ω02
Unit Step Response. By definition, the unit step response g(t) of a circuit is the zero-state
response (ZSR) to the input vS (t) = u(t). For the 2’nd order LPF considered here the unit
step response is of the form (if ζ 6= 1)
g(t) = K1 es1 t + K2 es2 t + 1 ,
t≥0
=⇒
1
g (1) (t) = s1 K1 es1 t + s2 K2 es2 t ,
t≥0,
with initial conditions g(0) = 0 and g (1) (0) = 0. The values of s1 and s2 are the solutions of
the characteristic equation
p
s2 + 2ζω0 s + ω02 = 0
=⇒
s1,2 = − ζ ± ζ 2 − 1 ω0 ,
and the properties of g(t) change fundamentally depending on whether ζ > 1, ζ = 1, or
ζ < 1.
Overdamped Case, ζ > 1. In this case the characteristic equation has two real solutions
p
p
and
s2 = −α2 , α2 = (ζ + ζ 2 − 1) ω0 .
s1 = −α1 , α1 = (ζ − ζ 2 − 1) ω0 ,
Note that α1 < α2 . The unit step response is of the form
g(t) = K1 e−α1 t + K2 e−α2 t + 1 ,
t≥0
g (1) (t) = −α1 K1 e−α1 t − α2 K2 e−α2 t ,
=⇒
t ≥ 0.
Using initial conditions g(0) = 0 and g (1) (0) = 0 yields
"
#" # " #
K1 + K2 = −1 ,
1 1
K1
−1
=⇒
=
.
α1 K1 + α2 K2 = 0 .
α1 α2
K2
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 −
Note that
g (1) (t) =
α2 e−α1 t − α1 e−α2 t
,
α2 − α1
α1 α2
(e−α1 t − e−α2 t ) ,
α2 − α1
t≥0.
t≥0,
and therefore g (1) (t) = 0 requires that e−α1 t = e−α2 t which can only happen at t = 0 or
t = ∞ if α1 6= α2 . This implies that the extrema of g(t) occur at t = 0 (where g(0) = 0) and
at t = ∞ (where g(∞) = 1) and thus g(t) has no overshoot.
Critically Damped Case, ζ = 1. In this case the characteristic equation has one real
double 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 ,
(1)
−αt
=⇒ g (t) = −αK1 e
−αt
+ K2 e
t≥0,
− αK2 t e−αt ,
Using initial conditions g(0) = 0 and g (1) (0) = 0 yields
K1 = −1 ,
K2 = −α ,
2
t≥0.
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 that
g (1) (t) = α2 t e−αt ,
t≥0,
and therefore g (1) (t) = 0 requires either t = 0 or t = ∞, which implies that the extrema of
g(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 complex
solutions which are conjugates of each other
p
s1 = −α + jβ , and s2 = s∗1 = −α − jβ ,
where
α = ζω0 , β = 1 − ζ 2 ω0 .
The unit step response is of the form
g(t) = K1 es1 t + K2 es2 t + 1 ,
t≥0
=⇒
g (1) (t) = s1 K1 es1 t + s2 K2 es2 t ,
t≥0.
Substituting s1,2 = −α ± jβ and using initial conditions g(0) = 0 and g (1) (0) = 0 yields
"
#" # " #
K1 + K2 = −1 ,
1
1
K1
−1
=⇒
=
.
(−α+jβ)K1 − (α+jβ)K2 = 0 .
−α+jβ −α−jβ
K2
0
From this K1 and K2 = K1∗ are obtained as
K1 =
−β + jα
= ρ ejφ ,
2β
K2 =
−β − jα
= ρ e−jφ ,
2β
where
p
ρ=
α2 + β 2
1
= p
,
2β
2 1 − ζ2
φ = π − tan−1
and
α
ζ
= π − tan−1 p
.
β
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) ,
or, with ρ and φ substituted
ζ
e−αt
cos βt − tan−1 p
,
g(t) = 1 − p
1 − ζ2
1 − ζ2
t≥0.
To obtain a formula for g (1) (t) easily, first note that s2 K2 = (s1 K1 )∗ and
−β + jα
α2 + β 2
s1 K1 = (−α + jβ)
=
.
2β
2jβ
3
t≥0,
Then use g (1) (t) = s1 K1 ejβt + s2 K2 e−jβt e−αt to obtain
α2 + β 2 −αt ejβt − e−jβt α2 + β 2 −αt
g (t) =
e
=
e
sin βt ,
β
2j
β
(1)
t≥0.
To find the times where the extrema of g(t) occur, set g (1) (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
√
−πζ/ 1−ζ 2
e
ζ
π
= 1 + 2ρ e−πα/β cos(π + φ) = 1 + p
gmax = g
cos tan−1 p
.
β
1 − ζ2
1 − ζ2
Using the identity
cos tan−1 √
√
x
= 1 − x2 ,
1 − x2
the final result is
√
2
gmax = 1 + e−πζ/ 1−ζ
√
2
Overshoot in %: 100 e−πζ/ 1−ζ .
=⇒
The following graph shows the unit step response g(t) for several values of ζ.
Unit Step Responses of Second Order LPF, ω =3141.5927
0
1.6
1.4
1.2
g(t)
1
0.8
0.6
0.4
ζ=0.2
ζ=0.5
ζ=0.70711
ζ=1
ζ=2
0.2
0
0
0.5
1
1.5
2
2.5
t in msec
4
3
3.5
4
4.5
5
Frequency Response. From the phasor analysis the system function of the LPF is obtained
as
VO
ω02
.
H=
= 2
VS
ω0 − ω 2 + j 2ζω0 ω
The magnitude and the phase of H are
ω02
|H| = p
(ω02
− ω 2 )2 + (2ζω0 ω)2
,
∠H = − tan−1
and
2ζω0 ω
.
ω02 − ω 2
Note that, at ω = ω0 ,
1
,
2ζ
|H|ω0 =
∠Hω0 = −90◦ ,
and
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
+ ◦
•
vS (t)
◦ +
•
L
C
− ◦
•
•
vO (t)
◦ −
Using phasor analysis, vO (t) ⇔ VO is computed as
VO =
R
jωL
(jω)2 LC+1
+ (jω)jωL
2 LC+1
VS =
1
jω RC
1
(jω)2 + jω RC
+
1
LC
VS .
√
Setting ω0 = 1/ LC and 2ζω0 = 1/(RC) yields
VO =
(jω)2
jω 2ζω0
VS
+ jω 2ζω0 + ω02
(2)
(1)
(1)
vO (t) + 2ζω0 vO (t) + ω02 vO (t) = 2ζω0 vS (t) .
⇐⇒
The blockdiagram that represents this differential equation is (after integrating both sides
(1)
so that the input is vS (t) rather than vS (t))
vO (t)
vS (t)
2ζω0
+
(1)
+
−
vO (t)
R
•
vO (t)
2ζω0
+
+
+
5
R
R
vO (τ )dτ
ω02
Frequency Response. From the phasor analysis the system function of the BPF is obtained
as
VO
j 2ζω0 ω
H=
.
= 2
VS
ω0 − ω 2 + j 2ζω0 ω
The magnitude and the phase of H are
2ζω0 ω
|H| = p 2
,
(ω0 − ω 2 )2 + (2ζω0 ω)2
and
∠H = π/2 − tan−1
2ζω0 ω
.
ω02 − ω 2
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
p
ω02 − ω32− = 2ζω0 ω3−
=⇒
ω3− =
1 + ζ 2 − ζ ω0 .
Similarly, the upper half-power (or -3dB) frequency ω3+ of the BPF is obtained from
p
ω32+ − ω02 = 2ζω0 ω3+
1 + ζ 2 + ζ ω0 .
=⇒
ω3+ =
Thus, the half-power (or -3dB) bandwidth of the BPF is equal to 2ζω0 , and ζ and ω0 can
therefore be determined from the magnitude and phase of H.
c
2003–2007,
P. Mathys.
Last revised: 2-12-07, PM.
6