EE 380:
Linear Control
Systems
III. System
Responses
EE 380: Linear Control Systems
Time
Response
III. System Responses
Frequency
Response
Summary
Department of Electrical Engineering
Pennsylvania State University, University Park, USA
Instructor: Ji-Woong Lee
September 16, 2009
III. System Responses
EE 380:
Linear Control
Systems
III. System
Responses
1 Time Response of Systems
Time
Response
Frequency
Response
Summary
2 Frequency Response of Systems
3 Summary
III. System Responses
EE 380:
Linear Control
Systems
III. System
Responses
1 Time Response of Systems
Time
Response
Frequency
Response
Summary
2 Frequency Response of Systems
3 Summary
Time Response of Systems
Transient Response Vs. Steady-State Response
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Time Response:
Performance evaluation is based on time responses.
C (s) = G (s)R(s) ⇒ c(t) = ct (t) + css (t) , t ≥ 0
| {z }
| {z }
transient
steady-state
Transient Response ct :
limt→∞ ct (t) = 0
Transient response reflects “dynamics.”
Steady-State Response css :
Steady-state error: css (t) − r (t), large t ≥ 0
Steady-state response reflects final “accuracy.”
Performance Objective:
Make transient response and steady-state error “small.”
Time Response of Systems
Test Signals
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Unit Impulse
1
0
1
1
0
Unit Ramp
Unit Step
0
0
0
0
1
Unit Impulse R(s) = 1:
Infinite amplitude, infinitesimal width, and unit area
Impulse response = L−1 of transfer function
Unit Step R(s) = 1/s:
Jump discontinuity
Useful in evaluating “quickness” of response
Unit Ramp R(s) = 1/s 2 :
Useful in evaluating ability to follow “linear changes”
Time Response of Systems
Impulse Response
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Unit Impulse δ:
If δ is unit impulse, then
Z ∞
f (σ)δ(t − σ) dσ = f (t)
−∞
for all t and for all continuous f .
Unit Impulse Response g :
If r = δ, then R(s) = 1, so
C (s) = G (s)R(s) = G (s)
⇒
c = L−1 G (s) = g .
System Response for General Input r :
If g is unit impulse response, then G (s) = L(g ), so
Z t
C (s) = G (s)R(s) ⇒ c(t) =
g (σ)r (t − σ) dσ.
0
Time Response of Systems
Unit Step Response
EE 380:
Linear Control
Systems
III. System
Responses
Final Value Theorem
If L(f ) = F (s) and if limt→∞ f (t) exists, then
lim f (t) = lim sF (s).
t→∞
Time
Response
Frequency
Response
Summary
s→0
R∞
R∞
lim 0 e st dfdt(t) dt = 0 lim e −st dfdt(t) dt =
Proof. lim L df
dt = s→0
s→0
s→0
R ∞ df (t)
dt
=
lim
f
(t)
−
f
(0).
Thus
lim
f
(t)
= lim L df
+ f (0).
dt
dt
0
t→∞
t→∞
s→0
df
However, L dt = sF (s) − f (0), so the result follows.
System DC Gain:
If r is unit step, then R(s) = s −1 , so system dc gain (or
steady-state gain) is
lim c(t) = lim sG (s)R(s) = lim G (s) = G (0).
t→∞
s→0
s→0
Time Response of Systems
Step Response of First-Order Systems
EE 380:
Linear Control
Systems
III. System
Responses
Standard Form of First-Order Systems:
K
K /τ
G (s) =
=
τs + 1
s + 1/τ
where K is steady-state gain, τ time constant.
Time
Response
Frequency
Response
Summary
Steady-State Gain K :
If r is unit step, then R(s) = s −1 , so
K
−K
K /τ 1
=
+
s + 1/τ s
s
s + 1/τ
−t/τ
−t/τ
⇒ c(t) = |{z}
K + −Ke
, t ≥ 0.
| {z } = K 1 − e
C (s) = G (s)R(s) =
=css (t)
=ct (t)
Time Constant τ :
If steady-state gain K = 1, then initial decay rate of ct is
dct (t) d
1
=
− e −t/τ = .
dt t=0 dt
τ
t=0
Time Response of Systems
Step Response of First-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Step Response of 1st−Order System
K
0
τ
c(t) = K (1 − e −t/τ )
t
e −t/τ
0
τ
2τ
3τ
4τ
1
0.3679
0.1353
0.0498
0.0183
Time Constant τ (Cont’d):
Consider e −t/τ ≈ 0 ( within 2% ) for t ≥ 4τ .
Compare decay of terms in partial-fraction expansion:
K2
K1
G (s) =
+
+ ···
τ1 s + 1 τ2 s + 1
Example:
5
2.5
G (s) =
=
⇒ c(t) ≈ 2.5 for t ≥ 4τ = 2 sec
s +2
0.5s + 1
Time Response of Systems
Step Response of Second-Order Systems
EE 380:
Linear Control
Systems
Standard Form of Second-Order Systems:
G (s) =
III. System
Responses
Time
Response
Frequency
Response
Summary
K ωn2
s 2 + 2ζωn s + ωn2
where ζ is damping ratio, ωn natural (or undamped) frequency,
and K dc (or steady-state) gain.
(ωn ≥ 0)
Transient and Steady-State Step Responses:
If r is unit step, so that C (s) = G (s)R(s) = G (s)s −1 , then
C (s) =
K ωn2
=
s(s 2 + 2ζωn s + ωn2 )
K
s
|{z}
steady-state
+
−K (s + 2ζωn )
.
s 2 + 2ζωn s + ωn2
|
{z
}
transient
Two Poles of Transfer Function G (s):
Characteristic equation s 2 + 2ζωn s + ωn2 = 0 has two roots
p
s = −ζωn ± jωn 1 − ζ 2 .
Time Response of Systems
Step Response of Second-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
Root Locus of Characteristic Equation s 2 + 2ζωn s + ωn2 = 0:
III. System
Responses
ζ=0
0<ζ<1
Time
Response
Frequency
Response
Summary
ωn
ζωn
ωn
p
1 − ζ2
ζ>1
ζ>1
0
0
ζ=1
ζ=0
ζ = −1
ζ < −1 ζ < −1
0 > ζ > −1
Natural Undamped Frequency ωn :
Under zero damping (ζ = 0), roots of characteristic equation
are imaginary
⇒ Step response is purely sinusoidal with frequency ωn .
Time Response of Systems
Step Response of Second-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Damping Ratio ζ:
(Assume K = 1)
ζ > 1 (Overdamped):
c(t) = 1 + k1 e −t/τ1 + k2 e −t/τ2 , t ≥ 0,
1√
1√
where τ1 =
and τ2 =
2
2
ζωn +ωn
ζ −1
ζ = 1 (Critically damped):
ζωn −ωn
c(t) = 1 + (k1 + k2 t)e −t/τ ,
where τ = 1/ωn
ζ −1
t ≥ 0,
0 < ζ < 1 (Underdamped):
1
c(t) = 1 − e −ζωn t sin(βωn t + θ), t ≥ 0,
β
p
2
where β = 1 − ζ and θ = tan−1 (β/ζ)
ζ = 0 (Undamped): c(t) = 1 − cos(ωn t), t ≥ 0
ζ < 0 (Negatively damped): c(t) grows without bound.
Time Response of Systems
Step Response of Second-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
Classification of 2nd-Order System Dynamics:
ζ>1
ζ=1
0<ζ<1
III. System
Responses
1
1
1
Time
Response
Frequency
Response
0
0
Summary
1
0
−1 < ζ < 0
ζ=0
ζ < −1
1
0
1
0
0
ζ ≥ 1: No overshoot (i.e., output never exceeds reference)
ζ > 0: Stable (i.e., output eventually reaches reference)
Time Response of Systems
Step Response of Second-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
Pole Locations of 2nd-Order Transfer Function:
ζ>1
ζ=1
jω
jω
III. System
Responses
0
Time
Response
Frequency
Response
Summary
σ
jω
σ
−1 < ζ < 0
ζ=0
jω
0
0
0<ζ<1
0
σ
ζ < −1
jω
jω
σ
0
σ
0
σ
ζ ≥ 1: No overshoot (i.e., output never exceeds reference)
ζ > 0: Stable (i.e., output eventually reaches reference)
Time Response of Systems
Time Response Specifications
EE 380:
Linear Control
Systems
III. System
Responses
Typical Step Response
Mpt
1+d
css
1.0
0.9
Time
Response
1−d
Frequency
Response
Summary
0.1
0
Tr
Tp
Ts
Time Constant τ : In the case of 2nd-order step response,

1√

Overdamped (ζ > 1):
τ=


ζωn ±ωn ζ 2 −1
Critically damped (ζ = 1):


Underdamped (0 < ζ < 1):
τ=
τ=
1
ωn
1
ζωn
Rise Time Tr : Time to rise from 10% to 90% of final value css
Time Response of Systems
Time Response Specifications (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Peak Time Tp : Time to reach peak value Mpt
(e.g., First nonzero time dc(t)
dt becomes zero)
In the case of 2nd-order step response,
h
i
dc(t)
d
1 −ζωn t
=
1
−
e
sin(βω
t
+
θ)
n
dt
dt
β
ζωn −ζωn t
n −ζωn t
sin(βωn t + θ) − βω
cos(βωn t +
β e
β e
n −ζωn t
= ζω
[sin(βωn t) cos θ + cos(βωn t) sin θ]
β e
n −ζωn t
− βω
[cos(βωn t) cos θ − sin(βωn t) sin θ] .
β e
p
cos θ = ζ, sin θ = β, and β = 1 − ζ 2 to obtain
=
Frequency
Response
Summary
Use
dc(t)
dt
=
ωn −ζωn t
β e
sin(βωn t).
Putting βωn Tp = π yields
Tp =
π
π
= p
.
ωn β
ωn 1 − ζ 2
θ)
Time Response of Systems
Time Response Specifications (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Peak Response Mpt :
(Assume: K = 1)
In the case of 2nd-order step response,
sin(βωn Tp + θ) = sin(π + θ) = − sin θ = −β
⇒ Mpt = c(Tp ) = 1 − β1 e −ζωn Tp sin(βωn Tp + θ)
√
−ζπ/ 1−ζ 2
=1+e
.
Percent Overshoot:
Mpt − css
× 100.
css
In the case of 2nd-order step response,√
2
percent overshoot = e −ζπ/ 1−ζ × 100.
Settling Time Ts : Time to settle within 2% of css .
Approximately four time constants are required for 2nd-order
step response c(t) = 1 − β1 e −ζωn t sin(βωn t + θ) to settle to
4
within 2% of css , so
Ts = 4τ =
.
ζωn
percent overshoot =
Time Response of Systems
Example
EE 380:
Linear Control
Systems
Closed-Loop System:
e
+
Time
Response
r
Gc
u
Gp (s) =
Gp
y
{
Frequency
Response
Summary
Motor
Amplifier
III. System
Responses
⇒
0.5
s(s+2)
Gc (s) = Ka
Gp (s)Gc (s)
Y (s)
0.5Ka
=
= 2
R(s)
1 + Gp (s)Gc (s)
s + 2s + 0.5Ka
Control Objective:
Fastest step response (i.e., smallest Ts ) without overshoot
Compensator Design: Choose critical damping ζ = 1.
⇒ s 2 + 2s + 0.5Ka = s 2 + 2ζωn s + ωn2 = s 2 + 2ωn s + ωn2
⇒ 2ωn = 2 and ωn2 = 0.5Ka ⇒ ωn = 1 and hence Ka = 2.
Achieved Settling Time: Ts = 4τ =
4
ωn
= 4 sec
Time Response of Systems
Time Response and Pole Locations
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Settling Time Specification:
For
( 2nd-order systems,
Characteristic equation: s 2 + 2ζωn s + ωn2 = 0
p
Pole locations:
s = −ζωn ± jωn 1 − ζ 2
4
4
⇒ Ts =
=
ζωn
− real(s)
Summary
jω
−ζωn
0
−4/Ts,max
Speed of response is
increased by moving
poles to left in s-plane.
σ
If Ts ≤ Ts,max is
required, then place
poles s.t.
ζωn ≥ 4/Ts,max .
Time Response of Systems
Time Response and Pole Locations (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Overshoot Specification:
p
For 2nd-order systems with poles s = −ζωn ± jωn 1 − ζ 2 , let
p
1 − ζ2
−1 imaginery(s) −1
α = tan =
tan
real(s) ζ
√ 2
⇒ percent overshoot = e −ζπ/ 1−ζ × 100 = e −π/ tan α × 100
Summary
jω
p
jωn 1 − ζ 2
αmax
−ζωn
0
σ
Decreasing angle α
reduces percent
overshoot.
If α ≤ αmax is
required, then place
poles s.t.
ζ ≥ √ 12
.
1+tan αmax
Time Response of Systems
Example: RLC Circuit
EE 380:
Linear Control
Systems
III. System
Responses
Closed-Loop System:
L
+
r(t)
Time
Response
Frequency
Response
Summary
{
+
C
R
y(t)
{
Gp (s) =
VC (s)
IL (s)
=
1/C
s+1/RC
Gc (s) =
IL (s)
VL (s)
=
1/L
s
Y (s)
Gp (s)Gc (s)
1/LC
K ωn2
=
= 2
= 2
R(s)
1 + Gp (s)Gc (s)
s + (1/RC )s + 1/LC
s + 2ζωn s + ωn2
Design Specifications:
Percent overshoot no more than e −π × 100% = 4.32%
Settling time no more than 2 sec
Goals:
Given R choose C that meets specifications.
Determine range of L that meets specifications.
Time Response of Systems
Example: RLC Circuit (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Choice of Capacitance C :
p
System poles s = −ζωn ± jωn 1 − ζ 2 ⇒
Settling time Ts = ζω4 n = 8RC ≤ Ts,max = 2
2ζωn = 1/RC
1
⇒ C ≤ 4R
Range of Inductance L:
% overshoot e −π/ tan α × 100 ≤ e −π × 100 ⇒ αmax = 45◦
Damping ratio ζ ≥ √ 1 2
= √12 = 0.707 (common spec)
Natural freq ωn =
1+tan αmax
1
√1 =
√1
2ζRC ≤ 2(1/ 2)RC
LC
⇒
L ≥ 2R 2 C
jω
j2
−2
0
−j2
σ
Ts,max = 2 is met if
4
real(s) ≤ − Ts,max
= −2.
Percent overshoot ≤ 4.32%
if ζ ≥ √12 , which holds if
α ≤ 45◦ .
Time Response of Systems
Step Response of Higher-Order Systems
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
A Third-Order System:
K ωn2 /τ
(s + 1/τ )(s 2 + 2ζωn s + ωn2 )
K
K2 ωn2
K3 ωn2 s
+ 2
+ 2
=
2
τ s + 1 s + 2ζωn s + ωn
s + 2ζωn s + ωn2
G (s) =
3rd-Order Step Response: c(t) = c1 (t) + c2 (t) + c3 (t) where
c1 (t): Step response of a 1st-order system
c2 (t): Step response of a standard 2nd-order system
c3 (t): Derivative of 2nd-order step response
Estimate of Settling Time for 3rd-Order Responses:
Ts ≈ 4 max {τ, 1/ζωn }
Estimate of Percent Overshoot for 3rd-Order Responses:
√
2
Percent overshoot ≈ e −ζπ/ 1−ζ × 100
Time Response of Systems
System Response to Faster Test Signals
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Unit Ramp Response of First-Order Systems:
If r is unit ramp, then R(s) = 1/s 2 , so
K /τ 1
Kτ
K
−K τ
C (s) = G (s)R(s) =
+
= 2+
2
s + 1/τ s
s
s
s + 1/τ
⇒
−t/τ
c(t) = |Kt −
{z K τ} + |K τ e{z }, t ≥ 0
=css (t)
=ct (t)
If K = 1, then css (t) = t − τ
⇒ τ = time constant for step response
= steady-state delay in ramp response
To track fast input, system is necessarily of high order.
Relation Between Test Signals:
Unit impulse response: Derivative of unit step response
Unit step response: Derivative of unit ramp response
III. System Responses
EE 380:
Linear Control
Systems
III. System
Responses
1 Time Response of Systems
Time
Response
Frequency
Response
Summary
2 Frequency Response of Systems
3 Summary
Frequency Response of Systems
Complex Exponential Function
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Representations of Complex Numbers
(a) If x, y ∈ R, then x + jy = Re jφ = R 6 φ with

p
Magnitude: R = 
|x + jy | = x 2 + y 2 ,






if

tan−1 |y /x|




π − tan−1 |y /x|
if

Angle:
φ=

−1


− tan |y /x|
if







−π + tan−1 |y /x| if
x, y ≥ 0;
x < 0 ≤ y;
y < 0 ≤ x;
x, y < 0
(b) If R, φ ∈ R, then Re jφ = x + jy with
Real part: x = R cos φ;
Imaginary part: y = R sin φ.
p
p
p
2 x/ x 2 + y 2 + jy / x 2 + y 2 =
Proof.
(a) x + jy = x 2 + yp
p
x 2 + y 2 (cos φ + j sin φ) = x 2 + y 2 e jφ . (b) Re Re jφ = Re jφ
+Re −jφ /2 = R cos φ; Im Re jφ = Re jφ − Re −jφ /2j = R sin φ.
Frequency Response of Systems
Sinusoidal Response
EE 380:
Linear Control
Systems
Sinusoidal Test Signal:
r (t) = A cos ω1 t, t ≥ 0
III. System
Responses
Time
Response
R(s) =
s2
As
+ ω12
Sinusoidal Response of Linear System:
If r is sinusoidal, and if C (s) = G (s)R(s), then
C (s) = G (s)
Frequency
Response
Summary
⇔
As
k1
k2
=
+
+ Cg (s)
(s − jω1 )(s + jω1 )
s − jω1 s + jω1
Assumption: limt→∞ cg (t) = 0 (i.e., lims→0 sCg (s) = 0)
⇒
k1 =
A
2 G (jω1 ),
k2 = k1∗ =
A
2 G (−jω1 )
Sinusoidal Steady-State Response:
If G (jω1 ) = Re jφ and G (−jω1 ) = Re −jφ , then
css (t) = k1 e jω1 t + k2 e −jω1 t =
= AR cos(ω1 t + φ).
AR j(ω1 t+φ)
e
+ e −j(ω1 t+φ)
2
Frequency Response of Systems
Sinusoidal Response (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Frequency Response of Linear Systems
If C (s) = G (s)R(s), then G (jω), as a function of ω ∈ [0, ∞),
is called the frequency response function. For r (t) = A cos ω1 t,
t ≥ 0, the steady-state response is given by
css (t) = AR cos(ω1 t + φ),
where steady-state gain R and phase shift φ satisfy
Summary
G (jω1 ) = Re jφ .
Example: G (s) =
5
s+2
and r (t) = 7 cos 3t, t ≥ 0.
5(2−j3)
5
10
15
6 φ
G (j3) = 2+j3
= (2+j3)(2−j3)
= 10−j15
22 +32 = 13 − j 13 = R
q 2
2
10
R=
+ 15
= 1.387; φ = tan−1 −15/13
× 180
13
13
10/13
π
⇒
⇒
= −56.3◦
css (t) = 7R cos(3t + φ) = 9.707 cos(3t − 56.3◦ )
c(t) ≈ css (t) after 4τ = 2 sec.
Frequency Response of Systems
Frequency Response of First-Order Systems
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Standard Form of First-Order Systems:
K
G (s) =
τs + 1
Frequency Response of Standard 1st-Order Systems:
K
K (1 − jτ ω)
G (jω) =
=
= |G (jω)|e jφ(ω) ,
1 + jτ ω
1 + τ 2 ω2
where
K |1 − jτ ω|
K
|G (jω)| =
=√
,
2
2
1+τ ω
1 + τ 2 ω2
φ(ω) = tan−1 (−τ ω) = − tan−1 τ ω
System Bandwidth ωB :
√
Frequency at which gain is 1/ 2 times gain at zero frequency.
1
|G (jωB )| = √ |G (j0)|
2
K
K
1
⇒ √
= √ = 0.707K ⇒ ωB =
2
2
τ
2
1+τ ω
Frequency Response of Systems
Frequency Response of First-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Typical Frequency Response of 1st-Order Systems:
Magnitude of G(jω)
Phase Angle of G(jω)
0
K
0.707K
−45
Frequency
Response
Summary
0
−90
ωB
ω
Two plots completely describe frequency response.
Here, frequency ω (rad/sec) is in “log” scale.
Phase angles are in degrees (radians ×180/π).
τ = time constant for 1st-order step response
= steady-state delay in 1st-order ramp response
= inverse of 1st-order system bandwidth
ω
Frequency Response of Systems
Frequency Response of Second-Order Systems
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Standard Form of Second-Order Systems:
K ωn2
K
G (s) = 2
=
2
2
s + 2ζωn s + ωn
(s/ωn ) + 2ζ(s/ωn ) + 1
Frequency Response of Standard 2nd-Order Systems:
K
G (jω) =
= |G (jω)|e jφ(ω) ,
1 − (ω/ωn )2 + j2ζ(ω/ωn )
where
K
|G (jω)| = q
2 ,
2 1 − (ω/ωn )2 + 2ζ(ω/ωn )

ω
− tan−1 2ζ(ω/ωn )2 ,
ωn < 1;
1−(ω/ω
n)
φ(ω) =
−π + tan−1 2ζ(ω/ωn )2 , ω > 1.
ωn
1−(ω/ωn )
System Bandwidth ωB :
2 2
|G (jωB )| = √12 |G (j0)| ⇒ 1−(ωB /ωn )2 + 2ζ(ωB /ωn ) = 2
Frequency Response of Systems
Frequency Response of Second-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
Typical Frequency Response of 2nd-Order Systems:
III. System
Responses
Magnitude of G(jω)
ζ
ζ
ζ
ζ
Time
Response
Frequency
Response
K
Phase Angle of G(jω)
= 0.1
= 0.25
= 0.5
= 0.707
0
−90
ζ = 0.1
Summary
ζ=1
0
ζ=1
ωB /ωn
2
−180
0
ω/ωn
1
2
ω/ωn
Here, normalized frequency ω/ωn is in linear scale.
Given a ζ, increasing ωn increases bandwidth ωB .
We know, given a ζ, increasing ωn decreases peak time Tp .
⇒ Increasing ωn also decreases Tr , as well as Ts .
Frequency Response of Systems
Frequency Response of Second-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Bandwidth–Rise Time Product ωB Tr :
1.6 ≤ ωB Tr ≤ 2.2
for ζ ≥ 0
If ζ = 0, then c(t) = K (1 − cos ωn t) and
2 2 2
1 − (ωB /ωn )2 + 2ζ(ωB /ωn ) = 1 − (ωB /ωn )2 = 2.
Tr = t2 − t1pwhere c(t1 ) = 0.1K and c(t2 ) = 0.9K .
√
⇒ ωB Tr = 1 + 2ωn (t2 − t1 ) = 1.6.
If ζ ≫ 1, then c(t) = K + K1 e −t/τ1 + K2 e −t/τ2 with
1√
1√
τ1 =
and τ2 =
.
2
2
ζωn +ωn
ζ −1
e −t/τ2 )
⇒ c(t) ≈ K (1 −
⇒ ωB Tr ≈ ln 9 = 2.2.
ζωn −ωn
for ζ ≫ 1.
ωB Tr ≈ constant ≈ 2 for all ζ ≥ 0
ζ −1
Frequency Response of Systems
Frequency Response of Second-Order Systems (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Peak Frequency ωr :
|G (jω)| = q
1 − (ω/ωn )2
K
2
2
+ 2ζ(ω/ωn )
= 0 yields (ω/ωn ) (ω/ωn )2 − (1 − 2ζ 2 ) = 0.
( p
√
ωn 1 − 2ζ 2 , ζ < 1/ 2;
√
⇒ ωr = peak frequency =
0,
ζ ≥ 1/ 2.
d
d(ω/ωn ) |G (jω)|
Peak Frequency Response Mpω :

√
 √K
,
ζ
<
1/
2;
2
Mpω = |G (jωr )| = 2ζ 1−ζ
√
K ,
ζ ≥ 1/ 2.
Resonance: Peaking in frequency response
ωr = peak frequency = resonance frequency
Peak frequency response is directly related to overshoot in
step response.
Frequency Response of Systems
Frequency Response of Higher-Order Systems
EE 380:
Linear Control
Systems
III. System
Responses
Bandwidth–Rise Time Product:
ωB Tr ≈ constant
Time
Response
To increase speed of response, it is necessary to increase
system bandwidth.
Frequency
Response
How to increase ωB will be covered in later chapters.
Summary
Peak Frequency Response:
The higher peak frequency response, the more overshoot
in step response.
A control system should not have significant resonance.
A common performance specification is given by maximum
allowable Mpω .
Frequency Response of Systems
Example: A Third-Order System
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Transfer Function:
G (s) =
s3
1.524
s + 2.5
1.524
=
s + 2.5
=
8
8
=
+ 9s + 10
(s + 2.5)(s 2 + 2s + 4)
1.008e −j2.428
1.008e j2.428
√ +
√
+
s +1−j 3
s +1+j 3
−1.524s + 0.762
+
s 2 + 2s + 4
+ 4.5s 2
1st-Order Term: Time constant τ1 = 0.4
2nd-Order Term: Natural freq ωn = 2, damping ratio
ζ = 0.2, time constant τ2 = 1/ζωn = 1
Settling Time:
Estimate: Ts ≈ 4 max{τ1 , τ2 } = 4τ2 = 4 sec
Actual value (obtained via simulation): Ts = 3.2 sec
Frequency Response of Systems
Example: A Third-Order System (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Percent Overshoot:
√
2
Estimate: Percent overshoot ≈ e −ζπ/ 1−ζ × 100 = 16%
Actual value (obtained via simulation):
Percent overshoot = 11%
Bandwidth–Rise Time Product:
Actual bandwidth (obtained via simulation): ωB = 2.13
rad/sec
Actual rise time (obtained via simulation): Tr = 1.05 sec
Actual bandwidth–rise time product: ωB Tr = 2.24
Rise Time Specification:
Estimated bandwidth–rise time product: ωB Tr ≈ 2
If Tr ≤ 0.5 is required, then achieve ωB ≥ 4.
Frequency Response of Systems
Example: A Third-Order System (Cont’d)
EE 380:
Linear Control
Systems
G (s) As Parallel Connection of Two Subsystems:
III. System
Responses
G1(s)
1.524
s + 2.5
−1.524s + 0.762
G2 (s) =
s 2 + 2s + 4
G1 (s) =
+
R(s)
C(s)
Time
Response
G2(s)
+
Frequency
Response
Summary
Frequency and Time Responses:
Step Response of G(s)
Frequency Response |G(jω)|
0.8
0.8
0.566
0
ωB
ω
0
0
2
4
6
8
t
Frequency Response of Systems
Time and Frequency Scaling
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Time Scaling: Scaling factor σ > 0 gives scaled time tν by
t = σtν .
Basic Effects of Time Scaling:
1 Time scaling results in scaled time constant τν = τ /σ.
2 Time scaling results in frequency scaling ων = σω.
Example: For damped sinusoidal signal
c(t) = Ke −t/τ cos(ωt + φ),
the scaled signal is
t ≥ 0,
cs (tν ) = c(t)|t=σtν = c(σtν )
= Ke −σtν /τ cos(ωσtν + φ) = Ke −tν /τν cos(ων tν + φ).
Noneffects of Time Scaling:
Time scaling does not affect amplitude or phase angle.
Time scaling does not affect damping ratio.
Frequency Response of Systems
Time and Frequency Scaling (Cont’d)
EE 380:
Linear Control
Systems
III. System
Responses
Effect of Time Scaling on Derivatives:
d k c(t) d k c(σtν )
1 d k cs (tν )
=
=
dt k t=σtν
d(σtν )k
σ k dtνk
⇒
s k c = σ −k sνk cs
Effect of Time Scaling on Transfer Functions:
Time
Response
Frequency
Response
Summary
C (s)
bm s m + bm−1 s m−1 + · · · + b1 s + b0
=
R(s)
s n + an−1 s n−1 + · · · + a1 s + a0
The scaled transfer function is
Cs (sν )
bm σ −m sνm + bm−1 σ −(m−1) sνm−1 + · · · + b1 σ −1 sν + b0
=
Rs (sν )
σ −n sνn + an−1 σ −n−1 sνn−1 + · · · + a1 σ −1 sν + a0
σ n−m bm sνm + σ n−m+1 bm−1 sνm−1 + · · · + σ n−1 b1 sν + σ n b0
=
sνn + σan−1 sνn−1 + · · · + σ n−1 a1 sν + σ n a0
The effect of time scaling t = σtν is to replace ai , bi with
σ n−i ai , σ n−i bi , respectively.
Gs (jων ) = Cs (jων )/Rs (jων ) = G (jων /σ) with ων = σω.
III. System Responses
EE 380:
Linear Control
Systems
III. System
Responses
1 Time Response of Systems
Time
Response
Frequency
Response
Summary
2 Frequency Response of Systems
3 Summary
System Responses
Summary
EE 380:
Linear Control
Systems
III. System
Responses
Time
Response
Frequency
Response
Summary
Time response
≈ Zero-state response to step input
= Step response under zero initial conditions
(Transient response plus steady-state response)
= Sum of lower-order step responses
Frequency response
= Zero-state steady-state responses to sinusoidal inputs
(Steady-state response over all frequencies)
Performance specifications:
Time domain: Settling time, percent overshoot, rise time,
steady-state error, etc.
Freq domain: Bandwidth, peak frequency response, etc.
Reduced-Order Models:
Q: Given a high-order model, how do you obtain a
low-order (e.g., 1st- or 2nd-order) approximation of it?
A: Disregard terms with relatively small time constants.