Chapter 4 : System Response
Objectives
✓ Systems and System Models
✓ Dynamic Responses of Systems
✓ Frequency Responses of Systems
4. 1 Systems & System Models
✓ What is a System?
Any physical device that responds to a stimulus or force.
✓ Examples
• Electrical systems
• Mechanical systems (translation/rotation motion)
• Hydraulic systems
• Thermal systems, or
• Combination of the above: Mechatronics systems
2
Electrical Systems
✓ Building Blocks or System Elements
VR
• Resistor R
VR = iR R
R
VR2
▪ Dissipate power PR: PR = iR R =
R
iR
Vc
• Capacitor C
dVc i
= ;
dt
C
C
t
1
Ec =  Pc dt = CVC2
2
0
ic
▪ Store energy EC in the form of electrical field
• Inductor L
VL
t
di
1
V L= L L ; EL =  PL dt = Li L2
dt
2
0
▪ Store energy EL in the form of magnetic field
L
iL
3
✓ Construction of Electrical Systems:
• Consists of a structure of a number of R, C and L.
• As a simple example: RC or RLC circuits
RC Systems
R
• Input Vin & output Vc
• By the KVL law
Vin
ic
C
dVc
Vin = RC
+ Vc
dt
• RC system model: a first order linear differential equation
• The output, change of output and input are related in the system
model.
4
Vc
RLC Systems
R
• Input Vin & output Vc
L
• By KVL law
ic
2
Vin
dVc
d Vc
Vin = RC
+ LC 2 + VC
dt
dt
Vc
C
• RLC system model:
A second-order linear differential equation
5
✓ Why do we need those system models?
• To study the system response to a stimulus
• Example
▪ Apply a DC voltage V at some time to study the output and change of
output of the system (Dynamic Response) to this stimulus, or
▪ Apply a AC voltage of different frequency V = Asin(ωt), to predict the
change of system’s output due to the frequency change of the input
(Frequency Response)
6
Mechanical Systems
✓ Building Blocks or System Elements
• Spring stiffness k
Applied Force F
▪ Input-output relationship
F = kx
Change in length x
▪ Stored elastic energy
1 F2 1 2
E=
= kx
2 k
2
Input F
Output x
Spring
7
• Mass m
▪ Newton’s second law
Input F
dx 2
F = ma = m 2
dt
Output x
Mass
▪ Stored kinematic energy
1
1  dx 
2
Em = mv = m 
2
2  dt 
2
▪ Analogy to electrical system:
➢ Mass m ~ inductance L
➢ Speed dx/dt ~ current i
8
• Damper b
▪ Represent damping force that tries to slow down a moving object
▪ Resistive force examples:
Force F
➢ Frictional force
➢ Viscosity force
➢ All proportional to but against speed
▪ Input Force F and output displacement x:
Change in
position x
dx
F =b
dt
▪ Dissipate power Pb : Pb = b dx 
 dt 
▪ Analogy to electrical system:
➢ Damper b ~ resistor R
2
Output x
Input F
Damper
✓ Building Blocks for Rotational Systems
• Torsional spring k
T = k ,
▪ Stored elastic energy Ek :
• Rotary damper b
1 T2
Ek =
2 k
T = b = b
▪ Dissipated power Pb :
• Moment of inertia J
d
dt
Pb = b 2
d
d 2
T = J = J
=J 2
dt
dt
▪ Stored kinematic energy Ej: E j =
1 2
J
2
10
✓ Construction
• Consist of a combination of any
number of spring k, mass m and
damper b.
Mass of the Vehicle
b
• Example: a vehicle
k
Suspension
Mass of the suspension
Tyre
Applied Force F
(bumping road)
11
A Spring-mass-damper system
k
Applied Force F
b
m
➢ Free-body diagram of mass m:
Displacement x
kx
F
➢ Newton’s second law:
bv
d 2x
dx
m 2 + b + kx = F
dt
dt
➢ System model: 2nd order differential equation
12
Systems and System Models
✓ More complex systems
• Using free-body diagram to build a system model for each mass element
▪ Obtain a set of differential equations for the whole system
• For non-linear systems, using linearization procedures to convert a nonlinear system to a linear system around the points where the system
operates.
▪ This course will not cover the topic of linearization
✓ With system models, we may systematically study
• Dynamic Responses of systems
• Frequency Responses of Systems
13
4.2 Dynamic Responses of Systems
Dynamic Responses of Systems:
✓ To predict what the output will be for a particular input
✓ To predict how the output will change with time if the input changes
with time
✓ Differential equations are used as mathematical tools
14
Dynamic Response
A general linear system model is of the form
• Differential equations
M
d n X out
d m X in
An
=  Bm

n
m
dt
dt
n =0
m =0
N
• Order of the system: N
• Will study systems with N = 0, 1, 2, i.e. zero-order, first-order
and second-order systems (assume all cases M = 0)
15
Dynamic Response: Zero-order System
Zero-order system: When N = 0
A0 X out = B0 X in
B0
X out =
X in = KX in
A0
• Output Xout is proportional to the input Xin , K is the gain of the system.
Ideal measurement system: zero-order system
• Example: Potentiometer
Rx
Vs
Vout =
Vs =
X in
Rp
L
▪ L is the maximum amount of wiper travel
▪ Xin the displacement of the potentiometer wiper
• Other examples
▪
▪
Incremental Encoder: number of pulses proportional to speed
16
LVDT: induced voltage magnitude is proportional to the linear displacement that the iron core
travels
Dynamic Response: First-order System
First-order system: N = 1
A1
dX out
+ A0 X out = B0 X in
dt
• Static sensitivity
B0
K =
A0
• Time constant
A1
=
A0
or
B
A1 dX out
+ X out = 0 X in
A0 dt
A0
• System model can be rewritten as
dX out

+ X out = KX in
dt
17
First-order system response to an step input
• Homogeneous solution or the transient solution
▪ of the form Cest,
▪ the characteristic equation
s + 1 = 0;
s=-
1
X outh = Ce −t /

• Particular or steady state solution has the form as the particular input
 0, t  0
X in = 
 Ain , t  0
X out p = KAin (t  0)
• The general solution is the sum of the homogeneous and particular
solutions
t
−
X out = Ce  + KAin
18
1-order System Response to an Step-input
• From the zero initial condition of the system output
X out (0) = 0
• The constant C can be determined C = − KAin
• So, the system step-response is
−
t
X out (t ) = KAin (1 − e  )
19
Graphical display
Vin
• Step input
• System response
t
−
t
Vout (t ) = KVin (1 − e  )
▪ It takes some time (4τ) for the
system to reach the steady value
KVin
Vout(t)
KVin
98.2%
95%
86%
Vout
63%
▪ Time constant τ: measure how fast
the system responses to the input
Vin
 2 3 4
t
Note : e −1 = 0.37
20
R=1kΩ
Example: 1st order system
• RC System
Vin=5V
dVc
Vin = RC
+ Vc
dt
Vc
C=0.1µF
• Compare equation with the general form of 1st order differential
equation
dX out

+ X out = KX in
dt
▪ Static gain K=1
▪ Time constant τ = RC = 1k×0.1µ= 0.1ms
▪ System output (voltage cross C) for an step input (a DC 5V) is
Vc (t ) = Vin (1 − e
−
t
RC
)
t
−

−4 
1

10

Vc (t ) = 51 − e



21
Example cont.
• It takes around 4τ, or 0.4ms for the capacitor to be charged up to 4.91V.
• However if the capacitance is 10 times bigger, say 1µF, then time
constant τ =1k×1µ=1ms, or it takes 4ms for the capacitor to be charged
up to 4.91V.
Vin
VC
small RC,  = RC
large RC,  = RC
t
22
Dynamic Responses: Second-order Systems
• A general 2nd-order differential equation:
d 2 X out
dX out
A2
+
A
+ A0 X out = X in
1
2
dt
dt
• Example:
▪ A spring-mass-damping system
d 2x
dx
m 2 + b + kx = F
dt
dt
▪ An RLC system
d 2Vc
dVc
LC
+ RC
+ VC = Vin
dt
dt
Consider a spring-mass-damping
system
b
b
k
k
• At t = 0 → Static balanced
• Apply a force by a hook of 8N
weight, x(t) = ?
d 2x
dx
m 2 + b + kx = F
dt
dt
m
x(t)
m
t=0
F=8N
Free-body
diagram
kx
bv
m
F=8N
24
• The displacement x(t) consists of two components:
▪ Unforced (or transient) response and
▪ A typical response due to the typical force (in this case a hook)
Unforced (or transient) Response: of the form Cest
• Homogeneous equation
d 2x
dx
m 2 +b
+ kx = 0
dt
dt
• Characteristic equation
• Two roots:
ms 2 + bs + k = 0
− b  b 2 − 4mk
s1, 2 =
2m
• The transient response depends on the radicand: zero, positive, or
negative.
25
• Case 1: Without damping, b = 0 → double imaginary roots
s1, 2 =  j
k
m
 k 
 k 
xh (t ) = A cos
t  + B sin 
t 
 m 
 m 
▪ Natural frequency or resonant frequency n =
k
m
▪ Undamped system: Any disturbance to the system will cause it to
oscillate forever at the nature frequency with the same amplitude as the
disturbance, even if no external forces exert on the system (F = 0)
26
▪ Physical meaning
➢ Infinite conversion cycle between elastic energy stored in spring k
and kinematic energy stored in mass m
➢ No energy dissipation: up-and-down oscillation
k
k
x(t)
m
0
m
Disturbance:
t
F(t = 0) = 1; F(t ≠ 0) = 0
27
➢ For the case of a vehicle with zero damping
Magnitude of bump
Vertical displacement x(t)
t
A single bump
on the road
If car has zero damping, b=0
Then uncomfortable riding: the car
bumps up and down forever (energy
will not dissipate away with zero
damping)
28
• Case 2: Critical damping
▪ With the damping b equals the critical damping constant bc
b = bc = 4mk
▪ The radicand = 0, double real roots s1,2= ωn
▪ Transient homogeneous solution
xh (t ) = ( A + Bt)e −nt
▪ Critically damped system
▪ System energy stored in the spring and mass will gradually dissipate.
▪ Any disturbance to the system will be exponentially die out.
29
▪ Both elastic energy and kinematic energy are reducing with time
b
b
b
k
k
k
m
m
xh(t)=0 after a while
m
Pull x(t)
xh (t)
0
t
30
• Case 3: Under damping
▪ With the damping b smaller than the critical damping constant or
damping ratio ξ smaller than 1,
 =
b
1
bc
▪ Negative radicand gives two complex-conjugate roots
s1, 2 = −n  jn 1 −  2
▪ Transient homogeneous solution
xh (t ) = e −nt
A cos( 1 −  t ) + B sin( 1 −  t )
2
n
2
n
▪ Under-damped system: Any disturbance to the system will cause an
oscillation with exponentially decaying amplitude
31
• Case 4: Over damping
▪ With the damping b bigger than the critical damping constant bc or
damping ratio ξ bigger than 1
▪ Positive radicand gives two distinct real roots
▪ Transient homogeneous solution
xh (t ) = Ae
 − +  2 −1  t

 n
+ Be
 − −  2 −1   t

 n
▪ Over-damped system: Any disturbance to the system will die out very
fast (faster than the critical damping case).
32
• Unforced (transient) response
Nodamping
Dynamic Responses: 2nd-order Systems
Forced Response
• Exterior force (step input):
0 t  0
F =
 F0 t  0
• Particular solution or the steady state solution
F0
x p (t ) =
k
• General solution: sum of homogeneous transient solution and the
particular solution due to the force.
34
x(t ) = xh (t ) + x p (t )

 k 
 k 
t  + B sin 
t ;
b=0
 A cos

 m 
 m 


e −nt A cos( 1 −  2 t ) + B sin(  1 −  2 t ) ; b  b
n
n
c

F 
= 0 +
k 
− n t
(
A
+
Bt
)
e
; b = bc


 
 − −  2 −1   t
− +  2 −1  n t

 Ae 

 n
+ Be 
; b  bc





• Dynamic response to step input
Steady
response
36
• Analogy of electrical systems to mechanical systems
d 2Vc
dVc
LC
+
RC
+ VC = Vin
dt 2
dt
d 2x
dx
m 2 + b + kx = F
dt
dt
Electrical System
Mechanical System
L
R
1/C
m
b
k
37
Example: 2nd order Systems
Consider the RLC system with R=100 Ω,
L=2.0 H, and C = 20 µF:
R
L
ic
d 2Vc
dVc
LC
+
RC
+ VC = Vin
2
dt
dt
Vin
Vc
C
Questions:
• What is the natural frequency?
• Is this system over damping or under
damping?
• What is the dynamic response of this
system, i.e., Vc(t) = ?
38
Solution
• In analogy m=LC, b = RC, k = 1.
• Natural frequency
n =
1
1
=
= 158 Hz
−6
LC
2  20 10
• Damping coefficient ξ
b
RC
100  20 10 −6
= =
=
= 0.16  1
−
6
bc
4 LC
4  2  20 10
Under-damped
• The damped frequency ωd
d = n 1 −  2 = 158  1 − 0.162 = 156 Hz
39
• The output:
Vc (t ) = e −nt A cos(td ) + B sin(td ) + Vin
Vc (t ) = e −0.16158 t A cos(156t ) + B sin(156t ) + Vin
= e −25.3 t A cos(156t ) + B sin(156t ) + Vin
• Constants (A, B) can be determined by initial condition
Vc(0) = 0;
→ A = -Vin
dVc (t )
= 0 = − 25.3  e − 25.3t A cos(156t ) + B sin(156t )
t =0
dt t =0
+ e − 25.3t − 156 A sin(156t ) + 156 B cos(156t )
t =0
− 25.3 A + 156 B = 0  B = −0.16Vin
Vc (t ) = Vin − Vin e −25.3t cos(156t ) + 0.16 sin(156t )
Performance measures
• Steady state value: the value the system reaches after all transients dissipate
• Rise time: the time required for the system to reach 90% of its steady state
value.
--- Measure how fast system response to input
41
Performance Measures
• Overshoot: the amount the output exceeds the steady state value
• Setting time (with a given tolerance): the time required for the system
to settle within the tolerance of the steady state value
Summary: When the system model is given, all the above performance
measurement parameters can be calculated.
42
4.3 Frequency Response of Systems
Outline
• Frequency Domain Representation of Signals
▪ Fourier series (periodic inputs)
▪ Fourier transform (general input)
• Frequency Response of Systems
▪
Fourier transform of system models
▪
Obtain frequency response of systems from system models
Frequency Domain Representation of Signals
Idea
• Rational: the input signal to the system is a sum (or integration) of a
bunch of sinusoidal signals of different frequency & different
amplitude.
• Fourier Series & Fourier Transform are the tools.
44
Fourier Series Representation of Signals
• Any periodic signal (e.g., the input to a system) can be represented as
the sum of a series of complex exponential (or sine and cosine waves)
of different amplitude and frequency.
• Mathematically, a periodic signal F(t) of period T can be represented

F (t ) =  An e
jn 0 t
n = −
;
1
0 = 2 = 2f 0
T0
▪ ω0 -- fundamental or first harmonic frequency
▪ Coefficient An
T
1
An =  F (t )e − jn0t dt
T 0
•
A plot of An vs nω0 is called the spectrum.
45
Example: a square wave of period T
F(t)
T

1
,
0

t



2
F (t ) = 
− 1, T  t  T

2

• Fundamental frequency
• A0 (n = 0)
• An (n ≠ 0)
0 =
1
0.5T T
t
2
T
T
1 T
1  0.5T
A0 =  F (t )dt =  
dt −  dt  = 0
0.5T 
T 0
T 0
T
0.5T
T


1
1
− jn0t
− jn0t
− jn0t
A n =  F ( t )e
dt =   e
dt −  e
dt 
T 0
T 0
0.5T

 2

;
n
=
odd
1 2e
− 2 1 − ( −1)


=
=
=  jn

T
− j0 n
jn
0;
n = even 

− j0 n0.5T
n
46

F (t) =
jn0t
n
−1
2 jn0t
e
=
+

n=−,n=odd jn
 A e4sin( n t)
= 
n

n=−

2 jn0t
 jn e
n=1,n
=odd
0
n=1,3
,5,...
Fourier Series Representation of Sign1
• Spectrum of signal: A plot of the coefficients of exponential FS vs frequencies.
F(t)
Time Domain
Representation
1
Frequency Domain
Representation
Amplitude
0.5T T
2/π
0
▪ Knows spectrum
t
ω0 3ω0 5ω0 7ω0 9ω0 11ω0 13ω0
Frequency
knows the signal.

F (t ) =  An e jn 0 t
n = −
▪ From the above, the input signal to a system is the sum of a bunch of
sinusoidal waves of different frequencies
The system may attenuate some frequency components
and amplify other frequency components of the input.
This different responses to different frequency components
of an input signal is called System Frequency Response
49
Frequency Response of Systems
Frequency Response
• Definition: amplitude ratio of output and input
K ( ) =
X out ( )
A ( )
= out
X in ( )
Ain ( )
dB:
dB = 20  log 10
X out ( )
X in ( )
50
Frequency Response (by Bode Plot)
• Bode Plot: A plot in 20log10 scale or in dB.
▪ 20dB means a magnitude of 10;
▪ 40dB means 100;
▪ -60dB means 1/1000;
• Bandwidth: Frequency range between 3dB cutoffs.
Amplitude ratio in dB
0
ωL
ωH
frequency
-3dB
51
System Frequency Response
• Importance: if the system
frequency response is known,
then we may predicate its
response to any input.
• It is advantageous to do this in
the frequency domain.
52
Example
What is the output for a system described by its frequency response to the
input Xin described by its spectrum ?
X in = A1 sin( 0t ) + A2 sin( 20t ) +
A3 sin( 30t ) + ... + A10 sin(100t )
+ A11 sin(110t ) + ...
X out = 0.25 A2 sin( 20t ) + A3 sin( 30t )
+ ... + 0.5 A10 sin(100t )
53
From the above,
▪ A periodic wave can be represented by its Fourier Series.
▪ The importance of the frequency response of a system
• Questions:
▪ How to obtain the frequency response of a system?
▪ How to use Fourier Transform to represent a non-periodic signal
(similar to the case of using FS to represent periodic signals)?
Fourier Transform (FT)
• A signal x(t) can be represented in the frequency domain as
1 
j t
x(t ) =
X
(
j

)
e
d

−

2
• Fourier Transform or frequency domain representation

X ( j ) =  x(t )e − j t dt
−
• X(jω) and x(t) form a Fourier Pair
F
x(t )  X ( j )
55
Example: Given a signal
−7 t
x(t ) = e u (t )
x(t)
1
Its Fourier Transform
t
X ( j )

= e
−

= e
−7 t
|X(jω)|
u (t )e
− ( 7 + j ) t
0
=
1
7 + j
− j t
Magnitude
1/7
dt
dt =
e
− ( 7 + j ) t 
t =0
− (7 + j )
ω=
2πf
<X(jω)
π/2
ω
-π/2
Phase
56
Fourier Transform: Time Derivative Property
• Time derivative once corresponding to the multiplication of jω once
in the frequency domain
 1 

j t
d   X ( j )e d 
dx(t )
1 
2 −


jt
=
= j
X
(
j

)
e
d = j x(t )

−

dt
dt
2
• Or
d F
⎯→ j
dt
• Further
dk
k
F
⎯→( j )
k
dt
57
FT of 1st Order System Models
• Consider the system model of a RC circuit
R
dV (t )
Vin (t ) = RC c + Vc (t )
dt
• Its Fourier Transform of both sides
Vin
ic
Vc
Vin ( j ) = RCj Vc ( j ) + Vc ( j )
• The output/input ratio for a particular frequency component
H ( j ) =
Vc ( j )
1
=
Vin ( j ) 1 + jRC
58
• System Frequency Response
Magnitude of H(jω)
–
c =
H ( j ) =
1
RC
1
 

1 + 
 c 
2
– The higher frequency, the
smaller H(jω)
– Low pass filter
– Bandwidth: [0, 1/RC]
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FT of 2nd Order System Models
• Consider a spring-mass-damping system
d 2 x(t )
dx(t )
m
+b
+ kx(t ) = Fext (t )
2
dt
dt
• Rewrite it using natural frequency ωn and damping ratio ξ
Fext (t)
1 d 2 x(t ) 2 dx(t )
+
+ x(t ) =
2
2
n dt
n dt
k
• Taking Fourier Transform of both sides
 ( j ) 2 2

Fext (j )
j + 1 X ( j ) =
 2 +
n
k
 n

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• The output/input ratio
H ( j ) =
X ( j )
1/ k
1/k
=
=
2
Fext ( j )     2 
1
-r
+ jr 2ξ

1 −    + j
2
n
  n  
where: Frequency ratio r:

r=
n
K multiply the magnitude:
k H ( j ) =
1
1 − r  + (2r )
2 2
2
Phase response:
 2r 
−1  2 r 
H ( j ) = 0 − tan 
= − tan 
2
2
1
−
r
1
−
r




−1
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• Magnitude
k H ( j ) =
1
1 − r  + (2r )
2 2
2
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• Phase response
Example (Problem 4.21 of the textbook)
For a spring-mass-damper system with Fext = 20sin(0.75t) N, m = 10 kg,
k =12 N/m, and b = 10 Ns/m. What is the equation for the steady state
sinusoidal response of x(t)?
Solution
From:
x(t ) = H ( j )  Fext (t ) = H ( j )20 sin( 0.75t )
Input frequency ω = 0.75rad/s
Frequency response H(j0.75) = ?
n =
k
12
b
10

0.75
=
= 1.095rad / sec; =
=
= 0.456, r =
=
= 0.685
m
10
n 1.095
4mk
4 *10 *12
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H ( j )  =0.75 =
1/ k
1 − r  + (2r )
2 2
2
=
1/12
1 - 0.685  + (0.685  2  0.456)
2 2
= 0.1017
2
 2 r 
−1 2  0.685  0.456
H ( j )  =0.75 = −tg −1 
=
−
tan
= − tan −1 1.1767 = −0.866 rad
2
1 − 0.685  0.685
1 − r 
x(t ) = H ( j )  Fext (t ) = H ( j ) e jH ( j )  Fext (t )
= 2.03 sin( 0.75t − 0.866) meter
Steady state output is a delayed sinusoidal displacement of mass.
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