System Response
LAPLACE DOMAIN
TRANSFER FUNCTION
Y (S) = H(S) • F(S )
y (t) = h(t) ⊗ f (t)
RESPONSE
MODELS
Y ( jω) = H( jω) • F( jω)
FRF
IMPULSE
t
y ( t ) = ∫ h (τ)f ( t − τ)dτ
0
TIME DOMAIN
Peter Avitabile
Mechanical Engineering Department
University of Massachusetts Lowell
FREQUENCY DOMAIN
22.451 Dynamic Systems – System Response
1
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – First-order System
τy& + y = f ( t ) τ > 0 I.C. y(0) = y o
Free Response
No forcing function – Only I.C.
Laplace Transform
yo
τy o
τ[sY (s) − y o ] + Y(s) = 0 ⇒ Y (s) =
=
τs + 1 s + 1
Inverse Laplace
y( t ) = y o e
τ
( τ)
− t
start at yo and exponentially
∞
decay to zero as t
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
2
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – First-order System
Forced Response – Step Response
f ( t ) = A if t > 0 otherwise = 0
Laplace Transform
A
τy o s + A
τ[sY (s) − y o ] + Y (s) =
⇒ Y (s) =
s
s(τs + 1)
break up using partial fraction expansion
A
y os +
c1
c2
τ
=
+
Y (s) =
1
s s+1

s s + 
τ
τ

22.451 Dynamic Systems – System Response
3
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – First-order System
where
c1 = sY (s) s = 0
τy o s + A
=
τs + 1
1

c 2 =  s + Y (s) s = − 1
τ
τ

s=0
=A
A
y os +
τ
=
s
s=− 1
τ
= yo − A
A
1
+ (y o − A )
∴ Y (s) =
1
s
s+
τ
22.451 Dynamic Systems – System Response
4
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – First-order System
Inverse Laplace
−t
y STEP ( t ) = A + ( y o − A)e τ
−t 

if y o = 0 → y STEP ( t ) = A1 − e τ 


y(τ) = 0.632A
y(2τ) = 0.865A
y(3τ) = 0.950A
y(4τ) = 0.982A
y(5τ) = 0.995A
Scan Fig. 7.2 p. 337
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
5
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – First-order System
Forced Response – Ramp Response
f ( t ) = At if t > 0 otherwise = 0
Laplace Transform
A
τ[sY (s) − y o ] + Y (s) = 2
s
A
+ y os 2
B 2 B1
C
τ
= 2 +
+
Y (s) =
1
s s+1
2
s
s s + 
τ
τ

where B = A; B = − τA; C = y + τA
2
1
o
22.451 Dynamic Systems – System Response
6
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – First-order System
Inverse Laplace
y RAMP ( t ) = At − Aτ + Aτe
−t
τ
Note that the SS
response is At − Aτ.
After the transient
portion decays, the
response will track the
ramp but with an error
of Aτ.
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
7
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – Second-order System
&x& + 2ζω n x& + ωn 2 x = f ( t ) I.C. x o ; x& o
Free Response – No forcing function – Only I.C.
[
Laplace Transform
2
]
2
&
s X (s) − sx (0) − x (0) + 2ζω n [sX (s) − x (0)] + ωn X (s) = 0
(
s + 2ζω )x
X(s) =
•
− x0
2
s 2 + 2ζωn s + ωn
n
0
Solution form depends on the poles of the characteristic equation
s 2 + 2ζω n s + ωn 2 = 0
∴ s1, 2 = −ζω n ±
(ζω n )2 − ωn 2
= −ζω n ± ωn ζ 2 − 1
22.451 Dynamic Systems – System Response
8
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – Second-order System
Three cases of damping
•Case 1 – Overdamped (ζ>1)
•Case 2 – Critically damped (ζ=1)
•Case 3 – Underdamped (0<ζ<1)
•Case 4 – Undamped (ζ=0)
Scan Fig. 7.5 p. 343
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
9
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – Second-order System
Case 1 – Overdamped (ζ>1) – Two Real Roots
(s + 2ζω n )x 0 + v 0 (s + 2ζω n )x 0 + v 0
X(s) = 2
=
2
(s − s1 )(s − s 2 )
s + 2ζω n s + ωn
2
s1 = −ζω n + ωn ζ − 1; s 2 = −ζω n − ωn ζ 2 − 1
Partial Fraction Expansion
A1
A2
X(s) =
+
s − s1 s − s 2
22.451 Dynamic Systems – System Response
ωn  ζ + ζ 2 − 1  x 0 + v 0


A1 =
2
2ω n ζ − 1
− ωn  ζ − ζ 2 − 1  x 0 + v 0


A2 =
2
2ω n ζ − 1
10
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – Second-order System
Case 1 – Overdamped (ζ>1) – Two Real Roots
 ω  ζ + ζ 2 − 1  x + v 
0
0
 n
−ω

x(t) = 
e

2
2ω n ζ − 1




n
(ζ −
 ω  ζ − ζ 2 − 1  x + v 
0
0
 n
−ω

e
−

2
2ω n ζ − 1




22.451 Dynamic Systems – System Response
11
)
ζ 2 −1 t
n
(ζ +
)
ζ 2 −1 t
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – Second-order System
Case 2 – Critically damped (ζ=1) – Two Real Repeated Roots
s1, 2 = −ωn ( two real repeated roots)
(s + 2ωn )x 0 + v 0 (s + ωn )x 0 + ωn x 0 + v 0
=
X(s) = 2
2
(s + ωn )2
s + 2ω n s + ω n
x0
ωn x 0 + v 0
X(s) =
+
s + ωn
(s + ωn )2
Inverse Laplace
x(t) = x 0e
−ω n t
−ω t
(
)
+ ωn x 0 + v 0 te
22.451 Dynamic Systems – System Response
n
12
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – Second-order System
Case 3 – Underdamped (0<ζ<1) – Complex Conjugate Pair
Define:
σ = ζω n
ωd = ω n 1 − ζ 2
Then
s 2 + 2ζω n s + ωn 2 = (s + ζω n )2 − ζ 2 ωn 2 + ωn 2
= (s + σ ) + ωd
2
2
Then
(s + 2ζω n )x 0 + v 0 (s + 2σ )x 0 + x 0 + v 0
X(s) = 2
=
2
(s + σ )2 + ωd 2
s + 2ζω n s + ωn
22.451 Dynamic Systems – System Response
13
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Transient Response – Second-order System
which can be rearranged into
(s + σ )x 0
σx 0 + v 0
X(s) =
+
2
2
2
2
(s + σ ) + ωd (s + σ ) + ωd
(s + σ )x 0
(σx 0 + v 0 )
ωd
=
+
2
2
ωd
(s + σ ) + ωd
(s + σ )2 + ωd 2
Inverse Laplace
x(t) = x 0e
OR
=e
− σt
σx 0 + v 0 − σt
cos ωd t +
e sin ωd t
ωd
− σt 

σx 0 + v 0
sin ωd t 
 x 0 cos ωd t +
ωd


22.451 Dynamic Systems – System Response
14
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Impulse
&x& + 2ζω n x& + ωn 2 x = f ( t ) = δ( t )
with I.C.
x o ; x& o
Laplace Transform
[s X(s) − sx(0) − x& (0)]+ 2ζω [sX(s) − x(0)] + ω
2
n
Grouping terms
X(s) =
F(s)
2
s + 2ζω n s + ωn
2
+
n
2
X (s) = F(s)
(s + 2ζω n )x o + v o
s 2 + 2ζω n s + ωn 2
If I.C. are zero, then the result is the system transfer function
X (s)
1
H(s) =
= 2
F(s) s + 2ζω n s + ωn 2
The inverse Laplace of this yields the impulse response
of the system. (Note that I.C. are zero)
22.451 Dynamic Systems – System Response
15
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Impulse
Case 1 – Overdamped
x(t) =
1
2ω n ζ 2 − 1
e
(
)
− ω n ζ − ζ 2 −1 t
−
1
2ω n ζ 2 − 1
 ω  ζ + ζ 2 − 1  x + v 
0
0
 n
−ω

e
+

2
2
ω
ζ
−1


n


 ω  ζ − ζ 2 − 1  x + v 
0
0
 n
−ω

e
−

2
2
ω
ζ
−1


n


22.451 Dynamic Systems – System Response
16
n
(ζ −
e
(
)
− ω n ζ + ζ 2 −1 t
)
ζ 2 −1 t
I.C.
n
(ζ +
)
ζ 2 −1 t
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Impulse
Case 2 – Critically Damped
x ( t ) = te − ω t + x o e − ω t + (ωn x o + v o )te − ω
n
n
n
t
I.C.
22.451 Dynamic Systems – System Response
17
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Impulse
Case 3 – Underdamped

(σx o + v o )
1 − σt
− σt 
x(t) =
e sin ωd t + e  x o cos ωd t +
sin ωd t 
ωd
ωd


I.C.
Case 4 – Undamped
vo
1
x(t) =
sin ωn t + x o cos ωn t +
sin ωn t
ωn
ωn
22.451 Dynamic Systems – System Response
18
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Impulse
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
19
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Impulse
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
20
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Step
&x& + 2ζω n x& + ωn 2 x = f ( t ) = u ( t )
with I.C.
x o ; x& o
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
21
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Step
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
22
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Step
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
23
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response – Unit Step
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
24
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Frequency Response Function
The FRF can be obtained from the Fourier Transform of
Input-Output Time Response (and is commonly done so in
practice)
The FRF can also be obtained from the evaluation of the
system transfer function at s=jω.
For a 1st order system
1
1
H(s) =
⇒ H (s) s = jω = H( jω) =
τs + 1
1 + jωτ
For a 2nd order system
H(s) =
1
s 2 + 2ζω n s + ωn 2
⇒ H(s) s = jω = H( jω) =
22.451 Dynamic Systems – System Response
1
− ω2 + 2ζω n jω + ωn 2
25
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Frequency Response Function
The FRF for a mechanical system
H(s) =
1
2
ms + cs + k
⇒ H( jω) =
1
− mω2 + jcω + k
This is normally presented in a LOG-MAG and PHASE plot
called a BODE DIAGRAM.
22.451 Dynamic Systems – System Response
26
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
FRF – Bode Diagram – 1st Order
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
27
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
FRF – Bode Diagram – 2nd Order
Source: Dynamic Systems – Vu & Esfandiari
22.451 Dynamic Systems – System Response
28
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Fourier Series
A periodic function is a function that satisfies the relationship
where
f (t ) = f (t + T )
T is the period of the function
The fundamental angular frequency is defined by
2π
ωo =
(rad/sec)
T
22.451 Dynamic Systems – System Response
29
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Fourier Series
Consider a family of waveforms
f1 (t ) = 1
f 2 (t ) = sin (2πt )
1
f 3 (t ) = sin (6πt )
3
1
f 4 (t ) = sin (10πt )
5
1
f 4 (t ) = sin (14πt )
7
Summed as
N
g (t ) = ∑ f n (t )
1< N ≤ 5
n =1
22.451 Dynamic Systems – System Response
30
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Fourier Series
The summed waveforms can be plotted in Matlab to
approximate a square wave
22.451 Dynamic Systems – System Response
31
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Fourier Series
()
A Fourier Series representation of a real periodic function f t
is based upon the summation of harmonically related sinusoidal
components.
If the period is T , then the harmonics are sinusoids with
frequencies that are integer multiples of ωo ; This is written
as
f n (t ) = a n cos(nωo t ) + b n sin (nωo t )
= A n sin (nωo t + φn )
Note: These two are equivalent – one is written with sine and
cosine terms – the other is written as a sin with amplitude and
phase
22.451 Dynamic Systems – System Response
32
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Fourier Series
The Fourier Series representation of an arbitrary periodic
waveform f t is an infinite sum of harmonically related
sinusoids and is commonly written as
()
∞
1
f (t ) = a 0 + ∑ [a n cos(nωo t ) + b n sin (nωo t )]
2
n =1
∞
1
= a o + ∑ [A n sin (nωo t + φn )]
2
n =1
Or
(
)
(
a n jnω t
b n jnω t
− jnω t
f (t ) =
e
+e
+
e
+ e − jnω t
2j
2
1
1
jnω t
= (a n − jb n )e
+ (a n + jb n )e − jnω t
2
2
o
o
o
∞
[
= ∑ Fn e jnω t
−∞
o
o
o
)
o
]
22.451 Dynamic Systems – System Response
33
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Fourier Coefficients
The derivation of the expression for computing the
coefficients in a Fourier Series is beyond the scope of this
course. Without proof, the coefficients Fn are
1
Fn =
T
t1 +T
− jnω t
(
)
f
t
e
dt
∫
o
t1
which is evaluated over any periodic segment or in sinusoidal
form
1
an =
T
1
bn =
T
t1 +T
∫ f (t )cos(nωo t )dt
t1
t1 +T
∫ f (t )sin(nωo t )dt
t1
22.451 Dynamic Systems – System Response
34
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Consider a linear SDOF system with a steady state periodic
input (assume that the initial conditions and transients have
decayed to zero)
The input excitation can be described by a Fourier Series
representation as
∞
f (t ) = ∑ Fn e jnω t
o
−∞
∞
1
= a o + ∑ A n sin (nωo t + φn )
2
n =1
22.451 Dynamic Systems – System Response
35
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
The system output response is related to the input excitation
through the system frequency response function as
Y( jω) = H( jω) ⋅ F( jω)
The nth real harmonic input component of the Fourier Series
f n (t ) = A n sin (nωo t + φ)
Generates an output sinusoidal component y n (t ) with a
magnitude and phase determined by the nth component of H( jω)
y n (t ) = H( jnωo ) A n sin[nωo + φn + ∠H( jnωo )]
22.451 Dynamic Systems – System Response
36
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
From superposition, the total output y n (t ) is the sum of all the
n components as
y(t ) =
∞
∑ y n (t )
n =0
∞
1
= a o H( jo ) + ∑ A n H( jnωo ) sin[nωo t + φn + ∠H( jnωo )]
2
n =1
Note: This is also a Fourier Series with the same fundamental
and harmonic frequencies as the input
In complex form, this is written as
y n (t ) = H( jnωo )Fn e jnω t
o
The complete output Fourier Series is
y(t ) =
n =∞
∑ y n (t ) =
n = −∞
n =∞
jnω t
(
)
H
jn
ω
F
e
∑
o n
o
n = −∞
22.451 Dynamic Systems – System Response
37
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Consider a simple mass spring dashpot system that is under
steady state sinusoidal excitation (but not a ωn of a SDOF
system
Time Domain
h (t )
f (t ) = sin (ωt )
System
FFT
h ( jω)
Frequency Domain f ( jω)
22.451 Dynamic Systems – System Response
y(t ) = A sin (ωt + φ)
38
IFT
y ( jω)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Now consider excitation at the natural frequency, ωn ,of the
SDOF system
22.451 Dynamic Systems – System Response
39
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Now consider a periodic input which is not sinusoidal.
However, the Fourier Series representation is nothing more
than the sum of different amplitude sine waves.
22.451 Dynamic Systems – System Response
40
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.7
The first-order electric network shown in Fig. 15.7 is excited
with the sawtooth function. Find an expression for the series
representing the output Vout(t).
22.451 Dynamic Systems – System Response
41
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.7 (cont.)
The electric network has a transfer function
1
H(s ) =
RCs + 1
and therefore has a frequency response function
1
H ( j ω) =
(ωRC )2 + 1
∠H( jω) = tan −1 (− ωRC )
the input function u(t) may be represented by the Fourier Series
2(− 1)n +1
sin (nωo t )
u (t ) = ∑
nπ
n =1
∞
22.451 Dynamic Systems – System Response
42
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.7 (cont.)
At the output the series representation is
2(− 1)n +1
y ( t ) = ∑ | H ( jω ) |
sin[nωo t + ∠H( jnωo )]
nπ
n =1
∞
n +1


(
)
2
1
−
 sin nω t + tan −1 (− nω RC )
y(t ) = ∑ 
o
o
 nπ (nω RC )2 + 1 
n =1
o

[
∞
]
As an example consider the response if the period of the
input is chosen to be T = πRC so that ω = 2 ,then
o
RC
∞
2(− 1)n +1
 2n

sin 
t + tan −1 (− 2n )
2
 RC

n =1 nπ (2n ) + 1
y (t ) = ∑
22.451 Dynamic Systems – System Response
43
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.7 (cont.)
Figure 15.8 shows the computer response found by summing
the first 100 terms of the Fourier Series.
T = πRC
22.451 Dynamic Systems – System Response
ωo =
44
2
RC
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.8
A cart shown in Figure 15.9a, with a mass m=1.0 kg is supported on low
friction bearings that exhibit a viscous drag B = 0.2 N-s/m and is coupled
through a spring with stiffness K=25 N/m to a velocity source with a
magnitude of 10 m/s but which switches direction every π seconds as shown
in Figure 15.9b.
Vin (t ) =
10 m / s
0≤t<π
− 10 m / s
π ≤ t < 2π
The task is to find the resulting velocity of the mass v m (t )
22.451 Dynamic Systems – System Response
45
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.8 (cont.)
The input Ω(t ) has a period of T = 2π and a fundamental
frequency of ωo = 2π / T = 1 rad/s. The Fourier Series for the
input contains only odd harmonics:
20 ∞ 1
u (t ) = ∑
sin[(2n − 1)ωo t ]
π n =1 2n − 1
=
20 
1
1

(
)
(
)
(
)
sin
ω
t
+
sin
3
ω
t
+
sin
5
ω
t
+
...
o
o
o

π 
3
5
22.451 Dynamic Systems – System Response
46
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.8 (cont.)
The frequency response of the system is
H ( jω ) =
25
25 − ω2 + j0.2ω
(
)
which when evaluated at the harmonic frequencies of the input
nωo = n radians/s is
H( jnωo ) =
25
25 − n 2 + j0.2n
[(
)
22.451 Dynamic Systems – System Response
47
]
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.8 (cont.)
The following table summarizes the first five odd spectral
components at the system input and output.
22.451 Dynamic Systems – System Response
48
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Response of Linear System Due to Periodic Inputs
Example 15.8 (cont.)
Figure 15-10a shows the computed frequency response
magnitude for the system and the relative gains and phase
shifts (rad) associated with the first five terms in the series.
22.451 Dynamic Systems – System Response
49
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
22.451 Dynamic Systems – System Response
50
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory