Ch. 3: Forced Vibration of 1-DOF System
3.0 Outline
„ Harmonic Excitation
„ Frequency Response Function
„ Applications
„ Periodic Excitation
„ Non-periodic Excitation
3.0 Outline
Ch. 3: Forced Vibration of 1-DOF System
3.1 Harmonic Excitation
Force input function of the harmonic excitation is the
harmonic function, i.e. functions of sines and cosines.
This type of excitation is common to many system
involving rotating and reciprocating motion. Moreover,
many other forces can be represented as an infinite
series of harmonic functions. By the principle of
superposition, the response is the sum of the individual
harmonic response.
It is more convenient to use the frequency domain
technique in solving the harmonic excitation problems.
This is because the response to different
excitation frequencies can be seen in one graph.
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Let us focus on the particular solution of
mx + cx + kx = F0 cos ωt
normalize the equation of motion
x + 2ζωn x + ωn2 x = f 0 cos ωt , f 0 = F0 / m
f ( t ) = f 0 Re ⎡⎣eiωt ⎤⎦
∴ solve for z ( t ) from z + 2ζωn z + ωn2 z = f 0 eiωt
and the solution is the real part of z ( t ) ; x ( t ) = Re ⎡⎣ z ( t ) ⎤⎦
Assume the solution to have the same form as the forcing function
z ( t ) = Z ( iω ) eiωt ( same frequency as the input w/ different mag. and phase )
( −ω
2
+ i 2ζωωn + ωn2 ) Z ( iω ) eiωt = f 0 eiωt
f 0 / ωn2
f0
=
Z ( iω ) = 2
2
ωn − ω + i 2ζωωn 1 − (ω / ωn )2 + i 2ζω / ωn
=
F0
2
k ⎡1 − (ω / ωn ) + i 2ζω / ωn ⎤
⎣
⎦
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
z (t ) =
F0
iωt
iωt
=
ω
e
H
i
F
e
(
)
0
k ⎡⎣1 − r 2 + i 2ζ r ⎤⎦
⎡
⎤
F0
iωt
∴ x ( t ) = Re ⎢
e ⎥ , r = ω / ωn
2
⎢⎣ k ⎡⎣1 − r + i 2ζ r ⎤⎦
⎥⎦
1
iθ
=
If H ( iω ) =
is the frequency response
ω
H
i
e
(
)
2
k ⎡⎣1 − r + i 2ζ r ⎤⎦
∴ x ( t ) = F0 H ( iω ) cos (ωt + θ )
1
where H ( iω ) =
k
(1 − r ) + ( 2ζ r )
2 2
= magnitude
2
−2ζ r
= phase
2
1− r
The system modulates the harmonic input by
θ = tan −1
the magnitude H ( iω ) and phase
H ( iω )
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
total response = homogeneous soln. + particular soln.
Recall the homogeneous solution of the underdamped system
xh = Ce −ζωnt cos (ωd t − φ ) or xh = e −ζωnt ( A1 sin ωd t + A2 cos ωd t )
∴ x ( t ) = Ce −ζωnt cos (ωd t − φ ) + F0 H ( iω ) cos (ωt + θ )
or x ( t ) = e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) + F0 H ( iω ) cos (ωt + θ )
The initial conditions will be used to determine C , φ or A1 , A2
They will be different from those of free response
because the transient term now is partly due to the excitation force
and partly due to the initial conditions
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Ex. 1 Compute and plot the response of a spring-mass
system to a force of magnitude 23 N, driving
frequency of twice the natural frequency and i.c.
given by x0 = 0 m and v0 = 0.2 m/s. The mass
of the system is 10 kg and the spring stiffness
is 1000 N/m.
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
ωn = k / m = 1000 /10 = 10 rad/s
ζ = c / ( 2mωn ) = 0
ω = 2 ×10 = 20 rad/s
H ( iω ) =
1
1
−3
0.333
10
=
=
−
×
k ⎣⎡1 − r 2 + i 2ζ r ⎤⎦ 1000 × (1 − 22 )
x ( t ) = A1 sin ωnt + A2 cos ωnt − 23 × 0.333 ×10−3 cos ωt
x ( t ) = ωn A1 cos ωnt − ωn A2 sin ωnt + ω × 23 × 0.333 × 10−3 sin ωt
i.c. x ( 0 ) = 0 = A2 − 23 × 0.333 ×10−3 , A2 = 7.667 ×10−3
x ( 0 ) = 0.2 = 10 × A1 , A1 = 0.02
∴ x ( t ) = 0.02sin10t + 7.667 ×10−3 ( cos10t − cos 20t ) m
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Ex. 2 Find the total response of a SDOF system with
m = 10 kg, c = 20 Ns/m, k = 4000 N/m, x0 = 0.01 m,
v0 = 0 m/s under an external force F(t) = 100cos10t.
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
ωn = k / m = 20 rad/s
ζ = c / ( 2mωn ) = 0.05
r = ω / ωn = 0.5
H ( iω ) =
1
1
=
= 332.6 E − 6 − 0.0666
2
2
k ⎣⎡1 − r + i 2ζ r ⎦⎤ 4000 × (1 − 0.5 + i 2 × 0.05 × 0.5 )
x p ( t ) = F0 H ( iω ) cos (ωt + θ ) = 33.26 E − 3cos (10t − 0.0666 )
xh ( t ) = e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) , ωd = ωn 1 − ζ 2 = 19.975 rad/s
x ( t ) = e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) + F0 H ( iω ) cos (ωt + θ )
x ( t ) = −ζωn e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) + e −ζωnt (ωd A1 cos ωd t − ωd A2 sin ωd t )
− ω F0 H ( iω ) sin (ωt + θ )
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
ωresponse finally becomes ω, and in phase
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
ωresponse finally becomes ω, and out of phase
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
F0ωnt/(2k)
In case of ζ = 0 and ω = ωn , the guess solution of the form
x ( t ) = X ( iω ) eiωt = A cos ωt + B sin ωt is invalid. This is because
it has the same form as the homogeneous solution.
Fω t
The correct particular solution is x p ( t ) = 0 n sin ωnt.
2k
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Beat when the driving frequency is close to natural freq.
The total solution can be arranged in the form
x (t ) =
v0
ωn
sin ωnt + x0 cos ωnt +
f0
( cos ωt − cos ωnt )
ωn2 − ω 2
x02ωn2 + v02
⎛
xω ⎞
2f
⎛ ω − ω ⎞ ⎛ ωn + ω ⎞
t ⎟ sin ⎜
t⎟
sin ⎜ ωn t + tan −1 0 n ⎟ + 2 0 2 sin ⎜ n
v0 ⎠ ωn − ω
ωn
⎝ 2
⎠ ⎝ 2
⎠
⎝
If the system is at rest in the beginning,
=
x (t ) =
2 f0
⎛ ω n − ω ⎞ ⎛ ωn + ω ⎞
t ⎟ sin ⎜
t⎟
sin
⎜
2
2
ωn − ω
⎝ 2
⎠ ⎝ 2
⎠
The response oscillates with frequency
ωn + ω
inside
2
2f
⎛ ω −ω ⎞
t⎟
the slowly oscillated envelope 2 0 2 sin ⎜ n
ωn − ω
⎝ 2
⎠
∴ The beat frequency is ωn − ω
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Beat when the driving frequency is close to natural freq.
3.1 Harmonic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.2 Frequency Response Function
The core of the particular solution to the harmonic function is
H ( iω ) =
1
; frequency response function
2
k (1 − r + i 2ζ r )
It specifies how the system responds to harmonic excitation.
As a standard, we normalize the frequency response function
1
and then study how it varies as the
2
1 − r + i 2ζ r
excitation frequency ω and system parameters ζ ,ωn vary.
G ( iω ) =
It is indeed more convenient since we already normalized the frequency;
r = ω / ωn . So we can now study its variation to ζ and r.
For the fixed damping ratio, we plot G ( iω ) with r varies.
G ( iω ) has both magnitude and phase ⇒ magnitude and phase plot.
Then we repeatedly evaluate G ( iω ) by varying ζ .
3.2 Frequency Response Function
Ch. 3: Forced Vibration of 1-DOF System
H ( iω ) =
1
(1 − r
)
2 2
+ ( 2ζ r )
2
Frequency response plot
(Bode diagram)
⎛ −2ζ r ⎞
2 ⎟
1
r
−
⎝
⎠
θ = tan −1 ⎜
3.2 Frequency Response Function
Ch. 3: Forced Vibration of 1-DOF System
Resonance is defined to be the vibration response at
ω=ωn, regardless whether the damping ratio is zero.
At this point, the phase shift of the response is –π/2.
The resonant frequency will give the peak amplitude for
the response only when ζ=0. For0 < ζ < 1/ 2 ,the peak
amplitude will be at ω = ωn 1 − 2ζ 2 , slightly before ωn.
For ζ ≥ 1/ 2 , there is no peak but the max. value of the
output is equal to the input for the dc signal (of course,
for this normalized transfer function).
3.2 Frequency Response Function
Ch. 3: Forced Vibration of 1-DOF System
Ex. 3 Consider the pivoted mechanism with k=4x103 N/m,
l1=0.05 m, l2=0.07 m, l=0.10 m, and m=40 kg.
The mass of the beam is 40kg which is pivoted
at point O and assumed to be rigid. Calculate c
so that the damping ratio of the system is 0.2.
Also determine the amplitude of vibration of the
steady-state response if a 10 N force is applied
to the mass at a frequency of 10 rad/s.
3.2 Frequency Response Function
Ch. 3: Forced Vibration of 1-DOF System
l − l1 ⎞
⎡⎣ ∑ M O = I Oθ ⎤⎦ Fl − mglθ − Mg ⎛⎜
⎟ θ − c l2θ l2 − k ( l1θ ) l1
2
⎝
⎠
( )
2
2
⎡ 2
l + l1 )
(
⎛ l − l1 ⎞ ⎤
= ⎢ ml + M
+M⎜
⎟ ⎥θ
12
⎝ 2 ⎠ ⎥⎦
⎢⎣
0.1×10 cos10t = 0.5θ + 0.0049cθ + 59.05θ
ωn = 15.37, ζ = 0.2 =
0.0049c
, c = 627.3 Ns/m
2 × 0.5 × ωn
ω = 10, r = 0.6506
H ( iω ) =
1
= 0.02677 − 24.268°
59.05 ( 0.5767 + i 0.26 )
θ ss = 0.02677 cos (10t − 0.424 )
3.2 Frequency Response Function
Ch. 3: Forced Vibration of 1-DOF System
Ex. 4 A foot pedal for a musical instrument is modeled
as in the figure. With k=2000 kg/s2, c=25 kg/s,
m=25 kg, and F(t)=50cos2πt N, compute the
steady-state response assuming the system starts
from rest. Use the small angle approximation.
3.2 Frequency Response Function
Ch. 3: Forced Vibration of 1-DOF System
(
)
⎡⎣ ∑ M O = I Oθ ⎤⎦ F × 0.15 − k ( 0.05θ ) × 0.05 − c 0.05θ × 0.1 = m × 0.152 θ
5
100
θ = 50 cos 2π t , positive CW
3.75θ + θ +
6
3
Find the parameters
ωn = 2.98, ζ = 0.0373, ω = 2π , r = 2.108
H ( iω ) =
1
= 0.0087 − 177.4°
k (1 − r 2 + i 2ζ r )
since ζ ≠ 0, the transient response will die out
θ ss = F0 H ( iω ) cos (ωt + θ ) = 0.435cos ( 2π t − 3.096 )
3.2 Frequency Response Function
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
2
measured acc. ωn z
10
=
=
=
y
actual acc.
9.81
(1 − r ) + ( 2ζ r )
2 2
2
1
(1 − r 2 ) + ( 2ζ r )
2
2
= 0.962
From the problem statement, ω = 628 rad/s, ωd = ωn 1 − ζ 2 = 628 rad/s
r
ω
=
=1
2
ωd
1− ζ
ωn =
ωd
1− ζ 2
= 758 rad/s =
k
c
, ζ = 0.56 =
m
2mωn
k = 5745.6 N/m, c = 8.49 Ns/m
3.3 Applications
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
A periodic function is any function that repeats itself in
time, called period T.
f (t ) = f (t + T )
It is more general than the harmonic function. Here, we
will find the response to the input that is a periodic
function. The idea is to decompose that periodic input
into the sum of many harmonics. The response, by the
superposition principle of linear system, is then the sum
of the responses of individual harmonic. The response
of a harmonic function was studied in section 3.1
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Fourier found the way to decompose the periodic
function into sum of harmonic functions (sine & cosine)
whose frequencies are multiples of the fundamental
frequency. The fundamental frequency is the frequency
of the periodic function.
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Fourier series
Fourier series in real form:
a0 ∞
2π
f ( t ) = + ∑ ( an cos nω0t + bn sin nω0t ), ω0 =
2 n =1
T
Fourier coefficients:
T
2
an = ∫ f ( t ) cos nω0t dt , n = 0,1, 2,…
T 0
T
2
bn = ∫ f ( t ) sin nω0t dt , n = 1, 2,3,…
T 0
Fourier series in complex form:
2π
T
n =−∞
Fourier cofficients (complex):
f (t ) =
∞
∑ Cneinω0t , ω0 =
T
1
Cn = ∫ f ( t ) e − inω0t dt , n = … , −2, −1, 0,1, 2,…
T 0
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Some properties of Fourier series
1) If f ( t ) is an even function, bn = 0.
2 ) If f ( t ) is an odd function, an = 0.
a0
is the average value of f ( t ) over one period.
2
⎛ ∞
⎞
4 ) If f ( t ) is real, Ck = C− k ⇒ f ( t ) = C0 + 2 Re ⎜ ∑ Cn einω0t ⎟
⎝ n =1
⎠
3)
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Frequency spectrum tells how much each harmonic
contributes to the periodic function f ( t ) .
In real form, the harmonic at nω0 has the amplitude an2 + bn2
In complex form, the harmonic at nω0 has the amplitude 2 Re ( Cn )
Plot of the amplitude of each harmonic vs. its frequency
is the (discrete) frequency spectrum.
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Superposition principle of linear system
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Response to harmonic excitation
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
mx + cx + kx = F ( t ) = Cn einω0t
From section 3.1, xss = Cn H ( inω0 ) einω0t
where H ( inω0 ) =
1
⎛ ⎛ nω ⎞2
k ⎜1 − ⎜ 0 ⎟ + i 2ζ
⎜ ⎝ ωn ⎠
⎝
⎛ nω0 ⎞ ⎞
⎜
⎟ ⎟⎟
⎝ ωn ⎠ ⎠
⎛ ∞
⎞
mx + cx + kx = F ( t ) = C0 + 2 Re ⎜ ∑ Cn einω0t ⎟
⎝ n =1
⎠
C0
⎞
⎛ ∞
by superposition, xss =
+ 2 Re ⎜ ∑ Cn H ( inω0 ) einω0t ⎟
k
⎠
⎝ n =1
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
response frequency spectrum
excitation frequency spectrum
system frequency response
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Ex. Calculate the response of a damped system to
the periodic excitation f(t) depicted in the figure
by means of the exponential form of the Fourier
series. The system damping ratio is 0.1 and the
driving frequency is ¼ of the system natural freq.
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Expand f ( t ) as sum of harmonic series
f (t ) =
∞
∑Ce ω
in
n =−∞
0t
n
T
T /2
T
⎤
1
1⎡
2π
− inω0t
− inω0t
Cn = ∫ f ( t ) e
dt = ⎢ ∫ Ae
dt + ∫ − Ae − inω0t dt ⎥ , ω0 =
T 0
T⎣0
T
T /2
⎦
⎧ 0, n = even
iA ⎡
n
⎪
Cn =
1 − ( −1) ⎤ = ⎨ i 2 A
⎦
nπ ⎣
⎪⎩− nπ , n = odd
⎛
π⎞
i 2 A inω0t
2 A i⎜⎝ nω0t − 2 ⎟⎠ 4 A ∞ 1
∴ f ( t ) = ∑ n =odd −
×e
= ∑ n =odd
×e
=
∑ sin nω0t
nπ
nπ
π n =1,3,… n
Gn ( iω ) =
Gn ( iω ) =
Gn =
ω nω0 n
1
r
ζ
,
0.1,
=
=
=
=
ωn ω n 4
1 − r 2 + i 2ζ r
1
1 − ( n / 4 ) + i 0.05n
2
1
2
,
Gn = tan −1
⎡1 − ( 0.25n )2 ⎤ + ( 0.05n )2
⎣
⎦
4A ∞ 1
∴ xss ( t ) =
∑ G sin ( nω0t + Gn )
π n =1,3,… n n
−0.05n
1 − ( 0.25n )
2
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Ex. The cam and follower impart a displacement y(t)
in the form of a periodic sawtooth function to the
lower end of the system. Derive an expression
for the response x(t) by means of Fourier analysis.
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
FBD and assume y > x
⎡⎣ ∑ Fx = max ⎤⎦ mx = k2 ( y − x ) − k1 x − cx
mx + cx + ( k1 + k2 ) x = k2 y, ωn =
k1 + k2
c
, ζ =
m
2mωn
Write y ( t ) in the Fourier series expansion
y (t ) =
∞
∑Ce ω
in
n =−∞
n
0t
, ω0 =
2π
A
, y (t ) = B + t, 0 ≤ t ≤ T
T
T
T
T
T
1
1
1 A
Cn = ∫ y ( t ) e − inω0t dt = ∫ Be − inω0t dt + ∫ te − inω0t dt
T 0
T 0
T 0T
e ax
Integration formula: ∫ e dx =
+ c and
a
ax
e ax
∫ xe dx = a 2 ( ax − 1) + c
ax
T
⎡
⎤
2π
⎡ − in 2π t ⎤
in
t
−
⎢
⎥
2π
B⎢e T ⎥
A⎢ e T
iA
⎛
⎞⎥
1
, n≠0
Cn = ⎢
in
t
+
−
−
=
⎥
⎟⎥
2 ⎜
2
π
T ⎢ −in 2π ⎥ T 2 ⎢ ⎛
T
n
⎠
2π ⎞ ⎝
−
in
⎢
⎥
⎜
⎟
⎣
T ⎦0
T ⎠
⎣⎝
⎦0
A
C0 = B +
2
T
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
A
⎡ ∞ iA
⎤
∴ y ( t ) = B + + 2 Re ⎢ ∑
( cos nω0t + i sin nω0t )⎥
2
⎣ n =1 2π n
⎦
A A ∞ 1
y ( t ) = B + − ∑ sin nω0t
2 π n =1 n
Frequency response H n ( iω ) =
H n ( iω ) =
Hn =
1
( k1 + k2 ) ⎡⎣1 − r 2 + i 2ζ r ⎤⎦
1
⎡ ⎛ nω ⎞ 2
⎛ nω0 ⎞ ⎤
0
( k1 + k2 ) ⎢1 − ⎜ ⎟ + i 2ζ ⎜ ⎟ ⎥
⎢⎣ ⎝ ωn ⎠
⎝ ωn ⎠ ⎥⎦
1
2
,
⎛ nω ⎞
−2ζ ⎜ 0 ⎟
⎝ ωn ⎠
H n = tan −1
2
⎛ nω0 ⎞
1− ⎜
⎟
⎝ ωn ⎠
⎛ ⎛ nω ⎞ ⎞ ⎛ ⎛ nω ⎞ ⎞
( k1 + k2 ) ⎜⎜1 − ⎜ 0 ⎟ ⎟⎟ + ⎜ 2ζ ⎜ 0 ⎟ ⎟
ω
ω
⎝ ⎝ n ⎠ ⎠ ⎝ ⎝ n ⎠⎠
k2 ⎛
A ⎞ k2 A ∞ 1
xss ( t ) =
∑ H n sin ( nω0t + H n )
⎜B+ ⎟−
k1 + k2 ⎝
2 ⎠ π n =1 n
2
2
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.5 Non-periodic Excitation
Harmonic and steady-state excitation and response are
conveniently described in the frequency domain. For
deterministic non-periodic excitation and response, time
domain technique is more suitable.
We cannot find the repeated pattern that lasts forever
(both in the past & future) for the non-periodic excitation.
System response to the unit impulse, called the impulse
response, will be first studied. Then, this fundamental
response will be used to synthesize the response of the
LTI system to arbitrary excitation.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Impulse
The unit impulse, or Dirac delta function, is defined as
δ ( t − a ) = 0 for t ≠ a
∞
∫ δ ( t − a ) dt = 1
−∞
This means that the unit impulse is zero everywhere
except in the neighborhood of t=a. Since the area
under the graph δ-t is 1, the value of δ ( t − a ) is very
large in the vicinity of t=a.
The impulse of magnitudeF̂ , which may represent a
large force acting over a short period, can be written as
F ( t ) = Fˆ δ ( t − a )
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
The unit impulse has a useful property called the
“sampling property”. Multiplying a continuous function
f ( t ) by δ ( t − a ) , and integrating w.r.t. time:
∞
∞
−∞
−∞
∫ f ( t ) δ ( t − a ) dt = f ( a ) ∫ δ ( t − a ) dt = f ( a )
which is just the value of f(t) at t=a. This is a way in
evaluating integrals involving with impulse.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Impulse response
The impulse response, h(t), is the response to the unit
impulse, δ(t), applied at t=0 with zero initial conditions.
The impulse response is very important since it contains
all the system characteristics and can be used to find
the response to arbitrary excitation of LTI system via the
convolution integral theorem.
The impulse response of a 1 DOF MBK system must
satisfy
mh ( t ) + ch ( t ) + kh ( t ) = δ ( t )
subject to i.c.
h ( 0 ) = 0, h ( 0 ) = 0
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Get rid of the impulse function by
integrating over the duration ( 0, ε ) of the impulse
ε
ε
0
0
∫ ( mh + ch + kh ) dt = ∫ δ ( t ) dt = 1
Take limit as ε → 0 and apply the i.c.
to evaluate the integral on the left hand side:
ε
lim ∫ mh ( t ) dt = lim mh ( t ) = mh ( 0+ ) ≠ 0, assuming h ( t ) is not continuous
ε →0
0
ε
ε →0
ε
0
lim ∫ ch ( t ) dt = lim ch ( t ) 0 = ch ( 0+ ) = 0, assuming h ( t ) is continuous
ε →0
0
ε
ε →0
ε
ε
lim ∫ kh ( t ) dt = lim gh ( 0 ) t 0 = 0, assuming h ( t ) is continuous
ε →0
0
ε →0
∴ mh ( 0+ ) = 1
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Therefore, the effect of a unit impulse at t=0 is to
produce equivalent initial velocity (impulse-momentum)
h ( 0+ ) = 1/ m
Now, we are ready to find the impulse response. The
equivalent system is a homogeneous system with i.c.
h ( 0 ) = 0, h ( 0 ) = 1/ m
If the system is underdamped, the impulse response is
⎧ 1 −ζωnt
e
sin ωd t , t ≥ 0
⎪
h ( t ) = ⎨ mωd
⎪
0, t < 0
⎩
Note that the above i.c. is not the actual i.c.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Impulse response of underdamped system
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Linear Time Invariant (LTI) system has the characteristic
that the shape of the response will not be influenced by
the time the input is applied to the system. That is
f (t )
f (t − a )
LTI system
LTI system
x (t )
x (t − a )
Hence if the impulse is applied at t=to, the response is
⎧ 1 −ζωn ( t −t0 )
e
sin ωd ( t − t0 ) , t ≥ t0
⎪
h ( t ) = ⎨ mωd
⎪
0, t < t0
⎩
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Total response of underdamped MBK with i.c. x(a)=x0
and v(a)=v0 subject to the impulse force F̂δ ( t − a )
x ( t ) = xh + x p
=e
−ζωn ( t − a )
=e
−ζωn ( t − a )
x ( t ) = −ζωn e
+e
Fˆ −ζωn (t − a )
sin ωd ( t − a )
( A1 sin ωd ( t − a ) + A2 cos ωd ( t − a ) ) + mω e
d
⎧⎪⎛
Fˆ
⎨⎜ A1 +
mωd
⎩⎪⎝
−ζωn ( t − a )
−ζωn ( t − a )
⎫⎪
⎞
⎟ sin ωd ( t − a ) + A2 cos ωd ( t − a ) ⎬ , t ≥ a
⎠
⎭⎪
⎧⎪⎛
Fˆ
⎨⎜ A1 +
mωd
⎪⎩⎝
⎫⎪
⎞
⎟ sin ωd ( t − a ) + A2 cos ωd ( t − a ) ⎬
⎪⎭
⎠
⎧⎪ ⎛
⎫⎪
Fˆ ⎞
⎟ cos ωd ( t − a ) − ωd A2 sin ωd ( t − a ) ⎬
⎨ωd ⎜ A1 +
ω
m
⎪⎩ ⎝
⎪⎭
d ⎠
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Total response of underdamped MBK with i.c. x(a)=x0
and v(a)=v0 subject to the impulse force F̂δ ( t − a )
Apply i.c. x ( a ) = x0 and x ( a ) = v0 to solve for A1 and A2 :
x0 = A2
⎛
Fˆ ⎞
v0 = −ζωn A2 + ωd ⎜ A1 +
⎟
ω
m
d ⎠
⎝
1 ⎛
Fˆ ⎞
∴ A2 = x0 and A1 =
⎜ ζω x + v − ⎟
ωd ⎝ n 0 0 m ⎠
⎫
−ζωn ( t − a ) ⎧ 1
∴ x (t ) = e
⎨ (ζωn x0 + v0 ) sin ωd ( t − a ) + x0 cos ωd ( t − a ) ⎬ , t ≥ a
⎩ ωd
⎭
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Arbitrary Excitation
Ideally, arbitrary excitation can be expressed as linear
combinations of simpler excitations. The simpler
excitations are simple enough that the response
is readily available. This concept is exactly used by
Fourier.
Now, the idea is to regard the arbitrary excitation as a
superposition of impulses of varying magnitude and
applied at different times. It is used when the excitation
can be easily described in time domain.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Consider the excitation F(t). We can imagine that it is
constructed from infinite impulses at different times.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Convolution integral theorm
Focus on the time interval τ < t < τ + Δτ , at which
the impulse of magnitude F (τ ) Δτ is acting. This
shifted impulse can be written as F (τ ) Δτδ ( t − τ ) .
The response of the LTI system to this particular
impulse is
Δx ( t ,τ ) = F (τ ) Δτ h ( t − τ )
Since by sampling property F ( t ) = ∑ F (τ ) Δτδ ( t − τ ),
τ
and the system is linear, the response to F ( t ) is
x ( t ) = ∑ F (τ ) Δτ h ( t − τ )
τ
t
In the limit as Δτ → 0, x ( t ) = ∫ F (τ ) h ( t − τ ) dτ .
0
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Convolution integral theorm
The response of the arbitrary excitation is the
superposition of shifted impulse responses.
Interpretation
for the whole
range of time; t
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
To obtain h ( t − τ ) from h (τ ) , we need to carry out
two operation; shifting and folding. This is another interpretation
for the specific time t. The figures show the steps in evaluating the convolution.
If we define a new variable λ = t − τ , then τ = t − λ
and dτ = −d λ. With the change of the integration limits,
0
t
t
0
x ( t ) = ∫ F ( t − λ ) h ( λ )( −d λ ) = ∫ F ( t − τ ) h (τ ) dτ
That is the convolution is symmetric in F ( t ) and h ( t ) .
To decide which formula to use depends on the nature of F ( t ) and h ( t ) .
It is obvious that if the excitation F ( t ) or the impulse response h ( t )
is too complicated, we may be unable to evaluate the closed form
solution of the convolution integral. The excitation may not at all
be written as functions of time. In these cases, the integration must
be carried out numerically.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Interpretation for the specific time; t
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Ex. Determine the response of the underdamped MBK
to the unit step input.
1
0
u(t)
t
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
t
x ( t ) = ∫ F (τ ) h ( t − τ ) dτ
0
F (τ ) = u (τ ) and h ( t − τ ) is the system impulse response
shifted by t and mirrored about the vertical axis.
If t < 0, F (τ ) h ( t − τ ) = 0 because of no overlap
If t > 0, F (τ ) h ( t − τ ) = h ( t − τ )
∴ x ( t ) = 0, t < 0
t
∴ x ( t ) = ∫ h ( t − τ ) dτ , t > 0
0
Let t − τ = λ. Hence dτ = − d λ
t
t
1 −ζωn λ
sin ωd λd λ
e
mωd
0
∴ x (t ) = ∫ h ( λ ) d λ = ∫
0
eiωd λ − e − iωd λ
e ax
ax
Substitute sin ωd λ =
and use ∫ e dx =
+c
2i
a
⎛
⎞⎤
ζω
1⎡
∴ x ( t ) = ⎢1 − e −ζωnt ⎜ cos ωd t + n sin ωd t ⎟ ⎥ , t > 0
k⎣
ωd
⎝
⎠⎦
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Ex. Find the undamped response for the sinusoidal
pulse force shown using zero i.c.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
t
x ( t ) = ∫ F (τ ) h ( t − τ ) dτ
0
⎛ 2π ⎞
⎛π ⎞
F (τ ) = F0 sin ⎜
τ ⎟ = F0 sin ⎜ τ ⎟
T
2
⎝ 0 ⎠
⎝ T0 ⎠
1
h (τ ) =
sin ωnτ
mωn
h ( t − τ ) is the system impulse response shifted by t
and mirrored about the vertical axis.
If t < 0, F (τ ) h ( t − τ ) = 0 because of no overlap
∴ x ( t ) = 0, t < 0
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
⎛π ⎞ 1
If 0 < t < T0 , F (τ ) h ( t − τ ) = F0 sin ⎜ τ ⎟ ×
sin ωn ( t − τ )
⎝ T0 ⎠ mωn
t
t
⎛π ⎞
F0
sin
x ( t ) = ∫ F (τ ) h ( t − τ ) dτ =
⎜ τ ⎟ sin ωn ( t − τ ) dτ
∫
mωn 0
⎝ T0 ⎠
0
using the relation sin α sin β =
∴ x (t ) =
1
⎡cos (α − β ) − cos (α + β ) ⎤⎦ and some arrangements
2⎣
F0
π
ω
2
ω
ω
t
r
t
t
T
ω
ω
−
<
<
=
,
r
=
,
k
=
m
sin
sin
,
0
where
(
)
n
n
0
ωn
T0
k (1 − r 2 )
t
T0
0
0
If t > T0 , x ( t ) = ∫ F (τ ) h ( t − τ ) dτ = ∫ F (τ ) h ( t − τ ) dτ =
∴ x (t ) =
{
t
∫ F ( t − τ ) h (τ ) dτ
t −T0
}
F0
[sin ωt − r sin ωnt ] − ⎡⎣sin ω ( t − T0 ) − r sin ωn ( t − T0 )⎤⎦ , t > T0
2
k (1 − r )
superposition of the out-of-phase shifted sine trains
3.5 Non-periodic Excitation