complex analysis, differential equations, and laplace transform
TRANSCRIPT
1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Complex Analysis, Differential Equations, and Laplace Transform
Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell
2 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
iyxjyxz +=+=
1ji −==x)zRe( =
Where:
Complex Analysis
y)zIm( =Real part of the functionImaginary part of the function
111 jyxz +=
222 jyxz +=( ) ( )212121 yyjxxzz +++=+
Addition
3 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Complex Numbers – Multiplication and Division
)y x y j(x) y – yx (x · zz 12212121 21 ++=2
12
111 y x * · zz +=onjugate complex c* where: z1 =
jy x zif: x – jy * and: z
1
1
+==
Multiplication
Division
( ) ( )2
22
2
12212121
22
22
22
11
22
11
yxxyxyjyyxx
jyxjyx
jyxjyx
jyxjyx
+
−++=
−−
∗++
=++
4 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – Complex Number Multiplication
Perform the following multiplication and express the result in rectangular form.
( )( )2j35j2 −+−
Solution: Treating the two complex numbers as binomials, the product is obtained as
10j15j4j6 2−++−=( )( )2j35j2 −+−
1019j6 ++−= 19j4+=
5 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – Complex Number DivisionPerform the following division of complex numbers and express the result in rectangular form.
5j23j1
++−
Solution:
2911j
2913
2911j13
5j25j2
5j23j1
+=+
=−−
⋅++−
6 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – Complex Conjugate
Given: Evaluate: j21z +−= *zz
Solution:
( )( )j21j21*zz −−+−=
( ) ( ) 521 22 =+−=
7 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Location of Complex Conjugates in the Complex Plane
jyx ±−
Complex Conjugates
8 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
jyxzsinrycosrx
+=θ=θ=
Complex Numbers in Polar Form
9 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
( ) θ
θ
=θ+θ=
θ+θ=+=θ+θ=
j
j
ersinjcosrz
sinrjcosrjyxzsinjcose
Polar Form - Euler’s Formula
{ } { }
( )directionccwinpositivexytan
yxzImzRezr
1
2222
θ
=θ
+=+==
−
10 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example
Express the complex number
in polar form.
3j3z +=
3212z ==63
3tan 1 π=
=θ −
Solution:
π
= 6j
e32z
11 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
ExampleDetermine the location and the phase angle of the complex number:
j12+−
Solution: Express this number in standard rectangular form by multiplying its numerator and denominator by the conjugate of the denominator.
j111
j22j1j1
j12
−−=+−−
=−−−−
⋅+−
Location = third quadrant Phase Angle = +225º or -135º
12 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Multiplication and Division (Polar Form)
Multiplication
( )21
21
j2121
j22
j11
errzz
erz;erzθ+θ
θθ
=
==
Division( )21j
2
1
2
1 err
zz θ−θ=
Complex Conjugate
2jj
j1
rrere*zz
re*z11
1
=⋅=⋅
=θ−θ
θ−
13 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
( )nmwhere)s(D)s(N
bsbsbsasasasK)s(G
js
n1n1n
1n
m1m1m
1m
<=
++++++++
=
ω+σ=
−−
−−
K
K
Complex FunctionsComplex Variable, s
14 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Complex FunctionThe complex function can be expressed in POLE-ZERO form as:
( )( ) ( )( )( ) ( )pnspsps
zszszsK)s(G21
m21−−−−−−
=K
K
Often this can be written is partial fraction form as:
n
n
2
2
1
1ps
aps
aps
a)s(G−
+−
+−
= K
The roots of the numerator are referred to as ZEROS.The roots of the denominator are referred to as POLES.
15 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
ExampleExpress the given complex function in pole-zero form. Identify the zeros and the poles, as well as the multiplicity of each.
( ) ( )3s102ss1s2)s(G 2 ++
+=
Solution: G(s) can be written in pole-zero form as:
( )( ) ( )3.0s2ss10
5.0s2)s(G 2 ++
+=
Simple zero: s=-0.5Two simple poles: s=0 and s=-0.3Double pole: s=-2
( ) ( )3.0s2ss5.0s
51
2 ++
+
=
16 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
( )0for)t(fx1x >τ=τ
+&
Differential Equations1st Order ODE
τ is the time constant
ζ - damping ratiocc – critical dampingωn – natural frequency
)t(fxx2x 2nn =ω+ςω+ &&&
mk;m2c;
cc 2
nncc
=ωω==ζ
2nd Order ODE
17 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
2nd order – 2 integrals/derivatives – 2 constants
Initial Value Problem
Constant Coefficients – time independent
x is dependent variable – t is independent
If coefficients do not depend on x, then the equations are linear (linear superposition possible)
Has Xh and Xp – homogeneous & particular
2nd Order ODE
18 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – Types of ODEs
( )( )
( ) 0x9x1xxtxx2
t3sinxtsinx0x2x1t2xtsin2x9x3x
0xx2
2
=+++=+
=+=+−+
=++=+
&&&
&
&
&&&
&&&
&
If the coefficients are constants or functions of t, the ODE is linear. Otherwise it is non-linear.
*Highest derivative identifies the order
First-order, linear with constant coefficientsSecond-order, linear with constant coefficientsSecond-order, linear
First-order, linear
First-order, nonlinear
Second-order, nonlinear
19 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1st Order ODE
Consider the single-tank, liquid-level system shown in the figure below. The mathematical model of this system is given by the following first-order, linear ODE with constant coefficients.
)t(qgRhh
gRA
i=+&
FlowrateIN
FlowrateOUTLevel
20 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1st Order ODE
The ODE can easily be expressed in the standard form as:
)t(qA1h
RAgh i=+&
As a result, the system’s time constant is identified as:
gRA
=τ
21 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response – First Order
Homogeneous Solution
00x1x >τ=τ
+&
Characteristic Equation
τ−=λ⇒=
τ+λ
101
The solution becomes
τ−=
tce)t(x
22 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Step Response – First Order
Consider a 1st order system described as:
)t(fBxxA =+&
And subjected to a step input.
Bf(t)x
dtdx
=+τ
In standard form:
Af(t)x
τ1
dtdx
=+
23 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Step Response – First Order
First Order homogeneous response is in the form of an exponential function
th e)t(y λ−= where
τ=λ
1
The solution is
where C is determined from initial condition for a particular solution and yp indicates the particular solution.
)t(yCe)t(y pt+= τ−
24 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Step Response – First Order
If the system is subjected to a step change
>=≤=
µ=0t10t0
)t()t(f s
The particular solution can be found to be
1Ce)t(yt+= τ−
Given the initial conditions when the system is initially at rest (t=0, ys(0)=0 requires that C=-1 which then gives
τ−−=
te1)t(y
25 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Step Response - First Order
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (sec)
Perc
enta
ge o
f Res
pons
e
τ 2τ 3τ
63.2
86.595.0
Step Response – First Order
26 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Step Response – First OrderResponse is broken up into two regions:1. Transient region where system is still responding
dynamically2. Steady-state region where system has reached its final
valueNote: There is no clear break point; often, 3τ, 4τ, or 5τ ischosen based on desired accuracy
The initial slope of the response may be found by differentiating y(t) to give
τ=
1dtdy
27 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Step Response – First Order
Using the following equation
HOMEWORK ASSIGNMENT -----
)t(fBxxA =+&
Calculate the response by hand and plot by hand. Let A = month and B = day of your birthday
Use MATLAB to confirm your results
28 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Transfer Function of Step Response
)t(fA1x1x
)t(fB1xx
BA
)t(fBxxA
BA =
+
=+
=+
&
&
&
( )ABs
A1
f(s)x(s)
+
=
Diff. Eq.
reduces to
or
Transfer Function is
Transferring to the LaPlace domain:
29 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Step Response – First Order
Using MATLAB and the equivalent Laplace form, the system transfer function is described as
+
=
+
−
ABS
A1
ABS
A1 1
› NUM = [0 1/A];› DEN = [1 B/A];› step(NUM,DEN)
Step Response - First Order
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (sec)
Perc
enta
ge o
f Res
pons
e
τ 2τ 3τ
63.2
86.595.0
30 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response – Second Order
Homogeneous Solution
0xx2x 2nn =ω+ζω+ &&&
Characteristic Solution
02 2nn
2 =ω+λζω+λ
( )
12nn
2n
2nn2,1
−ζω±ζω−=
ω−ζω±ζω−=λ
mk;m2c;
cc 2
nncc
=ωω==ζ
31 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response – Second Order
For purposes of development of these general equations a simple mass, spring, dashpot system will be used.
Equation of motion is obtained from Newton’s second law
)t(fkxdtdxc
dtxdm 2
2=++
With I.C.’s
ox)0(x = 0x)0(x && =and
k
c
m
x(t)
f(t)
32 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response – Second Order
A solution of the form fitstex λ=
( ) 0ekcm t2 =+λ+λ λ
0kcm 2 =+λ+λ
Characteristic equation
m2
mk4cc 2
−±−
=λ⇒
Solution has 2 roots and 3 possible solutions depending on the term under the √.
33 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response - Case 1 ( )1mk4c2 <ζ<
( )m2
cmk4c 2
−−±−
=λ
jm2
cmk4m2c 2
2,1−
±−=λ
jβ±α−= (Two complex conjugate roots)
The solution is
( ) ( )tjtjh BeAex β−α−β+α− +=
34 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Factor out: te α−
( )tjtj ββ −−+=⇒ BeAeex αt
h
Recall that:
θ−θ=θ+θ= θ−θ sinjcose;sinjcose jj
Then,
( ) ( )[ ][ ]tjctce
tjtBtjtAext
th
ββ
ββββα
α
sincos
sincossincos
21 +=
−++=−
−
Free Response - Case 1
35 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response - Case 1
Using
equation can be written as
ysinxcosycosxsin)yxsin( +=+
=α
m2c( )φβα += − tcex t
h sin
φ+
−=
−t
m2cmk4sincex
2tm2c
h
36 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response - Case 1
Now if we divide through by m, then
0mfxx2x 2
nn ==ω+ζω+ &&& (homogeneous)
whereζ = damping ratio =
ωn = natural frequency =
km2c
cc
c=
mk
( )φ+ω=⇒ ζω− tsinex dtn
where 2nd 1 ζ−ω=ω
37 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response - Case 1
1. For small ζ ---› ωn ≈ ωd2. ωn is independent of damping3. If c = 0, then ωn ≡ ωd4. Solution response is always of the form of a
damped exponentially decaying sinusoidal formFree Response - Damped Exponential Decay
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 0.5 1 1.5 2 2.5 3
Time (sec)
Mag
nitu
de
38 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response – Case 2 ( )1mk4c2 >ζ>
m2mk4c
m2c 2 −
±=λ Two REAL Roots
t2
t1
21 ececx λλ +=Recall
θ−θ=θ+θ= θ−θ coshsinhe;coshsinheThen
tcoshctsinhcx 2413 λ+λ=
Solution of this type will always be of the form of a lag in the system.
39 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response – Case 2
Response will have an exponential envelope but will not have oscillatory motion about steady-state.
Damping ζ > 1
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (sec)
Am
plitu
de
40 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response – Case 3 ( )14mkc2 == ζ
m2c
11 −=λ=λ Two REAL REPEATED Roots
t2
t1
21 tececx λλ +=
Solution of this type will also be in the form of a lag to the system, but this system will return to steady state faster than any other damping without overshoot.
This is the break point between structural dynamics and controls problems.
41 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Free Response - Summary
1
1
1
>ζ
=ζ
<ζ
42 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
S-plane Representation
dn1 jω+ζω−=λ
dn*
1 jω−ζω−=λ
n2,1 ζω−=λ1
1
1
>ζ
=ζ
<ζX – Conjugate Poles
– Repeated Roots
– Real Roots
43 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
jω
σ
UNSTABLESTABLE
TIME
FRF
FRF
TIME
FRF
TIME
FRF
TIME
TIME
TIME
> 1.0ζ
= 1.0ζ
= 0.7ζ
= 0.3ζ
= 0.1ζ = 0ζ
S-PLANE PLOTS FOR IMPULSE RESPONSE
OF A SINGLE DEGREE OF FREEDOM
MECHANICAL SYSTEM
44 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.12
Example from Dynamic Systems by Vu and Esfandiari
45 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.12 (cont)
Example from Dynamic Systems by Vu and Esfandiari
46 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.13
Example from Dynamic Systems by Vu and Esfandiari
47 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.14
Example from Dynamic Systems by Vu and Esfandiari
48 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.16
Example from Dynamic Systems by Vu and Esfandiari
49 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.16 (cont)
Example from Dynamic Systems by Vu and Esfandiari
50 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.17
Example from Dynamic Systems by Vu and Esfandiari
51 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.17 (cont)
Example from Dynamic Systems by Vu and Esfandiari
52 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.18
Example from Dynamic Systems by Vu and Esfandiari
53 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example –1.18 (cont)
Example from Dynamic Systems by Vu and Esfandiari
54 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.19
Example from Dynamic Systems by Vu and Esfandiari
55 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example 1.19 (cont)
Example from Dynamic Systems by Vu and Esfandiari
56 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Using Matlab to Solve Differential Equations
Matlab’s “dsolve” command is a common alternative to solving complicated differential equations by hand.
Example 1.17 will be solved again using Matlab
57 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Using Matlab to Solve Differential Equations
The same solution can be obtained with Matlaband compared to the solution from Example 1.17.
( ) t4t2t e31ee
35tx −−− +−=
58 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Differential Equations with force
or I.C. in time domain
Algebraic Equationin
Laplace domain
Rearrange terms intogood form
Closed-Form Solution or
Numerical Solution
Laplace Transform using I.C.
Inverse Laplace
Alternative: Convolution Integral
(very difficult)
Laplace Transform
Frequency domain X(s)Time domain x(t)
59 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Laplace Transform Equation
{ }
{ }{ } )0()0()()(
)0()()(
)()()(
2
0
ggssGstggsGstg
dttgetgsG st
&&&
&
−−=
−=
⋅== ∫∞ −
L
L
L
Derivatives
60 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Laplace Transform of several functions
ExponentialExponential
SINSIN
COS
Unit SinusoidUnit Sinusoid
1Unit Impulse (Dirac Unit Impulse (Dirac delta function)delta function)
Unit PulseUnit Pulse
Unit RampUnit Ramp
Unit StepUnit Step
Laplace TransformLaplace TransformFunction f(t)
ssµ
s12s1
22
22
ωssωs
ω
+
+
as1+
61 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Laplace Transform
{ } ∫∞ −==
0)()()( dtetusutu st
sssL
su
sue
su sssts =
−−−=−= ∞− 0|0
•Unit Step
•Unit Ramp
{ } ∫∞ −==
0)()( dtetsutu st
rsL
202
stst
00
st
s1|
sedt
se|
set =
−=
−−
−= ∞
−−∞∞−
∫
Requires manipulation of Laplace domain equation to get in a suitable form to apply L -1.
•Inverse Laplace
62 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Convolution Integral
{ } { } ))(()()()( 11 thgsHsGsF ⊗=•= −− LL
•If G(s) and H(s) have known inverses g(t) and h(t), then the product of GH can be obtained by the convolution integral.
))(()()(0
thgdtthgt
∫ ⊗=−= ττ
))(()()(0
tghdttght
∫ ⊗=−= ττ
63 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Unit Impulse Response – First Order
)t(x1x δ=τ
+&
•The Equation
1)s(X1)s(Xs =τ
+
0x0 =
•Laplace
1)s(X)1s( =τ
+
τ+
= 1s
1)s(X τ−=
te)t(x∴
look up inverse laplace
64 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Laplace – First Order ODE
•The basic first order ODE can be expressed in the Laplace Domain as for unit impulse and can be recast as
•This can be stated as follows:The basic value of is multiplied by A -- This
value is equal to 1 minus B times the integral of
•Normalize the equation so the coefficient on =1,
sAB
A1s −=∴ &
1BssA =+&Bs1sA −=&
s&s&
s&
65 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Block Diagram – First Order ODE
s1 SCOPE
integrate
Multiply byB/A
Normalizeto A coefStep
= 1/A
1
-
+ s& s
66 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
MATLAB/SIMULINK – (see tutorial)
•Simulink•File New Model (workspace appears)Select the following items and place them on the worksheet
Unit Step (from Sources) – change amplitudeSum (from Math) – need + and –Gain (from Math) – change gain valueIntegrator (from Continuous)Scope (from Sinks)
Double click items to view or change property GAIN block can be rotated by format.T branch – mouse online/CTRL and right mouse button to extend line.
67 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
SIMULINK – First Order Step Response
68 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
SIMULINK – First Order Impulse Response
69 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Unit Impulse Response – Second Order
•For unit impulse, 0x0,xδ(t),f(t) 00 === &
Then,
∴
δ(t)xωx2ζx 2nn =++ &&& ω
with 00 =x& 0x0 =and
•Laplace with I.C. = 0
1)s(X)s2s( 2nn
2 =ω+ζω+
2nn
2 s2s1)s(X
ω+ζω+=
70 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Unit Impulse Response - Second Order
•Note that (assume ζ<1)
2d
2)s(1)s(X
ω+σ+=So that
2n
2n
2n
2nn
2 )()s(s2s ω+ζω−ζω+=ω+ζω+
)1()s( 22n
2n ζ−ω+ζω+=
2d
2)s( ω+σ+=
71 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
++⋅= −
221
)(1)(
d
d
d stx
ωσω
ωL
•The inverse Laplace
te dt
d
ωω
σ sin1 −=
te dt
d
ωω
ζω sin1 −=
If I.C. ≠ 0, then a more involved (but possible solution) exists.
Unit Impulse Response - Second Order
72 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Unit Step Response – Second Order
•For unit step
)t(uxx2x 2nn =ω+ζω+ &&&
•Laplace with IC=0 gives
s1)s(x)s2s( 2
nn2 =ω+ζω+
∴
with
0x,0x),t(u)t(f 00 === &
0x,0x 00 ==&
2nn
2 s2s1
s1)s(X
ω+ζω+⋅=
73 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Unit Step Response – Second Order
•Again assume ζ<1 - then
+++
−= 222 2211
nn
n
n ωsζωsζωs
sωX(s)
•Then the inverse Laplace
++
+−= −
2221
2211)(
nn
n
n ωsζωsζωs
sωtX L
−+−= − ttetX dd
t
n
n ωζ
ζωω
ζω sin1
cos11)(22
74 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.40
Example from Dynamic Systems by Vu and Esfandiari
75 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
Example – 1.40 (cont)
Example from Dynamic Systems by Vu and Esfandiari
76 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
General Laplace Formulation
•Laplace
)t(fxx2x 2nn =ω+ζω+ &&&
[ ] F(s)x(s)ωx(0)sx(s)2ζ(0)xsx(0)x(s)s 2nn
2 =+−+−− ω&
)()()()(
)0()()()0()0()()( 2
sFtFsXx
xssxxxsxsxsx
==
−=−−=
LLLL
&
&&&
77 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
General Laplace Formulation
If initial conditions are zero, then the system transfer function is
)s(x)s2s( 2nn
2 ω+ζω+
2nn
2 s2s1)s(H
)s(F)s(x
ω+ζω+==
Applied Force and Initial Conditions
(0)x)x(0)ζω2(sF(s) n &+++=
2nn
2n
2nn
2 ωsζω2s(0)x)x(0)ζω2(s
ωsζω2sF(s)x(s)
++++
+++
=&
inputoutput
Many books use G(s)!
kcsms1aka 2 ++
78 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
SIMULINK – 2nd Order Impulse Response (or RAMP)
)t(Fx1000x40x100 =++ &&&
OR…
79 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
SIMULINK – 2nd Order Impulse Response (or RAMP)
)t(Fx1000x40x100 =++ &&&
In alternate form (using the Transfer Function Block)
80 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics
SIMULINK – 2nd Order Impulse Response (or RAMP)
This example is
General Equation is
)t(Fkxxcxm =++ &&&
[ ]xckx)t(Fm1x &&& −−=
)t(Fx1000x40x100 =++ &&&