Identification Methods for Structural Systems · multiple degree of freedom systems (mdof). We will use an approximation of a structural system as the main example for our demonstration

Identification Methods for Structural Systems

Prof. Dr. Eleni Chatzi

Lecture 4 - Part 1

Institute of Structural Engineering Identification Methods for Structural Systems 1

Fundamentals

Overview

What have we seen so far?

Properties of the response of dynamic systems and how theseare visible in the time and frequency domain

The FRF H(ω) and the TF H(s) for a SDOF system

What is shown in this class?

Continuous State-space Formulation

Multiple dof Systems

Solution of mdof Systems

Continuous versus Discrete Formulations


Fundamentals

Overview










Fundamentals

Overview










Fundamentals

Overview










State Space Formulation

In the previous lecture, we discussed the usefulness of a system’s transfer,or frequency response, function. However, as systems become morecomplex, e.g. when their dimension increases, representing them withdifferential equations or transfer functions becomes cumbersome. Thisissue is more critical for multiple degree of freedom systems, where multipleinputs and outputs may exist. To tackle this problem, we will introducehere the continuous state-space form of a system, which will degenerate thegeneral nth order differential equation, describing the system dynamics,with a first order matrix differential equation.The continuous state-space form is described as:

x = Ax + Bu (state Eqn)

y = Cx + Du (observation Eqn)

for more info, see here


http://lpsa.swarthmore.edu/Representations/SysRepSS.html


Continuous State-space form

x = Ax + Bu (state Eqn)

y = Cx + Du (observation Eqn)

wherex is the state vector ∈ Rn, it is a function of timeA is the state matrix ∈ Rnxn, a constantB is the input matrix ∈ Rnxp, a constantu is the input ∈ Rp, a function of timeC, is the output matrix ∈ Rnxm, a constantD is the direct transition (or feedthrough) matrix ∈ Rmxp, a constanty is the output r ∈ Rm, a function of time and it represents the quantity weobserve or measure



http://lpsa.swarthmore.edu/Representations/SysRepSS.html


From Differential Equations to Continuous State-space form

There exist more than one methods for bringing a differential equation intoa continuous state space form. We will here describe the controllablecanonical form.We will use a simple illustrative example to describe the process.Assume a single degree of freedom (sdof) Single Input Single Output(SISO) dynamical system described by the following equation of motion indifferential form:

d3y

dt3+ 2

d2y

dt2+ 3

dy

dt+ 4y = 5u




d3y

dt3+ 2

d2y

dt2+ 3

dy

dt+ 4y = 5u

Step 1:Define the state-space vector. We define new variables, the so-calledsystem states. The state variables are the smallest possible subset ofsystem variables that can represent the entire state of the system at anygiven time. The number of state variables n, is typically chosen equal tothe order of the system’s differential equation. In this case we have a 3rdorder equation, hence we define 3 states, corresponding to the derivativesof the original system response variable y :

x1 = y , x2 =dy

dt=

dx1

dt, x3 =

d2y

dt2=

dx2

dt

x = [x1 x2 x3]T




d3y

dt3+ 2

d2y

dt2+ 3

dy

dt+ 4y = 5u (1)

Step 2:Rewrite Equation (1), using the state variables:

dx3

dt+ 2x3 + 3x2 + 4x1 = 5u (2)

Notice how the above expression contains 3 variables, but is a first order

differential equation.




Step 3:Rewrite Equation (2) in matrix form:

x =d

dt

x1

x2

x3

=

0 1 00 0 1−4 −3 −2

︸︷︷︸

A

x1

x2

x3

+

005

︸︷︷︸

B

u

This equation is of the form x = Ax + Bu

Notice how the last row of the state matrix A occurs by solving equation

(2) in terms ofdx3

dt.




Step 4:Write the observation Equation, which delivers the actually desiredresponse quantity, i.e., y :

y =[

1 0 0]︸︷︷︸

C

x1

x2

x3

+[

0]︸︷︷︸

D

u

This equation is of the form y = Cx + Du.



C Tackling derivatives of the input u

The standard state space form, admits no derivatives of the input (load) uon the right hand side. However, the original equation of motion couldinclude such derivatives, as for instance the case of a base-excited structurein absolute coordinates (see Lecture 3, slide 11).For instance, assume the same SISO dynamical system described by thefollowing equation of motion in differential form:

d3y

dt3+ 2

d2y

dt2+ 3

dy

dt+ 4y = 2

d2u

dt2+ 5u

In order to bring this system into a canonical state-space form, we use atransformation to a new variable y :

y = 2d2y

dt2+ 5y



Tackling derivatives of the input

By substituting Equation (4) into Equation (3) we obtain:

2d2

dt2

(d3y

dt3+ 2

d2y

dt2+ 3

dy

dt+ 4y

)+ 5(d3y

dt3+ 2

d2y

dt2+ 3

dy

dt+ 4y

)=

2d2u

dt2+ 5u

which implies that y satisfies:

d3y

dt3+ 2

d2y

dt2+ 3

dy

dt+ 4y = u

This is now in regular form, and we may continue with the previous steps

for deriving the state-space form.


Multiple DOF Systems

1m1 1xk

1 1c x

( )1 tF

2m

( )2 2 1x xk −

( )2 2 1x xc −

( )2 tF

( )2 2 1x xk −

( )2 2 1x xc −

FBD

(Lumped Mass System)

1m

1k

1c( )1 tx

( )1 tF

2m

2k

1c( )2 tx

( )2 tF

The equations of motion can be written as

m1x1 + (c1 + c2)x1 − c2x2 + (k1 + k2)x1 − k2x2 = F1(t)

m2x2 + c2x2 − c2x1 + k2x2 − k2x1 = F2(t)

The system can be written in matrix form as follows:[m1 00 m2

] [x1

x2

]+

[c1 + c2 −c2

−c2 c2

] [x1

x2

]+

[k1 + k2 −k2

−k2 k2

] [x1

x2

]=

[F1(t)F2(t)

]Eq. (1)


Multiple DOF Systems

From sdof to modf

We have so far seen the general steps for converting a sdof systemdescribed by an ordinary differential equation into state-space form.

We will now demonstrate how the same context can be applied intomultiple degree of freedom systems (mdof).

We will use an approximation of a structural system as the main example

for our demonstration (see next slide)


State Space Equation Formulation for MDOF systems

2dof Mass Spring System

or otherwise more compactly, in matrix form:

Mxd + Cxd + Kxd = u

where xd =[x1 x2

]Tand u =

[F1 F2

]TWe now introduce the augmented state vector:

x =[

xd xd]T

=[x1 x2 x1 x2

]T. Then,

x =

x1

x2

x1

x2

=

0 0 1 00 0 0 1[−M−1K

] [−M−1C

]

x1

x2

x1

x2

+

0 00 0[M−1

][ F1

F2

]


State Space Equation Formulation for MDOF systems

State Equation

This is rewritten as:

x =

[��O2x2 I2x2[

−M−1K]

2x2

[−M−1C

]2x2

]x +

[02x2[M−1

] ]u

We therefore obtain the following equivalent state-space formulation:

⇒ x = Ax + Bu

where it is reminded that xd =[x1 x2

]T, u =

[F1 F2

]TFor a general n-dimensional system (n dofs), matrices A and B areobtained as:

A =

[��Onxn Inxn[

−M−1K]nxn

[−M−1C

]nxn

]B =

[0nxn[M−1

] ]Institute of Structural Engineering Identification Methods for Structural Systems 15

State Space Equation Formulation

Observation EquationThe observation equation contains the quantities we “observe”, i.e.measure, using relevant sensors. Assume we monitor (measure)both displacements x1, x2, e.g. via a laser sensor. Then the“observation vector” is:

y =

[1 0 0 00 1 0 0

]︸︷︷︸

C

x1

x2

x1

x2

+��O2×2u(t)

Assume we monitor (measure) the 2nd dof velocity, x2, e.g. via ageophone sensor. Then the “observation vector” is:

y =[

0 0 1 0]︸︷︷︸

C

x1

x2

x1

x2

+��O1×2u(t)



Observation EquationAssume we monitor both accelerations, x1, x2 via accelerometers:

y =[ [−M−1K

]2x2

[−M−1C

]2x2

]︸︷︷︸C

x1

x2

x1

x2

+[

M−12x2

]︸︷︷︸D

u

If we only monitor the 1st dof acceleration, x1, the “observation vector”is:

y =[−(k1 + k2)/m1 k2/m1 −(c1 + c2)/m1 c2/m1

]︸︷︷︸C

x1

x2

x1

x2

+ 1/m1︸︷︷︸D

F1

The observation vector is written in matrix form as y = Cx + Du



Note

Using the state space representation we have converted a 2nd orderODE into an equivalent 1st order ODE system.

We can now use any of the aforementioned 1st order ODEintegration methods in order to convert the continuous system intoa discrete one and obtain an approximate solution

For instance MATLAB’s ode45, which is a Runge Kutta integrationscheme may be used.

What are other integration schemes that may be utilized?


Numerical Integration for 1st order ODEs

Using these methods a continuous system is brought into an equivalentdiscrete formulation and an approximative solution is sought. 1st orderODE Integration Methods

Assumedy

dt= f (t, y(t)), y(t0) = 0

Forward Euler Method

yn+1 = yn + hf (tn, yn)

where h is the integration time step. This explicit expression isobtained from the truncated Taylor Expansion of y(tn + h). More infohere

Backward Euler Method

yn+1 = yn + hf (tn+1,yn+1)

This implicit expression (since yn+1 is on the right hand side) isobtained from the truncated Taylor Expansion of y(tn+1 − h). where his the integration time step. This explicit expression is obtained fromthe truncated Taylor Expansion of y(tn + h).


http://lpsa.swarthmore.edu/NumInt/NumIntFirst.html

Numerical Integration for 1st order ODEs

2nd Order Runge Kutta (RK2)

k1 = hf (tn, yn), k2 = hf (tn +1

2h, yn +

1

2k1)

yn+1 = yn + k2 + O(h3)

More info here

4th Order Runge Kutta (RK4) - MATLAB ode45funcction

k1 = hf (tn, yn), k2 = hf (tn +1

2h, yn +

1

2k1)

k3 = hf (tn +1

2h, yn +

1

2k2), k4 = hf (tn + h, yn + k3)

yn+1 = yn +1

6k1 +

1

3k2 +

1

3k3 +

1

6k4 + O(h5)

More info hereInstitute of Structural Engineering Identification Methods for Structural Systems 20

http://lpsa.swarthmore.edu/NumInt/NumIntSecond.html

http://lpsa.swarthmore.edu/NumInt/NumIntFourth.html

Solution of mdof systems

Laplace Transform for MDOF SystemsAssume the previous 2-DOF system, which we bring to a state-space form,i.e., the system may be summarized by the following 1st order ODE asfollows:

x(t) = Ax(t) + Bu(t) state-space or process equationy(t) = Cx(t) + Du(t) measurement or observation equation

Since we are now dealing with higher dimensions, let us now define theLaplace Transform for a vector of for instance 4 components:

L

x1(t)x2(t)x3(t)x4(t)

=

L{x1(t)}L{x2(t)}L{x3(t)}L{x4(t)}

= X (s)

Also, the Laplace property in relation to differentiation:

L{x(t)} = sX (s)− X (0) applies now in vector form.



Laplace Transform for MDOF Systems

Let us therefore return to the original state-space formulation and apply theLaplace Transform therein:

x(t) = Ax(t) + Bu(t)y(t) = Cx(t) + Du(t)

⇒ sX (s)− X (0) = AX (s) + BU(s)Y (s) = CX (s) + DU(s)

(sI− A)X (s) = BU(s) + X (0)⇒X (s) = (sI− A)−1BU(s) + (sI− A)−1X (0)

Assuming D = 0 and 0 Initial Conditions the observation equation yields

X (s) = (sI− A)−1BU(s)



Laplace Transform for MDOF Systems

The Input-Response Transfer FunctionThe previous relationships provides the connection, in the form of a transferfunction, H(s), between the input (load), U(s), and the response (output),X (s):

H(s) = (sI− A)−1B

H(s) can reveal significant information about the system as we havealready seen for the case of SDOF systems.However, we are interested in obtaining the solution in a time domain form,x(t). In doing so, we may use the Inverse Laplace transform:

x(t) = L−1{

(sI− A)−1BU(s)}



Reminder: Impulse Response FunctionThe IRF is provided through the Inverse Laplace of the FRF (fromfrequency to time domain):

h(t) = L−1H(ω)

The Input-Measurement Transfer FunctionOf additional interest is the connection, in the form of a transferfunction, HY (s), between the input (load), U(s), and themeasurement vector Y (s)This may be obtained as:

Y (s) = {C(sI− A)−1B}U(s)⇒HF (s) = C(sI− A)−1B



Note: In calculating the TF the inverse of a matrix needs to be calculated:

by definition: (sI− A)−1 =adj(sI− A)

det(sI− A)

where adj(Z) =[εij |Z

�ij

], Z

�ij: det(Z) without row i, column j

Then, det(sI− A)Y (s) = C [adj(sI− A)B]U(s)

where det(sI− A) = sn + an−1sn−1 + · · ·+ a0 is the characteristic

polynomial with roots λ1, · · · , λn

These roots are directly related to the eigenvalues of the original eigenvalue

problem: (K−Mω2)φ = 0



mdof Structural Systems

As an alternative, we can obtain the Transfer Function (TF) for theMDOF, directly from the Laplace Transform of the dynamicequation of motion (in matrix form):

mx + cx + kx = f ⇒[ms2 + cs + k

]X(s) = F(s)⇒

TF: H(s) =[ms2 + cs + k

]−1

For underdamped systems with symmetric matrices like the above,using partial fraction expansion the TF can be written as:

H(s) =m∑

k=1

Ak

s − λk+

A∗k

s − λ∗k* denotes complex conjugate

where

λk = −ζkωk + iωk

√1− ζ2

k



For s = iω the TF is evaluated on the Imaginary axis yielding theFrequency Response Function (FRF) with the following terms

Hij(ω) =m∑

k=1

kAij

iω − λk+

kA∗ijk

iω − λ∗k* denotes complex conjugate

The term Hij(ω) corresponds to a particular output response at point i dueto an input force at point j . This follows from the property L{δ(t)} = 1.

Since for a structural system m, c, k are symmetric we have that Hij = Hji

(reciprocity). This means that each term could be experimentally evaluatedby applying impact at a point i and recording the response at point j .The terms αijk are named the residues and they are related to the modeshape (eigenvector) matrix Φ as:

kAij = qkφikφjk



Solving MDOF systems via the continuous state-space form:

Other than the inverse Laplace transform, we may solve mdofsystems by using the state-space formulation:Examine the state-space system with no input, u = 0:

x = Ax given I.C. at x(0)

Assume the above system has a solution of the type x = Cestφ,where φ is a vector.Plug this in the differential equation to obtain:

sestφ = Aφest ⇒ [sI− A]φ = 0



This is a standard eigenproblem with eigenvalues s obtained fromdet(sI− A) = 0 and φ are the corresponding eigenvectorsThen, the total solution of the system is obtained throughsuperposition as

x(t) = C1φ1es1t + C2φ2e

s2t + · · ·+ Cnφnesnt

Hence, at t = 0⇒ x(0) = C1φ1 + C2φ2 + · · ·+ Cnφn =

=[φ1 φ2 · · · φn

]

C1

C2...Cn

⇒ C1

C2...

Cn

= Φ−1x(0)

where Φ is the matrix of eigenvectors, which is directly related to thestructure’s modal shape matrix.



and therefore, x(t) =[φ1 φ2 · · · φn

] C1es1t

...Cne

snt

=

= Φ

es1t · · · 0...

. . ....

0 · · · esnt

C1

C2

...Cn

⇒ x(t) = Φ

es1t · · · 0...

. . ....

0 · · · esnt

Φ−1x(0)



We define the State Transition Matrix M(t) as

M(t) = Φ

es1t · · · 0...

. . ....

0 · · · esnt

Φ−1 and x(t) = M(t)x(0)

If Initial Conditions are specified at a generic point t0 we get:

M(t − t0) = Φ

es1(t−t0) · · · 0...

. . ....

0 · · · esn(t−t0)

Φ−1

and x(t) = M(t − t0)x(t0)



State Transition Matrix PropertiesA. Assume we would like to obtain the connection of the response betweenthree successive time interval, t0,t1,t2.

x(t1) = M(t1 − t0)x(t0)x(t2) = M(t2 − t1)x(t1)

}⇒

x(t2) = M(t2 − t1)M(t1 − t0)x(t0)

Which by definition means that

M(t2 − t0) = M(t2 − t1)M(t1 − t0)

B. Now consider t2 = t0 (so we move 1 step back in time) thenI = M(t0 − t1)M(t1 − t0)⇒

M(t1 − t0)−1 = M(t0 − t1)

Comparing the above to the properties of an exponential we observe:

eα(t2−t1)eα(t1−t0) = eα(t2−to)

(eα(t1−t0))−1 = eα(t0−t1)

}⇒ similar traits



The Matrix ExponentialKnowing that M(t) behaves like an exponential, let us assume A isan m × n square matrix and

define: eAt = I + At +1

2!A2t2 +

1

3!A3t3 · · ·

If we plug this in x = Ax , where A is the state-space matrix:

A +2

2!A2t +

3

3!A3t2 · · · = A(I + At +

1

2!A2t2 +

1

3!A3t3 · · · )

we notice that eAt is a solution to our original problem! Indeed, wemay write the solution as:

x = Ax ⇒{

x(t) = eAtx(0), for given x(0)

x(t) = eA(t−t0)x(t0), for given x(t0)

}



The Matrix ExponentialThis implies that the state transition matrix actually is:

M(t) = eAt

In fact, the full solution to x = Ax + Bu is proved to be (for IC x(0)):

x(t) = eAtx(0) +

∫ t

0eA(t−τ)Bu(τ)dτ

Notice the resemblance to Duhamel’s Integral

It is interesting to therefore note the equivalence between SDOF andMDOF systems.



from Continuous to Discrete timeThe continuous state space form was derived as:

x(t) = Ax(t) + Bu(t) (continuous state Eqn)

y(t) = Cx(t) + Du(t) (observation Eqn)

Information from sensors (monitoring) comes at discrete time intervals,with measurements obtained usually at a fixed interval, e.g. every Tseconds. For such systems, it is more convenient to use a discretizedrepresentation (state-space) to describe the system:

x((k + 1)T ) = G(T )x(KT ) + H(T )u(KT ) (discrete state Eqn)

y(kT ) = Cx(kT ) + Du(KT ) (observation Eqn)

where k is an integer. The discretized state-space matrices G,H depend onthe value of the sampling interval T , while the observation equationremains the same in both the continuous and discrete form (since it issimply a linear equation, and not a differential equation).



from Continuous to Discrete timeThe discrete matrices may be obtained by using the solution of thecontinuous state space equation:

x((k + 1)T ) = eA(k+1)Tx(0) + eA(k+1)T

∫ (k+1)T

0

e−AτBu(τ)dτ

x(kT ) = eAkTx(0) + eAkT

∫ kT

0

eA(−τ)Bu(τ)dτ

Using the above equations, we express x((k + 1)T ) as a function of x(kT ):

x((k + 1)T ) = eATx(kT ) + eA(k+1)T

∫ (k+1)T

kT

e−AτBu(τ)dτ



from Continuous to Discrete time

By assuming u(t) = u(kT ) within the interval [kT (k + 1)T ] (zero-orderhold assumption), we may take this expression together with matrix B, outof the integral to obtain:

x((k + 1)T ) = eATx(kT ) + eA(k+1)T

∫ (k+1)T

kT

e−Aτdτ Bu(kT )

τ ∈ [kT , (k + 1)T ]

With an appropriate change of variables, it is proven that the discretestate-space matrices are obtained as:

G(T ) = eAT

H(T ) = A−1(eAT − I

)B



http://www.engr.iupui.edu/~skoskie/ECE595_f05/handouts/discretization.pdf

Documents

Identification Methods for Structural Systems · multiple degree of freedom systems (mdof). We will use an approximation of a structural system as the main example for our demonstration