1 Lavi Shpigelman, Dynamic Systems and control – 76929 – Linear Time Invariant systems definitions, Laplace transform, solutions, stability.

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Linear Time Invariant systems

definitions, Laplace transform, solutions, stability

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Ready to GO !?

Disclaimer: the following slides are a quick review of Linear, Time Invariant systems

If you feel a bit disoriented think:a. I can easily read up on it (references will be given)b. Its VERY simple if you’ve seen it before (i.e. intuitive).c. This part will be over soon.

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Lumpedness and causality

Definition: a system is lumped if it can be described by a state vector of finite dimension. Otherwise it is called distributed.Examples: • distributed system: y(t)=u(t- t)

• lumped system (mass and spring with friction)

Definition: a system is causal if its current state is not a function of future events (all ‘real’ physical systems are causal)

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Linearity and Impulse Response description of linear systems Definition: a function f(x) is linear if

(this is known as the superposition property)

Impulse response: Suppose we have a SISO (Single Input Single Output) system

system as follows:

where: y(t) is the system’s response (i.e. the observed output) to the

control signal, u(t) . The system is linear in x(t) (the system’s state) and in u(t)

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Linearity and Impulse Response description of linear systems Define the system’s impulse response, g(t,), to be the

response, y(t) of the system at time t, to a delta function control signal at time (i.e. u(t)=t) given that the system state at time is zero (i.e. x()=0 )

Then the system response to any u(t) can be found by solving:

Thus, the impulse response contains all the information on the linear system

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Time Invariance

A system is said to be time invariant if its response to an initial state x(t0) and a control signal u is independent of the value of t0.So g(t,) can be simply described as g(t)=g(t,)

A time linear time invariant system is said to be causal if

A system is said to be relaxed at time 0 if x(0) =0 A linear, causal, time invariant (SISO) system that is

relaxed at time 0 can be described bycausal

relaxed ConvolutionTime invariant

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LTI - State-Space Description

Every (lumped, noise free) linear, time invariant (LTI) system can be described by a set of equations of the form:

Linear, 1st order ODEs

Linear algebraic equations

Controllable inputs u

State xDisturbance

(noise) wMeasurement Error (noise) n

Observationsy

Plant

Dynamic Process

A

B+

ObservationProcess

C

D+x

u

1/s

Fact: (instead of using the impulse response representation..)

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What About nth Order Linear ODEs? Can be transformed into n 1st order ODEs

1. Define new variable:

2. Then:

Dx/dt = A x + B u y = [I 0 0 0] x

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Using Laplace Transform to Solve ODEs The Laplace transform is a very useful tool in the

solution of linear ODEs (i.e. LTI systems).

Definition: the Laplace transform of f(t)

It exists for any function that can be bounded by aet (and s>a ) and it is unique

The inverse exists as wellLaplace transform pairs are known for many

useful functions (in the form of tables and Matlab functions)

Will be useful in solving differential equations!

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Some Laplace Transform Properties Linearity (superposition):

Differentiation

Convolution

Integration

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Some specific Laplace Transforms (good to know) Constant (or unit step)

Impulse

Exponential

Time scaling

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Using Laplace Transform to Analyze a 2nd Order system Consider the unforced (homogenous) 2nd order system

To find y(t):

Take the Laplace transform (to get an algebraic equation in s)

Do some algebra

Find y(t) by taking the inverse transform

characteristic polynomial determined by Initial condition

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2nd Order system - Inverse Laplace The solution of the inverse depends on the

nature of the roots 1,2 of the characteristic polynomial p(s)=as2+bs+c:• real & distinct, b2>4ac • real & equal, b2=4ac • complex conjugates b2<4ac

In shock absorber example:a=m, b=damping coeff., c=spring coeff.

We will see: Re{} exponential effectIm{} Oscillatory effect

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Real & Distinct roots (b2>4ac)

Some algebra helps fit the polynomial to Laplace tables.

Use linearity, and a table entry To conclude:

• Sign{} growth or decay• || rate of growth/decay

p(s)=s2+3s+1y(0)=1,y’(0)=01=-2.622=-0.38

y(t)=-0.17e-2.62t+1.17e-0.38t

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Real & Equal roots (b2=4ac)

Some algebra helps fit the polynomial to Laplace tables.

Use linearity, and a some table entries to conclude:

• Sign{} growth or decay• || rate of growth/decay

p(s)=s2+2s+1y(0)=1,y’(0)=01=-1

y(t)=-e-t+te-t

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Complex conjugate roots (b2<4ac)

Some algebra helps fit the polynomial to Laplace tables.

Use table entries (as before) to conclude:

Reformulate y(t) in terms of and

Where:

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Complex roots (b2<4ac)

For p(s)=s2+0.35s+1 and initial condition y(0)=1,y’(0)=0 The roots are =+i=-0.175±i0.9846 The solution has form:

and the constants areA=||=1.0157r=0.5-i0.0889=arctan(Im(r)/Re(r)) =-0.17591

We see the solution is an exponentially decaying oscillation where the decay is governed by and the oscillation by

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The “Roots” of a Response

Stable

MarginallyStable

Unstable

Re(s)

Im(s)

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(Optional) Reading List

LTI systems: • Chen, 2.1-2.3

Laplace:• http://www.cs.huji.ac.il/~control/handouts/laplace_Boyd.pdf

• Also, Chen, 2.3 2nd order LTI system analysis:

• http://www.cs.huji.ac.il/~control/handouts/2nd_order_Boyd.pdf

Linear algebra (matrix identities and eigenstuff)• Chen, chp. 3• Stengel, 2.1,2.2