Presentation is loading. Please wait.

Presentation is loading. Please wait.

Linear Time Invariant systems

Similar presentations


Presentation on theme: "Linear Time Invariant systems"— Presentation transcript:

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

2 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) Lavi Shpigelman, Dynamic Systems and control – – We assume that the state vector contains all the information needed to describe the future of the system given the future control signal and disturbances. Systems that are not lumped may be rather simple as in the above example but do not yield to the tools made for the state space description to be introduced later.

3 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) Lavi Shpigelman, Dynamic Systems and control – – The linearity of a system allows us to treat it’s response to control inputs at every instant separately (as if the control input is a continuous series of impulses) and then simply sum them up in the integral.

4 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,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 Lavi Shpigelman, Dynamic Systems and control – – The linearity of a system allows us to treat it’s response to control inputs at every instant separately (as if the control input is a continuous series of impulses) and then simply sum them up in the integral. If a system is causal (as we will assume in a minute) then its current state is not a function of future inputs, thus for t<tau g(t,tau)=0.

5 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,0) A 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 by Lavi Shpigelman, Dynamic Systems and control – – In practice, the difference between time invariant to time varying systems is the time constants of system changes. For example a new car in its “running in” period is time varying on the scale of days (till its parts settle down) but is time invariant on a scale of minutes. After the running in period it is relatively time invariant on the scale of years for the next 2-5 years and then it begins to show time dependency again as the rate it which it breaks down accelerates. The changes from the previous integral are independent. Each can be undone if it’s corresponding assumption doesn’t hold. causal Time invariant Convolution relaxed

6 LTI - State-Space Description
Fact: (instead of using the impulse response representation..) 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 Lavi Shpigelman, Dynamic Systems and control – – Controllable inputs u State x Disturbance (noise) w Measurement Error (noise) n Observations y Plant Dynamic Process A B + Observation Process C D x u 1/s If the system is not lumped, the state vector is infinitely long (assuming it holds all the information about the system) and therefore can not generally be described by a finite set of equations. Later we will deal with additive Gaussian white noise in this LTI setting in the context of estimation and control.

7 What About nth Order Linear ODEs?
Can be transformed into n 1st order ODEs Define new variable: Then: Lavi Shpigelman, Dynamic Systems and control – – Any linear nth order ODE can similarly be transformed into a set of n 1st order ODEs as seen in the example. Dx/dt = A x B u y = [I 0 0  0] x

8 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 aet (and s>a ) and it is unique The inverse exists as well Laplace transform pairs are known for many useful functions (in the form of tables and Matlab functions) Will be useful in solving differential equations! Lavi Shpigelman, Dynamic Systems and control – –

9 Some Laplace Transform Properties
Linearity (superposition): Differentiation Lavi Shpigelman, Dynamic Systems and control – –

10 Remember integration by parts:
Using that and the transform definition: Lavi Shpigelman, Dynamic Systems and control – –

11 Some Laplace Transform Properties
Linearity (superposition): Differentiation Convolution Lavi Shpigelman, Dynamic Systems and control – –

12 Integration over triangle 0 <  < t
Using definitions Integration over triangle 0 <  < t Define  = t-t, then ds = dt and region is  > 0, t > 0 Lavi Shpigelman, Dynamic Systems and control – –

13 Some Laplace Transform Properties
Linearity (superposition): Differentiation Convolution Integration Lavi Shpigelman, Dynamic Systems and control – –

14 Switch integration order Plug  = t-
By definition: Switch integration order Plug  = t- Lavi Shpigelman, Dynamic Systems and control – –

15 Some specific Laplace Transforms (good to know)
Constant (or unit step) Impulse Exponential Time scaling Lavi Shpigelman, Dynamic Systems and control – –

16 Homogenous (aka Autonomous / no input) 1st order linear ODE
Solve: Do the Laplace transform Do simple algebra Take inverse transform Lavi Shpigelman, Dynamic Systems and control – – Known as zero input response

17 1st order linear ODE with input (non-homogenous)
Solve: Do the Laplace transform Do simple algebra Take inverse transform Lavi Shpigelman, Dynamic Systems and control – – Known as the zero state response

18 Example: a 2nd order system
Solve: Do the Laplace transform Do simple algebra Take inverse transform Lavi Shpigelman, Dynamic Systems and control – –

19 Using Laplace Transform to Analyze a 2nd Order system
Consider the autonomous (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 Lavi Shpigelman, Dynamic Systems and control – – characteristic polynomial determined by Initial condition

20 2nd Order system - Inverse Laplace
Solution of inverse transform depends on 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 effect Im{} Oscillatory effect Lavi Shpigelman, Dynamic Systems and control – –

21 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 Lavi Shpigelman, Dynamic Systems and control – – p(s)=s2+3s+1 y(0)=1,y’(0)=0 1=-2.62 2=-0.38 y(t)=-0.17e-2.62t+1.17e-0.38t

22 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 Lavi Shpigelman, Dynamic Systems and control – – p(s)=s2+2s+1 y(0)=1,y’(0)=0 1=-1 y(t)=-e-t+te-t

23 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: Lavi Shpigelman, Dynamic Systems and control – –

24 Complex conjugate roots (b2<4ac)
E.g. p(s)=s2+0.35s+1 and initial condition y(0)=1 , y’(0)=0 Roots are =+i=-0.175±i0.9846 Solution has form: with constants A=||= r=0.5-i =arctan(Im(r)/Re(r)) = Solution is an exponentially decaying oscillation Decay governed by  oscillation by . Lavi Shpigelman, Dynamic Systems and control – –

25 The “Roots” of a Response
Stable Marginally Stable Unstable Re(s) Im(s) Lavi Shpigelman, Dynamic Systems and control – –

26 (Optional) Reading List
LTI systems: Chen, Laplace: Also, Chen, 2.3 2nd order LTI system analysis: Linear algebra (matrix identities and eigenstuff) Chen, chp. 3 Stengel, 2.1,2.2 Lavi Shpigelman, Dynamic Systems and control – –


Download ppt "Linear Time Invariant systems"

Similar presentations


Ads by Google