### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# ANALYSIS OF LIN SYS E C E 801

Clemson

GPA 3.84

### View Full Document

## 82

## 0

## Popular in Course

## Popular in ELECTRICAL AND COMPUTER ENGINEERING

This 30 page Class Notes was uploaded by Eloy Ferry on Saturday September 26, 2015. The Class Notes belongs to E C E 801 at Clemson University taught by Staff in Fall. Since its upload, it has received 82 views. For similar materials see /class/214315/e-c-e-801-clemson-university in ELECTRICAL AND COMPUTER ENGINEERING at Clemson University.

## Similar to E C E 801 at Clemson

## Reviews for ANALYSIS OF LIN SYS

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/26/15

Chapter 7 Stability We are familiar with the intuitive concept of stability in the context of system poles of the transfer function KSZl szm H S sp1spn If the 19139s are The system is all in the left halfplane stable any in the right halfplane unstable in the left halfplane except marginally stable for possibly nonrepeated poles on the lmaxis These rules do not always account for different types of stability nor for statevariable descriptions Except for the fact that we know the eigenvalues of a SISO system39s Amatrix to be the poles of the transfer function through the relationship especially HS CSI A IB D nonlinear dayS1 A sYstems D 1 Same expression used to find eigenvalues Next we39ll define several kinds of useful types of stability and discuss new tests for these in terms of state variable descriptions DEFINITION For a system given by a general equation not necessarily linear x0 fltxltrgtultrgtrgt W 0 x6 is an equilibrium point if in the absence of any input xt xe for all t2 t0 Because xe is constant this implies that xe constant at 0 u O xt 0 fxe0t Or in discretetime xk 1 2 xk xe for all k 2 0 quotequilibrium pointquot quotcritical pointquot Note that for LTl Linear TimeInvariant systems this results in 0 So either xe 0 when A is full rank or xe 6 MA in which case A has at least one zero eigenvalue and there are an infinite number of actual equilibrium points lying in a subspace that must of course pass through the origin We will consider only A39s of full rank so the only equilibrium points considered are x620 It is the equilibrium point that we will classify as stable or unstable ie does a system stray from its equilibrium and if so how far xe unstable amp unstable x A l marginally stable stable stable Definitions of different types of stability ut O Denote the solution of a zeroinput system of equations as tx0 to which isjust CDtt0 x0 for our linear systems and let the origin x6 0 be the equilibrium point of interest DEFINITION The origin is a stable equilibrium if for any a gt 0 there exists a 88 to gt 0 such that if Hydro lt 5 then HxtH lt 8 for all tgtto gt stable quotin the sense of Lyapunovquot isL DEFINITION The origin is asymptotically stable if it is stable and there exists an 53900 gt 0 such that whenever Hxt0H lt 53900 then limetH 0 l gtoo A Picture for a two variable system unstable m x1 asymptotically stable stable isL Extra Conditions If 5 or 539 are independent of the initial time then the stability is called quotuniformquot If 5 or 539 can be chosen arbitrarily large then the equilibrium point is globally stable or stable in the large DEFINITION If there is a fixed constant M such that u M for every tor k then the input is bounded If for every bounded input and every initial condition 26 there exists a scalar NSMtoxfo gt 0 such that Ham 3 NS then the system is BIBS bounded input bounded state stable DEFINITION If input u is bounded above by constant M and there exists a scalar N such that for all time the system output satisfies HyHS N the system is BIBO bounded input bounded output stable Now we consider linear systems Recall that the solution of a linear system can be written xt 13010 xt0 j tTBTuTd C Z0 Ol39 yt Ctlt t0 xt0 jCtDt TBTuTdT Dtut I to P3 t i l i s f Ct1t t0xt0 jCrc1r 031 5 IDIurdr if Q Ct1tt0xt0 jHl CuTd C Z0 This is also known quotwe39ghtmg matr39Xquot as the impulse response matrix of the system Consider the zeroinput solution the the state variables xt c130 t0xt0 ut 0 By Cauchy39s Theorem llxmll S lllt1gtta toll llxtoll Ifthere is a bound K00 on the statetransition matrix CIgtUo such that IICDUJMII 3 K00 then by choosing 5008 g we can prove stability isL by the definition 0 This is a necessary and suf cient condition for stability isL of the zero equilibrium state Proof of necessity is by contradiction t gtOO If in addition H t t0H gt 0 then the equilibrium is asymptotically stable All linear systems that are asymptotically stable are also exponentially stable This means that the norm of the state transition matrix is bounded by an exponential function as it tends to zero Now reconsider the inputrelated types of stability BIBS and BIBO Theorem A system is BIBS stable iff there exists a finite fixed value N00 such that assume utl lt M lll lttaTBTHdI 3 N00 lt 00 to for all tZtO triangle amp Cauchy inequalities xt s cDtt0xt0 l cDt TBTu39c d39c sll ltnrogtllllxltrogtllllllt1gtltnrgtBltrgtllllultrgtlldr boundedness of the input S Dante H xtl J HCDU TBTHd CMlt oo t0 Stability isL is necessary for BIBS stability which we can see by setting MU 0 above so we know that As long as HCItt0Hlt 00 and that l mBrllcr 3 N00 lt oo l 0 then the state will be bounded For BIBO stability we consider any initial conditions as having resulted from some bounded past input so that we may write From slide 243 yl jHtTuTdT WithtOZ oo Then because we are considering only bounded inputs we can conclude that f lt M is necessary and sufficient for BIBO stability l HHW OO ldISNHltoo Recall that we can find the 2 norm of a matrix M by taking the square root of the largest eigenvalue of MTM In practice this integral test is not often used Instead it can be shown that A timeinvariant system is BIBO stable iff all of the poles of the reduced transfer function are in the open lefthalf complex plane or for discretetime systems inside the open unit circle Note that eigenvalues and poles are not necessarily the same thing ie for timevarying systems TimeInvariant svstems 13tt0 eAU tO continuous time Pk0 Aquot discrete time The behavior of these statetransition matrices is obvious from the forms above A summary of zeroinput stability properties in terms of the eigenvalues of the Amatrix is shown in the table Continuous time Discrete time xAx xk1Axk Any Revlgt0 orRemipo Any lxigt1 or M21 if x1 unStable If if Ki is repeated is repeated Stable isL ifA ReltiigtsolorReltligtlt0 A M31 or lt1 if 139 if Xi is repeated is repeated Asymptotically All R x 0 All A 1 stable If e lgtlt lzllt Modal Stability and Classification of Equilibria Recall that each eigenvector corresponds to an eigenvalue and that the eigenvectors span invariant subspaces in the statespace Then we can talk about the stability of these subspaces by considering the value of the eigenvalue corresponding to each such subspace Recall the phase portraits in the last chapter that illustrated the quotphase planesquot of some 2D systems Consider how they are constructed from their modes Eigenvectors invariant subspaces Large motion along 4 vertical eigenvector toward origin fast mode corresp 5 39 39 to O 4 2 V 1 O A 1 A e2 0 4 X2 0 e1 7L 2 1 4 1 39 39 1 O M 2 A O 1 3 Small motion along 4 horizontal eigenvector 394 392 0 2 4 toward origin X1 slow mode corresp to O1 Stable Node Eigenvectors invariant subspaces 2 3 4 ez 4 2 D 2 4 X1 Stable Node Large motion along second eigenveotor toward origin fast mode corresp to O 4 Ali iii x 1 4 Mli ii Large motion along first eigenveotor toward origin slow mode corresp to O 1 Similar pictures can be drawn for 3D systems where the invariant subspaces might then be planes also Higher dimensions work the same way but are difficult to imagine Lvapunov Stabilitv Methods This can be a powerful stability testing technique It applies to many nonlinear systems as well Consider the unforced mechanical system K iiikm l B M56BXKx0 The energy stored in this system is Vx39c Mx2 Kx2 Normally we expect a passive system to have positive energy which would require sufficiently that MKgt0 If the system is quotstablequot this energy should dissipate so the rate of change of energy over time should be nega ve Vx X KXx Evaluating this quotalong the trajectories of the systemquot ie according to the differential equation g Mx V39xx max Kxxlx x M Bx2 Kxx Km 2 Bx2 So if we expect this passive physical system to dissipate positive energy then we must have B gt 0 All three of the constants should therefore be positive in order for the energy to be positive and the change in energy to be negative This is the idea behind Lyapunov39s Direct Method What would we ask as conditions on the coefficients of MSZ BSK Quadratic Mjng gt 0 Formula for a stable system Hint RouthHurwitz Usually though the state equations have little intuitive sense of quotenergyquot to them so we generalize the terminology Notation A function of a state vector Vx is positive de nite if in a neighborhood of an equilibrium point V0 0 and Vx gt 0 forx 7 0 Vx 2 0 for semidefinite reverse the signs to get negative semidefinite Lvaounov Theorem For any system at fx with equilibrium point f0 0 if a positive de nite function Vx can be found such that Vx is negative semide nite then the system is stable in the sense of Lyapunov The function Vx is then called a quotLyapunov Functionquot If the derivative Vx is negative definite then the origin is asymptotically stable Theorem If a system is globally asymptotically stable then a Lyapunov function exists that is valid for all x i 0 Notes What these theorems do and don39t say 1 The neighborhood in which the Lyapunov function is defined is the only neighborhood in which the initial condition can lie for which stability is guaranteed A system might be unstable if we leave this neighborhood 2 The theorems do not say howto find such Lyapunov functions Doing so generally takes lots of experience and trial and error especially for nonlinear systems 3 If we cannot nd a Lyapunov function our candidates do not work this does not imply the system is unstable It just means we haven39t shown it is When linear systems are stable they are globally stable Lets s examine x Z Ax Lyapunov s method Linear timeinvariant systems give some interesting results we choose a quadratic form as a quotcandidatequot Lyapunov function Vx xTPx Obvious VX iS PD with P being real symmetric and positivedefinite Now compute the derivative Vx XTPx xTPX xTATPx xPAx xTATPPA1x We want this to be a negative definite function so ATPPA Q where Q is some positive definite matrix This is a famous equation called a Lyapunov equation So we can guess a P matrix and compute Q to see if it is positive definite or semidefinite for isL stability A more common and reliable technique is choose a positive definite Q and solve the Lyapunov equation for P which is often not easy Anwaill do so Q I is often used Theorem A system with system matrix A is asymptotically stable iff the solution P of the Lyapunov equation is positive definite whenever Q is positive definite Note that any pd Q will show whether a system is stable in the reverse approach whereas choosing a P and solving for Q will show neither stability or instability if Q turns out not to be positive definite Interestingly the Lyapunov equation has a closedform solution but it is of little help because it requires knowledge of the statetransition matrix for the LTl system P IeATthAtdt 0 Example Prove that the nonlinear system 561 xz 322 a1xz Clle b1xz 923602362 a1 a2 gt 0 is asymptotically stable Consider the candidate Lyapunov function 2 Vx 6123612 x2 So Vx is obviously positive definite First find equilibra O 2 X2 2 0 a1x2 02x1 51952 52x1 952 The only equilibrium is x1e xze 0 NOW Vx 2a2x1xl 2362362 2 2612x1362 2x2 czle czle b1x2 b2x12 x2 2 2a1x W2 2x b1x2 529502 2a1x 2x b1x2 b2x12 lt0 362750 This is obviously So at any time that x20 72 0 the system is asymptotically stable If x20 2 0 the system is stable isL This is the quotregion of stability Stability of timevarying svstems Interestingly if we freeze time and compute the eigenvalues ofAt at that instant the results do not tell us whether the system is stable except in special cases wherein the eigenvalues are changing quotslowlyquot Also when a timevarying system has an unstable eigenvalue that does not necessarily imply that the system is unstable overallll unless all of them are unstable Lyapunov techniques are better for timevarying systems but are still very difficult

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.