Today’s goalsState space overviewState space solution to the uncompensated 2.004 Tower modelPoles are the eigenvalues of A | StabilityEigenvectors and modesState space representation as block diagramState feedbackExampleLecture 28 – Wednesday, Nov. 142.004 Fall ’07 Today’s goals• State space so far– Definition of state variables– Writing the state equations– Solution of the state equations in the Laplace domain– Phase space and phase diagrams• Today– Stability in state space– State feedback controlLecture 28 – Wednesday, Nov. 142.004 Fall ’07 State space overviewFrom the Equation of Motion to the State—Space representation:m¨x(t)+b ˙x(t)+kx(t)=w(t) →µx˙x¶≡ q(t)=µq1q2¶state,y(t) ≡ ˙x(t)output⇒ ˙q(t)=µ˙q1˙q2¶=µ01−k/m −b/m¶µq1q2¶+µ01¶w(t); y( t)=(0 1)µq1q2¶≡ cq.k1b1m1x1towerswaywA=µ01−k/m −b/m¶b =µ01¶Solution to the state equations:sˆq(s)=Aˆq(s)+bW (s) ⇒ˆq(s)=(sI − A)−1bW (s).Y (s)=cˆq(s)=c (sI − A)−1bW (s).Lecture 28 – Wednesday, Nov. 142.004 Fall ’07 State space solution to the uncompensated 2.004 Tower modelˆq(s)=(sI − A)−1bW (s)=1s2+(b1/m1) s +(k1/m1)µ1/m1s/m1¶W (s).From this result we can obtain transfer functions for position, velocity:for p osition choose c =(1 0),X(s) ≡ Y (s)=c (sI − A)−1bW (s) ⇒X(s)W (s)=1/m1s2+(b1/m1) s +(k1/m1).for velocity choose c =(0 1),V(s) ≡ Y (s)=c (sI − A)−1bW (s) ⇒V (s)W (s)=s/m1s2+(b1/m1) s +(k1/m1).positionvelocityphase diagramvelocitypositionLecture 28 – Wednesday, Nov. 142.004 Fall ’07 Poles are the eigenvalues of A | StabilityConsider the eigevalue problem for the matrix A:Aξ = μξ,where the solutions for μ are the eigenvalues and ξ are the eigenvectors.To solve the eigenvalue problem, we set det ( μI − A)=0.That is, the eigenvalues are the roots of the determinantof the matrix (μI − A).Recall that the state—space solution wasˆq(s)=(sI − A)−1bW (s)=adj (sI − A)det (sI − A)bW (s)==1s2+(b1/m1) s +(k1/m1)µ1/m1s/m1¶W (s),where adj(.) denotes the adjoint. The same denominator det (sI − A)appears in the transfer functions for both velocity and position.This denominator is also referred to as characteristic equation.Therefore, the polesof the system are the roots of the determinantof the matrix (s I − A), i.e., the eigenvalues.The uncompensated 2.004 Tower is a 2nd order system withω2n=k1m1; ζ =b12√k1m1Therefore the eigenvalues/poles ares±= −ωn³ζ ±pζ2− 1´.The system represented by A isstable if A’s eigenvalueshave negative real part(i.e., are on the left—hand half—plane.)The phase diagram is then orientedtowards the origin (“sink.”)velocitypositionThe system represented by A isunstable if A’s eigenvalueshave positive real part (phase diagramexplodes outwards — “source”)and marginally stable if A’s eigenvalueshave zero real part (phase diagramrotates around the originwithout either approaching or moving away.)Lecture 28 – Wednesday, Nov. 142.004 Fall ’07 Eigenvectors and modesLet’s do a specificexample:m1=1,b1=1,k1=1,that is ωn=1,ζ =1/2, s±= −(1/2) ± jp3/2.This system is stable and, in fact, underdamped,consistent with ζ < 1andpolesoff the real axis.Now let’s compute the eigenvectors, starting with ξ(+)corresponding to the eigenvalue s+:µ01−1 −1¶ξ(+)= s+ξ(+)⇒(ξ(+)2=¡−1+j√3¢ξ(+)1/2ξ(+)1+ ξ(+)2=¡−1+j√3¢ξ(+)2/2.It can be verified that the two equations are equivalent.Therefore, the eigenvector corresponding to s+isξ(+)=⎛⎜⎜⎝1√2−1√2+ j√3√2⎞⎟⎟⎠α(+),where α(+)is any arbitrary real number. By convention,the eigenvector is writte n so that if α(+)=1⇒¯¯ξ(+)¯¯=1.Similarly we can find the eigenvector ξ(−)corresponding to the eigenvalue s−:ξ(−)=⎛⎜⎜⎝1√2−1√2− j√3√2⎞⎟⎟⎠α(−).The two eigenvectors ξ(+), ξ(−)are referred to as the modes of the system.The imaginary parts of the corresponding polesare the eigenfrequencies of the modes.In the uncompensated 2.004 Tower,the two modes are degeneratebecause the two poles are conjugate(i.e., they have the same imaginaryparts with ± signs.)This is true for any 2nd order system.The compensated 2.004 Toweris a 4th order system, and so it hastwo non—degenerate modes.The significance of a modeis that if the system is excitedwith a sinusoid of frequency equalto the mode’s eigenfrequency,then the response of the system willbe the mode itself (i.e., the eigenvector.)At other freque ncies, the responseis a mixture of modes.Lecture 28 – Wednesday, Nov. 142.004 Fall ’07 State space representation as block diagramwBqACy++q.Figure by MIT OpenCourseWare.˙q(t)=Aq(t)+Bw(t),y( t )=Cq(t) .C (sI − A)−1BW (s)Y ( s)Equivalent block diagram representation as transfer function:Lecture 28 – Wednesday, Nov. 142.004 Fall ’07 State feedbackw(t)=r( t) − Kq(t) ⇒˙q(t)=Aq (t)+B³r(t) − Kq´,y(t)=Cq(t).˙q(t)=³A − BK´q(t)+Br(t),y(t)=Cq(t ).⇒Closed—Loop TF:Y (s)R(s)= C (sI − A + BK)−1BwrBqAKCy+++-q.Figure by MIT OpenCourseWare.Lecture 28 – Wednesday, Nov. 142.004 Fall ’07 ExampleDesign problem: We are given transient response requirements of 9.5% over-sho ot and 0.74sec settling time. Moreover, we wou ld like t o approximatelycancel the zero in the opn—loop transfer function.To meet these goals, we select two closed—loop poles at −5.4 ± j7.2. Thesemeet the transient reponse requirements. Moreover, we select an additionalclosed—lo op pole at −5.1 to approximately cancel the open—loop zero. Therefore,the desired closed—loop transfer function should be proportional to(s +5)(s +5.1)(s +5.4 − j7.2)(s +5.4+j7.2)=(s +5)s3+15.9s2+ 136s +413.Next we convert the given transfer function to a state—space representation.This is done by first considering a system without the open—loop zeros, i.e. ofthe formX(s)W (s)=1s(s +1)(s +4)=1s3+5s2+4s⇔ x(3)+5¨x +4˙x = w.The state variable are selected asq =⎛⎝q1q2q3⎞⎠=⎛⎝x˙x¨x⎞⎠⇒ A =⎛⎝01 000 10 −4 −5⎞⎠, b =⎛⎝001⎞⎠.This choise of state variables is also known as phase—variable form,becauseit agrees with the phase diagram representation that we saw earlier (except inthis case we have a 3rd order system, and hence three state/phase variables:position, velocity, acceleration.)We also need to determine the observation matrix C .Sincetheopen—looptransfer function has a zero, the response includes a derivative
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