DESIGN OF CONTROLLERS WITH ARBITRARY POLE PLACEMENT AND STATE OBSERVERS Dr. C. Vivekanandan Professor and Vice Principal Dept. of EEE Jan. 9, 20161SNSCE

+ _ SECOND ORDER SYSTEMS R(S)C(S) Jan. 9, 20162SNSCE

TIME DOMAIN SPECIFICATIONS Jan. 9, 20163SNSCE

DOMINANT POLES System may be of higher order Poles, closer to imaginary axis, contribute more during Transients If only a pair of complex conjugate pair is closer and all other poles are sufficiently away from imaginary axis…..????? Jan. 9, 20164SNSCE

S = -2+/- j5 S = -3+/- j5 S = -10+/- j5 Jan. 9, 20165SNSCE

STATE SPACE ANALYSIS Jan. 9, 20166SNSCE

State Model Σ Σ CB A D u x y dx/dt Jan. 9, 20167SNSCE In General for all practical applications Matrix D = [0]

Linear Transformation Jan. 9, 20168SNSCE

Linear Transformation Characteristic Polynomial Let P is an n x n, non-singular non-unique matrix. Then Invariance property Jan. 9, 20169SNSCE

Controllabiliy If it is possible to transfer the system from any initial state to any other desired state by means of an unconstrained control vector – Controllable. Controllablity Matrix: R ank of (N) = n for complete controllability i.e. B, AB …. are linearly independent. Jan. 9, SNSCE

Contorllable Canonical Form B Jan. 9, SNSCE

Linear Transformation Define and Then if the system is controllable Jan. 9, SNSCE

Observability If it is possible to estimate the states by observing the output over finite period of time – Observable. Observable Matrix Rank of (N) = n for complete Observability Jan. 9, SNSCE

Observable Canonical Form Jan. 9, SNSCE

Linear Transformation Define and Then though linear transformation Jan. 9, SNSCE

Design by Pole Placement Settling Time = 2 Sec Peak Overshoot = 16% Damping factor ζ = 0.5 Natural frequency of oscillation ω n = 4 rad/sec Poles at S 1,2 = -2 ± j3.464 All other poles with real part between -10 and -20 Jan. 9, SNSCE

Design by Pole Placement by State Feeback Let the control signal u be (Control Law) So, or Characteristic Eqn.: Solution Vector Jan. 9, SNSCE

Closed Loop system with Control Law u = -Kx Σ Σ CB A D u dx/dt x y -K Jan. 9, SNSCE

Jan. 9, 2016SNSCE19 System Model Kontrol Law -K x u Full-state feedback

Regulator Problem Problem converted in to regulator problem Roots of are regulator poles By proper selection of K, the matrix can be made asymptotically stable matrix for all Jan. 9, SNSCE

Pole Placement Let the characteristic equation of the given system be Let the poles of the desired system be The characteristic equation of the desired system: Jan. 9, SNSCE

Pole Placement Let K be the feedback gain matrix such that The system, represented in regulator form will have its poles at and will comply with the specifications. Jan. 9, SNSCE

Pole Placement Let the characteristic equation of the given system be Transform the given system into controllable canonical form Jan. 9, SNSCE

Pole Placement As, feedback gain matrix will take the form Hence, The poles of and are Jan. 9, SNSCE

Pole Placement Let * Jan. 9, SNSCE

Pole Placement As It may be arrived that And so Jan. 9, SNSCE

Example: Consider the third-order system with the differential equation Dorf and Bishop, Modern Control Systems Given Specifications: Peak Overshoot 1.5% (Approx.) Settling Time: 1 Sec

We can select the state variables as x 1 =y, x 2 =dy/dt, x 3 =d 2 y/dt 2. (Phase variables) and

If the state variable matrix is and then the closed-loop system is The state feedback matrix is Dorf and Bishop, Modern Control Systems

and the characteristic equation is If we seek a rapid response with a low overshoot, we choose a desired characteristic equation such that

If we want a settling time (with a 2% criterion) equal to 1 second, then If we choose ω n =6 rad/s, the desired characteristic equation is Comparing two characteristic equations yields Therefore,

Difficulties in Pole Placement All State Variables shall be available for feedback to achieve arbitrary pole placement – State Variables may not be available for measurement – Available state variables may be corrupted by noise – Measurement of state variables and subsequent conditioning may not be economical Jan. 9, SNSCE

STATE OBSERVERS A Computer system that estimates the state variables Takes input signal u and output y as inputs The estimated state variables is the output Jan. 9, SNSCE