Modern Control System EKT 308 SteadyState and Stability

Modern Control System EKT 308 Steady-State and Stability

Quick Review 1. Laplace transform 2. Poles and zeros , transfer function 3. Simplification complex block diagram, signal flow diagram 4. State space modeling 5. Modeling physical systems 6. Time response of first and second order systems Topic to cover 1. Steady State Error 2. Routh Hurwitz Stability Criterion

Review : Transient and Steady-State Response Analysis Step response of a control system

Review: Performance Measures for step response ü Delay Time üRise Time : Time to reach half of the final value for the first time. : Time to rise to the final value. Underdamped (0% 100%) Overdamped (10% 90%) ü Peak Time : Time to reach the first peak of the overshoot. ü Percentage overshoot,

Steady State Error R(s) + E(s) B(s) - Y(s) From the diagram Consider and Use the FINAL VALUE THEOREM and define steady state error, that is given by

Unit step input, From Steady state error, We define step error coefficient, Thus, the steady state error is By knowing the type of open-loop transfer function, We could determine step error coefficient and thus the steady state error

For open-loop transfer function of type 0: For open-loop transfer function of type 1: , For open-loop transfer function of type 2: ,

STEADY STATE ERROR EXAMPLE A first order plant with time constant of 9 sec and dc gain of 5 is negatively feedback with unity gain, determine the steady state error for a unit step input and the final value of the output. Solution: The block diagram of the system is. R(s) + Y(s) As we are looking for a steady state error for a step input, we need to know step error coefficient, Knowing the open-loop transfer function, then And steady state error of Its final value is

Unit Ramp As in the previous slide, we know that , while its Laplace form is Thus, its steady state error is Define ramp error coefficient, Which the steady state error is For open-loop transfer function of type 0: For open-loop transfer function of type 1: For open-loop transfer function of type 2: Just like the unit step input we can conclude the steady state error for a unit ramp through the type of the open-loop transfer function of the system.

Example: A missile positioning system is shown. (i) Find its closed-loop transfer function (ii) Determine its undamped natural frequency and its damping ratio if (iii) Determine the steady state error, if the input is a unit ramp. Compensatot + - DC motor

Solution: (a) By Mason rule, the closed-loop transfer function is , (b) If Comparing with a standard second order transfer function

Comparing Thus undamped natural frequency rad. s-1 and damping ratio of (c) To determine the ramp error coefficient, we must obtain its open-loop transfer function As it is a type 1, the system will have a finite ramp error coefficient, putting Hence steady state error of

Steady State Error of Feedback Control System

Stability

Stability

Stability Routh-Hurwitz Stability Criteria If a polynomial is given by Where, are constants and Necessary condition for stability are: (i) . All the coefficients of the characteristic polynomial are of the same sign. If not, there are poles on the right hand side of the s-plane (ii) All the coefficient should exist except for the For the sufficient condition, we can now form a Routh-array,

Routh’s Array

Routh Array elements Routh-Hurwitz Criteria states that the number of roots of charateristic equation with positive real parts is the same as the number of sign changes of the first column.

Case 1: No zero on the first column After the Array has been tabled, all the elements on the first column are not equal to zero. If there is no sign changed, all the poles are in the LHP. While the number of poles on the RHP is equal to number of sign change on the first column of the Routh’s array. Example: Consider a fourth order characteristic equation

Example Solution: Form the Routh’s array 2 12 1 8 2 2 2 There are two sign change on rows 2 and 3. Hence, there two poles on the RHP (Right-half f s-palne).

Scilab solution

Case 2: Coeffiecient of the first column is zero but not the others. Change the zero element by a small positive number, . The number of pole on the RHP will depend on the number of sign change. Example: Consider a fifth order characteristic equation Solution: Form the Routh’s array 1 3 5 2 6 3 , =1 3 3 If , is a small positive number there are two sign change at row 3 and 4, and also at row 4 and 5 . Hence, there two poles on the RHP.

Scilab Solution -->CE=poly([3 5 6 3 2 1], 's', 'c') -->roots(CE) ans = 0. 3428776 + 1. 5082902 i 0. 3428776 - 1. 5082902 i - 1. 6680888 - 0. 5088331 + 0. 7019951 i - 0. 5088331 - 0. 7019951 i

Case 3: All the coefficients on a row are zeros. Form an auxiliary equation from the row above it and replace the coefficient of the row with the differentiated coefficient of the auxiliary equation. For this case, if there is no sign change, the characteristic equation has a pair of poles with opposite sign of real component or/and a pair of conjugate poles on the imaginary axis. Example: Consider this fifth order characteristic equation Formed Routh’s array 1 6 8 7 42 56 84 28 21 9. 3 56 56

Form the auxillary equation on the second row: Differentiate the equation: As there is no sign change, there is a pair of conjugate poles on the axis and/or a pair of poles with opposite sign of real component. To be sure we can use Scilab -->CE=poly([56 8 42 6 7 1], 's', 'c') -->roots(CE) ans = - 7. - 8. 049 D-16 + 2. i - 8. 049 D-16 - 2. i 1. 4142136 i - 1. 4142136 i

Use of Routh Hurwitz Criteria Main use is to determine the position of the poles, which in turns can determine the stability of the response. j STABLE UNSTABLE

Example A closed-loop transfer function is given by Determine the range for K for the system to be always stable. Solution: Charactristic equation is Expand the equation Form the Routh’s array 1 3 3 2 K K K

To ensure that there is no poles on the RHP of the s-plane, there must be no sign change on the first column of the Routh’s Array, therefore for no sign change: Refering to row 4: which gives and row 5 Hence its range

Concept of Root Locus Fig 1 Characteristic equation (above fig),

Concept of Root Locus (contd…) The root locus is the path of the roots of the characteristic equation traced out in the s-plane as the system parameter varies from zero to infinity. Root Locus Procedure Step 1:

Root Locus Procedure (contd…) Locate poles and zeros in the s-plane (‘x’ for poles, ‘o’ for zeros Note: Let us vary K from 0 to infinity. When K = 0, So, when K=0, values of s coincide with poles of P(s).

The locus of the roots of the characteristic equation 1+KP(s)=0 begins at the poles of P(s) and ends at the zeros of P(s) as K increases from 0 to infinity. If n > M, (n-M) branches of root locus approach the (n-M) zeros at infinity. Step 2: Locate the segments of the real axis that are root loci. The root locus on the real axis lies in a segment of the real axis to the left of an odd number of poles and zeros.

Example fo step 1 an 2.