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ChE 170 Lecture 8 Stability of control system Response of control system for a unit-step change in set point For a unit-step change in R, the transform of the response is As K c increases, the system response becomes more oscillatory Beyond a certain value of Kc , the successive amplitudes of the response grow rather than decay; this type of response is called unstable. Definition of stability (linear systems) Stable system - one for which the output response is bounded for all bounded inputs. Unstable system - exhibiting an unbounded response to a bounded input is unstable. STABILITY CRITERION *Let open-loop TF Basic single-loop control system. Characteristic equation of the control system **Where is the open-loop transfer function because it relates the measured variable B to the Stability criterion: A linear control system is unstable if any real parts of the set point R if the feedback loop is disconnected from the comparator (i.e., if the loop is opened). roots of its characteristic equation are on, or to the right of, the imaginary axis. Otherwise the system is stable. At RHP: Variation exponentially grow. The farther from the origin, the faster is the growth At LHP: Variation exponentially decay. The farther from the origin, the faster is the decay Limit of stability!!! Is when the pole is purely imaginary (no real part) CRITERION OF STABILITY Characteristic equation of the control system: 1 + G = 0 Characteristic equation of the control system: 1 + G = 0 Equivalent to: Where, *The stability of a control system is determined solely by its open-loop transfer function through the roots of the characteristic equation. *If the real part of the roots of the characteristic equation is negative (-), the system is stable. HW Consider the liquid-level control system shown .The tanks are noninteracting. The following information is known: 1. The resistances on the tanks are linear. These resistances were tested separately, and it was found that, if the steady-state flow rate q cfm is plotted against steady-state tank level h ft, the slope of the line dq/dh is 2 ft2/min. 2. The cross-sectional area of each tank is 2 ft2. 3. The control valve was tested separately, and it was found that a change of 1 psi in pressure to the valve produced a change in flow of 0.1 cfm. 4. There is no dynamic lag in the valve or the measuring element. (a) Draw a block diagram of this control system, and in each block give the transfer function, with numerical values of the parameters. ROUTH TEST FOR STABILITY a purely algebraic method for determining how many roots of the characteristic equation have positive real part if there are no roots with positive real parts, the system is stable Is limited to systems that have polynomial characteristic equations, which means it cannot be used to test the stability of a control system containing a transportation lag When and How to apply the ROUTH TEST FOR STABILITY 1. Write the characteristic equation in the form: Where ao is positive. (If ao is originally negative, both sides are multiplied by -1). Coefficients of equation: ao , a1 , a2 , . . . , an-1 , an *If any coefficient is negative, the system is definitely unstable, NO need to apply Routh test *If all the coefficients are positive, the system may be stable or unstable, apply Routh test ROUTH TEST FOR STABILITY 1. Arrange the coefficients of characteristic equation (in standard form) into the first two rows of the Routh array 2. Determine the remaining rows from the following: Coefficient of characteristic equation in standard polynomial form 3. Apply Routh test criterion: STABLE if (Theorem 1) all elements of the first column of the Routh array are positive and nonzero (Theorem 2) there is NO sign changes in the first column. Note: the number of sign changes is equal to the number of roots with positive real parts Apply Routh test criterion: STABLE if (Theorem 3) If one pair of roots is on the imaginary axis, equidistant from the origin, and all other roots are in the left half plane, The location of the pair of imaginary roots can be found by solving the equation: where the coefficients C and D are the elements of the array in the (n - 1)th row as read from left to right, respectively. this last rule to be of value in the root-locus method n=7 Row of (n-1) Example 6 Given the characteristic equation: s4 + 3s3 + 5s2 + 4s + 2 = 0 Determine the stability by the Routh criterion. Since all the coefficients am positive, the system may be stable, therefor Routh test is applicable n=4 Coefficient of characteristic equation in standard form Calculated by equation Example 7 Given the characteristic equation: a. Determine the values of Kc for which the control system is stable. b. For the value of Kc for which the system is on the threshold of instability, determine the roots of the characteristic equation with the help of Theorem 3 The characteristic equation 1 + G ( s ) = 0 becomes: Rearrangement of this equation for use in the Routh test gives Routh array: <0 > -1 Theorem 1: Stable if elements 1st column are positive and non-zero >0 >0 Therefore system will be stable: -1 < < 10 Example 8 Determine the stability of the system shown below for which a PI controller is used. Limit of stability of a control system (Direct substitution method – Frequency response stability) To get the limit of stability let into the equation and solve of ultimate , Limit of stability of a control system (Direct substitution method – Frequency response stability) • The imaginary axis divides the complex plane into stable and unstable regions for the roots of characteristic equation, as indicated in Fig. 11.26. • On the imaginary axis, the real part of s is zero, and thus we can write s=jw. Substituting s=jw into the characteristic equation allows us to find a stability limit such as the maximum value of Kc. • As the gain Kc is increased, the roots of the characteristic equation cross the imaginary axis when Kc = Kcm. Example 9 Use the direct substitution method to determine Kc,u for the system with the characteristic equation given as: Solution: Let s jω Where gives 10 jω3 17ω2 8 jω 1 K c,u 0 and Group the real and imaginary term of the equation: is the sustained oscillation , ω ω If both the real and imaginary parts are identically zero: Solving: ω ω ω , **To get the location of these purely imaginary pole, apply theorem 3 of the Routh stability test