Stability Analysis PDF

O2 MIA Dymamics system Stability Analysis Stabilty Study and PID Controllers The goal of This chapter is to study the stability of linear systems and to learn about about the PID controller and a few basic tuning rules of it. After taking this lesson, you will be able t...

O2 MIA Dymamics system Stability Analysis Stabilty Study and PID Controllers The goal of This chapter is to study the stability of linear systems and to learn about about the PID controller and a few basic tuning rules of it. After taking this lesson, you will be able to 1. Study any stability of linear systems by using the criterion of Routh-Hurwitz 2. relate PID controller parameters to step response characteristics of the controlled system 3. To apply the famous Ziegler-Nichols tuning method to come up with an initial set of working PID parameters for an unknown system.  Stability Stability is an important concept. In this chapter, let us discuss the stability of system and types of systems based on stability. What is Stability? A system is said to be stable, if its output is under control. Otherwise, it is said to be unstable. A stable system produces a bounded output for a given bounded input. The figure shows the response of a stable system. This is the response of first order control system for unit step input. This response has the values between 0 and 1. So, it is bounded output. We know that the unit step signal has the value of one for all positive values of t including zero. So, it is bounded input. Therefore, the first order control system is stable since both the input and the output are bounded 3 Stability 4  Stability Tacoma Narrows at Puget Sound, 1940, USA 5  Types of Systems based on Stability We can classify the systems based on stability as follows. Absolutely stable system Conditionally stable system Marginally stable system Absolutely Stable System If the system is stable for all the range of system component values, then it is known as the absolutely stable system. The open loop control system is absolutely stable if all the poles of the open loop transfer function present in left half of ‘s’ plane. Similarly, the closed loop control system is absolutely stable if all the poles of the closed loop transfer function present in the left half of the ‘s’ plane. Conditionally Stable System If the system is stable for a certain range of system component values, then it is known as conditionally stable system. Marginally Stable System If the system is stable by producing an output signal with constant amplitude and constant frequency of oscillations for bounded input, then it is known as marginally stable system. The open loop control system is marginally stable if any two poles of the open loop transfer function is present on the imaginary axis. Similarly, the closed loop control system is marginally stable if any two poles of the closed loop transfer function is present on the imaginary axis 6  Stability Analysis Routh-Hurwitz Stability Criterion The necessary condition is that the coefficients of the characteristic polynomial should be positive. This implies that all the roots of the characteristic equation should have negative real parts. Necessary Condition for Routh-Hurwitz Stability Consider the characteristic equation of the order ‘n’ is Routh-Hurwitz stability criterion is having one necessary condition and one sufficient condition for stability. If any control system doesn’t satisfy the necessary condition, then we can say that the control system is unstable. But, if the control system satisfies the necessary condition, then it may or may not be stable. So, the sufficient condition is helpful for knowing whether the control system is stable or not n 1 a s a s n n n 1 ...  a s  a s  a  0 2 2 1 0 Note that, there should not be any term missing in the nth order characteristic equation. This means that the nth order characteristic equation should not have any coefficient that is of zero value. Sufficient Condition for Routh-Hurwitz Stability The sufficient condition is that all the elements of the first column of the Routh array should have the same sign. This means that all the elements of the first column of the Routh array should be either positive or negative. 7  Stability Analysis Routh Array Method where 8  Stability Analysis Example s 4  3s 3  4 s 2  2 s  1  0 Step 1- Verify the necessary condition for the Routh-Hurwitz stability. All the coefficients of the characteristic polynomial are positive. So, the control system satisfies the necessary condition. Step 2- Form the Routh array for the given characteristic polynomial S4 1 4 1 S3 3 2 0 S2 b1=(3x4-1x2)/3=10/3 b2=(3x1-1x0)/3=1 0 S1 c1=(b1x2-b2x3)/b1=1.1 c2=(b1x0-3x0)/b1=0 0 S0 d1=(c1xb2-c2xb1)/c1=1 0 Step 3 − Verify the sufficient condition for the Routh-Hurwitz stability. All the elements of the first column of the Routh array are positive. There is no sign change in the first column of the Routh array. So, the control system is stable. 9  Stability Analysis Example: Determine the values of K to give a system is stable for the following equation s  2s  4s  K  0 3 2 s3 1 4 s2 2 K s1 (8- 0 K)/2 s0 K 0 For a stable system, we require that : 0 < K < 8. 10  Stability Analysis Example: K is considered positive, which values of K gives the system stable s 4  s3  s 2  s  K  0 s4 1 1 K s3 1 1 0 s2 0 K 0 s1 (0-K)/0=-K/0=-infinty 0 0 s0 K 0 0 Therefore, for any value of K greater than zero, the system is unstable 11  Stability Analysis Example: A control system has the structure shown in the following figure Determine the gain at which the system will become unstable. 12  Stability Analysis Example: A control system has the structure shown in the following figure Determine the gain at which the system will become unstable. 13  Stability Analysis Example :A negative feedback system has an open loop transfer function K ( s  2) L( s )  s ( s  1) (a)Find the value of the gain when the damping ratio of the closed loop roots is equal to 0.707. (b)( Find the value of the gain when the closed-loop system has two roots on the imaginary axis. 14  Stability Analysis Example :Consider the feedback system in the following Figure Determine the range of Kp and KD for stability of the closed-loop system. Determine the range of Kp and KD for stability of the closed-loop system. 15  Exercises Determine whether the systems with the following characteristic equations are stable or unstable:  Exercises A cassette tape storage device has been designed for mass-storage. It is necessary to control the velocity of the tape accurately. The speed control of the tape drive is represented by the system shown in the following Figure a) Determinethe limiting gain for a stable system. (b) Determine a suitable gain so that the overshoot to a step command is approximately 5%. The design of a control system is concerned with the arrangement, or the plan, of the system structure and the selection of suitable components and parameters. Types of compensation. (a) Cascade compensation. (b) Feedback compensation. (c) Output, or load, compensation. (d) Input compensation. A compensator is an additional component or circuit that is inserted into a control system to compensate for a deficient performance.  BODE DIAGRAM A Bode diagram is a graph commonly used in control systems engineering to determine the stability of a control system. A Bode diagram maps the frequency response of the system through two graphs - the Bode magnitude diagram (expressing magnitude in decibels) - and the Bode phase diagram (expressing phase shift in degrees). bode(sys) creates a Bode plot of the frequency response of a dynamic system model sys. The plot displays the magnitude (in dB) and phase (in degrees) of the system response as a function of frequency. bode automatically determines frequencies to plot based on system dynamics. Create a Bode plot of the following continuous-time SISO dynamic system. 10 𝑇 𝑠 = 𝑠 = 𝑗𝜔 𝑠 + 10 10 1 T(𝑗𝜔)= = % Create a bode diagram 𝑗𝜔+10 𝑗𝜔+1 10 T=tf(10, [1 10]); 1 𝑗𝜔 𝜔 𝑇(𝑗𝜔) = < 1+ = 𝜑(𝜔)=-atan( ) Bode(T) 𝜔 10 10 ( )2 +1 10 grid 𝜔 𝑇(𝑗𝜔) 𝑑𝐵 =20𝑙𝑜𝑔1 − 20𝑙𝑜𝑔 ( )2 +1 10 𝜔 2 𝑇(𝑗𝜔) 𝑑𝐵 =0-10log((10) +1) 𝜔 ≪ 10 𝑇(𝑗𝜔) 𝑑𝐵 =0 𝜔 𝜔 ≫ 10 𝑇(𝑗𝜔) 𝑑𝐵 =-20log( ) 10 𝜔 = 10 𝑇(𝑗𝜔) 𝑑𝐵 =−10log(2) = −3dB Phase 𝜔 𝜔 < 𝑇 = ∠(1 + 𝑗 ) = 00 − tan−1 10 10 𝜔 𝜔 ≺≺ 10 ⇒ ≈ 0 ⇒ ∠𝑇 ≈ tan−1 0 = 0𝑜 10 𝜔 𝜔 ≻≻ 10 ⇒ ≈ ∞ ⇒ ∠𝑇 ≈ − tan−1 ∞ = −90𝑜 10  Drawing Bode Plot Second order 𝑇(𝑠) == 𝜔𝑛2 𝑠 2 + 2𝜉𝜔𝑛 𝑠 + 𝜔𝑛2 𝑠 = 𝑗𝜔 𝜔𝑛2 2𝜉𝜔𝑛 𝜔 𝑇(𝑗𝜔) = 𝐺(𝑗𝜔)𝐻(𝑗(𝜔) = 2 ∠𝑇(𝑗𝜔) = −𝑎tan( (𝜔𝑛 − 𝜔 2 ) + 2𝑗𝜉𝜔𝑛 𝜔 𝜔𝑛2 − 𝜔 2 𝜔 1 2𝜉 𝜔𝑛 𝑇(𝑗𝜔) = ∠𝑇(𝑗𝜔) = −𝑎tan( 2) 𝜔2 𝜔 𝜔 1 − 2 ) + 2𝑗𝜉 1− 𝜔𝑛 𝜔𝑛 𝜔𝑛    0 ,  1   n    ,  1 T ( j )    20 log( 2 ) , 1  0 0 n     n      1 T ( j )    900 , 1  180o  n  40 log( ) ,      n n ,  1 n  Drawing Bode Plot Second order T(𝑠) == 4 𝑠 2 + 1.2𝑠 + 4 ω𝑛 = 2 𝜉 = 0,3 𝑇(𝑗𝜔) 𝑑𝐵 = 𝜔 0 , ≺≺ 1 % bode diagram of econd order dzeta 0 We can check by Routh-Hurwitz In closed loop, the characteristic equation is : 𝑠 3 +2𝑠 2 +s+1=0 1 1 0 2 1 1 0 2 0 All coefficients of first column are positive, then The system is stable