Objectives

This chapter introduces the analysis of the time response of different control systems under different scenarios. Only first and second order systems will be considered in details using analytical and numerical methods.

Extension to higher order systems will be developed. Both transient and steady-state responses will be evaluated. Stability analysis will be analyzed for different kinds of feedback while investigating the effect of both proportional and derivative control actions on the performance of the closed-loop system.

Finally systems types and steady state errors will be calculated for unity feedback. Outcomes By the end of this chapter, students will be able to:  evaluate both transient/steady state responses for control systems,  analyze the stability of closed-loop LTI systems,  investigate the effect of P and I control actions on performance, and  understand dominant dynamics of higher order systems.

Dr. Ashraf Zaher Introduction 3 Test signals Transient response Steady state response Analytical techniques and Numerical (simulation) techniques. Stability (definition and analysis methods), Relative stability, and Effect of P/I control actions on stability and performance.

Summary of the used systems:

 First order systems,

 Second order systems, and

 Higher order systems. Dr. Ashraf Zaher Test Signals Dr. Ashraf Zaher 4 Impulse function:  Used to simulate shock inputs,

 Laplace transforms 1. Step function:

 Used to simulate sudden disturbances,  Laplace transform: 1/s. Ramp function:

 Used to simulate gradual Fundamentals Dr. Ashraf Zaher 5 Definitions:  Zeros (Z) of the TF

 Poles (P) of the TF  Transient Response (Natural)

 Steady-State Response (Forced)

 Total Response Limits:  Initial values  Final values Systems (?Zs):  First order (one P)

 Second order (two Ps)

 Higher order! More:  Stability and relative stability

 Steady state errors (unity feedback) First Order Systems Dr. Ashraf Zaher 6 TF: T: time constant Unit Step Response: 1 1 ( ) ( ) + = R s Ts C s (1/ ) 1 1 1 1 1 1 1 ( ) Ts s s T T s Ts s C s + = − + = − + = t T c t e / ( ) 1 − = − ( ) 1 0.632 1 = = − = − c t T e T e dt T dc t t T t ( ) 1 / 1 0 = = − = ( 0) 1 0 0 c t = = − e = ( = ∞) =1− =1 −∞ c t e First Order Systems Dr. Ashraf Zaher 7 Unit Ramp Response: (1/ ) 1 1 1 1 ( ) 2 2 s T T s T s Ts s C s + = − + + = c(t = 0) = 0 c(t = ∞) = t −T t T c t t T Te / ( ) − = − + Unit Impulse Response: (1/ ) 1 1 1 1 ( ) Ts T s T C s + = + = c(t = 0) =1/T c(t = ∞) = 0 t T e T c t 1 / ( ) − = First Order Systems Dr. Ashraf Zaher 8 Example: RC Circuit  T = RC  Input is ei and output is e0  Transient Response (Natural)  Steady State Response (Forced) Second Order Systems Dr. Ashraf Zaher 9 TF: ζ : damping ratio ωn: natural frequency 2 2 2 ( ) 2 ( ) n n n R s s s C s ζω ω ω + + = Characteristic equation: Effect of damping:  ζ = 0: no damping  0 < ζ < 1: under damping  ζ = 1: critical damping  ζ > 1: over damped  ζ < 0: unstable! 2 2 2 n n s + ζω s +ω 2nd order: under damped case Dr. Ashraf Zaher 10 Step Response: Time Response: 2 2 2 2 1 ( ) n n n s s s C s ζω ω ω + + = 2 2 2 2 2 2 ( ) ( ) 1 ( ) 2 1 2 ( ) n n d d n s s s s s s s s C s σ ω σ σ ω σ ζω ω ζω + + − + + + = − + + + = −         − − − = −         − + − = −         − = − + − − − − − 2 1 2 2 1 2 2 1 cos tan 1 1 1 sin tan 1 1 sin 1 ( ) 1 cos ζ ζ ω ζ ζ ζ ω ζ ω ζ ζ ω σ σ σ t e t e c t e t t d t d t d d t 2 ωd =ωn 1−ζ σ = ζωn 2nd order: under damped case Dr. Ashraf Zaher 11 Transient Response Specifications:  Delay time, td  Rise time, tr  Peak time, tp  Maximum overshoot, Mp  Settling time, ts ζ ζ π σ σ ω π ω π β       − − = = = = − = 2 / 1 (5%) 3 (2%) 4 M e t t t p s d p d r 2nd order: under damped case Dr. Ashraf Zaher 12 2nd order: critically damped case Dr. Ashraf Zaher 13 Step Response: Time Response: 2 2 ( ) 1 ( ) n n s s C s ω ω + = 2 2 ( ) ( ) 1 1 ( ) 1 2 ( ) n n n n n s s s s s s C s ω ω ω ω ω + − + = − + + = − c t e ( t) n t n ω ω = − + − ( ) 1 1 c(t = 0) = 0 c(t = ∞) =1 = 0 ωd Properties:  No overshoot  No damped frequency (no oscillation):  Initial and final values 2nd order: over damped case Dr. Ashraf Zaher 14 Step Response: Time Response: ( )( ) 1 ( ) s a s b ab s C s + + = s b a b a s a b b a s C s + − + + − = − 1 /( ) /( ) ( ) ( ) at bt be ae b a c t − − − − = − 1 ( ) 1 c(t = 0) = 0 c(t = ∞) =1 = 0 ωd Properties:  No overshoot  Sluggish (slower than the Dr. Ashraf Zaher 15 2nd order: Comparison Dr. Ashraf Zaher 16 2nd order: Comparison Dr. Ashraf Zaher 17 2nd order: Comparison Dr. Ashraf Zaher 18 2nd order: Comparison Dr. Ashraf Zaher 19 2nd order: Impulse Response Time Response: Response type:  Under damped  Critically damped  Over damped 2 2 2 2 ( ) n n n s s C s ζω ω ω + + = Dr. Ashraf Zaher 20 2nd order: Impulse Response Under damped:  Response:  Maximum: Critically damped:  Response:  Maximum: Over damped c t e t d n t ω ζ ω σ sin 1 ( ) 2 − − = t n n c t te ω ω − = 2 ( )         − − =        − + −      − − − t t n n n c t e e ζ ζ ω ζ ζ ω ζ ω 1 1 2 2 2 2 1 ( ) d t ω β max =         − − = 2 1 max ζ βζ c ω en n t ω 1 max = n n c e ω 0.368ω 1 max = = − tmax = ? cmax = ? Dr. Ashraf Zaher 21 Higher Order Systems Response C(s): Time Response:  Break down the system into smaller systems (1st and 2nd)  Dominant behavior (simplification)

 Use Laplace transform table  Computer Simulations  Matlab/Simulink ( )( )…( ) ( )( )…( ) 1 ( ) ( ) ( ) 1 2 1 2 n m s p s p s p K s z s z s z H s G s C s + + + + + + = + = Stability Dr. Ashraf Zaher 22 Definition: A system is stable if every bounded input produces a bounded output. In short, this is called BIBO stability. If a bounded input produces an unbounded output, the system is unstable. Stability condition(s) for LTI systems: The natural response approaches zero as time approaches infinity All exponential terms have negative exponents Poles of the TF should be in the left half-plane How to find the poles of the TF:

 Factorization (easy vs. difficult polynomials)  Using Matlab (symbolic variables!)  Routh-Hurwitz criterion Routh’s Stability Criterion Dr. Ashraf Zaher 23 Closed-loop TF: C/Cs equation:  Location of the system poles  Roots of the c/cs equation ( ) ( ) … … ( ) ( ) 1 1 0 1 1 1 0 1 A s B s a s a s a s a b s b s b s b R s C s n n n n m m m m = + + + + + + + + = − − − − … 0 1 1 0 + 1 + + − + = − n n n n a s a s a s a Routh’s Stability Criterion Dr. Ashraf Zaher 24 Polynomials with +ve coefficients: Number of roots of the c/cs equation with +ve real parts = number of changes in sign of the coefficients of the first column of the array. Routh’s Stability C Routh’s Stability Criterion Dr. Ashraf Zaher 26 Special case I: Zero only in the first column Two sign changes unstable! ( 1.6681)( 0.5088 1.7020 )( 3429 1.5083 ) 10 2 3 6 5 3 10 ( ) ( ) 5 4 3 2 s s i s i R s s s s s s C s + + ± − ± = + + + + + = Routh’s Stability Criterion Dr. Ashraf Zaher 27 Special case II: Entire row is zero Auxiliary polynomial: Roots of P(s) No sign changes stable! ( 7)( 2 )( 2 ) 10 7 6 42 8 56 10 ( ) ( ) 5 4 3 2 s s i s i R s s s s s s C s + ± ± = + + + + + = ( ) 6 8 4 2 P s = s + s + 4 12 0 ( ) 3 = s + s = ds dP s 6 8 ( 2 )( 2 ) 4 2 s + s + = s ± i s ± i Relative Stability Dr. Ashraf Zaher 28 Definition: To find how close is the closed-loop system to instability. How to find it?  Shift the imaginary axis of the s-plane into the left  Re-check for stability in the p-plane  Repeat until the system is found to be unstable  Find the final value of a s = p − a Or, simply  Factorize the closed-loop c/cs equation and find the real-part of the nearest pole to the imaginary axis. Routh’s Criterion Applications Dr. Ashraf Zaher 29 s4 1 3 k s3 3 2 0 s2 (7/3) k s1 2-(9/7)k s0 k C/Cs equation: s + 3s + 3s + 2s + k 4 3 2 Range of k: 9 14 0 < k < Routh’s Criterion Applications Dr. Ashraf Zaher 30 s3 1 77 s2 18 k s1 (1386-k)/18 s0 k C/Cs equation: s +18s + 77s + k 3 2 Range of k: 0 < k <1386 Routh’s Criterion Applications Dr. Ashraf Zaher 31 ( 2)( 3) ( 20) ( ) + + + = s s s K s G s s3 1 6+k s2 5 20k s1 6-3k s0 k C/Cs equation: s 5s (6 k)s 20k 3 2 + + + + Range of k: 0 < k < 2 Effect of I Control Action Dr. Ashraf Zaher 32 P control action for plants with no internal integrators results in an offset that can be eliminated by I control action. Figure (a) shows that the control signal has a steady state value, even after the error goes to zero. Thus, I control action is used to remove offsets. In contrast, without I control action, P control actions results in a zero control signal, after the error vanishes. I control actions increase the order of the closed-loop system; thus, it might lead to instability and/or oscillatory performance Effect of P Action (1st order systems) Dr. Ashraf Zaher 33 ( ) 1 1 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) ( ) 1 ( ) ( ) ( ) ( ) ( ) 1 ( ) R s Ts K R s G s E s R s G s C s R s R s C s R s E s Ts K G s + + = + = + = − = − ⇒ = + = Ts K K Ts e e t sE s Ts K s Ts E s t s s ss + = + + + = = = + + + = →∞ → → 1 1 1 1 lim ( ) lim ( ) lim 1 1 1 ( ) 0 0 Effect of I Action (1st order systems) Dr. Ashraf Zaher 34 s Ts K s Ts R s E s s Ts K K R s C s + + + = + + = ( 1) ( 1) ( ) ( ) ( ) ( 1) ( ) 0 1 1 ( 1) ( ) lim 2 0 = + + + = = → Ts K s s Ts e sE s s ss P, I, and PI Controllers (1) Dr. Ashraf Zaher 35 p p D s Js bs K C s D s E s D s Js bs K C s + + = − = − + + = 2 2 1 ( ) ( ) ( ) ( ) 1 ( ) ( ) p d d p s ss K T s T Js bs K s e sE s = − + + − = = → 2 0 ( ) lim p d ss ss K T c = −e = s T d t T D s r t R s d = d ⇒ = = ⇒ = ( ) ( ) ( ) 0 ( ) 0 P, I, and PI Controllers (2) Dr. Ashraf Zaher 36 ( ) ( / ) ( ) ( ) ( ) ( ) ( / ) ( ) 3 2 3 2 p p i P p i Js bs K s K T s D s C s D s E s Js bs K s K T s D s C s + + + = − = − + + + = 0 ( / ) ( ) lim 3 2 2 0 = + + + − = = → s T Js bs K s K T s e sE s d p p i s ss = − = 0 ss ss c e s T d t T D s r t R s d = d ⇒ = = ⇒ = ( ) ( ) ( ) 0 ( ) 0 P, I, and PI Controllers (3) Dr. Ashraf Zaher 37 D s Js bs K C s D s E s Js bs K s D s C s + + = − = − + + = 3 2 3 2 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) lim 0 3 2 2 0 = + + − = = → s T Js bs K s e sE s d s ss = − = 0 ss ss c e s T d t T D s r t R s d = d ⇒ = = ⇒ = ( ) ( ) ( ) 0 ( ) 0 Wrong Equations (the closed-loop system is unstable): Effect of D Control Action Dr. Ashraf Zaher 38 When P control action is used – alone – in the plant on the left figure, the resulting closed-loop will result in an oscillatory response. D control action has an anticipatory behavior; thus, it damps out the oscillations via responding to the rate of change of the error. D control action is used with either P, I control actions, or both, as it can’t remove offsets, as shown on the figure on the right. PD Control Action Dr. Ashraf Zaher 39 d p d p p d Js B K s K s Js B R s E s Js B K s K K K s R s C s + + + + = + + + + = ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 p ss ss K B r t t e r t e = ⇒ = = ⇒ = ( ) ( ) 1 0 p d p d JK B K ζ J K s J B K s 2 ( ) 2 + + ⇒ = + + Unity Feedback Systems Dr. Ashraf Zaher 40 1 ( ) ( ) lim ( ) lim ( ) lim 1 ( ) 1 ( ) ( ) 1 ( ) ( ) 1 ( ) ( ) ( ) ( ) 0 0 G s sR s e e t sE s R s G s C s R s E s G s G s R s C s t s s ss + ⇒ = = =        + = − = − + = →∞ → → Steady State Errors (unity feedback) Dr. Ashraf Zaher 41 System type: the value of N in the above equation ( 1)( 1)…( 1) ( 1)( 1)…( 1) ( ) 1 + 2 + + + + + = s T s T s T s K T s T s T s G s p N a b m lim ( ) lim ( ) lim ( ) (0) 2 0 0 0 K s G s K sG s K G s G s a s v s p → → → = = = =

Summary

Dr. Ashraf Zaher 42 First and second-order systems can be effectively used to model the majority of existing control systems.

Thus, understanding their performance for various input is very important. Stability analysis is very crucial for the design of control systems.

Routh’s criterion is a powerful mathematical tool that can be used for such purpose.

Transient and steady-state responses can be evaluated for higher order systems via breaking them down into smaller first and second order subsystems. Knowing the systems type can simplify the determination of steady-state errors for different control systems.

What is next?

 Analysis and design using root-locus,

 Analysis and design using frequency response,

 PID controllers.