# Dynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.

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1 Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 1 / 71

2 BIBO stability There are many classifications of stability in system analysis. Figure: Stability, marginal stability and instability For LTI systems, they all are basically same which is bounded input bounded output (BIBO) stability. If input is bounded (i.e., u (t) < K 1 ), then output is also bounded (i.e., y (t) < K 2 ). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 2 / 71

3 BIBO stability In the time domain, the output has the form + y (t) = h (τ) u (t τ) dτ. Bounding the output results in y (t) = + h (τ) u (t τ) dτ + h (τ) u (t τ) dτ + h (τ) u (t τ) dτ. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 3 / 71

4 BIBO stability After noting that u (t) < K 1, we obtain + y (t) K 1 h (τ) dτ. So if a system is BIBO stable, then + h (τ) dτ <. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 4 / 71

5 Dynamic response We can now model dynamic systems with differential equations. Q. What do these equations mean? Q. How does this system respond to certain inputs? Q. If we add dynamics (i.e., a controller), how will the system respond? Q. How should the system respond (i.e., design specifications)? A. Essential tool is Laplace transform. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 5 / 71

6 Some important input signals Several signals recur throughout this course. The unit step function: 1 (t) = { 1, t 0 0, otherwise 1 (t) 1 t Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 6 / 71

7 Some important input signals The unit ramp function: r (t) = { t, t 0 0, otherwise r (t) t Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 7 / 71

8 Some important input signals The unit parabola function: p (t) = { t 2 2, t 0 0, otherwise p (t) t Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 8 / 71

9 Some important input signals The cosine/sine functions: cos (t) t sin (t) t Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 9 / 71

10 Some important input signals The impulse function: The impulse function is a very strange generalized function, only defined under an integral δ (t) = 0 for t 0 }{{} zero duration and It satisfies the useful sifting property δ (t) dt = 1. } {{ } unit area u (τ) δ (t τ) dτ = u (t). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 10 / 71

11 Response of linear time invariant systems Let y (t) be the output of an LTI system with input u (t) y (t) = T [u (t)] = T u (τ) δ (t τ) dτ = u (τ) T [δ (t τ)] dτ. Let then h (t, τ) = T [δ (t τ)] y (t) = u (τ) h (t, τ) dτ. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 11 / 71

12 Response of linear time invariant systems Note that, for time invariant systems then y (t) = h (t, τ) = h (t τ) u (τ) h (t τ) dτ = u (t) h (t). The output of an LTI system is equal to the convolution of its impulse response with the input. This makes life easy. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 12 / 71

13 Finding an impulse response Example Consider a first order system ẏ (t) + ky (t) = u (t) with y (0 ) = 0 and u (t) = δ (t) where k is a positive constant. So for positive time ẏ (t) + ky (t) = 0. Recall from previous courses that y (t) = A exp (st) with A and s being constants. We need to solve for A and s. Time derivative of y (t) = A exp (st) is found as ẏ (t) = As exp (st). Substituting the above expression results in As exp (st) + ka exp (st) = 0 s + k = 0 s = k. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 13 / 71

14 Finding an impulse response Example We have solved for s, now we will solve for A 0+ 0 ẏ (t) dt }{{} y(t) k 0+ 0 y (t) dt }{{} 0 = A exp ( k0 +) = 1 0+ A = 1. So the response of the system to impulse is h (t) = exp ( kt), t > 0 = exp ( kt) 1 (t). 0 δ (t) dt }{{} 1 Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 14 / 71

15 Finding an impulse response Example Response of this system to general input: y (t) = = = = 0 u (τ) h (t τ) dτ u (t τ) h (τ) dτ u (t τ) exp ( kτ) 1 (τ) dτ u (t τ) exp ( kτ) dτ. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 15 / 71

16 Transfer function Response to impulse is called the impulse response h (t). Response to general input u (t) is found by the below convolution u (t) h (t). Response to sinusoid or cosinusoid A cos (ωt) = A (exp (jωt) + exp ( jωt)) 2 reduces to response to exponential. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 16 / 71

17 Transfer function Let u (t) = exp (st) with s being complex. The response of the system is found as y (t) = = = = exp (st) u (t τ) h (τ) dτ exp (s (t τ)) h (τ) dτ exp (st) exp ( sτ) h (τ) dτ = exp (st) H (s). exp ( sτ) h (τ) dτ Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 17 / 71

18 Transfer function An input of the form exp (st) decouples the convolution into two independent parts: a part depending on exp (st), a part depending on h (t). Recall that, a signal for which the system output is a (possibly complex) constant times the input is referred to as an eigenfunction of the system, and the amplitude factor is referred to as the system s eigenvalue. Complex exponentials are eigenfunctions of all LTI systems. Notice that H (s) (which is a constant for a specific value of s) is the eigenvalue associated with the eigenfunction exp (st). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 18 / 71

19 Transfer function Example Consider the following system model with u (t) = exp (st). We know that ẏ (t) + ky (t) = u (t) y (t) = H (s) exp (st) ẏ (t) = sh (s) exp (st). Substituting the above expression into the system model, results in sh (s) exp (st) + kh (s) exp (st) = exp (st) sh (s) + kh (s) = 1 H (s) = 1 s + k y (t) = exp (st) s + k. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 19 / 71

20 Response to cosinusoid For s = jω, u (t) = exp (jωt) y (t) = H (jω) exp (jωt). For s = jω, u (t) = exp ( jωt) y (t) = H ( jω) exp ( jωt). Thus, for u (t) = A cos (ωt) = A 2 (exp (jωt) + exp ( jωt)), y (t) = A [H (jω) exp (jωt) + H ( jω) exp ( jωt)]. 2 Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 20 / 71

21 Response to cosinusoid Consider a real h (t) with so we know that H (jω) = L exp (jφ) H ( jω) = L exp ( jφ). Substitute these into y (t) to obtain y (t) = A [L exp (jφ) exp (jωt) + L exp ( jφ) exp ( jωt)] 2 = AL 2 [exp (j (ωt + φ)) + exp ( j (ωt + φ))]. The response of an LTI system to a sinusoid is a sinusoid (of the same frequency). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 21 / 71

22 Frequency response of a first order system For H (s) = 1 s + k H (s) s=jω = H (jω) = 1 jω + k. Magnitude of H (jω) Phase angle of H (jω) M = H (jω) = The output can be written as y (t) = 1 ω 2 + k 2. φ = H (jω) = tan 1 ( ω k A ( ω 2 + k cos 2 ). ( ωt tan 1 ω k )). Can we use these results to simplify convolution and get an easier way to understand dynamic response? Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 22 / 71

23 Laplace L transform We have seen that if a system has an impulse response h (t), we can compute its transfer function H (s) H (s) = h (t) exp ( st) dt. Since we deal with causal systems (possibly with an impulse at t = 0), we can integrate from 0 instead of negative infinity H (s) = 0 h (t) exp ( st) dt. This is called the one sided (uni lateral) Laplace transform of h (t). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 23 / 71

24 Laplace L transform Name Time function f (t) Laplace transform F (s) Unit impulse δ (t) 1 Unit step 1 (t) 1/s Unit ramp t1 (t) 1/s 2 nth order ramp t n 1 (t) n!/s n+1 Sine sin (bt) 1 (t) b/ ( s 2 + b 2) Cosine cos (bt) 1 (t) s/ ( s 2 + b 2) ) Damped sine exp ( at) sin (bt) 1 (t) b/ ((s + a) 2 + b 2 ) Damped cosine exp ( at) cos (bt) 1 (t) (s + a) / ((s + a) 2 + b 2 ( (s Diverging sine t sin (bt) 1 (t) 2bs/ + b 2) ) 2 ( Diverging cosine t cos (bt) 1 (t) s 2 b 2) ( (s / 2 + b 2) ) 2 Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 24 / 71

25 Properties of Laplace transform Superposition L {af 1 (t) + bf 2 (t)} = af 1 (s) + bf 2 (s). Time delay L {f (t τ)} = exp ( sτ) F (s). Time scaling L {f (at)} = 1 ( s ) a F. a Time scaling property is useful if original equations are expressed poorly in time scale. For example, measuring disk drive seek speed in hours. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 25 / 71

26 Properties of Laplace transform Differentiation { } L f (t) } L { f (t) { } L f (m) (t) Integration = sf (s) f ( 0 ) = s 2 F (s) sf ( 0 ) f ( 0 ). = s m F (s) s m 1 f ( 0 )... f (m 1) ( 0 ). t L f (τ) dτ = 1 s F (s). 0 Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 26 / 71

27 Properties of Laplace transform Convolution Y (s) = L {y (t)} = L {h (t) u (t)} t = L h (τ) u (t τ) dτ τ=0 = = t=0 t τ=0 t=τ τ=0 h (τ) u (t τ) dτ exp ( st) dt h (τ) u (t τ) exp ( st) dtdτ. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 27 / 71

28 Properties of Laplace transform Multiply the right hand side of the above expression with exp ( sτ) exp (sτ) to obtain Y (s) = h (τ) exp ( sτ) τ=0 t=τ u (t τ) exp ( s (t τ)) dtdτ. Let t = t τ Y (s) = h (τ) exp ( sτ) dτ τ=0 = H (s) U (s). t =0 ( u t ) ( exp st ) dt Laplace transform unwraps convolution for general input signals. This makes systems easy to analyze. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 28 / 71

29 Inverse Laplace transform Inverse Laplace transform converts F (s) to f (t). Once we get an intuitive feel for F (s), we won t need to do this often. The main tool for inverse Laplace transform is partial fraction expansion. For most physical systems F (s) = b 0s m + b 1 s m b m s n + a 1 s n a n with n m. If we write F (s) in the poles/zeros form F (s) = k m (s z i ). n (s p i ) i=1 i=1 Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 29 / 71

30 Inverse Laplace transform If p i are distinct c 1 F (s) = + c s p 1 s p 2 s p n (s p 1 ) F (s) = c 1 + c 2 (s p 1 ) c n (s p 1 ). s p 2 s p n Let s = p 1 then c 1 = [(s p 1 ) F (s)] s=p1. Similarly s = p i then c i = [(s p i ) F (s)] s=pi. Since, we know that then L {exp (kt) 1 (t)} = 1 s k f (t) = n c i exp (p i t) 1 (t). i=1 Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 30 / 71 c n

31 Inverse Laplace transform Example Find the inverse Laplace transform of F (s) = Using partial fraction expansion 5 s 2 + 3s c 1 = [(s + 1) F (s)] s= 1 = (s + 2) s= 1 = 5 c 2 = [(s + 2) F (s)] s= 2 = Then, f (t) is found as 5 (s + 1) s= 2 = 5. f (t) = [5 exp ( t) 5 exp ( 2t)] 1 (t). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 31 / 71

32 Inverse Laplace transform Repeated Roots If F (s) has repeated roots, we must modify the procedure. For example, consider below F (s) which includes a third order pole F (s) = = k (s p 1 ) 3 (s p 2 ) c 1,1 + c 1,2 s p 1 (s p 1 ) 2 + c 1,3 (s p 1 ) 3 + c 2 +. s p 2 The modified formula to compute the constants is [ ] c 1,3 = (s p 1 ) 3 F (s) s=p1 [ d ( c 1,2 = (s p 1 ) 3 F (s)) ] s=p1 ds [ d 2 ( c 1,1 = ds 2 (s p 1 ) 3 F (s)) ] s=p1 c 2 = [(s p 2 ) F (s)] s=p2. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 32 / 71

33 Inverse Laplace transform Example Find the inverse Laplace transform of Let F (s) = s + 3 (s + 1) (s + 2) 2. F (s) = c 1,1 s c 1,2 (s + 2) 2 + c 2 (s + 1). Using partial fraction expansion [ ] [ ] c 1,2 = (s + 2) 2 s + 3 F (s) s= 2 = s= 2 = 1 s + 1 [ d ( c 1,1 = (s + 2) 2 F (s)) ] [ ( )] d s + 3 s= 2 = s= 2 ds ds s + 1 [ 1 = s + 1 s + 3 ] (s + 1) 2 s= 2 = 1 1 = 2. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 33 / 71

34 Inverse Laplace transform Example Using partial fraction expansion c 2 = [(s + 1) F (s)] s= 1 = Then, f (t) is found as [ ] s + 3 (s + 2) 2 s= 1 = 2. f (t) = [2 exp ( t) 2 exp ( 2t) t exp ( 2t)] 1 (t). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 34 / 71

35 Using Laplace transform to solve problems Example We can use Laplace transform to solve both homogeneous and forced differential equations. Solve for y (t) ÿ (t) + y (t) = 0 with y (0 ) = α, ẏ (0 ) = β. Using Laplace transform yields Thus, Y (s) is found as s 2 Y (s) αs β + Y (s) = 0. From tables, Y (s) = αs + β s = α s s β 1 s y (t) = [α cos (t) + β sin (t)] 1 (t). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 35 / 71

36 Using Laplace transform to solve problems Example Solve for y (t) ÿ (t) + 5ẏ (t) + 4y (t) = u (t) with u (t) = 2 exp ( 2t) 1 (t) and y (0 ) = 0, ẏ (0 ) = 0. Using Laplace transform results in s 2 Y (s) + 5sY (s) + 4Y (s) = 2 s + 2. Thus, Y (s) is found as Y (s) = 2 (s + 2) (s 2 + 5s + 4) = 2 (s + 2) (s + 1) (s + 4) = 1 s /3 s /3 s + 4. From tables, y (t) = [ exp ( 2t) + 23 exp ( t) + 13 exp ( 4t) ] 1 (t). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 36 / 71

37 Time response vs. pole locations Once the transfer function is determined, we can start to analyze the response of the system it represents. When the system equations are ordinary differential equations, the transfer function will be a ratio of polynomials H (s) = b (s) a (s) where we assume that a (s) and b (s) have no common factors. The values of s such that a (s) = 0 represents points where H (s) is infinity and these s values are called poles of H (s). The values of s such that b (s) = 0 are points where H (s) = 0 and these s values are called zeros of H (s). Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 37 / 71

38 Time response vs. pole locations Impulse response is the time function corresponding to the transfer function. The impulse response is also called the natural response of the system. Poles and zeros are used to compute corresponding time response of the system. Poles qualitatively determine the behavior of the system. Poles identify the classes of signals contained in the impulse response. Zeros quantify this relationship. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 38 / 71

39 First order pole Consider a first order pole H (s) = 1 s + σ. From the tables, h (t) can be found as h (t) = exp ( σt) 1 (t). If σ > 0, pole is at s < 0, stable (i.e., impulse response decays, and any bounded input produces bounded output). If σ < 0, pole is at s > 0, unstable. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 39 / 71

40 First order pole h (t) exp ( σt) 1 e 0.2 Time (sec τ) Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 40 / 71

41 First order pole Figure: Unit step response of a stable first order pole Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 41 / 71

42 Second order systems without any zeros Standard form of a second order system without any zeros b 0 Kωn 2 H (s) = s 2 = + a 1 s + a 2 s 2 + 2ζω n s + ωn 2 where ζ is the damping ratio, ω n is the natural/undamped frequency. From the tables, h (t) can be found as h (t) = ω n 1 ζ 2 exp ( σt) sin (ω dt) 1 (t) where σ = ζω n and ω d = ω n 1 ζ 2 is the damped frequency. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 42 / 71

43 Second order systems without any zeros Figure: Damping ratios vs. pole locations Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 43 / 71

44 Second order systems without any zeros Figure: Envelope of sinusoid decays as exp ( σt) Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 44 / 71

45 Second order systems without any zeros Figure: Impulse responses of second order systems Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 45 / 71

46 Second order systems without any zeros Figure: Step responses of second order systems Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 46 / 71

47 Second order systems without any zeros Low damping, ζ 0, oscillatory. High damping, ζ 1, no oscillations. 0 < ζ < 1 underdamped. ζ = 1 critically damped. ζ > 1 overdamped. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 47 / 71

48 Second order systems without any zeros Figure: Impulse responses vs. pole locations Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 48 / 71

49 Second order systems without any zeros Figure: Step responses vs. pole locations Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 49 / 71

50 Time domain specifications We have seen impulse and step responses for first and second order systems. Our control problem may be to specify exactly what the response should be. Usually expressed in terms of the step response. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 50 / 71

51 Time domain specifications t r is the rise time and is equal to the time to reach vicinity of set point. t s is the settling time and is the time for transients to decay (to 5%, 1%). M p is the percent overshoot. t p is the time to peak. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 51 / 71

52 Rise time All step responses rise in roughly the same amount of time. Take ζ = 0.5 to be average. Time from 0.1 to 0.9 is approximately equal to t r 1.8 ω n. We could make this more accurate, but note that this is only valid for second order systems with no zeros. Use this as approximate design rule of thumb and iterate design until specifications are met. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 52 / 71

53 Peak time Step response can be found from ILT of H (s) /s as ( y (t) = 1 exp ( σt) cos (ω d t) + σ ) sin (ω d t) ω d where σ = ζω n and ω d = ω n 1 ζ 2. Peak occurs when ẏ (t) = 0 ( ẏ (t) = σ exp ( σt) cos (ω d t) + σ ) sin (ω d t) ω d So, exp ( σt) ( ω d sin (ω d t) + σ cos (ω d t)) ( σ 2 + ω 2 ) d = exp ( σt) sin (ω d t) = 0. ω d ω d t p = π t p = π ω d = π ω n 1 ζ 2. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 53 / 71

54 Percent overshoot vs. damping ratio Percent overshoot is M p = 100 exp ( ζπ/ ) 1 ζ 2 where common values are M p = 16% for ζ = 0.5 or M p = 5% for ζ = 0.7. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 54 / 71

55 Settling time Determined mostly by decaying exponential with ɛ = 0.01, 0.02, or For ɛ = 0.01, we obtain exp ( ω n ζt s ) = ɛ exp ( ω n ζt s ) = 0.01 ω n ζt s = 4.6 t s = 4.6 ω n ζ = 4.6 σ. For ɛ = 0.02, we obtain exp ( ω n ζt s ) = 0.02 ω n ζt s = 3.9 t s = 3.9 ω n ζ = 3.9 σ. For ɛ = 0.05, we obtain exp ( ω n ζt s ) = 0.05 ω n ζt s = 3.6 t s = 3.6 ω n ζ = 3.6 σ. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 55 / 71

56 Design synthesis Specifications on t r, t s, M p determine pole locations. ω n 1.8/t r. ζ function (M p ) (read from M p versus ζ graph). σ 4.6/t s (for ɛ = 0.01). Figure: Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 56 / 71

57 Converting specifications to s plane Example Specifications are: t r 0.6, M p 10%, t s 3 sec at 1%. ω n 1.8/t r = 1.8/0.6 = 3.0 rad/sec. From M p versus ζ graph, ζ 0.6. σ 4.6/t s = 4.6/3 1.5 sec for ɛ = Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 57 / 71

58 Converting specifications to s plane Designing a motor compensator Suppose a servo motor system for a pen plotter has transfer function 0.5K a s 2 + 2s + 0.5K a = ω 2 n s 2 + 2ζω n s + ωn 2. where there is only one adjustable parameter which is K a. So, we can choose only one specification (i.e., either of t r, t s or M p ). Let the design requirement be to allow no overshoot. Thus, M p = 0 results in ζ = 1. From the denominator of the transfer function, 2 = 2ζω n, thus ω n = 1. Thus 0.5K a = 1 resulting in K a = 2. Note that, for ɛ = 0.01, we obtain t s = 4.6 sec. We will need a better controller than this for a pen plotter! Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 58 / 71

59 Second order systems without any zeros We have looked at first order and second order systems without zeros, and with unity gain. Non unity gain: If we multiply by K, the dc gain becomes K. Note that, t r, t s, M p, t p are not affected. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 59 / 71

60 Second order systems with a zero Adding a zero to a second order system: Consider the following transfer functions H 1 (s) = H 2 (s) = 2 (s + 1) (s + 2) = 2 s s (s + 1.1) 1.1 (s + 1) (s + 2) = 0.18 s s + 2. Same dc gain (at s = 0). Coefficient of (s + 1) pole is greatly reduced. General conclusion: A zero near a pole tends to cancel the effect of that pole. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 60 / 71

61 Second order systems with a zero How about transient response? H (s) = s αζω n + 1 ( ) 2 s ω n + 2ζs ω n + 1 Zero at ασ. Poles at R (s) = σ. Large α, zero far from poles no effect. α 1, large effect. Overshoot goes up as α 0. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 61 / 71

62 Second order systems with a zero Figure: Step responses for a second order system with a zero Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 62 / 71

63 Second order systems with a zero Figure: Overshoot versus normalized zero location for a second order system with a zero Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 63 / 71

64 Second order systems with a zero Set ω n = 1 H (s) = s αζ + 1 s 2 + 2ζs + 1 = 1 s 2 + 2ζs s αζ s 2 + 2ζs + 1 = H o (s) + H d (s). where H o (s) is the original response (i.e., without the zero) and H d (s) is the added term due to the zero. Notice that H d (s) = 1 αζ sh o (s). The time response of H d (s) is a scaled version of the derivative of the time response of H o (s). If any of the zeros are in RHP, the system is called non minimum phase. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 64 / 71

65 Second order systems with a zero Figure: Step response for a second order minimum phase system Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 65 / 71

66 Second order systems with a zero Figure: Step response for a second order non minimum phase system Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 66 / 71

67 Adding a pole to a second order system Consider the transfer function 1 H (s) = ( ) ( ( ) 2 ). s αζω n + 1 s ω n + 2ζ ω n s + 1 Original poles are at R (s) = σ = ζω n. New pole is at s = αζω n. Major effect is an increase in rise time. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 67 / 71

68 Adding a pole to a second order system Figure: Step response for a pole added second order system (i.e., a third order system) Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 68 / 71

69 Adding a pole to a second order system Figure: Normalized rise time vs. normalized pole location Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 69 / 71

70 Summary of higher order approximations Extra zero in LHP will increase overshoot if the zero is within a factor of 4 from the real part of complex poles. Extra zero in RHP depresses overshoot, and may cause step response to start in wrong direction. Extra pole in LHP increases rise time if extra pole is within a factor of 4 from the real part of complex poles. Figure: Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 70 / 71

71 Summary of higher order approximations Matlab step and impulse commands can plot higher order system responses. Since a model is an approximation of a true system, it may be all right to reduce the order of the system to a first or second order system (if higher order poles and zeros are a factor of 5 or 10 times farther from the imaginary axis). Analysis and design are much easier. Numerical accuracy of simulations are better for lower order models. First and second order models provide us with great intuition into how the system works. May be just as accurate as higher order model, since higher order model itself may be inaccurate. Assoc. Prof. Enver Tatlicioglu EE362 Feedback Control Systems Chapter 3 71 / 71

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