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


 Bathsheba Mason
 1 years ago
 Views:
Transcription
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
STABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse
SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential
More informationThe Laplace Transform
The Laplace Transform Generalizing the Fourier Transform The CTFT expresses a timedomain signal as a linear combination of complex sinusoids of the form e jωt. In the generalization of the CTFT to the
More informationA system that is both linear and timeinvariant is called linear timeinvariant (LTI).
The Cooper Union Department of Electrical Engineering ECE111 Signal Processing & Systems Analysis Lecture Notes: Time, Frequency & Transform Domains February 28, 2012 Signals & Systems Signals are mapped
More informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22  Lecture 4 Step response and pole locations 4  Review Definition of transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More informationProfessional Portfolio Selection Techniques: From Markowitz to Innovative Engineering
Massachusetts Institute of Technology Sponsor: Electrical Engineering and Computer Science Cosponsor: Science Engineering and Business Club Professional Portfolio Selection Techniques: From Markowitz to
More informationNotes for ECE320. Winter by R. Throne
Notes for ECE3 Winter 45 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
More informationPoles, Zeros and System Response
Time Response After the engineer obtains a mathematical representation of a subsystem, the subsystem is analyzed for its transient and steady state responses to see if these characteristics yield the desired
More informationLaplace Transform Part 1: Introduction (I&N Chap 13)
Laplace Transform Part 1: Introduction (I&N Chap 13) Definition of the L.T. L.T. of Singularity Functions L.T. Pairs Properties of the L.T. Inverse L.T. Convolution IVT(initial value theorem) & FVT (final
More informationLab # 4 Time Response Analysis
Islamic University of Gaza Faculty of Engineering Computer Engineering Dep. Feedback Control Systems Lab Eng. Tareq Abu Aisha Lab # 4 Lab # 4 Time Response Analysis What is the Time Response? It is an
More informationDynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
More informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real
More informationTransient Response of a SecondOrder System
Transient Response of a SecondOrder System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a wellbehaved closedloop
More informationECE 3620: Laplace Transforms: Chapter 3:
ECE 3620: Laplace Transforms: Chapter 3: 3.13.4 Prof. K. Chandra ECE, UMASS Lowell September 21, 2016 1 Analysis of LTI Systems in the Frequency Domain Thus far we have understood the relationship between
More information2.161 Signal Processing: Continuous and Discrete Fall 2008
MIT OpenCourseWare http://ocw.mit.edu 2.6 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More information6.1 Sketch the zdomain root locus and find the critical gain for the following systems K., the closedloop characteristic equation is K + z 0.
6. Sketch the zdomain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)
More informationDynamic System Response. Dynamic System Response K. Craig 1
Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. NonLTI Behavior Solution of Linear, ConstantCoefficient, Ordinary Differential Equations Classical
More informationECEN 420 LINEAR CONTROL SYSTEMS. Lecture 2 Laplace Transform I 1/52
1/52 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 2 Laplace Transform I Linear Time Invariant Systems A general LTI system may be described by the linear constant coefficient differential equation: a n d n
More informationProcess Control & Instrumentation (CH 3040)
Firstorder systems Process Control & Instrumentation (CH 3040) Arun K. Tangirala Department of Chemical Engineering, IIT Madras January  April 010 Lectures: Mon, Tue, Wed, Fri Extra class: Thu A firstorder
More informationThe Laplace Transform
The Laplace Transform Syllabus ECE 316, Spring 2015 Final Grades Homework (6 problems per week): 25% Exams (midterm and final): 50% (25:25) Random Quiz: 25% Textbook M. Roberts, Signals and Systems, 2nd
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
More informationSTABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable
ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Boundedinput boundedoutput (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated
More informationC(s) R(s) 1 C(s) C(s) C(s) = s  T. Ts + 1 = 1 s  1. s + (1 T) Taking the inverse Laplace transform of Equation (5 2), we obtain
analyses of the step response, ramp response, and impulse response of the secondorder systems are presented. Section 5 4 discusses the transientresponse analysis of higherorder systems. Section 5 5 gives
More informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
More informationFirst and Second Order Circuits. Claudio Talarico, Gonzaga University Fall 2017
First and Second Order Circuits Claudio Talarico, Gonzaga University Fall 2017 Capacitors and Inductors intuition: bucket of charge q = Cv i = C dv dt Resist change of voltage DC open circuit Store voltage
More informationFirst and Second Order Circuits. Claudio Talarico, Gonzaga University Spring 2015
First and Second Order Circuits Claudio Talarico, Gonzaga University Spring 2015 Capacitors and Inductors intuition: bucket of charge q = Cv i = C dv dt Resist change of voltage DC open circuit Store voltage
More informationChapter 7: Time Domain Analysis
Chapter 7: Time Domain Analysis Samantha Ramirez Preview Questions How do the system parameters affect the response? How are the parameters linked to the system poles or eigenvalues? How can Laplace transforms
More informationControl Systems, Lecture04
Control Systems, Lecture04 İbrahim Beklan Küçükdemiral Yıldız Teknik Üniversitesi 2015 1 / 53 Transfer Functions The output response of a system is the sum of two responses: the forced response and the
More informationTime Response of Systems
Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) splane Time response p =0 s p =0,p 2 =0 s 2 t p =
More informationEE Control Systems LECTURE 9
Updated: Sunday, February, 999 EE  Control Systems LECTURE 9 Copyright FL Lewis 998 All rights reserved STABILITY OF LINEAR SYSTEMS We discuss the stability of input/output systems and of statespace
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More informationSolution of ODEs using Laplace Transforms. Process Dynamics and Control
Solution of ODEs using Laplace Transforms Process Dynamics and Control 1 Linear ODEs For linear ODEs, we can solve without integrating by using Laplace transforms Integrate out time and transform to Laplace
More informationExplanations and Discussion of Some Laplace Methods: Transfer Functions and Frequency Response. Y(s) = b 1
Engs 22 p. 1 Explanations Discussion of Some Laplace Methods: Transfer Functions Frequency Response I. Anatomy of Differential Equations in the SDomain Parts of the sdomain solution. We will consider
More informationEE 3054: Signals, Systems, and Transforms Summer It is observed of some continuoustime LTI system that the input signal.
EE 34: Signals, Systems, and Transforms Summer 7 Test No notes, closed book. Show your work. Simplify your answers. 3. It is observed of some continuoustime LTI system that the input signal = 3 u(t) produces
More informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steadystate error, and transient response for computercontrolled systems. Transfer functions,
More informationDynamic Behavior. Chapter 5
1 Dynamic Behavior In analyzing process dynamic and process control systems, it is important to know how the process responds to changes in the process inputs. A number of standard types of input changes
More informationEE/ME/AE324: Dynamical Systems. Chapter 7: Transform Solutions of Linear Models
EE/ME/AE324: Dynamical Systems Chapter 7: Transform Solutions of Linear Models The Laplace Transform Converts systems or signals from the real time domain, e.g., functions of the real variable t, to the
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More informationReview of Linear TimeInvariant Network Analysis
D1 APPENDIX D Review of Linear TimeInvariant Network Analysis Consider a network with input x(t) and output y(t) as shown in Figure D1. If an input x 1 (t) produces an output y 1 (t), and an input x
More informationSolution of ODEs using Laplace Transforms. Process Dynamics and Control
Solution of ODEs using Laplace Transforms Process Dynamics and Control 1 Linear ODEs For linear ODEs, we can solve without integrating by using Laplace transforms Integrate out time and transform to Laplace
More informationGATE EE Topic wise Questions SIGNALS & SYSTEMS
www.gatehelp.com GATE EE Topic wise Questions YEAR 010 ONE MARK Question. 1 For the system /( s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) s (C)
More informationControl Systems I. Lecture 5: Transfer Functions. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 5: Transfer Functions Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 20, 2017 E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/2017
More informationEE102 Homework 2, 3, and 4 Solutions
EE12 Prof. S. Boyd EE12 Homework 2, 3, and 4 Solutions 7. Some convolution systems. Consider a convolution system, y(t) = + u(t τ)h(τ) dτ, where h is a function called the kernel or impulse response of
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationThe Laplace Transform
The Laplace Transform Introduction There are two common approaches to the developing and understanding the Laplace transform It can be viewed as a generalization of the CTFT to include some signals with
More informationLECTURE 12 Sections Introduction to the Fourier series of periodic signals
Signals and Systems I Wednesday, February 11, 29 LECURE 12 Sections 3.13.3 Introduction to the Fourier series of periodic signals Chapter 3: Fourier Series of periodic signals 3. Introduction 3.1 Historical
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004 Revisit Temperature Control Problem τ dy dt + y = u τ = time constant = gain y ss =
More informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: SteadyState Error Unit 7: Root Locus Techniques
More informationLecture 7: Laplace Transform and Its Applications Dr.Ing. Sudchai Boonto
DrIng Sudchai Boonto Department of Control System and Instrumentation Engineering King Mongkut s Unniversity of Technology Thonburi Thailand Outline Motivation The Laplace Transform The Laplace Transform
More informationNotes on the Periodically Forced Harmonic Oscillator
Notes on the Periodically orced Harmonic Oscillator Warren Weckesser Math 38  Differential Equations 1 The Periodically orced Harmonic Oscillator. By periodically forced harmonic oscillator, we mean the
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationAutonomous Mobile Robot Design
Autonomous Mobile Robot Design Topic: Guidance and Control Introduction and PID Loops Dr. Kostas Alexis (CSE) Autonomous Robot Challenges How do I control where to go? Autonomous Mobile Robot Design Topic:
More information4.1. If the input of the system consists of the superposition of M functions, M
4. The ZeroState Response: The system state refers to all information required at a point in time in order that a unique solution for the future output can be compute from the input. In the case of LTIC
More informationController Design using Root Locus
Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers
More informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationEE 4343/ Control System Design Project LECTURE 10
Copyright S. Ikenaga 998 All rights reserved EE 4343/5329  Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phaselead and Phaselag compensators using
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More informationChapter 6: The Laplace Transform. ChihWei Liu
Chapter 6: The Laplace Transform ChihWei Liu Outline Introduction The Laplace Transform The Unilateral Laplace Transform Properties of the Unilateral Laplace Transform Inversion of the Unilateral Laplace
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationSecond Order and Higher Order Systems
Second Order and Higher Order Systems 1. Second Order System In this section, we shall obtain the response of a typical secondorder control system to a step input. In terms of damping ratio and natural
More informationReview: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system
Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system Plan of the Lecture Review: transient and steadystate
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
More informationSolutions to SkillAssessment Exercises
Solutions to SkillAssessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 505900 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
More informationEE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =
1. Pole Placement Given the following openloop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the statevariable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More informationCHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System
CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages
More informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e at, an exponential function s + a sin wt, a sine fun
More informationLaplace Transforms and use in Automatic Control
Laplace Transforms and use in Automatic Control P.S. Gandhi Mechanical Engineering IIT Bombay Acknowledgements: P.Santosh Krishna, SYSCON Recap Fourier series Fourier transform: aperiodic Convolution integral
More informationLecture 7:Time Response PoleZero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion
Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 FirstOrder Specs: Step : Pole Real inputs contain
More informationSchool of Mechanical Engineering Purdue University. ME375 Dynamic Response  1
Dynamic Response of Linear Systems Linear System Response Superposition Principle Responses to Specific Inputs Dynamic Response of f1 1st to Order Systems Characteristic Equation  Free Response Stable
More informationSinusoidal Forcing of a FirstOrder Process. / τ
Frequency Response Analysis Chapter 3 Sinusoidal Forcing of a FirstOrder Process For a firstorder transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A
More informationDr. Ian R. Manchester
Dr Ian R. Manchester Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the splane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus
More informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
More information( ) Frequency Response Analysis. Sinusoidal Forcing of a FirstOrder Process. Chapter 13. ( ) sin ω () (
1 Frequency Response Analysis Sinusoidal Forcing of a FirstOrder Process For a firstorder transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A tis: sin
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles
More informationCHAPTER 5: LAPLACE TRANSFORMS
CHAPTER 5: LAPLACE TRANSFORMS SAMANTHA RAMIREZ PREVIEW QUESTIONS What are some commonly recurring functions in dynamic systems and their Laplace transforms? How can Laplace transforms be used to solve
More informationFeedback Control of Linear SISO systems. Process Dynamics and Control
Feedback Control of Linear SISO systems Process Dynamics and Control 1 OpenLoop Process The study of dynamics was limited to openloop systems Observe process behavior as a result of specific input signals
More informationPractice Problems For Test 3
Practice Problems For Test 3 Power Series Preliminary Material. Find the interval of convergence of the following. Be sure to determine the convergence at the endpoints. (a) ( ) k (x ) k (x 3) k= k (b)
More informationEEE 184: Introduction to feedback systems
EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)
More informationTopic # Feedback Control Systems
Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that the
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : RouthHurwitz stability criterion Examples Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
More information9.5 The Transfer Function
Lecture Notes on Control Systems/D. Ghose/2012 0 9.5 The Transfer Function Consider the nth order linear, timeinvariant dynamical system. dy a 0 y + a 1 dt + a d 2 y 2 dt + + a d n y 2 n dt b du 0u +
More informationECE 380: Control Systems
ECE 380: Control Systems Course Notes: Winter 2014 Prof. Shreyas Sundaram Department of Electrical and Computer Engineering University of Waterloo ii Acknowledgments Parts of these course notes are loosely
More informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
More informationFATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY
FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY Senkottai Village, Madurai Sivagangai Main Road, Madurai  625 020. An ISO 9001:2008 Certified Institution DEPARTMENT OF ELECTRONICS AND COMMUNICATION
More informationIntroduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured
More informationOutline. Control systems. Lecture4 Stability. V. Sankaranarayanan. V. Sankaranarayanan Control system
Outline Control systems Lecture4 Stability V. Sankaranarayanan Outline Outline 1 Outline Outline 1 2 Concept of Stability Zero State Response: The zerostate response is due to the input only; all the
More informationControl System (ECE411) Lectures 13 & 14
Control System (ECE411) Lectures 13 & 14, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 SteadyState Error Analysis Remark: For a unity feedback system
More informationCourse Background. Loosely speaking, control is the process of getting something to do what you want it to do (or not do, as the case may be).
ECE4520/5520: Multivariable Control Systems I. 1 1 Course Background 1.1: From time to frequency domain Loosely speaking, control is the process of getting something to do what you want it to do (or not
More informationCDS 101/110: Lecture 3.1 Linear Systems
CDS /: Lecture 3. Linear Systems Goals for Today: Describe and motivate linear system models: Summarize properties, examples, and tools Joel Burdick (substituting for Richard Murray) jwb@robotics.caltech.edu,
More informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control DMAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
More information