Chapter 7. Digital Control Systems


 Emma Weaver
 1 years ago
 Views:
Transcription
1 Chapter 7 Digital Control Systems 1
2 1 Introduction In this chapter, we introduce analysis and design of stability, steadystate error, and transient response for computercontrolled systems. Transfer functions, representing compensators built with analog components, are now replaced with a digital computer that performs calculations that emulate the physical compensator. The digital computer (or microcontroller, microprocessor) receives the error or only the reference signal and performs calculations (program) in order to provide an output near or equals the desired signal. Figure (a) block diagram of a singleloop digital control system. Figure (b) Conversion of antenna aimuth position control system from: a. analog control to b. digital control 2
3 1.1 Advantages of Digital Computers The use of digital computers in the loop yields the following advantages 1. Reduced cost. A single digital computer can replace numerous analog controllers with a subsequent reduction in cost, analog controllers implied numerous adjustments and resulting hardware. 2. Flexibility in response to design changes. Any changes or modifications that are required in the future can be implemented with simple software changes rather than expensive hardware modifications. 3. Noise immunity. Digital systems exhibit more noise immunity than analog systems. The computer replaces the cascade compensator and uses digital signals Fig. (a) A device that converts analog signals to digital signals is called an analogtodigital (A/D) converter. a device that converts digital signals to analog signals is called a digitalto analog (D/A) converter Figure (a) a. Placement of the digital computer within the loop; b. detailed block diagram showing placement of A/D and D/A converters 3
4 1.2 DigitaltoAnalog Conversion (DAC) Digitaltoanalog conversion is simple and effectively instantaneous. Properly weighted voltages are summed together to yield the analog output. if the binary number is 110 2, the center and bottom switches are on, and the analog output is 6 volts Fig(a). the switches are electronic and are set by the input binary code. Figure (a) Digitaltoanalog converter 4
5 1.3 AnalogtoDigital Conversion (ADC) The analog signal is first converted to a sampled signal and then converted to a sequence of binary numbers, the digital signal. ADC is a twostep process (sampling and quantiation) with delay between the input analog voltage and the output digital word (not instantaneous). The sampling rate must be at least twice the bandwidth of the signal (Nyquist sampling rate), The analog signal is sampled and held at periodic intervals by a eroorderholder (ZOH) (staircase approximation to the analog signal.) The ADC converts the sample to a digital number. The analog signal's voltage is divided into discrete levels M 2B, and each level is assigned a digital number. M is the maximal value of the analog input signal and B is the length of the binary bits. In Fig (a) the analog signal is divided into eight levels 2 3. A threebit digital number (B=3) can represent each of the eight. The quantiation error e is due to the quantiation process that rounds off the analog voltage to the next higher or lower level. e 1 2 M M 2B = Figure (a) Steps in analogtodigital conversion: a. analog signal; 2 B+1 b. analog signal after sampleandhold; 5 c. conversion of samples to digital numbers
6 2 Modeling the Digital Computer 2.1 Modeling the Sampler The objective is to derive a mathematical model (transfer function of a subsystem) for the digital computer as represented by a sampler and eroorder hold. When signals are sampled, the Laplace transform can be replaced by another related transform called the transform. The sampling model is a switch turning on and off at a uniform sampling rate. Fig(a) Sampling can also be considered to be a product of the time waveform to be sampled f(t) and a sampling function s(t). Fig(b) The time equation of the sampled waveform f Tw t f Tw t = f t s t = f t + integer period of the pulse train [u t kt u t kt T w ] pulse width FIGURE (1) Two views of uniformrate sampling: a. switch opening and closing; b. product of time waveform and sampling waveform k= 6
7 Using the Laplace transform F Tw s = + k= f kt e kts s e kts T ws s = + k= f kt 1 e T ws s e kts = T w f t Replacing e T ws with its series expansion, we obtain F Tw s = + k= s = F Tw f kt 1 {1 T ws + T ws 2 } 2! s + k= f kt T ws s For small T w = e kts + k= e kts f kt T w e kts Laplace inverse transform [ LIT e kts = δ t kt ] Finally, converting back to the time domain, we have + f Tw t = T w k= Dirac Delta function f kt δ t kt = T w f t Hold ideal sampler Figure: Model of sampling with a uniform rectangular pulse train 7
8 2.2 Modeling the eroorder hold The final step in modeling the digital computer is modeling the eroorder hold, it follows the sampler and holds the last sampled value of f(t). Using an impulse input at ero time, the output is a step that starts at t = 0 and ends at t = T The output is: y(t)=h(t)=u(t) u(t T) Laplace transform is: L{u(t) u(t T)}= 1 s 1 s e Ts = 1 ets s ( s )ܪ = G h s = 1 ets s FIGURE (1) Ideal sampling and the eroorder hold The digital computer is modeled by cascading two elements: (1) an ideal sampler and (2) a eroorder hold. 8
9 3 The Transform The goal is to develop a transform that contains the information of sampling from which sampleddata systems can be modeled with transfer functions, analyed and designed Laplace transform of a sampled signal f t is: f t = + k= f kt δ t kt F s = + k= f kt δ t kt e st dt F s = + k= f kt δ t kt e st dt = + k= f kt e skt It s nonlinear. To overcome the nonlinearity problem, we transform Sdomain to another domain where the operator is linear: Z domain by setting = e st, The transform is defined as: F = + k= f kt k 9
10 Find the transform of a sampled unit ramp. Example 1 SOLUTION For a unit ramp, f kt = kt. The ideal sampled step: f t = + Z transform k T δ t kt F = + f kt k F = + kt k = T k=0 k=0 k=0 Multiplying by Z F = T F F = T We have the sum of geometric series (r < 1): 1 F = T s = 1 + r 1 + r 2 + r 3 +r N rs = r + r 2 + r 3 + r 4 +r N + r N+1 s rs = 1 r N+1 s = 1 rn+1 1 r s = 1 1 r for N Then, = F = T
11 Hint: use i=0 N a i = 1 an+1 1 a  Ztransform of unit impulse: δ = k=0 δ k k = 0 = 1  ZTransform of the unitstep: U = k=0 u k k = k=0 k = Ztransform of the exponential function: f t = r t F = k=0  If r = e a then F = e at r kt k = k=0 r T 1 k = 1 1 r T 1 = r T  For general sinusoidal form: f k = r k sin kθ F = rcos θ 2 2rcos θ +r 2 11
12 3.1 Partial table of  and stransforms 12
13 3.2 Some Properties 13
14 3.3 The Inverse Transform Three methods for finding the inverse transform (the sampled time function from its transform) will be described: 1. partialfraction expansion. 2. Residue. 3. the power series method. Since the transform came from the sampled waveform, the inverse transform will yield only the values of the time function at the sampling instants. 14
15 3.4 Inverse Transforms via PartialFraction Expansion The Laplace transform consists of a partial fraction that yields a sum of terms leading to exponentials, that is, A/(s + a). Knowing that: a akt or 1 e bt s+b e b k T F = N D F = A a + B A B + F = + b a b + The inverse is: f kt = A a k + B b k + 15
16 Example Given the function F() find the sampled time function. F = SOLUTION First we divide F() by and then we perform a partialfraction expansion F = = A B 0.7 = Next, multiply through by. F = = Thus the inverse transform f kt = k k The ideal sampled time function is: f t = k=0 f kt δ t Kt = k= k k δ t Kt Plot by hand the function f t for k = 0, 1,
17 4 Transfer Functions of a discrete system The objective is to find transfer functions of sampleddata systems. Consider the continuous system shown in Figure (a) the output is conceptually sampled in synchroniation with the input by a phantom sampler. The sampled input, r*(t), to the system of Figure(c) is: (convolution theorem) r t = c t = k=0 n=0 r nt δ(t nt) sampled input r nt g(t nt) system sampled input Time out put of G(s), discrete convolution Figure Sampleddata systems: a. continuous; b. sampled input; c. sampled input and output C = k=0 c kt k = let m = kn k = m + n C = n=0 m+n=0 n=0 k=0 Change t by kt r nt g[ k n T] k r nt g mt (m+n) C = C(t) for t = kt m=0 The transform of the sampled output is the product of the transforms of the sampled input and the pulse transfer function of the system. g mt m n=0 r nt n = G R() m+n changed to m because m+n=0 yields to m<0 for n>0 and g(mt) = 0 for m<0. 17
18 Example: Converting G1(s) in Cascade with.o.h. to G() Given a.o.h. in cascade with G 1 s = s + 2 s + 1 or G s = 1 e Ts s s + 2 s + 1 find the sampleddata transfer function, G(), if the sampling time, T, is 0.5 second. SOLUTION Knowing that Z 1 = e s T G s = 1 e Ts G 1 s s the impulse response (inverse Laplace transform) of G 1 s s G = 1 1 Z G 1 s s = 1 Z G 1 s s G 2 s = G 1 s s = s + 2 s(s + 1) = A s + B s + 1 = 2 s 1 s + 1 Taking the inverse Laplace transform g 2 t = 2 e t g 2 kt = 2 e kt From Table G 2 = 2 1 e T Substituting T = 0.5s G 2 = Z G 1 s s = = ( 0.607) G = 1 Z G 1 s s =
19 Available Commands for Continuous/Discrete Conversion The commands c2d, d2c, and d2d perform continuous to discrete, discrete to continuous, and discrete to discrete (resampling) conversions, respectively. sysd = c2d(sysc,ts) % Discretiation w/ sample period Ts sysc = d2c(sysd) % Equivalent continuoustime model sysd1= d2d(sysd,ts) % Resampling at the period Ts Step Response Amplitude Time (seconds) 19
20 5 Block Diagram Reduction Our objective here is to be able to find the closedloop sampleddata transfer function of an arrangement of subsystems that have a computer in the loop. When manipulating block diagrams for sampleddata systems, the rule is: Z G 1 s G 2 (s) G 1 G 2 () G 1 s G 2 (s) G 1 s G 2 (s) G 1 s G 2 (s) = G 1 s G 2 (s) Sampled data systems Their transforms C = R G() no sampler between G 1 s and G 2 (s) single transfer function G 1 s G 2 (s)=g 12 (s) C = R G 12 () cascaded of two subsystems each like in (a) C = R G 1 G 2 () the continuous signal entering the sampler is R s G 1 (s) RG 1 =Z{R s G 1 (s)} C = RG 1 G 2 () Figure Sampleddata systems and their transforms 20
21 Example Find the transform of the system shown in Figure SOLUTION The objective is to reduce the block diagram of Figure (a) and reducing it to the one shown in Figure (f). 1. place a phantom sampler at the output of any subsystem that has a sampled input (is not an input to other subsystem) 2. add phantom samplers S2 and S3 at the input to a summing junction whose output is sampled (synchronied samplers). 3. move sampler S1 and G(s) to the right past the pickoff point (to yield a sampler at the input of G(s)H(s)) G s H s with samplers S1 and S3 becomes GH() G s with samplers S1 and S4 becomes G Converting R s to R and C s to C Now we have the system represented totally in the domain 4. using the feedback formula, we obtain the first block (Fig(e)) 5. multiplication of the cascaded sampleddata systems yields the final result (Fig(f)) 21 FIGURE Steps in block diagram reduction of a sampleddata system
22 6 Stability The stability of digital system can be analyed in ZDomain or in SDomain. Changes in sampling rate not only change the nature of the response from overdamped to underdamped, but also can turn a stable system into an unstable one. 22
23 6.1 Digital System stability via ZPlane In the Splane, the region of stability is the left halfplane. If the transfer function, G(s), is transformed into a sampleddata transfer function, G(), the region of stability on the plane can be evaluated from Z = e Ts. Letting s = α + jω we obtain: Z = e Ts = e T(α+jω) = e αt e jωt = e αt cosωt + jsinωt = e αt ωt From the above equation, we can deduce that the stable domain that corresponds to <0, lies inside the unity circle, the jω ( =0) axis lies on the unity circle, and the unstable domain >0 lies outside the unity circle. Thus, a digital system is stable if and only if all poles of the closedloop transfer function T() are inside the unity circle. The digital system is marginally stable if poles of multiplicity one of the closedloop transfer function T() are on the unity circle and other are inside the unity circle. 23
24 Example Study the stability of the closedloop system in the figure. Where G 1 s = 1 and T=0.5s s+2 SOLUTION G = 1 1 G(s) s = G() T = 1 + G( ) = since the pole is inside the unity circle then the system is stable. MATLAB Code: 24
25 We can check the stability with regards to the sampling period T: G s = 1 e Ts s(s + 2) = e Ts ( 1 s 1 Taking the ZTransform s + 2 ) G = e 2T 1 e 2T = 0.5 e 2T T = 1 e G 0.5 2T 1 + G = e 2T = e 2T e 2T 0.5(1 e 2T ) (1.5e 2T 0.5) The pole is 1.5e 2T 0.5 The system is stable for all T >0. Let G s = 10 s+1 1 e T 10 1 e T G = 10 e T and T = 11e T 10 The system is stable for: 0 < T< 0.2. for T = 0 11e 0 10 = 1 and for T = e = 1 The pole is 11e T 10, monotonically decreases from +1 to 1 for 0 < T < 0.2. For 0.2 < T <, 11e T 10 monotonically decreases from  1 to 10. Thus, the pole of T() will be inside the unit circle, and the system will be stable if 0 < T < 0.2. In terms of frequency, where f= 1/T, the system will be stable as long as the sampling frequency is 1/0.2 = 5 hert or greater. 25
26 6.1 Stability via Splane (RouthHurwit criterion) We can study the stability of digital system using the RouthHurwit criterion. Indeed, we have just to find the transformation from Domain to sdomain. We have: Z = e st, we cannot use the transformation: G s = G(Z) =e st with a nonlinear operator. The most used transformation is the Bilinear Transformation, where: Z = s+1 sdomain to domain saw previously. s 1 This transformation verifies the mapping from Example: Let the characteristic equation of a system be: D = = 0 In sdomain for Z = s+1 s 1, this is equivalent to : s3 19 s 2 45 s 17 = 0. Thus system is unstable and has 1 pole outside the unity circle. No pole on the unity circle and two poles inside the unity circle. 26
27 7 SteadyState Errors We examine the effect of sampling upon the steadystate error for digital systems. the placement of the sampler changes the openloop transfer function. we assume the typical placement of the sampler after the error and in the position of the cascade controller, and we derive our conclusions accordingly about the steadystate error of digital systems. Consider the digital system in Figure, where the digital computer is represented by the sampler and eroorder hold. The transfer function of the plant is represented by G 1 (s). we have: E = R C, R() or: E = 1 + G() Using the final value theorem (for discrete signals see slide 13): e ss = e = lim E = R G E E + G E = R E [1 + G ] = R() E = lim 1 1 R() G() 27
28 7.1 Unit Step Input: R = 1 Thus, the steady state error is: Defining the static error constant K p as: e ss = lim G() = lim 1 G K p = limg() e ss = K p 7.2 Unit Ramp Input: R = T 1 2 The steady state error for this case is: e ss = e( ) = 1 K v Where K v = 1 T lim 1 lim 1 1 T G() lim 1 T G() 1 G() 1 lim 1 1 T T 0 1 G() 7.3 Unit Parabolic Input: R = T2 (+1) The steady state error for this case is: e ss = e( ) = 1 K a Where K a = 1 T 2 lim G() The equations developed above for e*( ), Kp, Kv, and Kaare similar to the equations developed for analog systems. Multiple pole placement at the origin of the Splane reduced steadystate errors to ero in the analog case. Multiple pole placement at = 1 reduces the steadystate error to ero for digital systems. s = 0 maps into = 1 under 28 = e Ts
29 Example For step, ramp, and parabolic inputs, find the steadystate error for the feedback control system shown in Figure if: G 1 s = 10 s s+1 SOLUTION First find G(s), the product of the.o.h. and the plant. The transform is then: G = Thus: G s = T e Ts s 2 s = 10 1 e Ts 1 s 2 1 s + 1 s + 1 e T = 10 T e T 1. For a step input: K p = lim 1 G = e ss = e = 1 1+K p = 0 2. For a ramp input: K v = 1 T lim 1 1 G = 10 e ss = e = 1 K v = For a parabolic input: K a = 1 T 2 lim 1 ( 1)2 G 1 = 0 e ss = e = 1 K a = 29
30 8 Transient Response on the ZPlane On the splane: vertical lines were lines of constant settling time, horiontal lines were lines of constant peak time, and radial lines were lines of constant percent overshoot. T r = 1.8 ω, T s = 4 σ, T p = π ω, %OS = e ξπ 1 ξ 2 The transformation used to domain is: = e st, for s = σ + jω we obtain = e σt e jωt = re jωt constant settling time are concentric circles of radius r. for σ is positive, the circle has a larger radius than the unit circle. if σ is negative, the circle has a smaller radius than the unit circle. lines of constant peak time are radial lines at an angle of θ 1. If σ is negative, that section of the radial line lies inside the unit circle. If σ is positive, that section of the radial line lies outside the unit circle. for s = σ + jω we obtain = e σt e jωt = e σt e jθ 1, ωt = θ 1 = π T T p Fig (a) Constant damping ratio, normalied (to the sampling interval) settling time, and normalied peak time plots on the 30 plane
31 For constant percent overshoot we obtain curves on the plane. radial lines in the Splane are represented by σ ω = tan ξω n sin 1 ξ = ω n 1 ξ = ξ 2 1 ξ 2 Transforming = e st = e ζω n T e jω n 1 ξ 2 T s = σ + jω = ω n ξ + jω n 1 ξ 2 to Zplane Radius and phase depend on ξ Thus, given a desired damping ratio, ξ, curves can then be used as constant percent overshoot curves on the plane through a range of ωt (see previous slide). Fig. The splane sketch of constant percent overshoot line 31
32 9 Design Gain (PController) Design via Root Locus The objective is to plot root locus and determine the gain required for stability as well as the gain required to meet a transient response requirement. the openloop and closedloop transfer functions for the generic digital system shown in Figure (a) are identical to the continuous system except for a change in variables from s to, we can use the same rules for plotting a root locus. the region of stability on the plane is within the unit circle and not the left halfplane. Thus, in order to determine stability, we must search for the intersection of the root locus with the unit circle rather than the imaginary axis. we derived the curves of constant settling time, peak time, and damping ratio. In order to design a digital system for transient response, we find the intersection of the root locus with the appropriate curves as they appear on the plane. Fig(a) Generic digital feedback 32
33 Example (Stability Design via Root Locus) Sketch the root locus for the system shown in Figure (a). Also, determine the range of gain, K, for stability from the root locus plot. SOLUTION Treat the system as if were s, and sketch the root locus. The result is shown in Figure (b) using Matlab. Fig(a) Digital feedback control search along the unit circle for 180. Identification of the gain, K, at this point yields the range of gain for stability. We find that the intersection of the root locus with the unit circle is The gain at this point is 0.5. Hence, the range of gain for stability is 0 < K < 0.5. if the openloop transfer function is given by: K G = K + γ + α + β The root locus is a circle with center 0 = γ, 0 with the radius r = γ α γ β Fig(b) Root locus for the system of fig(a) 33
34 Example (Transient Response Design via Gain Adjustment) For the system of the previous example, find the value of gain, K, to yield a damping ratio of 0.7. SOLUTION Figure (a) shows the constant damping ratio curves superimposed over the root locus for the system as determined from the last example. Draw a radial line from the origin to the intersection of the root locus with the 0.7 damping ratio curve (a line). We obtain the gain by searching along a line for 180, the intersection with the root locus. K = at j 0.215, the point where the 0.7 damping ratio curve intersects the root locus. We can now check our design by finding the unit sampled step response of the system Using our design, K = Input: R() = /(  1), sampled step R G() the sampled output C = 1 + G() = Since the overshoot is approximately 5%, the requirement of a 0.7 damping ratio has been met. Fig(a): Root locus for the system with constant 0.7 damping ratio curve Fig (b): Sampled step response of the system with K =
35 10 Cascade Compensation via the splane Rather than designing directly in the domain, we can design on the splane, using Splane analysis, and then convert the continuous compensator to a digital compensator using the bilinear. A bilinear transformation that yields a digital transfer function whose output response at the sampling instants is approximately the same as the equivalent analog transfer function is called the Tustin transformation. Tustin transformation is used to transform the continuous compensator, G c (s), to the digital compensator, G c (), by: Tustin transformation Inverse Tustin transformation As the sampling interval, T, gets smaller (higher sampling rate), the designed digital compensator's output yields a closer match to the analog compensator. 35
36 Example (Digital Cascade Compensator Design) For the digital control system of Figure(a), where the plant G p s is given, design a digital lead compensator, G c (), as shown in Figure (b), so that the system will operate with 20% overshoot and a settling time of 1.1 seconds. Create your design in the s domain and transform the compensator to the domain (Sampling period T=0.01 second). FIGURE a. Digital control system showing the digital computer performing compensation; b. continuous system used for design; SOLUTION Using Figure(b), design a lead compensator using the techniques described previously. The design was created as part of an Example, where we found that the lead compensator was (see previous chapters) 36
37 Using Tustin transformation : We have the analog compensator transfer function Using Tustin transformation with T = 0.01 Yields to the digital compensator TF The transform of the plant and eroorder hold, with T = 0.01 second, is The transformed digital system The time response in Figure (T = 0.01 s) shows that the compensated closedloop system meets the transient response requirements. The figure also shows the response for a compensator designed with sampling times at the extremes of Astrom and Wittenmark's guideline. FIGURE: Closedloop response for the compensated system of Example showing effect of three different sampling frequencies 37
Discrete 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 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 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 informationEE451/551: Digital Control. Chapter 3: Modeling of Digital Control Systems
EE451/551: Digital Control Chapter 3: Modeling of Digital Control Systems Common Digital Control Configurations AsnotedinCh1 commondigitalcontrolconfigurations As noted in Ch 1, common digital control
More information1 x(k +1)=(Φ LH) x(k) = T 1 x 2 (k) x1 (0) 1 T x 2(0) T x 1 (0) x 2 (0) x(1) = x(2) = x(3) =
567 This is often referred to as Þnite settling time or deadbeat design because the dynamics will settle in a Þnite number of sample periods. This estimator always drives the error to zero in time 2T or
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
More informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
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 informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
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 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 informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
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 informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
More informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
More informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
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 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 informationEC CONTROL SYSTEM UNIT I CONTROL SYSTEM MODELING
EC 2255  CONTROL SYSTEM UNIT I CONTROL SYSTEM MODELING 1. What is meant by a system? It is an arrangement of physical components related in such a manner as to form an entire unit. 2. List the two types
More informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and ClosedLoop Control Systems 3. Why ClosedLoop Control? 4. Case Study  Speed Control of a DC Motor 5. SteadyState Errors in Unity Feedback Control
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 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 Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
More informationTransient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n
Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
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 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 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 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 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 informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA EXPERIMENT NO : CS II/ TITLE : FAMILIARIZATION
More informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET  II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steadystate Error and Type 0, Type
More informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
More informationControl Systems. University Questions
University Questions UNIT1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
More informationRoundoff Noise in Digital Feedback Control Systems
Chapter 7 Roundoff Noise in Digital Feedback Control Systems Digital control systems are generally feedback systems. Within their feedback loops are parts that are analog and parts that are digital. At
More informationSoftware Engineering 3DX3. Slides 8: Root Locus Techniques
Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007
More informationEEE 188: Digital Control Systems
EEE 88: Digital Control Systems Lecture summary # the controlled variable. Example: cruise control. In feedback control, sensors and measurements play an important role. In discrete time systems, the control
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 informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies.
SET  1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies..
More informationContents. PART I METHODS AND CONCEPTS 2. Transfer Function Approach Frequency Domain Representations... 42
Contents Preface.............................................. xiii 1. Introduction......................................... 1 1.1 Continuous and Discrete Control Systems................. 4 1.2 OpenLoop
More informationEEE 184 Project: Option 1
EEE 184 Project: Option 1 Date: November 16th 2012 Due: December 3rd 2012 Work Alone, show your work, and comment your results. Comments, clarity, and organization are important. Same wrong result or same
More informationControl Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho kwangho@hansung.ac.kr Tel: 027604253 Fax:027604435 Introduction In this lesson, you will learn the following : The
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 informationDigital Control System Models. M. Sami Fadali Professor of Electrical Engineering University of Nevada
Digital Control System Models M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline Model of ADC. Model of DAC. Model of ADC, analog subsystem and DAC. Systems with transport
More informationIndex. Index. More information. in this web service Cambridge University Press
Atype elements, 4 7, 18, 31, 168, 198, 202, 219, 220, 222, 225 Atype variables. See Across variable ac current, 172, 251 ac induction motor, 251 Acceleration rotational, 30 translational, 16 Accumulator,
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 informationChapter 5 HW Solution
Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, timeinvariant system. Let s see, I
More informationSchool of Engineering Faculty of Built Environment, Engineering, Technology & Design
Module Name and Code : ENG60803 Real Time Instrumentation Semester and Year : Semester 5/6, Year 3 Lecture Number/ Week : Lecture 3, Week 3 Learning Outcome (s) : LO5 Module Coordinator/Tutor : Dr. Phang
More information7.1 Introduction. Apago PDF Enhancer. Definition and Test Inputs. 340 Chapter 7 SteadyState Errors
340 Chapter 7 SteadyState Errors 7. Introduction In Chapter, we saw that control systems analysis and design focus on three specifications: () transient response, (2) stability, and (3) steadystate errors,
More informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response
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 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 informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
More informationTest 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010
Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the
More informationCHAPTER 5 : REDUCTION OF MULTIPLE SUBSYSTEMS
CHAPTER 5 : REDUCTION OF MULTIPLE SUBSYSTEMS Objectives Students should be able to: Reduce a block diagram of multiple subsystems to a single block representing the transfer function from input to output
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationEC6405  CONTROL SYSTEM ENGINEERING Questions and Answers Unit  I Control System Modeling Two marks 1. What is control system? A system consists of a number of components connected together to perform
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 informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 OpenLoop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
More informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24Jan15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time
More informationAutomatique. A. Hably 1. Commande d un robot mobile. Automatique. A.Hably. Digital implementation
A. Hably 1 1 Gipsalab, GrenobleINP ahmad.hably@grenobleinp.fr Commande d un robot mobile (Gipsalab (DA)) ASI 1 / 25 Outline 1 2 (Gipsalab (DA)) ASI 2 / 25 of controllers Signals must be sampled and
More informationDepartment of Electronics and Instrumentation Engineering M. E CONTROL AND INSTRUMENTATION ENGINEERING CL7101 CONTROL SYSTEM DESIGN Unit I BASICS AND ROOTLOCUS DESIGN PARTA (2 marks) 1. What are the
More informationAN INTRODUCTION TO THE CONTROL THEORY
OpenLoop controller An OpenLoop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, nonlinear dynamics and parameter
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 informationLaplace Transform Analysis of Signals and Systems
Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.
More informationControl System. Contents
Contents Chapter Topic Page Chapter Chapter Chapter3 Chapter4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of
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 informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS
ENG0016 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS MODULE NO: BME6003 Date: Friday 19 January 2018
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 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 informationMethods for analysis and control of. Lecture 6: Introduction to digital control
Methods for analysis and of Lecture 6: to digital O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.lag.ensieg.inpg.fr/sename 6th May 2009 Outline Some interesting books:
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
More information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More informationCONTROL SYSTEMS ENGINEERING Sixth Edition International Student Version
CONTROL SYSTEMS ENGINEERING Sixth Edition International Student Version Norman S. Nise California State Polytechnic University, Pomona John Wiley fir Sons, Inc. Contents PREFACE, vii 1. INTRODUCTION, 1
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 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 informationControl Systems. EC / EE / IN. For
Control Systems For EC / EE / IN By www.thegateacademy.com Syllabus Syllabus for Control Systems Basic Control System Components; Block Diagrammatic Description, Reduction of Block Diagrams. Open Loop
More information100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =
1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot
More informationEE C128 / ME C134 Fall 2014 HW 8  Solutions. HW 8  Solutions
EE C28 / ME C34 Fall 24 HW 8  Solutions HW 8  Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot
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 informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #4 Monday, January 13, 2003 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Impulse and Step Responses of ContinuousTime
More informationEE C128 / ME C134 Final Exam Fall 2014
EE C128 / ME C134 Final Exam Fall 2014 December 19, 2014 Your PRINTED FULL NAME Your STUDENT ID NUMBER Number of additional sheets 1. No computers, no tablets, no connected device (phone etc.) 2. Pocket
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
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 informationCONTROL SYSTEMS LECTURE NOTES B.TECH (II YEAR II SEM) ( ) Prepared by: Mrs.P.ANITHA, Associate Professor Mr.V.KIRAN KUMAR, Assistant Professor
LECTURE NOTES B.TECH (II YEAR II SEM) (201718) Prepared by: Mrs.P.ANITHA, Associate Professor Mr.V.KIRAN KUMAR, Assistant Professor Department of Electronics and Communication Engineering MALLA REDDY
More informationIt is common to think and write in time domain. creating the mathematical description of the. Continuous systems using Laplace or s
It is common to think and write in time domain quantities, but this is not the best thing to do in creating the mathematical description of the system we are dealing with. Continuous systems using Laplace
More informationRoot Locus Techniques
4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since
More informationECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
More information7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM
ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)
More informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
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 informationBasic Procedures for Common Problems
Basic Procedures for Common Problems ECHE 550, Fall 2002 Steady State Multivariable Modeling and Control 1 Determine what variables are available to manipulate (inputs, u) and what variables are available
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 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 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 informationDigital Control System
Digital Control Sytem  A D D A Micro ADC DAC Proceor Correction Element Proce Clock Meaurement A: Analog D: Digital Continuou Controller and Digital Control Rt  c Plant yt Continuou Controller Digital
More informationDIGITAL CONTROLLER DESIGN
ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steadystate accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some
More information