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Chapter 5 On-Line Computer Control – The z Transform
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Analysis of Discrete-Time Systems 1. The sampling process 2. z-transform 3. Properties of z-transforms 4. Analysis of open-loop and closed-loop discrete time systems 5. Design of discrete-time controllers
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Continuous signal and its discrete-time representation with different sampling rates 3 t (sec) y*y* 6 9 12 3 t (sec) 1 y 79115 3 t (sec) 1 y*y* 6 912 yy*y* Continuous signal Disontinuous signal tt t = nTt y*y* tt t y*y* t y*y* From the response of a real sampler to the response of an ideal impulse sample (a) (b) (c) (a)(b) (c) T = 1 sec T = 3 sec
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The Sampling Process 1. At sampling times, strength of impulse is equal to value of input signal. 2. Between sampling times, it is zero. Impulse Sampler y (t)y*y* or Laplacing
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The Hold Process : From Discrete to Continuous Time Zero – Order Hold : m * (t)m (t) Continuous output discrete impulses Hold Device t m * (t) T m (t) 1 Transfer Function : Response of an impulse input : (t)
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First Order Hold Response to an impulse input 1 1 2 T 2T Transfer function:
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First Order versus Zero Order Hold Comparison of reconstruction with zero-order and first-order holds, for slowly varying signals. Comparison of reconstruction with zero-order and first-order holds, for rapidly changing signals. 02T4T6T8T10Tt m * (nT) 02T4T6T8T10Tt m (t) 02T4T6T8T 10T t m (t) 0 1T3T5T7T t m (t) 0 1T3T5T7T t m (t) 0 1T3T5T7T t m * (nT) (a) (b) (c)
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Z-Transforms y(t) y z (t) Remarks 1. z-transform depends only on the discrete values y(0), y(ז),y(ז)..etc. If two continuous functions have the same sampled values, then z-transform will be the same. 2. It is assumed that the summation exists and is finit. 3. We can also view t in the form Z[ y (s) ] = ŷ(z) Sample
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Z-Transforms of Basic Functions 1. Unit Step Function 2. Exponential Function
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Z-Transforms of Basic Functions - Continued 3. Ramp Function 4. Trigonometric Functions
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Z-Transforms of Basic Functions - Continued 5. Translation
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Z-transform for Numerical Derivative z -1 is like a back shift operator
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Properties of z-Transforms 1. Linearity 2. Final Value Theorem
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Numerical Integration in z-transform Using Trapezoidal Rule or solving
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1.Partial fraction expansion λ 1, λ 2,… λ n are low-order polynomials in z -1 compute c 1,c 2,…c n. Invert each part separately, we able Inversion of z-transforms
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y(nT) = -1/2 + 1/2 e 11n y:0,1,4,13,0,… 2.Inversion by Long-Division 1z -1 +4z -2 +13z -3 1-4z -1 +3z- 2 z -1 z -1 -4z -2 +3z -3 4z -2 +3z -3 4z -2 -16z -3 +12z -4 13z -3 -12z -4 y(0) = 0 y(T) = 1 y(2T) = 4 y(3T) = 13 From Tables of z-transforms
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z-transforms of various functions Function Lalpace transform z-transform in time domain unit impluse 1 unit step 1/s ramp: f(t) = at a/s 2 f(t) = t n n!/s n+1 f(t) = e -at 1/s+a f(t) =te -at 1/(s+a) 2
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z-transforms of various functions Function Lalpace transform z-transform in time domain f(t) = sinωt f(t) = cosωt f(t) = 1-e -at f(t) = e -at sinωt f(t) = e -at cosωt
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Discrete-Time Response of systems In computer control: measurements are taken periodically and control actions implemented periodically, This results in a discrete input/discrete output dynamic system. Discrete System enen cncn
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Example of Discrete Systems Let a discrete time approximation is Taking z-transform
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Z-transform for a given continuous system with transfer function G(s) and a ZOH
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Example: Pure Integrator with Hold c*(s)y*(s) Step response Hence of which impulse a ramp response
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Example : First order lag system
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Step Response for 1st order lag system y(t) time * * * * * ******* Note: Compare with discrete approximation to First-order system From tables, for
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Generalization or D (z)=Transfer function relating e and c Analogous to Laplace transfer Discrete time input/output model Remark : Note that D(z) is the z-transform of the response of the system to an impulse input
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Z-transform of a continuous process with Sample and Hold Hold H (s) Process Gp (s) discrete input c*(s) continuous variables y (s)y*(s) discrete output we seek a relationship (Z-transfer function) between c and y. Consider a impulse input c*(z)=1 c*(s)=1 HGp(z) called the pulse transfer function (since it represents the z-transform of the pulse response of Gp (s) ) Then
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Properties of pulse Transfer Function 1. 2.An impulse input is converted into a pulse input by the first order hold element. Hence HG(z) is the pulse response of G(s) sampled at z internals of T. 3.The pulse transfer function of two systems in series can be combined if there is a sample and hold in between. G 1 (z)G 2 (z) c1c1 T c3c3 c2c2
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Closed-Loop System D (z)H (s)Gp (s) set point Hold Process disturbance ysp (z) + - T m (s) T y(2) y1(s) sampled output or 1. Roots of the Characteristic equation 1+HGp(z)D(z)=0 Determine stability of the closed-loop system 2. Note similarity to continuous system.
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Example: closed-loop response of a first-order system For proportional control where
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For a unit step change in set point and
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The response is very similar to continuous control. The steady state value of y(t) is Hence the offset is
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Stability of Discrete Systems A system is consider to be stable if output remains bounded for bounded Inputs. Consider a discrete system with transfer function Where P 1,P 2,…,P n are n roots of:
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Im Unstable roots real Unit circle STABLE REGION
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Example: Stability of closed-loop
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Example: Stability of closed- loop - Continued
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Digital Feedback - Control
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1. No initialization is necessary. [ C s is not needed ] Bumpless transfer from manual / automatic 2. Automatic ‘reset-windup’ protection. 3. Protection in case of computer failure 1. Since different modes are indistinguishable, on-line tuning methods will not work. 2. Difficult to put constraints on integral and / or derivative term. 1. Ziegler – Nichols 2. Cohen – Coon settings 3. Time - integral performance criteria Disadvantages: Tuning Digital Controllers: Advantages of velocity Form
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Y(t) actual time ideal
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Derivation of Deadbeat Controller- Continued
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Deadbeat
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0~10 sec
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Deadbeat control for (1/(s+1) 3 ) Sampling time: 2
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Ringing and Pole-placement Ringing refers to excessive value movement caused by a widely oscillating controller output. Caused by negative poles in D(z). Hence avoid poles near -1. Change controller design such that poles are on the side or near zero on negative side
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SYS = TF(1,[1 3 3 1]) Transfer function: 1 --------------------- s^3 + 3 s^2 + 3 s + 1 >> sysd=c2d(SYS,2) Transfer function: 0.3233 z^2 + 0.3073 z + 0.01584 -------------------------------------- z^3 - 0.406 z^2 + 0.05495 z - 0.002479 Sampling time: 2 0.3233 0.6306 0.3231 0.0158 >> p1=[1 -1];p2=[0.3233 0.3073 0.01584] p2 = 0.3233 0.3073 0.0158 >> c=conv(p1,p2) c = 0.3233 -0.0160 -0.2915 -0.0158
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Canceling the ringing pole at z=-0.8958 ans = 1.0000 -0.8958 -0.0547 >> p1=[1 0];p2=[1 -0.99];p3=[1 0.0547]; >> c=conv(p1,p2) c = 1.0000 -0.9900 0 >> c=conv(c,p3) c = 1.0000 -0.9353 -0.0542 0 Warning: Using a default value of 1 for maximum step size. The simulation step size will be limited to be less than this value. >>
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Smoothing the Control Action p=[0.3233 -0.016 -0.2915 -0.01584]; r=roots(p) r = 1.0001 -0.8959 -0.0547 Delete the unstable pole z=-0.8959 p1=[1 -0.99]; p2=[1 0.0547]; c=conv(p1,p2) c = 1.0000 -0.9353 -0.0542
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Reconstruct the Control Loop
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The treatment of unstable poles sysd=c2d(SYS,1) Transfer function: 0.0803 z^2 + 0.1544 z + 0.01788 ----------------------------------- z^3 - 1.104 z^2 + 0.406 z - 0.04979 >> p2=[ 0.0803 0.1544 0.01788] p2 = 0.0803 0.1544 0.0179 >> p1=[1 -1] p1 = 1 -1 >> c=conv(p1,p2) c = 0.0803 0.0741 -0.1365 -0.0179
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The treatment of unstable poles >> roots(c) ans = -1.7990 1.0000 -0.1238 >> p1=[1 0];p2=[1 -0.99];p3=[1 0.1238]; >> c=conv(p1,p2) c = 1.0000 -0.9900 0 >> c=conv(c,p3) c = 1.0000 -0.8662 -0.1226 0
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3.Dahlin’s Method Require that the closed–loop system behave like a first-order system with dead-time. 1.Choose , such that D(z) is realizable 2.Lot of algebra Solving for D we want for
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Dahlin’s Method
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0~10sec 0~50 sec
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Dahlin’s Method for for (1/(s+1) 3 ) Sampling time: 2
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Regulatory Control Consider a process described by y n = a 1 y n-1 + a 2 y n-2 + … + b 1 m n-1 + … + b k m n-k In regulatory control, we want to keep y close to zero in presence of disturbances. Ideally choose m n such that y n ≡ y sp y sp = a 1 y n + a 2 y n-2 + … + b k y n-k+1 + b 1 m n + b 2 m n-1 + b k m n-k+1 Orm n = -1/b 1 [ y sp – a 1 y n – a y n-1 - a k y n-k+1 – b 2 m n-1 + … - b k m n-k+1 ] Remark 1. Problems can arise in practice if model parameters are not known 2. The above choice is equivalent to minimizing 3. If dead-time is present control will be unrealizable
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