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Asamenew NigusseiDepartment of Electrical Engineering

Defense University CollegeBishoftu, EthiopiaNovember 13, 201412/30/2014DCS, EP-5511, by Asamenew1DESIGN OF CONVENTIONAL DTCS Closed loop stability analysisIn this course we consider linear time-invariant discrete-time SISO control systems.Consider the following closed-loop pulse-transfer function

Then the stability of the system (or other discrete-time control systems) may be determined from the location of the closed loop poles in the z-plane, or the roots of the characteristics equationP(z) = 1 + G(z)H(z) = 0

12/30/2014DCS, EP-5511, by Asamenew2For the system

To be stable, the closed-loop poles or the roots of the characteristic equation must lie within the unit circle in the z-plane. Any closed-loop pole outside the unit circle makes the system unstable.If a simple pole lies at z = 1 or a single pair of conjugatecomplex poles lies on unit circle the system becomes critically stable.Any other multiple closed-poles on the unit circle makes the system unstable.Closed-loop zeros do not affect the absolute stability.

12/30/2014DCS, EP-5511, by Asamenew3Example 1: Determine the stability of the closed-loop transfer function of a system with

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12/30/2014DCS, EP-5511, by Asamenew5Methods for testing absolute stabilityStability tests can be applied directly to the characteristic equation without solving for the roots.Schur-Cohn stability testJury stability testBilinear transformation coupled with Routh-stability criterionThe Schur-Cohn and Jury stability tests only reveal the existence of any unstable roots. They neither give the locations of the unstable roots nor the effects of parameter changes on the system stability, except for the simple case of low-order systems.The Jury test will be the focus of this course.12/30/2014DCS, EP-5511, by Asamenew6The Jury stability testConsider a given characteristic equation P(z) = 0 of a system,where

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Note that;The last row in the table consists of three elements.The elements in any even-numbered row are simply the reverse of the immediately preceding odd-numbered row.12/30/2014DCS, EP-5511, by Asamenew8

Stability criterion by the Jury test12/30/2014DCS, EP-5511, by Asamenew912/30/2014DCS, EP-5511, by Asamenew10Example -1: Jury stability test construct the jury stability table for the following characteristics equation


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Example -2: Jury stability test

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12/30/2014DCS, EP-5511, by Asamenew13Example -3: Jury stability test

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12/30/2014DCS, EP-5511, by Asamenew15Example- 4: Jury stability test

12/30/2014DCS, EP-5511, by Asamenew16Example -5: Jury stability test

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12/30/2014DCS, EP-5511, by Asamenew1812/30/2014DCS, EP-5511, by Asamenew19Transient and Steady-State Response Analysis

Transient response refers to that portion of the response due to closed-loop poles of the system.Steady-state response refers to that portion of the response due to the poles of the input or forcing function.

The response of systems to arbitrary inputs may be estimated from the response of the systems to standard inputs (step inputs, ramp inputs, or sinusoidal inputs).12/30/2014DCS, EP-5511, by Asamenew20Transient Response Specifications

In almost all practical cases the performance of systems is specified in-terms of time-domain quantities, because systems with energy storage cannot respond instantaneously and will always exhibit transient response whenever they are subjected to inputs or disturbances.

12/30/2014DCS, EP-5511, by Asamenew21Frequently, the performance characteristics of a control system are specified in-terms of the transient response to a unit-step input, since the unit-step input is easy to generate and is sufficiently drastic to provide useful information for both the transient response and the steady-state response characteristics of the system.

The transient response of a system to a unit-step input depends on the initial conditions.

The transient response of a practical control system, where the output signal is continuous time, often exhibits damped oscillations.Transient Response Specifications (cont.)12/30/2014DCS, EP-5511, by Asamenew22

Figure: Unit-step response of control systems.

Like the continuous-time control systems, digital control systems are also characterized by the transient response specifications:

Transient Response Specifications (cont.)12/30/2014DCS, EP-5511, by Asamenew23

1) Delay time td2) Rise time tr3) Peak time tp4) Maximum overshoot Mp5) Settling time tsTransient Response Specifications (cont.)12/30/2014DCS, EP-5511, by Asamenew24Delay time td : The time required for the response to reach half the final value the very first time.

Rise time tr : The time required for the response to rise from 10% to90%, or 5% to 95%, or 0% to 100% of the final value depending on the situation.

Peak time tp: The peak time is the time required for the response to reach the first peak of the overshoot.Transient Response Specifications (cont.)12/30/2014DCS, EP-5511, by Asamenew25Maximum overshoot Mp: The maximum peak value of the response curve measured from unity. If the final steady-state value of the response differs from unity, then it is common to use the maximum percent overshoot. i.e.,

Setlting time ts : The time required for the response curve to reach and stay within a range about the final value of a size specified as an absolute percentage of the final value, usually 2%. The settling time is related to the largest time constant of the control system.

Transient Response Specifications (cont.)12/30/2014DCS, EP-5511, by Asamenew26Note thatThe control system being designed must be modified until the transient response is satisfactory.All transient response specifications does not apply to any given case.12/30/2014DCS, EP-5511, by Asamenew27Steady-State Error AnalysisThe steady-state performance of a stable control system is generally judged by the steady-state error due to step, ramp, and acceleration inputs.Discrete-time control systems can be classified according to the number of open-loop poles at z = 1.

Where: B(z)/A(z) contains neither a pole nor a zero at z = 1. Then the pulse transfer function is called Type-N system, where N = 0, 1, 2, . . .

12/30/2014DCS, EP-5511, by Asamenew28Steady-State Error AnalysisConsider the discrete-time control system

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Steady-State Error Analysis (cont.)12/30/2014DCS, EP-5511, by Asamenew30Summary: Steady-State Analysis

Steady-State Error Analysis (cont.)12/30/2014DCS, EP-5511, by Asamenew31Example 2. Steady-state error constantConsider a discrete-time control system

12/30/2014DCS, EP-5511, by Asamenew32Design Based On Root Locus Method

In addition to the transient response characteristics of a given system, it is often necessary to investigate the effects of the system gain and/or sampling period on the absolute and relative stability of the closed-loop system. For such purposes the root-locus method proves to be very useful.

While using the root-locus method, for the discrete-time control systems the stability boundary is the unit circle in the z-plane.12/30/2014DCS, EP-5511, by Asamenew33Definition:Root-locus is the curve traversed by the roots of the characteristic equation of a control system as the gain (or a parameter) of the system varies. Design based on root locus methodConsider the discrete-time control system

The characteristic equation is1 + G(z)H(z) = 1 + F(z) = 0,where F(z) = G(z)H(z)Design based on root locus method(cont.)12/30/2014DCS, EP-5511, by Asamenew341 + G(z)H(z) = 1 + F(z) = 0, where F(z) = G(z)H(z)Implies that F(z) = 1Leading to the Angle and Magnitude conditions

Thus, the values of z that satisfy the angle and magnitude conditions are the roots of the closed-loop system. As we vary some parameter of the angle and magnitude conditions, the values of z that satisfy themagnitude and angle conditions also vary. The curve along which the values of z traverse is then the root-locus.

Design based on root locus method(cont.)12/30/2014DCS, EP-5511, by Asamenew35Procedure rules for constructing root lociStep 1. Obtain the characteristic equation 1 + F(z) = 0 and rearrange the equation so that the parameter of interest, such as gain K > 0,appears as the multiplying factor in the form:

This implies that(z + p1)(z + p2) . . . (z + pn) + K(z + z1)(z + z2) . . . (z + zm) = 0Then, when K = 0, the poles of F(z) are the roots of the characteristic equation. When K =infinity, the zeros of F(z) are the roots of the characteristic equation

12/30/2014DCS, EP-5511, by Asamenew36Step 2. Find the starting and terminating points of the root loci.Step 3. Determine the root loci on the real axis. To construct the root loci on the real axis choose a test point on it. If the the total number of the real poles and real zeros liying to the right of the test point is odd, the the test point lies on a root locus.Step 4. Determine the asymptotes of the root lociProcedure rules for constructing root loci (cont.)12/30/2014DCS, EP-5511, by Asamenew37Design based on root locus Example 1: Plot the root locus of feedback control system with

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12/30/2014DCS, EP-5511, by Asamenew40Example 2: Determine the critical value of K and the poles of the closed-loop poles for K = 2 for the system described in Example (1).

12/30/2014DCS, EP-5511, by Asamenew41Bode plotThe Bode-plot is the amplitude and plots of a transfer function as a function of the frequency. This can be treated as the scale of the magnitude and the phase shift a transfer function can introduce to sinusoidal signals of different frequencies.

The Bode-plot of discrete time systems can also be plotted as follows:1) Obtain the pulse transfer function of the system.2) Transform the pulse transfer function in the z-plane to -plane using the bilinear transformation.12/30/2014DCS, EP-5511, by Asamenew423) Substitute a fictitious frequency j instead of to obtain G().4) Plot the bode plot of the G().5) Remember the frequency relation of the fictitious frequency ( )and is

Example 1: Draw the Bode-plot of the following pulse transfer function

were the sampling period is T = 0.1 s.

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12/30/2014DCS, EP-5511, by Asamenew47Individual assignments: From K. Ogata, DTCS, Second editionUse MATLAB to generate the Bode plot of the transfer functions given in problems A-4-11, A-4-12, and B-4-3 (MATLAB commands: tf, bode)12/30/2014DCS, EP-5511, by Asamenew48Solving DT state-space equations, Example 1:Consider the following state-matrix of a DT system

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12/30/2014DCS, EP-5511, by Asamenew51Pulse-transfer-function matrix, Example 2:Consider Example 1, with

determine the pulse-transfer-function matrix

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12/31/2014DCS, EP-5511, by Asamenew53CHAPTER SIXPole Placement and Observer Design

Pole placement: MotivationBy now it is clear that the stability of DT LTI system can be determined by the location of the poles of the system pulse-transfer function.

Given F(z) as the pulse-transfer function of a DT system, the system is stable if the poles of F(z) are within the unit circle in the complex z-plane.

12/31/2014DCS, EP-5511, by Asamenew54Thus, for a pulse-transfer-function matrix F(z) the poles are the values of z such (zI A) = 0, which are the eigen values of the state matrix A.

The eigen values of the state matrix A can also be determined by solving |zI A| = 0.

Assignment: Practice determination of the eigen values of a square matrix (or the poles of a pulse-transfer-function matrix).12/31/2014DCS, EP-5511, by Asamenew55Pole placementAs the location of the poles of pulse-transfer-function matrix (or the eigen values of the state-matrix A) determines the stability of a DT LTI system, a system can be stabilized or the performance can be changed by relocating the poles to the desired location.

However, it is not always possible to relocate the poles to the desired locations. It is important to check if the system is full-state controllable and observable.What does controllability and observability mean?12/31/2014DCS, EP-5511, by Asamenew56ControllabilityThe notion of controllability is concerned with the problem of whether is it possible to steer a system from a given initial state at a given time to an arbitrary state.

If it is possible to steer a system from any initial state at any time to any arbitrary state within a finite time, the system is called controllable (specifically full-state controllable).

It means that one can design a controller to the system such that the system can operate as it is desired to.12/31/2014DCS, EP-5511, by Asamenew57Sometimes it is possible that the system is not complete-state controllable, and the system states can be steered to limited state-space.This means that some of the states can be steered to any desired state while other states cannot be. Example 1Determine if the following system is controllable

12/31/2014DCS, EP-5511, by Asamenew58Solution: Observe how the control input u(k) can steer all the other states.We see that the control input can directly affect the fourth state x4(k). x4(k) can steer directly x3(k + 1), which implies that the control input can steer x3(k) indirectly. x3(k) can steer x2(k + 1) and x2(k) can steer x1(k + 1).

The analysis shows the control input u(k) can steer all the states.Therefore it is possible to steer the states from any initial value to any desired value in finite time steps. Hence, the given system is CONTROLLABLE.12/31/2014DCS, EP-5511, by Asamenew59The analysis in Example 1 is simple, as the state-space representation is in controllable canonical form. In cases where the state-space representation is not easy to analyze, if the state-space representation ofa DT LTI system has to satisfy the following condition it is called complete state controllable.

rank{C} = rank{[B AB A2B . . . A n1B]} = nNote also that: If rank{C} = n, then |zI A| 0. Hence, checking the determinant of (zIA) is sufficient to check complete state controllability of a DT LTI system.12/31/2014DCS, EP-5511, by Asamenew60Class-Work.: Determine the controllability of Example 1 using the above controllability condition.

Solution: rank{C} = 4 or |zI A| = z4 3z3 2z2 + 2 0, hence the system is complete state controllable.

Note that |zI A| = 0 only at its roots (i.e., eigen values of A).12/31/2014DCS, EP-5511, by Asamenew61Controllability, Output

Note that the notion a complete state controllability does not imply complete output controllability.

A system is complete output controllable only if it is possible to steer the output from any initial value to any arbitrary output value.Mathematically, a system is complete output controllable if the following rank condition is full-filled.rank{[D CB CAB CA2B . . . CA n1B]} = m12/31/2014DCS, EP-5511, by Asamenew62Observability

The notion of observability is concerned with the problem of whether is it possible to determine the initial states of a system from measured output of the system in a given time.

If it is possible to determine all of the initial states of a system from any the measurement of the output of the system in a finite time, the system is called observable (specifically complete-state observable). 12/31/2014DCS, EP-5511, by Asamenew63It means that one can design an observer that can determine the all the initial state of a system from the output measurements in a finite time.

Sometimes it is possible that the system is not full-state observable, and the system states can only be estimated from the system output measurements.

Consider the following DT LTI system:x(k + 1) = Ax(k) + Bu(k)y(k) = Cx(k) + Du(k)12/31/2014DCS, EP-5511, by Asamenew64Since the effect of the control input u(k) on the measurement y(k) is known, observability is mainly concerned on the effect of the states x(k) on the measured output.

Thus, it is sufficient to only consider an autonomous system (i.e., B = 0 and D = 0). A DT LTI system is complete state observable if

12/31/2014DCS, EP-5511, by Asamenew65Example 2.Determine if the following DT-LTI system is observable. If so, determine the values of x(0) given y(1) = 19, y(2) = 39.

Solution: Determine the observability matrix O

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The |O| = 5 0, thus it is full-rank system. i.e, the system is full-state observable. From the state-space model, we have x1(k + 1) = x1(k) andx2(k + 1) = 2x2(k)From the output equation we also have thaty(1) = x1(1) + 5x2(1) = 19y(2) = x1(2) + 5x2(2) = 39

12/31/2014DCS, EP-5511, by Asamenew67On substitution of x1(2) = x1(1) and x2(2) = 2x2(1) andsimplification of the equations we get x1(1) = 1 and x2(1) = 4Thus, from the state equation we get x1(0) = 1 and x2(0) = 2 as the initial value of the system states.12/31/2014DCS, EP-5511, by Asamenew68Design via Pole PlacementPole placement is a control design method where the poles of the closed-loop system is placed at a desired location to provide the required system performance.

The pole locations are defined based on the desired transient-response or frequency-response characteristics.In general, during the pole placement design method, two important assumptions are:1. All the states are observable (they are available for feedback)2. All the states are controllable (they can be steered to any state desired)12/31/2014DCS, EP-5511, by Asamenew69Let assume that based on the desired characteristics, the selected polesz1 = 1, z2 = 2, . . . , zi = i , . . . , zn = n results in the following closed-loop system:x(k + 1) = Acl x(k) + Bucl (k)y(k) = Cx(k) + Ducl (k)

Now let us derive the expression of Acl .Let the state-feedback matrix be K such that u(k) = Kx(k) + ucl (k)where x(k) is the state of the open-loop system12/31/2014DCS, EP-5511, by Asamenew70Thus, we have

Then the poles i are the eigen values of Acl = A BK or the roots of the characteristic equation |zI A + BK| = 0.

12/31/2014DCS, EP-5511, by Asamenew71Consider characteristic equations

12/31/2014DCS, EP-5511, by Asamenew72Example 3Given an LTI DT system

Design a state-feedback controller such that the poles of the closed-loop system will be z1,2 = 0.5 j0.5

12/31/2014DCS, EP-5511, by Asamenew73Solution: First determine the characteristic equation of the original system and the closed-loop system.

12/31/2014DCS, EP-5511, by Asamenew74Design via Pole Placement, Ackermanns FormulaThe characteristic equation

12/31/2014DCS, EP-5511, by Asamenew75Example 4, Ackremanns formulaConsider the system in Example 2, and design the state feedback controller using Ackremanns formula.

Solution: Compute (A): from the given pole locations, we get the characteristic equation to be z2 z + 0.5 = 0, thus we have: 0 = 1, 1 = 1, 2 = 0.5. Now

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End of the Course