formulation
I ho\\n Fig.
I.The system controllabk if for Jllh-order polynomia]CI
control 1/= polynollli'al of
And. from the of Section .3,
Example that an aClUator delay the to a
command input or a disturbance of the control implementation, This is an inhcrent of an aclUator delay for any system, of
estimator including the delay minimizes its Use of CUITent the best rcsponse to
Controllability and that describe features
of a dynamic These concepts explicitly identified and studied by Kalman (1960) Kalman. Ho. and (] 961). We will only
few of the known results for linear. input and one
output.
We encountered in conncction with design of
controlla\\'s and gains. We in Section 8.1.3 that if the
C by
The idea of pole that is abm'e to define controllability essentially
a time-domain definition is
8.7 *Controllability and Observability
is nonsingular. then a of can eli
description into the control canonical form control such that closed-loop equation can be arbitrary (rea]) coefficient',
We begin our of controllability definition first of
controllability definitions three l:
Time
t I
o
Idl
. ()
Ibl
.-10
,b,
O
I) I
,dl Time
\'
Time Time Figure 8.38
Responses to ImpulSive disturbance,w, for Example la) Ideal
case, no delay(K 93), 0 (b) claSSICal feedback
With delayiK 4), predictor estimator w/delay(K = 931, (d) current estimator w/delay = 93)
0
0
(eI
Figure 8.37 Responses to a step input,r,for Example 8.22 la) Ideal case, no delayIK 93), (bl classical feedback with delay(K 93), (c) predictor estimator
With delay, (dl classical . 0 I
feedback with delay lal Time
(K
-0 I
344
C
(k
+
I)='-''''1
(k)+
Yill.+
I) = +If we now define = - the equation in is
+ 1) - + I)= - +
which is the same as
1. Al1 characteristic of are distinct. and 2. No element of
r;
is zero.Observabilil) 347
corresponding input matrix with elementsVi'Then the structure is as shown In FIg. By the definition. the input must be connected to each mode so that noY, is zero. this is not enough if the roots are not distinct. Suppose.
tor Instance. = Then the equations in the first two states are
The point is that if two characteristic roots are equalilla
olleinput.we effectively a hidden mode that is not to the control. and the system is not controllable. Therefore. even in this simple case, we two condltlons for controllability:
let us consider the controllability matrix of this diagonal system. By direct computation. we obtain
Figure 8.39 Blockdiagram for a system With a diagonal
I
=
+
...
[II(J\':-
I)u(O)
We ha,'e assumed that the dimension of the state. and hence the number of ro\\, of the coefficient matrix of these equations. isII: the number of columns is If is less than II. we cannot possibly find a solation for every x,. If. on other hand.N is greater thanII. we will add a column and so on. But.
the Cayley-Hamilton theorem. isee Appendix CJ. is a linear combination lower powers of and the new columns add no new rank. Therefore we solution. and our system is controllable by definitionIIif and only if the rank C isII.exactly the same condition found for pole
Our final definition is closest to the structural character of controllability III.The system controllable if every mode in is connected to control input.
Because of the of modes. we will treat only the case of systems which can to diagonal form. (The double-integrator model for the satellite does lIotqualify.) Suppose a diagonal matrix Ifx( is to equalx,.then we must be able to solve the equations
and. solving for a few steps. find that if X(O)= then
x(1)= +
=
+
=
+ +
ru(1).II.The system is controllable for andXI there is a finite Nand a of II(11 ... IIiN)such that if the has at =O. it is forced to x,atk=
In this definition we are considering the direct action of the control II on the
Xand not concerned explicitly with modes or characteristic
develop a test for controllability for definitionII.The equations are x(k + I) = + l'ulk}.
346
(8.961
(8.97)
and 349
- -I
1
u
and
Because the natural mode at = disappears. it not connected to the input.
if we consider the RHP
C
= rr =
which is clearly singular. If we compUle the transfer function fromIIto we find
this allegation by computing the determinant of the controllability matrix. The system matrices are
and = then we must test the rank of
is clearly less than which means. again. uncontrollable. In conclusion.
we have three of controllability: pole assignment. state reachability.
and mode coupling to the input. The definitions are equivalent. and tests for of these properties are found in the rank of the controllability matrix or in the rank of the input system matrix
-We thus far discussed on Iy controllability. concept of Figure 8.40
Blockdiagram of a simple uncontrollable system
observability (iiJ = (].
(]
(i) =
[
(]or there is110nonzero such that
Of for that for
- = 0'
and so on, Thus we O'has a nontrivial solution. C is singular. and the system is not controllable. To show that a nontri\'ial if C is singular requires a bit more work and is omitted. See Kailath(1979).
We have given two pictures of an uncontrollable system. Either the input is not connected to a dynamic pan physically or else two parallel parts identical characteristic roots. The engineer should be aware of the existence third simple situation. illustrated in Fig. Here the problem is that the at = appears to be connected to the input but is by the zero in preceding member: the result is an uncontrollable system. First we will conflrm Then. multiplying by we find
= = (].
The controllability matrix a product of two tenns. and C nonsingular if only if each factor is nonsingular. The first term has a determinant that product of theY,' and the term is nonsingular if and only if the
I So once again we fi nd that our definition of controllability leads to same test: The matrix C must be nonsingular.IfC is nonsingular. then we assign the by state feedback. we can the to any pan the space in finite time. and we know that every is connected to the input.' .
As our final remark on the topic of controllability we present a test that an alternative to testing the rank (or detenninant) of C. This is the
Rosenbrock-(RHP) test Isee Rosenbrock(1970).Kailath(1979)].The system rJis controllable if the system of equations
This test is equivalent to the test.Itis easyto show that if such then C is singular. For if a nonzero such that = (]by (i). then.
multiplying (il byf on the right. we find
- = II.
has only the trivial = 0'. equivalently 348 Chapter Design
8.8 351
• For any controllable system of oruer11.there exists a discrete full state feedback control (K that will place thenclosed-loop poles at arbitrarY locations. acker.m or place.m using perform this function.
• C=
r ... J.
the controllability matrix. must be of rank11.the order of the system. for the system to be controllable. ctrb. m performs this calculation.• The general rule in selecting the desired pole locations is to move existing open-loop poles as little as possible in order meet the system specifications.
• For any observable system ) of order/1.there exists a discrete estimator with gain L that will place the11estimator error equation poles at arbitrary locations. acker.m or place.m using calculates
• = [H ... ]. the observability matrix. must be of rankn. the order of the system. for the system to be observable. obsf. m performs this calculation.
• Feedback via K using the estimated state elements results in system poles that consist of the control poles plus then estimator poles.
Estimator poles are usually selected to be approximately twice as fast as the controller poles in order for the response to be dominated by the control design. However. in order to smooth the effects of measurement noise. it is sometimes useful to select estimator poles as slow or slower than the control poles.
• Calculation of Nand N via refi.m and their usao-e with the structure in Figs.8.15 or 8. 16: the command structure. the best response to command inputs.
• Integral control is achieved by implementing the desired error integral and including this as part of an augmented plant model in calculating The estimator is based on the non-augmented model. Integral control eliminates steady state errors due to command inputs and input disturbances.
As we saw in the discussion of state controllability. rows in these equations cannot be independent of previous rows if N > n because of the Cayley-Hamilton theorem. Thus the test for observability is that the matrix
must be nonsingular. we take the transpose of and let HT= rand = then we find the controllability matrix of ( another manifestation of duality.
8.8 Summary
x(O)= x(k
+
1)== Hx(k):
and successive outputs fromk= 0 are
We will consider the development of a test for observability according definition OIl. The system is described
= Hxo•
= Hx(1)= H xo_
=
Hx(2)==
N - I) = In matrix form. these equations are
if that all of in
If inputs. a we a different
01. The system H) is observable if for any nth-order polynomial there an estimator gain L such that the characteristic equation of state error of the estimator is
011. The system H) is observable if for any x(O). there is a finite N such that x(O) can be computed from observation of I) I l.
0111. The system H) is observable if every dynamic mode in is to the output via H.
is parallelto that of controllability. and most of the results thus far discussed can be transferred statements about observability by the simple expedient01 substituting the transposes for and HTfor
r.
The result of these substi-tutions is a "dual"' system. We have already seen an application of duallty when we noticed that the conditions for the ability to select an observer gain L to. . . h HT
the state-error dynamics an arbitrary charactenstlc equatIOn were t at . I
must be controllable-and we were able to use the same Ackermann formula tor estimator gain that we used for control gain. The other properties that are dual to controllability are
observability definitions
350 8 State-Space \lethods
352 Design Stale-Space
this form for andrcalled canonical 8.10 (al For
For in Problem
(a) Find the predictor estimator equarions and thc of the L that the
estimator < ...
(b) Verify that is plotting the responsc of an initial \'aluc.
For the open-loop
= =
,-preceded ZOHand a sample rate of
Find the gainKso that control hale an s-planew
10rad/sec
Find the estimator gainLj, rhat the estimator poles an equilalent I-plane
=
and=
0.7.(e) Determine transfer function of the compensatioll.
(d) Design a lead compensation techniques so that thc equi"alent s-plane natural frequency 10 and 0.7. either locus or frequency
Comparc the compensation transfer functions from Iand Id) and the dlfferences.
For the inProblem 8.6. design the controller and estimator that the
unIt a command mput a time < and an o\'ershoot
< ..-hen
(a) the state structure.
(bl the output error command structure.
For both cases. (hat the specifications are plotting the step response.
For open-loop
= = [
find transform T that if x= then the equations in be in control canonical form.
= = _
+4001
design the controller and estimator that the unit responseto
com-mand mput has a rise time r. 200 and an M when
the state command specifications arc (he
response. ...
8.5
8.6
8.7
8.8
8.9 Compute I from for
[ -a,
= I
0 0r= H=[", a I
[)• Disturbance estimation is accomplished by augmenting the model in the estimator to include the unknown disturbance a state element.
disturbance could be an unknown constant or a sinusoid with unknown magnitude and phase. The disturbance estimate can then be used in control law to reject its effect. called disturbance rejection or to cause system to track the disturbance with no error. called disturbance
o Delays in the actuator or sensor can be modeled in the estimator so that no estimation errors occur. Sensor delays will cause a delay in the to a disturbance but there need not be any delay in the response to com-mand inputs. However. actuator delays will cause a delayed response from command and disturbances.
:l r=[l]
K so that the poles of the closed-loop with full
are at = ±
8.3 For the open-loop
= - = I
II(S) +
tal Find the discrete state-space modeL assuming there aZOHand the sample T 100
(bl Pick poles that the time r, I sec and find the K that will produce poles with full feedback.
(e) that satisfied plotting the response to an initial \'alue of
8.4 For thc in Prohlem find the estimator equations and the ,'alue of the gain L by hand so that = 0,6± for
(al a predictor estimator,
(b) current and
(e) a reduced-ordcr estimator =
G(s)
=
u(s)=
I(a) Find the state-space representation there aZOHand the sample periodT= sec.
(b) find the full state digital fcedback that I-plane at = with =
(e) the of closed-loop to an initial of
that the response is with the desired 8.2 For the open-loop
Problems
8.1 For the open-loop
8.9
Problems 355
) Gisl= .'-(51
te) Introduce a command reference with feedforward that the of.i: not forced r. or compute the frequency response from r to svstem error r xand the highest frequency for which the error amplitude is than of the command amplitude,
8.15 Derive Eg. (8.63) from Eq. (8.57).
8.16 For the open-loop
= = = + 0.9669J
(a) Find the control feedbackK gainL that will place control poles at
= 0.8 estimator poles at =0.6±
IbJ Plot the of for a unit step input Jcommand the state command structure. Would there be a steady-state error if 0"
(el Determine what the state value of would be if there an input disturbance.
(d) Detennine an integral control gain and show the block with the integral control term Set the control pole =
(e) Demonstrate thaI the will no steady state error for disturbance.u·.or an mput command. even whentv
8.17 For the open-loop
0.88150.4562] H=ll
-0.4562 - r, - - 0].
(a) the control feedbackK estimator gainL
pthat place control poles at
= 0.6± and estImator poles = 0.3 jO.3.
(b) Plot the response of for unit step of the disturbance
(e) Determine integral control gain and the system block diagram with the Integral control term included. Set the e.xtra control at = 0.9.
(d) Demonstrate that the system no steady-state error for a disturbance.
8.18 For the open-loop from Problem 8.17
(a) on some unknown but value. construct an estimator that mcludes an estimate of that disturbance. Place poles of the as in Problem
8.17. place the eSlimatorpole = 0.9. ofK and
Lpand sketch the block diagram showing the various quantities are used in the control. Indude the command inputrin the diagram using the command structure.
(b) Plot the response nf ." andii' to unit step inu'withr 0. State whether the meet your
(e) Plot the response of to a unit inr with O.State whether the meet expectations.
8.19 A disk drive read has the open-loop transfer function
Franklin.Powell.and EmamI (19941.
= lOOOx+2011.
r=[ ]
check the observability for:
(a) 10 11·
(b) [I 0].
(e) Rationalize your results to (a) and (bi. why the observability or lack of it occurred.
8.13 Design the antenna in Appendix A.2 by state-"ariable pole
(a) Write the equations in form withx,= and = the matricesF. G, and H. Let a = 0.1.
(b) LetT= Iand designKfor equivalent poles ats = -1/2±j(J3/21.Plotthestcp response of the resulting design.
Ie) a prediction estimator with selected that = that both poles are at the origin.
(d) Use the estimated states for computing the control and introduce the reference input as to the state estimate undisturbed. Plot the unit step response from thi, reference input and from a wind gust (step) disturbance that acts on the antenna just like the comrol force (but not on the estimator).
(e) Plot the root locus of the closed-loop with respect to the plant gain mark the of closed-lOOp poles.
8.14 In Problem we described an experiment in magnetic levitation described by equations
Let the time.T.be 10 msec.
(a) Use pole placement to design this system to meet the specifications that time is than 0.25 sec to an initial offset in x is less than (b) Design a reduced-order estimator for for this system such that the error-settling
time will be less than 0.08 sec,
Plot step of x . . and for an initial x displacement
(d) Plot the root locus for changes in the plant gain and mark the design pole location,.
= - L,.1',. A= (I - - [KJ. B= AL,' C = -K. and
D=-KL,'
Ib) Use the of Eg. 14.65) to show that the controller based on a current estimator has a zero = 0 for choice of control lawKor estimator law L, . 8.12 For the open-loop system
(b) .the gain. such ifII= - K the characteristic equation will
+0.7. •
(e) UseTfrom part (aJ to computeK,.the gain in the x-states.
8.11 (a) Show that the equations for the current estimator can be wriuen in standard form
354 8 Design
356
where = and =
(a) a digital that no to an input
command nor to a constant di,turbance The plant gain of 1000 not known
so it not acceptable to the error due to
can be eliminated feedforward with an
are that the rise time be than and 0\ < l
a rate of kHz.
(bl The disk at30L)Orpm. There a between the
circular on the and the rotation.
in location of the that be followed the read head. Determine
to the a
the wobble magnitude.
Ie) Emhellish from Ia) that the error to the eliminated
vou Plot of the tracking error in Fig.
the i;lput to track wobble. tbe that
spin rpm.
8.20 Apendulum with a torquer at hinge
[ ] = [ ] [ ] + [ ] [ ] lJ.
\ = [I OJ ]
89 357
_ 1'(1)
G,lsi= - -= -'- = Design digital controller (K and L
r) T= control at
= 0.75± and estimator poles = 0.3±jO.3.Vcrify by simulation it response to a unit-step command using the state-command input
that is with the selected poles.
controller IK and obtained in part lal in ,imulation of
infinite spring replaced with the one the output d.that
a fourth-order plant with controller. Examine and
it qualitatiwly an of dosed-loop of this combined system.
(e) Repeat panlbJ. but replace the plant output.d.
(d) roots of would be if measured and directly
(no and them back using K pan la,.
Ie) Design a founh-order controller with control =0.75± ±jO.6.
estimator at =0 0± with the measurement.
yenfy simulation that it provides the to a unit-step usmg the state-command input structure.
Plot the of the compensation Icontrol plus estimatorl from paI1
(e). State you think kind of referred as alIolch
.filrer.
(bl Find K to obtain equivalent s-plane control = 0.5 and = rad/sec.
Ie) Find L
"
to ohtain equivalent.\ -plane poles at = 0.5 and = 8 (d) Determine and then sketch a block diagram specifying controller
the reference input.
(el On you there will be error for this fnr input"
Plot the responsc and confirm your to (e).
Repeat the of controller and estimator) for 8 and but
place all four pules = 0. (This often referred to asafinile or the will senle in a finitc number of sample periods.I (a) Kand determine the time response of the to unit-step
the input
(bJ Determine compensation in function form.D and construct a roOi
the de g,ltn of the correspond, to the
\ forKand L
8.24 The described in Appendix of many
resonances. Placing the it measures
called it that it d called the
the designer not aware initiallv that a resonance in
the a situation problem. .
For = kg.III= Ikg. = and = we obtain
frequency of 5 rad/sec with damping ratio. =
To the case where the did not know about the resonance.
the coupling rigid. infinite. The function is then
Glsl=
tn form withT= 0.1
whereXl = = angle ofthc pendulum from = u'= IOrquc
and = Answer output
sampled T= 100mseC and control(II+ I/' applied through a ZOH.
(a) With no torque =0 I. augment so that the
bias a clement. this
(bl With no =0,. sO that the torquc
is state
(e) Augment the that both biases
a (control plus for the 111
add a of output and
measurement to the computer.
la) Compute K for = 0.8 O. O.
la) Compute K for = 0.8 O. O.