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Direct Computation of thePerformance Index for anOptimally Controlled ActiveSuspension with Preview Appliedto a Half-Car ModelA.G. Thompson & C.E.M. Pearce

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To cite this article: A.G. Thompson & C.E.M. Pearce (2001): Direct Computation of thePerformance Index for an Optimally Controlled Active Suspension with Preview Applied toa Half-Car Model, Vehicle System Dynamics: International Journal of Vehicle Mechanics andMobility, 35:2, 121-137

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Vehicle System Dynamics 0042-3114/01/3502-121$16.002001, Vol. 35, No. 2, pp. 121±137 # Swets & Zeitlinger

Direct Computation of the Performance Index for an

Optimally Controlled Active Suspension with

Preview Applied to a Half-Car Model

A.G. THOMPSON1 and C.E.M. PEARCE2

SUMMARY

A quadratic integral performance index is de®ned for a linear active preview±type suspensionand a series of matrix expressions derived for its evaluation by means of MATLAB or somesimilar computer program in the case of a unit step input to the system. The computation, whichis both fast and accurate compared to simulation, requires the solution of Lyapunov± andRiccati±type equations. Some examples of numerical computation are given and these showexcellent agreement with published results. The conclusion features a useful computer program.

1. INTRODUCTION

The ride and road-holding characteristics of two-dimensional half-car modelshave been investigated at various times in the past with the object ofdetermining the design factors for optimal performance [1±4]. For the case ofa half-car model excited by a unit step input but without preview, an explicitsolution for the optimal performance index has been given by Sharp andWilson [5]. In an earlier solution by Thompson and Pearce [1] the control isrestricted to the feedback of a weighted sum of the system state variables,which is more easily realised physically but less optimal. In a more recentpaper by Thompson and Pearce [6] a quite compact formula is obtained for theperformance index of a quarter-car model with preview. It seems natural toattempt to extend this analysis to the case of a half-car model. This is the

1Department of Mechanical Engineering, The University of Adelaide, Adelaide SA 5005,Australia.2Corresponding author. Department of Applied Mathematics, The University of Adelaide,Adelaide SA 5005, Australia. Tel.: 61 8 8303 5412; Fax: 61 8 8303 3696; E-mail:cpearce@maths.adelaide.edu.au

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content of the present paper. Although the formul� here are more complicated,they work just as well. As in the case of the quarter-car model, we have notallowed for an integral constraint, which would need to be included for thepurpose of a self-levelling system. Speci®cally, such a constraint serves toeliminate static body de¯ections due to passenger and baggage loading forces.As in the case of a quarter-car model, however, we believe that the inclusion ofthe additional state variables would not affect the expressions derived for thecomputation of the performance index.

As previously we have also restricted the input to a unit step in the beliefthat this is about the worst discrete form of road disturbance likely to beencountered and one which is analogous to an integrated white noise typerandom road input, in the sense that, due to the identical frequencydistributions of these inputs, a suspension system which is optimal for onewill also be optimal for the other. Here we are dealing with a theoreticalsuspension with direct feedback of the system state variables, which are not alldirectly measurable. In practice, either a state observer or an equivalentsystem based on direct measurements would be employed as in [7].

The numerical results obtained for the example in Section 9 are in accurateagreement with those in previously published references and can easily bederived by use of the MATLAB program given in the conclusion. We have notattempted to produce trade-off performance plots for different vehicle speeds,road conditions etc. as this has been done comprehensively by Hrovat [8, 9]based on a discrete-time method of design similar to that employed by Louam,Sharp and Wilson [10].

2. THE SYSTEM ANALYSIS

The half-car model with preview is shown in Figure 1. The car is dividedlongitudinally by a vertical plane so that M is half the total body mass and J ishalf the pitch moment of inertia about a transverse axis through the centre ofgravity. The front and rear unsprung masses M1 and M3 each correspond to asingle wheel, while S1 and S3 represent the tyre spring rates. In consideringvertical motions, the control forces u1 and u3 are assumed to be appliedbetween the wheels and the body at the front and rear, respectively. Thewheelbase L � a� b, where a and b are the distances of the centre of gravityof the body from its support points and Ps is the distance to the preview sensor.For a vehicle forward velocity V, the preview time Tp � Ps=V and the delay

122 A.G. THOMPSON AND C.E.M. PEARCE

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time Td � L=V . The road input xr � U�t� is a unit step which is encounteredby the sensor at time t � 0. The ensuing disturbance inputs at the tyre contactpoints may then be represented by

w1 � xa � U�t ÿ Tp�; w3 � xb � U�t ÿ Tp ÿ Td�:The vertical displacements x1 to x4 are measured from equilibrium on a levelroad and together with their corresponding velocities x5 to x8 form a set ofeight state variables which are all zero initially. The state equations for thehalf-car model may then be derived as

_x1 � x5; _x2 � x6; _x3 � x7; _x4 � x8;

_x5 � S1�xa ÿ x1�=M1 ÿ u1=M1;

_x6 � �1=M � a2=J�u1 � �1=M ÿ ab=J�u3 � �u1 � �u3;

_x7 � S3�xb ÿ x3�=M3 ÿ u3=M3;

_x8 � �1=M ÿ ab=J�u1 � �1=M � b2=J�u3 � �u1 � u3:

In accordance with previous references [1, 3], the performance index isde®ned as

� �Z 1

0

��1u21 � �3u2

3 � q1�x1 ÿ xa�2

� q2�x1 ÿ x2�2 � q3�x3 ÿ xb�2 � q4�x3 ÿ x4�2� dt: �1�With the above notation, the state vector x � �x1 x2 x3 x4 x5 x6 x7 x8�0,

the control force vector u � �u1 u3�0 and the disturbance input vector

Fig. 1. Half-car model with preview showing a unit±step road input.

DIRECT COMPUTATION OF THE PERFORMANCE INDEX 123

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w � �w1 w3�0 � �xa xb�0. A transformation to relative displacements is nowachieved by de®ning the new state variables

z1 � x1 ÿ xa; z5 � x3 ÿ xb;z2 � x2 ÿ x1; z6 � x4 ÿ x3;z3 � x5; z7 � x7;z4 � x6; z8 � x8:

The new state vector z � �z1 z2 z3 z4 z5 z6 z7 z8�0 and the matrix transfor-mation equation is

z � Cx� Dw �2�while the transformation matrices C and D may be written in full as

C �

1 0 0 0 0 0 0 0

ÿ1 1 0 0 0 0 0 0

0 0 0 0 1 0 0 0

0 0 0 0 0 1 0 0

0 0 1 0 0 0 0 0

0 0 ÿ1 1 0 0 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 1

266666666664

377777777775; D �

ÿ1 0

0 0

0 0

0 0

0 ÿ1

0 0

0 0

0 0

266666666664

377777777775: �3�

By differentiating (2) we obtain

_z � C _x� D _w � C _x� D� _xa _xb�0; �4�which introduces the delta functions

_xa � ��t ÿ Tp�; _xb � ��t ÿ Tp ÿ Td�:We require also the transformed state equations, which are

_z1 � z3 ÿ _xa;

_z2 � z4 ÿ z3;

_z3 � ÿ�S1z1 � u1�=M1;

_z4 � �u1 � �u3;

_z5 � z7 ÿ _xb;

_z6 � z8 ÿ z7;

_z7 � ÿ�S3z5 � u3�=M3;

_z8 � �u1 � u3

124 A.G. THOMPSON AND C.E.M. PEARCE

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In matrix form, the transformed state equations are represented by

_z � Az� Bu� D _w with z�0� � 0: �5�The 2±column matrix D is as in (3) while the matrices A and B may be writtenas

A �

0 0 1 0 0 0 0 0

0 0 ÿ1 1 0 0 0 0

ÿS1=M1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 ÿ1 1

0 0 0 0 ÿS3=M3 0 0 0

0 0 0 0 0 0 0 0

266666666666664

377777777777775;

B �

0 0

0 0

ÿ1=M1 0

� �

0 0

0 0

0 ÿ1=M3

�

266666666666664

377777777777775:

If D is partitioned into its columns D � �d1 d2�, the state equation (5)becomes

_z � Az� Bu� d1��t ÿ Tp� � d2��t ÿ Tp ÿ Td�and the performance index simpli®es to

� �Z 1

0

�u0Ru� z0Qz� dt:

The square symmetric matrices R and Q are de®ned by

Q � diag � q1 q2 0 0 q3 q4 0 0 � and R � diag � �1 �3 �:

DIRECT COMPUTATION OF THE PERFORMANCE INDEX 125

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3. THE CONTROL EQUATIONS

For the half-car model with preview, the optimal control forces have beenshown in [3, 4] to be the components of the 2-vector

u�t� � ÿRÿ1B0�Kz�t� � g�t��; �6�where g�t� is an 8-vector function and K is the 8� 8 positive-de®nite solutionof the algebraic Riccati equation

A0K � KAÿ KBRÿ1B0K � Q � 0: �7�In (6) the feedforward components of the control due to the preview arerepresented by the 2-vector

v�t� � �v1�t� v3�t��0 � ÿRÿ1B0g�t�:The preview function g�t� is in this case the solution of the vector equation

_g� A0cg� KD _w � 0; �8�in which Ac is the optimum closed-loop system matrix given by

Ac � Aÿ BRÿ1B0K: �9�The solution of (8), subject to the boundary condition that g�tf � � 0 as tf

tends to in®nity, is given by the integral

g�t� �Z 1

t

expÿA0c�� ÿ t��KD _w���d� for t � 0:

Making use of the fact that D _w��� � d1��� ÿ Tp� � d2��� ÿ Tp ÿ Td�, wewrite g�t� � g1�t� � g2�t�, where

g1�t� �Z 1

t

expÿA0c�� ÿ t��Kd1��� ÿ Tp

�d�;

g2�t� �Z 1

t

expÿA0c�� ÿ t��Kd2��� ÿ Tp ÿ Td

�d�:

We note that g1�t� � 0 if t > Tp, since in that case ��� ÿ Tp� is zero over theentire range of integration. Similarly g2�t� � 0 if t > Tp � Td. Hence we

126 A.G. THOMPSON AND C.E.M. PEARCE

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obtain

g1�t� �expÿA0c�Tp ÿ t��Kd1 for t � Tp

0 for t > Tp;

(�10�

g2�t� �expÿA0c�Tp � Td ÿ t��Kd2 for t � Tp � Td

0 for t > Tp � Td.

(�11�

4. REARRANGING THE PERFORMANCE INDEX

Since the optimal control is a vector u as given by (6) and Q is a symmetricmatrix satisfying the Riccati equation (7) we easily determine the scalarproducts

u0Ru � �z0K � g0�BRÿ1B0�Kz� g�;z0Qz � z0�KBRÿ1B0K ÿ KAÿ A0K�z:

We add these equations, noting that expansion gives two pairs of equal termswhich are differently expressed via transposition. This provides

u0Ru� z0Qz � 2�z0KBRÿ1B0Kz� z0KBRÿ1B0gÿ z0KAz�� g0BRÿ1B0g: �12�

Substituting for the optimal control force u in (5) yields

Az � _z� BRÿ1B0Kz� BRÿ1B0gÿ D _w:

This result, when substituted into (12), leads to the expression

� �Z 1

0

�2z0KD _wÿ 2z0K _z� g0BRÿ1B0g�dt

for the performance index.To evaluate � we separate the terms containing z from that containing g and

decompose � � �1 ��2, where we de®ne

�1 �Z 1

0

g0BRÿ1B0g dt

DIRECT COMPUTATION OF THE PERFORMANCE INDEX 127

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and

�2 � 2

Z 10

z0K�D _wÿ _z�dt: �13�

In the two sections following we address the determination of these twocomponent integrals.

5. THE INTEGRAL �1

Using g � g1 � g2, we derive

�1 �Z 1

0

�g1 � g2�0BRÿ1B0�g1 � g2�dt

�Z 1

0

�g01BRÿ1B0g1 � 2g01BRÿ1B0g2 � g02BRÿ1B0g2�dt:

Since g1 � 0 for t > Tp and g2 � 0 for t > Tp � Td, we have

�1 �Z Tp

0

�g01BRÿ1B0g1 � 2g01BRÿ1B0g2�dt �Z Tp�Td

0

g02BRÿ1B0g2dt: �14�

We may now substitute from (10) for g1 in the ®rst integral on the right toderive

I1 �Z Tp

0

fd01K expÿAc�Tp ÿ t��BRÿ1B0 exp�A0c�Tp ÿ t��Kd1gdt:

This may be simpli®ed via the substitution Tp ÿ t � � to yield

I1 � d01K

Z Tp

0

exp�Ac��BRÿ1B0 exp�A0c��d�� �

Kd1: �15�

Denote by G1 the well-known integral within the braces. This may bedetermined by

G1 � Pÿ exp�AcTp�P exp�A0cTp�; �16�where P is the solution of the Lyapunov matrix equation

AcP� PA0c � BRÿ1B0 � 0: �17�

128 A.G. THOMPSON AND C.E.M. PEARCE

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Our solution for I1 is then simply

I1 � d01KG1Kd1:

The second integral on the right of (14) may be written

I2 � 2

Z Tp

0

d01K expÿAc�Tp ÿ t��BRÿ1B0 exp

ÿA0c�Tp � Td ÿ t��Kd2

� dt:

Substituting Tp ÿ t � � in this case yields

I2 � 2d01K

Z Tp

0

exp�Ac��BRÿ1B0 exp�A0c��d�� �

exp�A0cTd�Kd2:

The integral within the braces is as in (15) and therefore the solution is

I2 � 2d01KG1 exp�A0cTd�Kd2:

The last integral on the right of (14) may be written

I3 �Z Tp�Td

0

d02K expÿAc�Tp � Td ÿ t��BRÿ1B0 exp

ÿA0c�Tp � Td ÿ t��Kd2

� dt:

Substituting Tp � Td ÿ t � � in this case yields

I3 � d02K

Z Tp�Td

0

exp�Ac��BRÿ1B0 exp�A0c��dt

� �Kd2:

The integral within braces will in this case be denoted by G3 and may bedetermined from

G3 � Pÿ exp�Ac�Tp � Td��P exp�A0c�Tp � Td��; �18�

where P is obtained from (17). Hence the solution for I3 is

I3 � d02KG3Kd2:

Adding I1; I2 and I3 yields

�1 � d01KG1Kd1 � 2d01KG1 exp�A0cTd�Kd2 � d02KG3Kd2: �19�

DIRECT COMPUTATION OF THE PERFORMANCE INDEX 129

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6. THE INTEGRAL �2

Determining an explicit expression for the integral (13) is more involved dueto discontinuities in the transformed state variables z�t� at times t � Tp andt � Tp � Td. However a reasonably simple exact solution can still be obtained.By employing the transformation equations (2) and (4), the integral �2 may bemodi®ed to

�2 � ÿ2

Z 10

z0K�_zÿ D _w�dt

� ÿ2

Z 10

�x0C0 � w0D0�KC _x dt

� ÿ2

Z 10

�x0C0KC _x� w0D0KC _x� dt:

Since the terms in the integrand are scalars, we have �2 � J1 � J2, where

J1 � ÿZ 1

0

�x0C0KC _x� _x0C0KCx�dt; �20�

J2 � ÿZ 1

0

2w0D0KC _x dt:

As there are no discontinuities in the original state variables x�t� due to thedisturbance inputs, (20) integrates to yield

J1 � ÿ�x0C0KCx�j10 :In the steady state as t!1, the displacements x1 to x4 are all equal to unityand the corresponding velocities x5 to x8 are all zero, and therefore

x�1� � �1 1 1 1 0 0 0 0 �0: �21�From (3) and (21), we obtain the product

Cx�1� � ÿ�d1 � d2�:Since x�0� � 0, the integral J1 evaluates to

J1 � ÿ�d1 � d2�0K�d1 � d2�: �22�

130 A.G. THOMPSON AND C.E.M. PEARCE

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Further w0 � �xa xb � � �0 0� over the interval t � 0 to Tp, and so

J2 �Z 1

Tp

�ÿ2w0D0KC _x�dt:

Over the interval �Tp;Tp � Td�, the term w0 � �1 0� , while over the interval�Tp � Td;1� we have w0 � �1 1� . Also, since Dw � d1xa � d2xb, we haveDw � d1 over the ®rst interval and Dw � d1 � d2 over the second. Hence

J2 � �ÿ2d01KCx�jTp�Td

Tpÿ 2�d1 � d2�0KCxj1Tp�Td

:

To revert to z-coordinates via (2) we may substitute Cx � zÿ d1 over the®rst interval and Cx � zÿ �d1 � d2� over the second to obtain

J2 � fÿ2d01K�zÿ d1�gjTp�TdÿTp� � fÿ2�d1 � d2�0K�zÿ d1 ÿ d2�gj1Tp�Td�:

Because z�t� is bi-valued at the instants t � Tp and t � Tp � Td, whichcorrespond to the integration limits, the notation z�Tpÿ� and z�Tp�� will beused respectively to represent the state vectors just before and just after theunit step at t � Tp. Similarly, z�Tp � Tdÿ� and z�Tp � Td�� represent the statevectors just before and just after the unit step at t � Tp � Td. With thisnotation

J2 � ÿ2d01Kfz�Tp � Tdÿ� ÿ d1g � 2d01Kfz�Tp�� ÿ d1gÿ 2�d1 � d2�0Kfz�1� ÿ d1 ÿ d2g � 2�d1 � d2�0Kfz�Tp � Td��ÿ d1 ÿ d2g:

Since z�Tp�� � z�Tpÿ� � d1; z�Tp � Td�� � z�Tp � Tdÿ� � d2 andz�1� � 0, we may collect terms to provide

J2 � 2d01Kz�Tpÿ� � 2d02Kz�Tp � Tdÿ� � 2d01Kd1 � 2�d1 � d2�0Kd2: �23�With the simplifying substitutions

Z1 � z�Tpÿ� and Z2 � z�Tp � Tdÿ�;we obtain from (22) and (23) that

�2 � 2d01KZ1 � 2d2KZ2 � d01Kd1 � d02Kd2: �24�The 8-vectors Z1 and Z2 may be computed by matrix operations as shown inthe next section.

DIRECT COMPUTATION OF THE PERFORMANCE INDEX 131

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7. THE SOLUTION FOR Z1

Over the preview time t � �0; Tpÿ�, the disturbance vector _w � �0 0�0. Hence(5), (6) and (9) lead to the closed-loop system equation

_z � Aczÿ BRÿ1B0g: �25�Since z�0� � 0, we obtain the solution

z�t� � ÿZ t

0

exp�Ac�t ÿ ���BRÿ1B0g��� d�

for t � Tp. Substituting g�t� � g1�t� � g2�t� and using (10) and (11) yields

z�t� � ÿZ t

0

exp�Ac�t ÿ ���BRÿ1B0fexp�A0c�Tp ÿ ���Kd1

� exp�A0c�Tp � Td ÿ ���Kd2g d�

� ÿZ t

0

exp�Ac�t ÿ ���BRÿ1B0 exp�A0c�t ÿ ���d��fexp�A0c�Tp ÿ t��Kd1

� exp�A0c�Tp � Td ÿ t��Kd2g: �26�With the substitution t ÿ � � �, the integral in (26) becomesZ t

0

exp�Ac��BRÿ1B0 exp�A0c�� d�:

For t � Tp, this last integral is the same as that in (15) and may be denoted byG1 as given by (16). Hence the required solution is

Z1 � z�Tpÿ� � ÿG1�Kd1 � exp�A0cTd�Kd2�: �27�

8. THE SOLUTION FOR Z2

Since z�Tp�� � z�Tpÿ� � d1 � Z1 � d1 may be taken as the initial conditionover the interval �Tp�; Tp � Tdÿ� and it is required to obtain a solution forZ2 � z�Tp � Tdÿ�, then again _w � �0 0�0 over this interval and (25) stillapplies and has the solution

z�t� � exp�Ac�t ÿ Tp���Z1 � d1�

�Z t

Tp

exp�Ac�t ÿ ���fÿBRÿ1B0g���g d�: �28�

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From (10) and (11) for Tp < t < Tp � Td,

g�t� � g2�t� � exp�A0c�Tp � Td ÿ t��Kd2

and therefore we obtain by substitution in (28) that

z�t� � exp�Ac�t ÿ Tp���Z1 � d1�

ÿZ t

Tp

exp�Ac�t ÿ ���BRÿ1B0 exp�A0c�Tp � Td ÿ ���Kd2 d�:

If we let t � Tp � Td, we have

z�Tp � Tdÿ� � exp�AcTd��Z1 � d1�

ÿZ Tp�Td

Tp

exp�Ac�Tp � Td ÿ ���BRÿ1B0

exp�A0c�Tp � Td ÿ ���Kd2 d�:

From the substitution � � Tp � Td ÿ � we obtain

Z2 � exp�AcTd��Z1 � d1�

ÿZ Td

0

exp�Ac��BRÿ1B0 exp�A0c�� d�Kd2:

The integral is similar to that in (15) and we conclude that

Z2 � exp�AcTd��Z1 � d1� ÿ G2Kd2; �29�where the matrix G2 is given by

G2 � Pÿ exp�AcTd�P exp�A0cTd� �30�and P is the solution of the Lyapunov equation (17).

9. EXAMPLE

The data employed here for the half-car model is the same as that in [3, 7]and the weighting factors used in the performance index are also the same.Thus in SI units, M1 � 28:58; M3 � 54:43;M � 505:1; J � 651:0; S1 � S3 �155900; a�1:0978; b�1:4676; L�2:5654; Ps�1:0; �1��3� 0:8 � 10ÿ9,

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q1 � q3 � 10; q2 � q4 � 1. Assuming a vehicle speed V � 10 m/s however,results in a preview time Tp � 0:1 sec and a delay time Td � 0:25654 sec.

The calculations can be done easily using MATLAB. The procedure isbrie¯y as follows: ®rst create the matrices A; B; D; Q and R. Then, solve theRiccati equation (7) for K and determine the closed-loop system matrix Ac

from (9). Use Ac in equation (17) and solve the Lyapunov equation for thematrix P. MATLAB has the Lqr and Lyap functions available to solve theseequations and also the matrix exponentiation functions required to computeG1; G2 and G3 from (16), (30) and (18) respectively, and hence, Z1 and Z2

from (27) and (29) respectively.Once G1 and G3 are determined, the integral �1 may be computed from

equation (19). The values obtained for the state vectors Z1 and Z2 can be usedto compute the integral �2 from equation (24) and the performance index � isthen obtained as �1 ��2. With the above data we obtain � � 0:346465. Thecalculation is very fast and takes less than a second on a PC. Gives the valuesof the performance index for various preview sensor distances.

The given values of � for V � 10 correspond fairly well with thoseobtained by integration over the transient response time, as tabulated in [7]. Itshould be noted that we have not included the factor of 1=2 in the performanceintegral (1). For V � 30 and Ps � 0 (zero look±ahead preview) the programyields � � 0:52956, in good agreement with the values given in [5, 10] andsubstantially less than the value � � 0:70184 obtained for a half-car modelusing only state feedback [1]. Figure 2 shows some plots of performance indexversus preview sensor distance for a range of vehicle speeds. These show thatat low speeds up to 10 m/s less than 1 m preview distance is adequate. Athigher speed more preview distance would be bene®cial but probablyimpracticable. Nevertheless even 1 m preview for the input to the front wheelsin this examples represents over 3.5 m preview for the input to the rear wheelsand at 30 m/s (108 km/h) the performance index for Ps � 1 would be less than0.5. That's much better than the value for a half-car model using only statefeedback. Some further improvements might be possible by manipulating the

Table 1. Performance index for various preview sensor distances.

V Ps 0 0.2 0.6 1.0 1.5 2.0 110 � 0.48432 0.44437 0.37542 0.34646 0.32638 0.31938 0.3161130 0.52956 0.50931 0.46906 0.46289 0.42562 0.38155 0.32090

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weighting factors �, q1 and q2 as in the global studies carried out by Hrovat[8, 9] but that is not within the scope of this paper.

10. CONCLUSION

The following, which represents a portion of a MATLAB m-®le forcomputation of the performance index, may assist in the setting up of aworking program. The % sign precedes a comment and the ellipsis (. . .)indicates that the foregoing statement continues on to the next line.Commented references at the ends of lines are to the relevant equationnumbers in the text.

% Kalman feedback gains k and a Riccati equation solution matrix

Kmat

[k,Kmat]=lqr(A,B,Q,R);

disp(-k), disp(Kmat)

R1=inv(R);

BRB=B*R1*B';

Ac=A-BRB*Kmat;

% Lyapunov equation solution matrix P

P=Lyap(Ac,BRB);

G1=P-expm(Ac*Tp)*P*expm(Ac0*Tp); %(16)

Fig. 2. Performance Index vs. Preview Sensor Distance for a Range of Vehicle Velocities.

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Z1=-G1*(Kmat*d1+expm(Ac0*Td)*Kmat*d2); %(27)

G2=P-expm(Ac*Td)*P*expm(Ac0*Td); %(30)

Z2=expm(Ac*Td)*(Z1+d1)-G2*Kmat*d2; %(29)

G3=P-expm(Ac*(Tp+Td))*P*expm(Ac0*(Tp+Td)); %(18)

PI1=d10*Kmat*G1*Kmat*d1+... %(19)

2*d10*Kmat*G1*expm(Ac0*Td)*Kmat*d2+...d20*Kmat*G3*Kmat*d2;

PI2=2*d10*Kmat*Z1+d10*Kmat*d1+... %(24)

2*d20*Kmat*Z2+d20*Kmat*d2;PI=PI1+PI2

% end Program

The intended application of the program is to design studies of the effects ofsystem parameters and weighting factors, etc. A similar program might beused to determine the optimum feedback gains for a real vehicle excited bystep inputs applied with a pre-determined time delay to the front and rear tyrecontact points. There is an obvious need to be able to set these gains withouthaving to measure all the physical parameters, such as body mass, suspensionspring rates and tyre radial rates. The practicality of this idea, however, is asyet untested.

The front and rear suspensions are not decoupled as in [8] by anyassumptions about the physical parameters of the system and tyre damping hasnot been included. The direct determination of the rms values for controlforce, tyre dynamic de¯ection and relative wheel travel for a half-car modelwith preview has not been undertaken in this paper. Some useful results havebeen obtained for a quarter-car model, which will be reported in due course.

REFERENCES

1. Thompson, A.G. and Pearce, C.E.M.: An Optimal Suspension for an Automobile on aRandom Road. SAE Paper 790478 (1979).

2. HaÂc, A.: Stochastic Optimal Control of Vehicles with Elastic Body and Active suspension.ASM J. Dynamic Systems, Measurement and Control 114 (1986), pp.556±562.

3. Thompson, A.G., Davis, B.R. and Pearce, C.E.M.: An Optimal Linear Active Suspensionwith Finite Road Preview. SAE Paper 800520 (1980).

4. Ha�c, A.: Optimal Linear Active Preview Control of Active Suspension. Vehicle SystemDynamics 21 (1992), pp.167±195.

5. Sharp, R.S. and Wilson, D.A.: On Control Laws for Vehicle Suspensions Accounting forInput Correlations. Vehicle System Dynamics 19 (1990), pp.353±363.

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6. Thompson, A.G. and Pearce, C.E.M.: Performance Index for a Preview Active Suspensionapplied to a Quarter-Car Model. Vehicle System Dynamics (to appear).

7. Thompson, A.G. and Pearce, C.E.M.: Physically Realisable Feedback Controls for a FullyActive Preview Suspension Applied to a Half-Car Model. Vehicle System Dynamics 30(1998), pp.17±35.

8. Hrovat, D.: Optimal Suspension Performance for 2-D Vehicle Models. J. of Sound andVibration 146(1) (1991), pp.93±110.

9. Hrovat, D.: Survey of Advanced Suspension Developments and Related Optimal ControlApplications. Automatica 10 (1997), 1781±1817.

10. Louam, N., Wilson, D.A. and Sharp, R.S.: Optimal Control of a Vehicle SuspensionIncorporating the Time Delay between Front and Rear Wheel Inputs. Vehicle SystemDynamics 17 (1988), 317±336.

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