Order reduction of linear systems using point-to-point step response matching technique

S. MUKHERJEE*, SATAKSHI**, R.C. MITTAL** *Department of Electrical Engineering; **Department of Mathematics I.I.T Roorkee, Roorkee-247667 INDIA

Abstract:- The proposed method of order reduction of linear systems is based on dominant pole retention coupled with step response matching technique for zero evaluation. At different points i.e., the time axis, of the unit step response of the original high order system (HOS), the output is equated with that of the reduced low order system (ROS). Depending on the order to be reduced to, the numbers of points are chosen. A set of linear equations equal in number to the unknowns in the numerator are solved to obtain these terms. A number of examples of different varieties are solved to check the validity of the proposed method and the results are encouraging.

Key-Words:- Order reduction, Steady state, Step response, Transient, Linear system, Dominant poles.

1. Introduction If the transfer function of the nth order HOS is G n (s) as Among a large number of methods proposed for shown below: obtaining low order model of a high order system, those c s n1  ......  c s  c G (s)  n1 1 0 using step response matching criteria occupy a premier n s n  ......  d s  d position. Although, there are several approaches in this 1 0 c s n1  ......  c s  c area, but these can be broadly put under two categories,  n1 1 0 (1) such as ones using the concept of response matching ‘a (s  1 )(s  2 )...... (s  n ) priori’ and the others as matching after obtaining the where  (i  1,2,.....,n) are the poles of HOS, that can low order model. Methods under ‘a priori’ approach in i be real or imaginary and distinct or repeated. fact uses error minimization between the step responses and consequently the unit step response of (1) is of HOS and ROS and some of the methods in this category can be mentioned as [3]-[7]. The other group of obtained as: methods uses several mathematical approaches but ultimately uses the error criteria between step responses n1 of HOS and ROS to evaluate the quality of ROS G n (s) c n1s  ......  c1s  c0 Tn (s)   (2) obtained using the method and some such methods can s s(s  1 )(s  2 )...... (s  n ) be located in literatures [1], [2], [8] and [9]. Present After taking the inverse Laplace of (2)  t  t  t attempt is a mixed approach, like both ‘a priori’ for the g (t)     e    e   ......   e n , (3) steady state part of the step response and exact matching n 0 1  n at several predefined points of the transient part of the then the reduced rth order ROS, G r (s) is put as, r1 step response. Later on, after obtaining the ROS, it is cˆ r1s  ......  cˆ1s  cˆ 0 G r (s)  compared with HOS graphically and also the value of s r  ......  dˆ s  dˆ the Integral Square Error (ISE) between the transient 1 0 cˆ s r1  ......  cˆ s  cˆ parts of the responses of HOS and ROS is calculated to  r1 1 0 ˆ ˆ ˆ (4) judge the quality of ROS. (s  1 )(s  2 )...... (s  n ) The problem statement and algorithm of the G r (s) proposed method is given in the following sections. and Tr (s)  s

where ˆ (i  1,2,....., r) are the poles of ROS. 2. Problem Formulation i having unit step response as: where g nss (t)   0 is known as steady state part of the ˆ t ˆ t ˆ t g (t)     e    e   ......   e r , (5) n r    r i t response and g ntr (t)   i e is called transient part is to be obtained such that the following conditions are i1 fulfilled. of the response. (i) Steady state parts of the responses shown in Similarly giving the unit step input to ROS    r1 (3) and (5) should be equal i.e., 0 0 G (s) cˆ s  ......  cˆ s  cˆ r  r1 1 0 (8) ˆ Tr (s)  (ii) i  i ,(i  1,2,...,r) are the dominant poles ˆ ˆ ˆ s s(s  1 )(s  2 )...... (s  r )

chosen from 1 ,2 ,...,n (r

3. Description of the Method Step3:Matching of steady state parts of The procedure of the order reduction methods responses of HOS and ROS. developed, based on the problem statement above is As mentioned above, in order to match steady state described in terms of steps to be followed as under: parts, following condition is required to be fulfilled: i.e.,    Step1:Selection of dominant poles to be retained 0 0 By definition, nearer a pole is to the origin, more Step4:Choice of ‘r’ matching points at dominant it is said to be. So, if Re(i )  Re(i1 ) , t  t , t ,..., t on the transient part of the (i=1,2,…,n-1) and Re( )  0 , then  is most 1 2 r n 1 unit step responses of HOS and ROS dominant and n is least dominant. In case of imaginary The ‘r’ matching points on system response are pole pairs, the real part will decide the dominance. In chosen in the transient part of the response curve only as this method, the dominant poles are to be selected as steady state matching has already been taken into under: account and hence no point is chosen on the steady state

If the poles of a nth order HOS are 1 ,2 ,...,n such part. These ‘r’ matching points are chosen using visual inspection of the response curve. that Re(i )  Re(i1 ) (i=1,2,…,n-1) and Re(n )  0 , then if it is to be reduced to rth order ROS(r

t  t n t g n (t)   0  1e    e  ......   n e , (7) ⋮ (10)

 g nss (t)  g ntr (t) g rtr (t r )  g ntr (t r )  f r where t1 t1 n t1 f1  g ntr (t1 )  1e   e  ......  n e Pr   r (r )(1  r )(2  r )...... (r1  r )  t  t  t  2  2 n 2 There are ‘r’ equations in (15) and each one is having r f 2  g ntr (t 2 )  1e   e  ......   n e cˆ ,cˆ ,...,cˆ ⋮ (11) unknowns as 0 1 r1 . So (15) is a system of r

t r t r n t r linear equations in r unknowns and hence can be solved f r  g ntr (t r )  1e   e  ......   n e to get cˆ ,cˆ ,...,cˆ . The value of cˆ determined earlier are constants as all  (i  1,2,..., n) and  (i  1,2,...,n) 0 1 r1 0 i i from the condition of exact steady state matching and are known from HOS. given in (13) coincides with the value obtained by So, ˆ ˆ ˆ solving the set of equations (15). Using the above steps t1 t1 r t1 e    e  ......   r e  f1 the proposed method is used to solve a number of ˆ t ˆ t ˆ t  2  2 r 2 examples and the results are shown in the next Section. e   e  ......   r e  f 2 ⋮ (12) ˆ t ˆ t ˆ t  e  r   e  r  ......   e r r  f   r r 4. Examples

Since f1 ,f 2 ,...,f r i.e., g ntr (t1 ),g ntr (t 2 ),...,g ntr (t r ) are all Example-1 known constants, (12) is a set of ‘r’ linear equations The transfer function of the original 5th order system, with ‘r’ unknowns i.e.,  i (i  1,2,..., r) and hence can be G 5 (s) is as under: solved. 10s 4  80s 3  264s 2  369s  156 G (s)  Step 6: Determination of unknown coefficients of 5 2s 5  21s 4  84s 3  173s 2  148s  40 cˆ ,cˆ ,...,cˆ nd th numerator of (4) i.e., 0 1 r1 . Transfer function of the reduced 2 and 4 order

 0 and  i (i  1,2,..., r) known from Step1 and Step3 systems using the proposed method are as under: respectively can be put as under: 3.321s 1.95 G2 (s)  2 cˆ 0 1 s 1.5s  0.5  0  sTr (s)   cˆ 0 K , where K  s0 02 12 ...r 12 ...r J = 7.21000 x 10 (13) 4.312s 3 19.33s 2  33.84s 15.6 cˆ   / K   / K  G 4 (s) = 4 3 2 So, 0 0 0 , is known as 0 and s  5.5s 14.5s 14.0s  4.0  (i  1,2,...,r) i are known from HOS J = 8.51625 x 1004 ˆ r1 ˆ r2 ˆ c r1 (1 )  c r2 (1 )  ...  c 0 The step responses between the original and reduced 1  (s  1 )Tr (s)  s1 (1 )(2  1 )(3  1 )...(r  1 ) order system are shown in Fig. 1. cˆ ( ) r1  cˆ ( ) r2  ...  cˆ   (s   )T (s)  r1 2 r2 2 0 1 2 r s Example-2 2 ( )( )( )...( ) th 2 1  2 3  2 r  2 The transfer function of the original 8 order system,

⋮ G 8 (s) is as under: cˆ ( ) r1  cˆ ( ) r2  ...  cˆ 7 6 5 4 3   (s   )T (s)  r1 r r2 r 0 35s  1086s  13285s  82402s  278376s  r r r s r ( )(   )(   )...(   ) 2 r 1 r 2 r r1 r 11812s  482964s  194480 G (s)  (14) 8 s 8  33s 7  437s 6  3017s 5  11870s 4  27470s 3  or, 2 r1 r2 37492s  28880s  9600 cˆ r1 (1 )  cˆ r2 (1 )  ......  cˆ 0  P1 Transfer function of the reduced 3rd and 4th order cˆ ( ) r1  cˆ ( ) r2  ......  cˆ  P r1 2 r2 2 0 2 systems using the proposed method are as under: ⋮ (15) 31.21s 2  59.26s  40.52 r1 r2 ˆ ˆ ˆ G 4 (z) = 3 2 c r1 (r )  c r2 (r )  ......  c 0  Pr s  3.0s  4.0s  2.0 where J = 6.70000 x 1002 P   ( )(   )(   )...... (   ) 1 1 1 2 1 3 1 r 1 33.66s3 151.2s 2  219.9s 121.5 P   ( )(   )(   )...... (   ) G (s) = 2 2 2 1 2 3 2 r 2 4 s 4  6.0s3 13.0s 2 14.0s  6.0 ⋮ (16) J = 1.40000 x 1003 The step responses between the original and reduced with different techniques of pole finding methods. order system are shown in Fig. 2. Distinct advantages of the method are like an assured stable reduced order system and perfect matching of the steady state part of the unit step responses of the original high order and reduced low order systems obtained using the proposed method.

References: [1] J.H. Anderson, Geometrical approach to the reduction of dynamical systems, Proc. IEE, Vol. 114, 1987, pp.1014-1018. [2] J.W.Bandler, N.D.Markettos and N.K.Sinha, Optimum system modelling using recent gradient methods, Int. Journal of System Science, Vol.4, 1973, pp.257-262. [3] S.Mukherjee and R.N.Mishra,Reduced order modelling of linear system using step response Fig.1: Step response comparison of original 5th and matching, Proc. ASMI International Conf. on reduced 2nd and 4th order models Modelling and Simulation, Sept.3-5, 1986, Williamsburg, USA. [4] S.Mukherjee and R.N.Mishra, Order reduction of linear system using error minimization technique, Journal of the Franklin Institute, USA, Vol.323, No.1, 1987, pp.23-32. [5] S.Mukherjee and R.N.Mishra, Reduced order modelling of linear multivariable system using an error minimization technique, Journal of the Franklin Institute, USA, Vol.325, No.2, 1988, pp.235-245. [6] S.Mukherjee and R.N.Mishra, Optimal order reduction of discrete systems, Journal of the Institution of Engrs(I), ET. Vol.68, 1988, pp.142- 144. Fig. 2: Step response comparison of original 8th and [7] S.Mukherjee, Order reduction of linear system using reduced 3rd and 4th order models eigenspectrum analysis, Journal of the Institution of Engrs(I), Vol.77, 1996, pp.76-79. [8] N.K.Sinha and W.Pille, A new method for reduction 5. Conclusion of dynamic systems, Int. Journal of Control, Vol.14, As can be seen in Figs (1-2), step response matching 1971, pp.111-118. between original high order and reduced low order [9] N.K.Sinha and G.T.Berezanai, Optimum system is nearly perfect. Though the poles of the new approximation of high order system by low order low order systems were obtained using the well-known models, Int. Journal of Control, Vol.14, 1971, dominant pole retention method, the approach of pp.951-959. obtaining the zeros is new in the proposed method and the results are encouraging. Different types of systems have been considered in the two examples solved using the proposed method. Further investigation is being undertaken by the authors towards application of the proposed method in variety of other systems as well as