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Problem 7.13PP
(a) Find the state transformation that will keep the description of the tapedrive system of Example 
7.10 in modal canonical form but will cause the poles to be displayed in Am in order of increasing 
magnitude.
(b) Use Matlab to verify your result in part (a), and give the complete new set of state matrices e 
. A ,. B, .C, and .D.
Example 7.10
EXAMPLE 7.10 Using M atlab to Find Poles and Zeros ofTape-Drive System
Find the dgeDvahies of die qrttem mMiix described below for the tape-drive 
conind (see Fig. 3.S0). Abo, compute the tramformatioa of the equations 
of the t^ie drive in their giwea fm n to modal canonical fionn. The system
contnd (see Hg. 3.S0). Abo, compute the tiansfonnatioa o f die equations 
of the t^ie drive in their given fnm to modal canmical fbnn. The system
A »
' 0 2 0 0 0 * '0'
- a i - 0 3 5 0.1 a i a75 0
0 0 0 2 0 0
0.4 a 4 - a 4 -1 .4 0 0
0 -0 4 B 0 0 -1 .1,
a.40)
C2 * [OD OlO 1.0 Oil Oil] Servomotor ontpm,
C3 s [(L5 0.0 OJ Oil Oil] Posidoii M leaiVwfite head as ootpal,
C r - [ - 0 .2 -O J 0.2 0.2 0.0] Ibnsion output,
D = 0 .0 .
The sMe vector b defined as
xi (tape positiao at capstan)
« i (speed of the drive wheel)
JK3 (posttioii of the tape at the head) 
(ouqwt speed)
The matrix C3 corresponds to malrii^ X3 (die position of the tape over the 
lead/write head) the output, and the matrix CTConespoods to making tension 
the output.
Sclution. To compute the eigenvalues using Mathb, we write
P««ig(A),
which results in
P s
-0 .6 3 7 1 + 0 .6669 i 
-0 .6 3 7 1 -0 .6 6 6 9 i 
0.0000 
-0 .5075 
-0 .9683
Notice that the system has all poles in the left half-plane (LHP) e x c ^ for 
one pole at the origin. This means that a stq> iiqNit will result in a ramp 
output, so we ctmclude that the system has Type 1 behavior.
To transform to modal fm n , we use the Matlab fimctitm canon:
sysG-ss(A,B,G,D);
[sysGm,Tl]-canon(sysG/modal');
[Am,Bm,Cm,Dm]ear in the 2 x 2 block in the uf îer-!
comer 5 (-0 .9 6 8 3 + O.OOOOi) > P, (-0 .6 3 7 1 + 0 .6 6 6 9 /)
> Pj (-0 .6371 -0 .6 6 6 9 /) > /J (-0 .5 0 7 5 + 0.0000/) >
P ,(-0.0000 + 0.0000/)
Hence, the matrix must be arranged as shown below.
r2= ['5 '2 U
Where,
Tj is the rearranged eigen vector T in the order of increasing magnitude.
/i,/j,/j,(j,and/jare the respective column of matrix T.
Therefore the rearranged eigen vector T in the order of increasing magnitude is given below.
■ 0.4714 0.3805 0.8697 1.3406 -2 .8284'
-0 .2282 -0 .4112 -0 .1502 -0 .3402 0.0000
0.5093 2.1334 -3 .0 3 3 0 2.3060 -2 .8284
-0 .2466 0.3317 1.6776 -0.5851 0.0000
0.2160 0.0130 -0 .0 1 1 4 0.0207 - 0.0000
Step 2 of 4
(b)
Write the matlab program to verify the result and to obtain the new set of state matrices, 
cic
A=[0 2 0 0 0;-0.1 -0.35 0.1 0.1 0.75;0 0 0 2 0;0.4 0.4 -0.4 -1.4 0;0 -0.03 0 0 -1];
B=[0:0;0:0;1]: 
c2=[0 0 1 0 0]: 
c3=[0.5 0 0.5 0 0]; 
cT=[-0.2 -0.2 0.2 0.2 O.Oj;
D=0;
[P]=eig(A);
sysG=ss{A,B,c3,D):
[sysGm,TI]=canon(sysG,'modal'):
[Am,Bm,Cm,Dm]=ssdata{sysGm):
TI=[-0.3439 -0.3264 0.3439 0.7741 0.4785; 0.1847 -0.7291 -0.1847 0.0969 -0.6247; -0.1844 
-1.3533 -0.1692 -0.3383 -1.0150;0.3353 -2.3627 -0.3353 -1.0161 -3.5980; -0.0017 0.2077 0.0017 
0.0561 4.9133];
T=inv(TI):
% [V,P]=eig{P)
[f,indices]=sort(abs(P));
T2=T(:,indices):
n=T2\B;
T3=T2*diag{n);
Am2=T3\A*T3
Bm2=T3\B
Cm2=c3*T3
dm2=0
Step 3 of 4
The output obtained on executing the code is given below; 
The Eigen values of matrix are displayed.
P =
-0.6371 + 0.66691 
-0.6371 - 0.66691 
0.0000 + O.OOOOi 
-0.5075 + O.OOOOi 
-0.9683 + O.OOOOi
Step 4 of 4
The new set of state matrices are shown below. 
Am2 =
0.0000 0.0002 0.0000 -0.0001 -0.0002 
-0.0000 -0.5075 -0.0000 -0.0000 -0.0000 
-0.0000 0.0002 -0.6371 -0.8707 -0.0040 
0.0000 0.0001 0.5108-0.6371 0.0016 
-0.0000 0.0000 -0.0000 -0.0000 -0.9683 
Bm2 =
1.0000
1.0000
1.0000
1.0000
1.0000 
Cm2 =
2.8705 -6.5593 0.6014 0.6755 2.4120 
dm2 =
0
Hence the results are verified new set of state matrices are obtained using matlab.