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Problem 7.41 PP
Consider the system in Fig. 
Figure Coupled pendulums
K=bt
A, - jr ( 0 , -0 2 )+ ^
^ = - « ^ + jT(0| - 02) - fhnl
K=kd
0, = - + /7ml 
^ + IT(0| - 02) - fhnl
(a) Write the state-variable equations for the system, using [ej $2 vector
and Fas the single input.
(b) Show that all the state-variables are observable using measurements of 01 alone.
(c) Show that the characteristic polynomial for the system is the product of the polynomials for 
two oscillators. Do so by first writing a new set of system equations involving the state-variables
■ y i ■ ■ 0 1 + ^ ■
y i . ^ O l-0 2 ,
Hint If A and D are invertible matrices, then
[^ •o r= [r‘ ^]■
(d) Deduce the fact that the spring mode Is controllable with F but the pendulum mode is not.
S tep -by-s tep s o lu tio n
step 1 of 4
A) The State vector
JT = [e . e, 6.
0 0
0 0
1 O'
0 1
0 0
K 0 0
X -F
Step 2 of 4
B) We have
r = [ l 0 0 0]X 
Observability
e =
H 1 0 0 0
HF 0 0 1 0
HF^ ~ - (® “ -i-j?:) i : 0 0
HF^ 0 0 -(® ^+j^r) K
The state is observable with
Step 3 of 4
Q Let
X = y i f
0 , + ^ ^ + 6 , 9 , - e , S i-S , ]
0 1 0 0
-“ + 2 ir )
The characteristic equation o f the system is 
det [ i f l - i? ) = (s ’ + «?) + o ’ + 2 ^ )
Step 4 of 4
D) There is no coi^ling between the spring mode (6 ̂— @2) and the pendulum mode
(S i+ a .)