24 CHAPTER 2. MATHEMATICAL MODELS IN CONTROL
in state-space form as
x ̇ 1 (t)
x ̇ 2 (t)
φ ̇ 1 (t)
φ ̇ 2 (t)
=
01 00
0
−(mL^2 +I)b
(M+m)I+MmL^2m^2 gL^2
(M+m)I+MmL^20
00 01(^0) (M+m−)mLbI+MmL (^2) (MmgL+m)(IM++MmLm) 2 0
x 1 (t)
x 2 (t)
φ 1 (t)
φ 2 (t)
+
0
(mL^2 +I)
(M+m)I+MmL^2
0
mL
(M+m)I+MmL^2
u(t)wherex ̇ 1 (t)=x ̇,x ̇ 2 (t)=®x,φ ̇ 1 (t)=φ ̇andφ ̇ 2 (t)=®φ. Since we are interested
in the position of the cart, as well as theangular position of the pendulum, the
output may be synthesized as
∑
y 1 (t)
y 2 (t)∏
=
∑
1000
0010
∏
x 1 (t)
x 2 (t)
φ 1 (t)
φ 2 (t)
+
∑
0
0
∏
u(t)For this example we will assume the following parameters:M =0.5kg
m =0.2kg
b =0.1N/m/s
l =0.3m
I =0.006 kg m^2To make the design more challenging, we will be applying a step input to the
cart. The cart should achieve its desired position within 5 seconds and have a
rise time under 0.5 seconds. We will also limit the pendulumís overshoot to 20
degrees (0.35 radians), and it should also settle in under 5 seconds. The design
requirements for the inverted pendulum are therefore
ïsettling time forxandθof less than 5 seconds,ïrise time forxof less than 0.5 seconds, andïovershoot ofθless than 20 degrees (0.35 radians).We useMatlabto perform several computations. First, we wish to obtain the
transfer function for the given set of parameters. Using them-fileshown below,
we can obtain the coefficients of the numerator and denominator polynomials.
M = .5;
m = 0.2;