Homework 1 - ECES 513 Fundamentals of Systems - III
Naresh Shah, Pranav Ram Kamarajugadda May 2, 2014
Problem 1. A serial robot can be described as a set of n coupled links parameterized by a vector of co-ordinates x = (x1 , x2 ...., xn ), with n µ(x)ij j=1 d2 xj dt2

n

= F (x, x, x)i + ˙ ¨ n j=1 n

Kij (xj − xj,0 ) Kij (xj − xj,0 )

τij xj = ¨ j=1 j=1

where µ(x) is a mass matrix, τ is an inertia matrix, F (x, x, x) is a vector of generalized ˙ ¨ resultant forces, x0 = (x1,0 , x2,0 ....xn,0 ) is a reference configuration and K is a matrix of joint stiffnesses. Select state variables for the system and write down the state equations. Under what conditions is the system linear? Proof. Assuming a two-link system: Let y1 = x1 , y2 = x1 , y3 = x1,0 , y4 = x1,0 be the state ˙ ˙ variables of the manipulator. Then, y 1 = x1 y2 = −µ−1 (x1 )[F (x1 , x1,0 ) + K(x1 − x1,0 )] y3 = x1,0 y4 = τ −1 K(x1 − x1,0 ) + τ −1 u where u is the input vector.

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Problem 2. Analyze the following equations graphically. In each case sketch the vector on the phase plane, find all the fixed points, classify their stability, and sketch the graph for different initial conditions.Attempt to obtain the analytical solution for x(t). • x = − sin(x) ˙ • x = x − x3 ˙ • x = e−x sin(x) ˙ Proof. .... 1. x = − sin(x) ˙

1 sin x sin x − sin x = − sin x2 = − 1−cos x2

⇒ Let u = cos x ⇒ du = − sin xdx ⇒ du 1−u2

→ −0.5 ln |u − 1| + 0.5 ln |u + 1|

Analytical Solution is ⇒ t = −0.5 ln | cos x − 1| + 0.5 ln | cos x + 1| Fixed points for this differential equation are the points where x = 0 are at x0 = nπ. ˙ Intuitively, when n = even, the solution is ‘stable’. When n=odd, the pendulum is inverted, and hence ‘unstable’.

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2. x = x − x3 ˙

1 x−x3 1 ⇒ (x + x ) 1−x2

=

1 x(1−x2 )

=

1 x

+

x 1−x2

= ln |x| + 0.5 ln |1 − x| − 0.5 ln |1 + x|

The analytical