HW2, MAE 270A. Linear systems I. Fall 2005. UCI

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HW2, MAE 270A. Linear systems I. Fall 2005. UCI

Nasser M. Abbasi

June 21, 2014

1 Problem 2.10

Consider the system described by

′′ ′ ′ y + 2y − 3y = u − u

What are the transfer function and the impulse response of the system?

Answer

The transfer function is deﬁned as the ratio of the Laplace transform of the output to the laplace transform of the input assuming zero initial conditions. i.e. assume that $′ − y (0 ) = 0$ , $− y (0 ) = 0$ , $− u (0 ) = 0$

Taking the laplace transform of the above diﬀerential equation we obtain

Hence

Y (s) 1 G (s) = ----- = ----- U (s) s + 3

Hence the impulse response is the inverse laplace transform of $G (s)$ , which is for this simple case can written directly

g (t) = e −3t

Note: the above solution is valid for $t ≥ 0$ . For $t < 0$ the impulse response is zero.

2 Problem 2.15 part(a)

Find state space equations to describe the pendulum system in following ﬁgure. Write down the linearized dynamic equations and the transfer function from $u(t)$ to $𝜃(t)$

Answer

The general state space representation for this system is

To simplify notations, I will not list time as an independent variables since it is implicit in $x$ and $u$ in this problem.

Now, assume we have a nominal solution $x0$ and a nominal input $u0$ and let the perturbation from these be $¯x$ and $¯u$ respectively. Hence (1) can be written as

Now pick a nominal solution when the system is in its stable equilibrium position (when the pendulum is hanging down at rest).

Hence $x0 = 0$ and $u0 = 0$ . For this state and input we obtain $h (x0,u0) = 0$ since $˙x0 (t) = 0$ as there is no state change with time, also we obtain that $f (x0, u0) = 0$ since $y0 (t)$ since the mass is not moving. Hence (2) becomes

Now since $¯x = x − x0$ and $x0 = 0$ then $¯x = x$

Similarly, $¯u = u − u 0$ and $u = 0 0$ then $u¯= u$ hence (3) can be written as

Hence we just need to evaluate $| | | | ∂h(x,u)|| , ∂h(x,u)|| , ∂f(x,u)|| , ∂f(x,u)|| ∂x x0,u0 ∂u x0,u0 ∂x x0,u0 ∂u x0,u0$ to obtain the linearized solution.

Since there are 2 states in this system and one input we obtain

Now we need to ﬁnd $h1, h2,f$ and substitute these into (5) and then into (4) to ﬁnd the solution.

First ﬁnd the dynamic equation for this system. The forces on the mass are

Applying Newton second law $F = ma$ , along the direction tangent to the motion we get

u(t)cos 𝜃(t) − mg sin𝜃 (t) = mL ¨𝜃(t)

Hence

¨ --1- g- 𝜃 (t) = mL u (t) cos𝜃 (t) − L sin𝜃 (t)

Now convert to state space. Let $x = 𝜃 1$ and $x = 𝜃˙ 2$ .

Hence

x˙1 = x2

and

Hence we can write

( ) ( ) ( ) || ˙x1 || || h1 (x1,x2,u) || || x2 || ( ) = ( ) = ( ) ˙x2 h2 (x1,x2,u) 1mLu (t)cosx1 − gL sinx1

(6)

Now for the output equation:

Hence

y = f (x,u) = (x1)

Now that we know $h1,h2,f$ , we can go back to (5) and evaluate that, we obtain

Now to obtain the solution (4) we need to evaluate (7) at the nominal solution $x0,u0$ and these are zero, i.e. $x1 = x1,0 = 0,x2 = x2,0 = 0,u = u0 = 0$ then (7) becomes

Hence, substitute (8) into (4) we obtain the ﬁnal linearized solution

The above is the linearized solution. Where

The transfer function is

but

( ) T ( ) | s − gL-| | s 1 | ( ) −1 |( |) |( |) 1 s − g- s || s − 1 || Adjoint(A-)- --|--------|-- -----L------- ( ) = Det(A ) = | | = s2 + g- gL- s || s − 1 || L | | || g- || | L s |

Hence