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Chapter 7

Q. 7.EX.2

Cruise Control Step Response
(a) Rewrite the equation of motion from Example 2.1 in state-variable form, where the output is the car position x.
(b) Use MATLAB to find the response of the velocity of the car for the case in which the input jumps from being u = 0 at time t = 0 to a constant u = 500 N thereafter. Assume that the car mass m is 1000 \text{ kg and } b = 50 \text{ N.sec/m}.

Step-by-Step

Verified Solution

(a) Equations of motion: First we need to express the differential equation describing the plant, Eq. (2.3), as a set of simultaneous first-order equations. To do so, we define the position and the velocity of the car as the state-variables x and v, so that x = [x v]^{T} . The single second-order equation, Eq. (2.3), can then be rewritten as a set of two first-order equations:

\ddot{x} + \frac{b}{m} \dot{x} = \frac{u}{m}.                (2.3)

\dot{x} = v,

\dot{v} = − \frac{b}{m}v + \frac{1}{m}u.

Next, we use the standard form of Eq. (7.3), \dot{ \pmb x} = \pmb{Fx} + \pmb Gu, to express these equations:

\left[\begin{matrix} \dot{x} \\ \dot{v}\end{matrix} \right] = \left[\begin{matrix} 0 & 1 \\ 0 & -b/m \end{matrix} \right] \left[\begin{matrix} x \\ v \end{matrix} \right] + \left[\begin{matrix} 0 \\ 1/m \end{matrix} \right] u .              (7.5)

The output of the system is the car position y = x1 = x, which is expressed in matrix form as

y= \left[\begin{matrix} 1 & 0 \end{matrix} \right] \left[\begin{matrix} x \\ v \end{matrix} \right] , ,

or

y = \pmb{Hx}.

So the state-variable-form matrices defining this example are

\pmb F = \left[\begin{matrix} 0 & 1 \\ 0 & -b/m \end{matrix} \right] , \ \ \ \pmb G = \left[\begin{matrix} 0 \\ 1/m \end{matrix} \right] , \ \ \ \pmb H = \left[\begin{matrix} 1 & 0 \end{matrix} \right] , \ \ \ J =0 .

(b) Time response: The equations of motion are those given in part (a), except that now the output is v = x_2. Therefore, the output matrix is

\pmb H = \left[\begin{matrix} 0 & 1 \end{matrix} \right] .

The coefficients required are b/m = 0.05 and 1/m = 0.001. The numerical values of the matrices defining the system are thus

\pmb F = \left[\begin{matrix} 0 & 1 \\ 0 & -0.05 \end{matrix} \right] , \ \ \ \pmb G = \left[\begin{matrix} 0 \\ 0.001 \end{matrix} \right] , \ \ \ \pmb H = \left[\begin{matrix} 0 & 1 \end{matrix} \right] , \ \ \ J =0 .

The step function in MATLAB calculates the time response of a linear system to a unit-step input. Because the system is linear, the output for this case can be multiplied by the magnitude of the input step to derive a step response of any amplitude. Equivalently, the \pmb G matrix can be multiplied by the magnitude of the input step.

The statements

MATLAB Verified Solution

Script Files

F = [0 1;0 −0.05];
G = [0;0.001];
H = [0 1];
J = 0;
sys = ss(F, 500*G,H,J); % step gives unit step response, so 500*G

gives u = 500 N.

step(sys); % plots the step response

calculate and plot the time response for an input step with a 500-N magnitude. The step response is shown in Fig. 7.2