A Simple System; Cruise Control Model 1. Write the equations of motion for the speed and forward motion of the car shown in Fig. 2.1 assuming that the engine imparts a force u as shown. Take the Laplace transform of the resulting differential equation and find the transfer function between the

Question

A Simple System; Cruise Control Model

1. Write the equations of motion for the speed and forward motion of the car shown in Fig. 2.1 assuming that the engine imparts a force [latex] u [/latex] as shown. Take the Laplace transform of the resulting differential equation and find the transfer function between the input [latex] u [/latex] and the output [latex] v.[/latex]
2. Use MATLAB to find the response of the velocity of the car for the case in which the input jumps from being [latex] u = 0 [/latex] at time [latex] t = 0 [/latex] to a constant [latex] u = 500 \ N [/latex] thereafter. Assume that the car mass [latex] m [/latex] is [latex] 1000 \ kg [/latex] and viscous drag coefficient, [latex] b = 50 \ N · sec/m.[/latex]

Accepted Answer

1. Equations of motion: For simplicity we assume that the rotational inertia of the wheels is negligible and that there is friction retarding the motion of the car that is proportional to the car’s speed with a proportionality constant, [latex] b . \ ^1 [/latex] The car can then be approximated for modeling purposes using the free-body diagram seen in Fig. 2.2, which defines coordinates, shows all forces acting on the body (heavy lines), and indicates the acceleration (dashed line). The coordinate of the car’s position [latex] x [/latex] is the distance from the reference line shown and is chosen so that positive is to the right. Note that in this case the inertial acceleration is simply the second derivative of [latex] x [/latex] (i.e., [latex] a = \ddot{x} ) [/latex] because the car position is measured with respect to an inertial reference. The equation of motion is found using Eq. (2.1). The friction force acts opposite to the direction of motion; therefore it is drawn opposite the direction of positive motion and entered as a negative force in Eq. (2.1). The result is

[latex] \pmb F = m \pmb a [/latex] (2.1)

[latex] u – b \dot{x} = m \ddot{x} [/latex] (2.2)

or

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

For the case of the automotive cruise control where the variable of interest is the speed, [latex] v (=\dot{x}), [/latex] the equation of motion becomes

[latex] \dot{v} + \frac{b}{m}v =\frac{u }{m} . [/latex] (2.4)

The solution of such an equation will be covered in detail in Chapter 3; however, the essence is that you assume a solution of the form [latex] v = V_0e^{st} [/latex] given an input of the form [latex] u = U_oe^{st}. [/latex] Then, since [latex] \dot{v} = sV_0e^{st}, [/latex] the differential equation can be written as

[latex] \left(s + \frac{b}{m}\right) V_0 e^{st} = \frac{1}{m} U_0 e^{st}. [/latex] (2.5)

The [latex] e^{st} [/latex] term cancels out, and we find that

[latex] \frac{V_0}{U_0} = \frac{\frac{1}{m}}{s+ \frac{b}{m}}. [/latex] (2.6)

For reasons that will become clear in Chapter 3, this is usually written as

[latex] \frac{V (s)}{U(s)} = \frac{\frac{1}{m}}{s+ \frac{b}{m}}. [/latex] (2.7)

This expression of the differential equation (2.4) is called the transfer function and will be used extensively in later chapters. Note that, in essence, we have substituted [latex] s [/latex] for [latex] d/dt [/latex] in Eq. (2.4).[latex] ^2[/latex]

2. Time response: The dynamics of a system can be prescribed to MATLAB in terms of row vectors containing the coefficients of the polynomials describing the numerator and denominator of its transfer function. The transfer function for this problem is that given in part (a). In this case, the numerator (called num) is simply one number since there are no powers of [latex] s, [/latex] so that num [latex] = \ 1/m \ = \ 1/1000.[/latex] The denominator (called den) contains the coefficients of the polynomial [latex] s + \ ^b/_m, [/latex] which are

[latex] den = \left[\begin{matrix} 1 & \frac{b}{m} \end{matrix} \right] = \left[\begin{matrix} 1 & \frac{50}{1000} \end{matrix} \right] . [/latex]

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, num can be multiplied by the magnitude of the input step.

[latex] ^1[/latex] If the speed is [latex] v, [/latex] the aerodynamic friction force is proportional to [latex] v^2. [/latex] In this simple model we have taken a linear approximation.

[latex] ^2[/latex] The use of an operator for differentiation was developed by Cauchy about 1820 based on the Laplace transform, which was developed about 1780. In Chapter 3 we will show how to derive transfer functions using the Laplace transform. Reference: Gardner and Barnes, 1942.

Question 2.EX.1: A Simple System; Cruise Control Model 1. Write the equation...

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