The implication from that is that the position at

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Unformatted text preview: n, , of the total population at the end of the previous day 365 (neglect leap-year effects). Every day 275 immigrants enter Freedonia. (a) Write a difference equation for the population at the beginning of the nth day after January 1, 2000 with the immigration rate as the excitation of the system. (b) By finding the zero-exctiation and zero-state responses of the system determine the population of Freedonia be at the beginning of the year 2050. Solutions 3-43 M. J. Roberts - 8/16/04 (b) The beginning of the year 2050 is the 18250th day. 61. A car rolling on a hill can be modeled as shown in Figure E61. The excitation is the force, f ( t) , for which a positive value represents accelerating the car forward with the motor and a negative value represents slowing the car by braking action. As it rolls, the car experiences drag due to various frictional phenomena which can be approximately modeled by a coefficient, k f , which multiplies the car’s velocity to produce a force which tends to slow the car when it moves in either direction. The mass of the car is m and gravity acts on it at all times tending to make it roll down the hill in the absence of other N⋅s forces. Let the mass, m, of the car be 1000 kg, let the friction coefficient, k f , be 5 m π and let the angle, θ , be . 12 (a) Write a differential equation for this system with the force, f ( t) , as the excitation and the position of the car, y( t) , as the response. (b) If the nose of the car is initially at position, y(0) = 0 , with an initial velocity, [y′(t)]t = 0 = 10 m , and no applied acceleration or braking force, graph the velocity of the car, s y′ ( t) , for positive time. (c) If a constant force, f ( t) , of 200 N is applied to the car what is its terminal velocity ? f(t) y(t) (θ) θ sin mg Figure E61 Car on an inclined plane (a) Summing forces, (b) The zero-excitation response can be found by setting the force, f ( t) , to zero. f ( t) − mg sin(θ ) − k f y′ ( t) = m y′′ ( t) − kf The homogeneous solution is y h ( t) = K h1 + K h 2e m . The particular solution must be in the form of a linear function of t, to satisfy the differential equation. t Solutions 3-44 M. J. Roberts - 8/16/04 t − 200 y( t) = 1.0346 × 10 1 − e − 507.28 t 5 t t − −t 1.0346 × 10 5 − 200 507.28 = 517.28e 200 − 507.28 = 517.28 e 200 − 1 + 10 y′ ( t) = e − 200 y’(t) 1000 t -550 (c) The differential equation is m y′′ ( t) + k f y′ ( t) + mg sin(θ ) = f ( t) We can re-write the equation as m y′′ ( t) + k f y′ ( t) = f ( t) − mg sin(θ ) treating the force due to gravity as part of the excitation. Then the impulse response is the solution of m h′′ ( t) + k f h′ ( t) = δ ( t) which is of the form, kf − t h( t) = K h1 + K h 2e m u( t) . The impulse response is − 1− e h( t) = kf kf m t u( t) . Now, if we say that the force, f ( t) , is a step of size, 200 N, the excitation of the system is x( t) = 200 u( t) − mg sin(θ ) . But this is going to cause a problem. The problem is that the term, − mg sin(θ ) , is a constant, therefore presumed to have acted on the system for all time before time, t = 0. The implication from that is that the position at time, t = 0, is at infinity. Since we are only interested in th...
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