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Simulation of Dynamic Systems with MATLAB and Simulink, 2nd by Harold, Allen, Randal Klee

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By Harold, Allen, Randal Klee

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4 T(t) Hot mh, Th water Simulation of Dynamic Systems with MATLAB® and Simulink® 38 A suitable model for this thermal system is based on the conservation of energy. 01375 Btu=8F=ft3). The heat flow terms on the right-hand side are given by Qh ¼ m_ h cp (Th À T) Qc ¼ m_ c cp (T À Tc ) Q0 ¼ 1 (T À T0 ) R where m_ h and m_ c are the mass flow rates (lb=min) of the hot and cold water cp is the specific heat of water (1 Btu=lb=8F) R is the thermal resistance (8F=Btu=min) of the chamber walls The expressions for Qh and Qc assume that the flow rates of the circulating fluids are great enough that both fluids exit at the same temperature at which they entered the chamber.

2002; Dorf and Bishop 2005). The unit step response assumes one of three forms depending on the location of the roots of the algebraic equation s2 þ 2zvn s þ v2n ¼ 0 (2:17) known as the characteristic equation of the system. 17 and are given by s1 , s2 ¼ Àzvn Æ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi z2 À 1vn (2:18) The natural modes of the second-order system are es1 t and es2 t . The step response depends on the value of the parameter z. There are three cases to consider. Case 1: z > 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi If we let s1 ¼ Àzvn À z2 À 1vn and s2 ¼ Àzvn À z2 À 1vn , then both roots are negative (assuming vn > 0) and s1 < s2 < 0.

18, resulting in the fund’s balance at the end of the 18th year. However, a general solution of the discrete-time model is preferable since it can be evaluated for any value of the discrete-time variable k. For the discrete-time model, y(k) ¼ y(k À 1) þ iy(k À 1) þ u(k), k ¼ 1, 2, 3, . . (1:38) ¼ (1 þ i)y(k À 1) þ A (1:39) ¼ ay(k À 1) þ A (1:40) where a ¼ 1 þ i and A is the constant net deposit each interest period. The first several outputs are y(1) ¼ ay(0) þ A Enhancer Apago PDF y(2) ¼ ay(1) þ A ¼ a[ay(0) þ A] þ A ¼ a2 y(0) þ aA þ A y(3) ¼ ay(2) þ A ¼ a[a2 y(0) þ aA þ A] þ A ¼ a[a3 y(0) þ a2 A þ aA] þ A suggesting the general expression for y(k) is & y(k) ¼ y(0), k ¼ 0 ak y(0) þ (1 þ a þ a2 þ a3 þ Á Á Á þ akÀ1 )A, k ¼ 1, 2, 3, .

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