PROCESS CONTROL - EXAMPLE 6.3 : The differential equation is 3 dy / dt + 2y = 1

PROCESS CONTROL - EXAMPLE 6.3 : The differential equation is 3 dy / dt + 2y = 1 with y(0) = 1. (a) The Laplace transformation, L for given terms are : L (dy / dt) = sY(s) - y(0), L(y) = Y(s), L(1) = 1 / s. Use such transformation to find Y(s). (b) The initial value theorem states that : When t approaches 0 for a function of y(t), it is equal to a function of sY(s) when s approaches infinity. Use the initial value theorem as a check to the answer found in part (a).

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PROCESS CONTROL - EXAMPLE 6.3 : The differential equation is 3 dy / dt + 2y = 1 with y(0) = 1. (a) ..

Answer / kangchuentat

PROCESS CONTROL - ANSWER 6.3 : (a) For the equation 3 dy / dt + 2y = 1, its Laplace transformation is 3 [ sY(s) - y(0) ] + 2Y(s) = 1 / s, 3 (sY - 1) + 2Y = 1 / s, 3sY - 3 + 2Y = 1 / s, 3sY + 2Y = 1 / s + 3, Y (3s + 2) = (1 + 3s) / s. Y = Y(s) = (1 + 3s) / [ s (3s + 2) ]. (b) Initial value theorem states that y(0) = sY(infinity), then 1 = s (1 + 3s) / [ s (3s + 2) ] = (1 + 3s) / (3s + 2) approaches 1 when the value of s approaches infinity. The checking is correct. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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