q0 pn 1 pn 2 p0 r dt n dt n 1 dt n 1 dt n 2 where

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Unformatted text preview: transform table and properties Solving the resulting algebraic transform of the variable of interest Refer to the Laplace transform table to find the inverse Laplace transform Use partial fraction expansion method if necessary Differential Equation dny d n −1 y d n −1r d n −2 r + qn −1 +.... + q0 = pn −1 + pn −2 + p0 r dt n dt n −1 dt n −1 dt n −2 Where y (t) is the response and r (t) the input or forcing function. Taking LT for both sides of the above equation we get s nY ( s ) + qn −1s n −1Y ( s ) + ... + q0Y ( s ) + some _ initial _ conditions = pn −1s n −1 R ( s ) + pn −2 s n −2 R ( s ) + ... + p0 R ( s ) + some _ initial _ conditions Assuming that all the initial conditions are equal to zero, then the transfer function can be written as [s n [ + qn −1s n −1 +... + q0 Y ( s ) = pn −1s n −1 + qn −2 s n −2 +... + q0 R ( s ) pn −1s n −1 + qn −2 s n −2 +... + q0 Y (s) G(s) = = R( s) s n + qn −1s n −1 +... + q0 Solving Differential Equations Example: Find the solution of the following differential equation d 2 y (t ) dy (t ) + 12 + 32 y (t ) = 32u (t ) 2 dt dt y (0) = y ' (0) = 0 Solving Differential Equations Exercise: Find the solution of the following differential equation ( t ) + 5 y( t ) + 4 y( t ) = 3 y y(0) = 1 y ( 0) = 2 Summary Laplace transform is just mathematical tools in control system....
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