EECE 301 Discussion 10 Laplace Transform Examples_2

# EECE 301 Discussion 10 Laplace Transform Examples_2 - EECE...

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1/19 EECE 301 Prof. Mark Fowler Discussion #10 • Laplace Transform Examples

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2/19 Examples of Solving Differential Equations using LT Notice how easy this is! -LT converts the differential equation into an algebraic equation -We can easily solve an algebraic equation for an output Y ( s ) -We can do partial fraction expansion -We can use the LT table But, this is where the hard part lies… although it is easy for certain inputs.
3/19 Example 6.29: 2 nd Order Differential Equation Given ) ( 2 ) ( 8 ) ( 6 ) ( 2 2 t x t y dt t dy dt t y d = + + Find the output y ( t ) for t 0 when the input is x ( t ) = u ( t ) 2 ) 0 ( 1 ) 0 ( = = y and y 1 ) ( t x t Input “starts” at t = 0 Assume that the system has ICs given by:

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4/19 Gives: [] [ ] ) ( 2 ) ( 8 ) 0 ( ) ( 6 ) 0 ( ) 0 ( ) ( 2 s X s Y y s sY y s y s Y s = + + Solving for Y ( s ) algebraically gives: [ ] ) ( 8 6 2 8 6 ) 0 ( 6 ) 0 ( ) 0 ( ) ( 2 2 s X s s s s y y s y s Y + + + + + + + = Using the specific IC’s gives: ) ( 8 6 2 8 6 8 ) ( 2 2 s X s s s s s s Y + + + + + + = IC part H ( s ) “Transfer Function” (notice how this comes directly from the D.E.) ) ( 2 ) ( 8 ) ( 6 ) ( 2 2 t x t y dt t dy dt t y d = + + Applying the LT to this D.E.
5/19 Notice how the IC part can be easily found using PFE and an LT table (For linear, constant coefficient differential equations it will always be like that!) ) ( 8 6 2 8 6 8 ) ( 2 2 s X s s s s s s Y + + + + + + = IC part H ( s ) “Transfer Function” (notice how this comes directly from the D.E.) Notice that the input part may not be easy to do PFE/ILT if the input X ( s ) is complicated. But in control systems and sometimes in electronics we are often interested in how

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EECE 301 Discussion 10 Laplace Transform Examples_2 - EECE...

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