Solution_to_problem_6.6

Solution_to_problem_6.6 - 1 and 2 are y t will contain for...

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6.6 – 1 Solution to problem 6.6 The first step is to write the overall transfer function ( ) G s in standard form before starting the analysis. Since the two processes (integrating element and first-order element) are operating in parallel, the overall transfer function is the sum of the individual transfer functions: ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 1 2 1 2 1 1 2 1 1 2 1 1 ( ) ( ) ( ) 1 1 1 1 K K s K K s K s K K s K K K K G s G s G s s s s s s s s s K τ + + + + + + = + = + = = = + + + + ( ) ( ) 1 2 2 1 1 1 1 1 1 ( ) 1 1 K K K K s K s K K G s s s s s + + + + = = + + Once we have the standard form we can give the answers: f ) For a given input ( ) U s , the output will be ( ) ( ) ( ) Y s G s U s = . In order to calculate ( ) y t we will have to apply the inverse Laplace Transform. Independently of the input ( ) U s , ( ) Y s will

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6.6 – 2 contain for sure the term ( ) 1 s s τ + in the denominator. That means that when calculating ( ) y t a partial fraction expansion will have to be performed involving the terms s and ( ) 1 s + . In other words, 1 2 ( ) ... 1 Y s s s α = + + + That means that whatever the values of
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Unformatted text preview: 1 and 2 are, ( ) y t will contain for sure the following terms: 1 1 2 1 2 ( ) ... ... 1 t y t e s s-- = + + = + + + L g ) If ( ) M U s s = then: ( ) ( ) 2 2 1 1 1 1 2 1 1 ( ) ( ) ( ) 1 1 K K K s MK s K K M Y s G s U s s s s s s + + + + = = = + + Proceeding as before, a partial fraction expansion of ( ) Y s would give: 3 1 2 2 ( ) 1 Y s s s s = + + + Therefore ( ) y t will contain the following terms: 1 3 1 2 1 2 3 2 ( ) 1 t y t t e s s s-- = + + = + ⋅ + + L Since 2 t ⋅ is not bounded, ( ) y t is not bounded for a step change. Note that the term 2 1 s is an integrating element....
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Solution_to_problem_6.6 - 1 and 2 are y t will contain for...

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