Tutorial 5 Laplace Transforms 2 Impedance Modelling

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Oct 30, 2023

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Tutorial 5 Laplace Transforms 2: Laplace Impedance Modelling Apply a forward Laplace transform to the following system and develop the s-domain transfer function for the system. What do we know? System is dynamic – capacitors & inductors 𝒊𝒊 = 𝑪𝑪 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅 = 𝑳𝑳 𝒅𝒅𝒊𝒊 𝒅𝒅𝒅𝒅 Map system to s-domain 𝒁𝒁 𝑹𝑹 = 𝑹𝑹 𝒁𝒁 𝑪𝑪 = 𝟏𝟏 𝑪𝑪𝑪𝑪 𝒁𝒁 𝑳𝑳 = 𝑳𝑳𝑪𝑪 Develop transfer function Algebraic model
We have now mapped to the s-domain 1. Simplify where possible 2. Develop transfer function How can we simplify? S + 2 Two parallel branches – solve separately!
How do we deal with parallel branches? Exactly as we would for resistive circuits… 1 𝑅𝑅 𝑇𝑇 = 1 𝑅𝑅 1 + 1 𝑅𝑅 2 + 1 𝑅𝑅 3 + … For the parallel branch AB we have… 1 𝑍𝑍 𝑇𝑇 = 1 𝑍𝑍 1 + 1 𝑍𝑍 2 + 1 𝑍𝑍 3 + … Inserting the values… Cross multiply by s+2 to yield common denominator… = = Invert to find total impedance… Thus the total impedance for the parallel network AB can be stated as… =
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For the parallel branch CD we have… = Thus multiplying through by 3s yields… = Therefore the total equivalent impedance for the parallel network CD can be stated as… Our system is now…
We now have two blocks in cascade (series) can add… Problem? Cross multiplying yields… Multiplying out the factorised terms yields… (s + 1) 2 = s 2 + 2s +1 (s 2 +2s +1) . (3s) = 3s 3 + 6s 2 + 3s (6s 2 + 1) . (s + 2) = 6s 3 + 12s 2 + s + 2 This process yields… Grouping like terms yields total impedance for the system…
TASK! Develop the s-domain transfer function for the following system…
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