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T5sol

# T5sol - ME360 Winter 2011 Tutorial#5 Solution 1 0.25 0.52...

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ME360 Winter 2011 Tutorial #5 Solution 1. a) Transfer function in standard form 𝐺𝐺 ( 𝑠𝑠 ) = 0.25 𝑠𝑠 2 + 0.35 𝑠𝑠 + 0.25 = 0.5 2 𝑠𝑠 2 + 2 0.35 0.5 𝑠𝑠 + 0.5 2 poles: 0.175 ± 0.468 𝑗𝑗 , 𝜍𝜍 = 0.35, 𝜔𝜔 𝑛𝑛 = 0.5, 𝜔𝜔 𝑑𝑑 = 0.468 (imaginary part of poles) 𝜃𝜃 = cos 1 𝜁𝜁 = cos 1 0.35 = 1.2132 𝑟𝑟𝑟𝑟𝑑𝑑 = 69.513 𝑜𝑜 -0.175 0.468j -0.468j Re Im 0.5 θ ζ =0.35 b) Step response plot 0 5 10 15 20 25 30 35 40 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Time (sec) µ ( t )

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ME360 Winter 2011 Tutorial #5 Solution 2. Time domain specifications a) Percentage overshoot (%OS) % 𝑂𝑂𝑂𝑂 = 𝑒𝑒𝑒𝑒𝑒𝑒 �− 𝜋𝜋𝜁𝜁 1 − 𝜁𝜁 2 × 100 = 𝑒𝑒𝑒𝑒𝑒𝑒 �− 𝜋𝜋 0.35 1 0.35 2 × 100 = 30.92 Peak time ( 𝑇𝑇 𝑒𝑒 ) 𝑇𝑇 𝑒𝑒 = 𝜋𝜋 𝜔𝜔 𝑛𝑛 1 − 𝜁𝜁 2 = 𝜋𝜋 0.5 1 0.35 2 = 6.7074 𝑠𝑠𝑒𝑒𝑠𝑠𝑜𝑜𝑛𝑛𝑑𝑑𝑠𝑠 Rise time ( 𝑇𝑇 𝑟𝑟 ): - first order approximation 𝑇𝑇 𝑟𝑟 = 0.8 + 2.5 𝜁𝜁 𝜔𝜔 𝑛𝑛 = 3.35
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T5sol - ME360 Winter 2011 Tutorial#5 Solution 1 0.25 0.52...

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