Tutorial 8 stability analysis

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Oct 30, 2023

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Tutorial 8: Pole-zeros & stability 3 𝑠𝑠 2 + 4𝑠𝑠 + 6 Determine the pole-zeros for the system What do we know? Denominator Second order so two poles Numerator no ‘s’ term so no zeros Denominator is quadratic −𝑏𝑏 ± 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 2𝑎𝑎 4 ± 4 2 (4 1 6) 2 4 ± 16 24 2 4 ± 8 𝑗𝑗 2 2 4 ± 2. 83𝑗𝑗 2 = 2 ± 1. 415𝑗𝑗 What is the stability of the system?
𝑠𝑠 − 3 3𝑠𝑠 2 9 Determine the stability of the system What do we know? Denominator is 2 nd order – two poles ‘s’ term in numerator – one zero No ‘s’ (b term in quadratic rule) −𝑏𝑏 ± 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 2𝑎𝑎 ± (4 3 ∗ − 9) (2 3) ±10.39 6 ± 1.732 To calculate poles… To calculate zero… Negate ‘s’ term thus zero at 3 What is the stability of the system? What type of poles are they?
𝑠𝑠 − 4 𝑠𝑠 2 + 0.64 Determine the stability of the system What do we know? Denominator is 2 nd order – two poles ‘s’ term in numerator – one zero No ‘s’ (b term in quadratic rule) −𝑏𝑏 ± 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 2𝑎𝑎 To calculate poles… ± (4 1 0.64) 2 What type of poles are they? ± 2.56 𝑗𝑗 2 2 ± 1. 6𝑗𝑗 2 = ±0. 8𝑗𝑗 To calculate zero… Negate ‘s’ term thus zero at 4 What is the stability of the system?
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𝑠𝑠 𝑠𝑠 2 − 4𝑠𝑠 + 5 Determine the stability of the system What do we know? Denominator is 2 nd order – two poles ‘s’ term in numerator – one zero Denominator is quadratic −𝑏𝑏 ± 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 2𝑎𝑎 To calculate poles… − − 4 ± 4 2 (4 1 5) 2 4 ± 2𝑗𝑗 2 = 2 ± 1𝑗𝑗 4 ± 16 20 2 4 ± 4 𝑗𝑗 2 2 To calculate zero… Negate ‘s’ term thus zero at 0 What is the stability of the system?
Is the previous system always unstable? A pole-zero plot (pzmap in Matlab) is only valid for ONE value of gain k 𝐺𝐺 𝑠𝑠 = 𝑘𝑘 𝑠𝑠 − 𝑧𝑧 1 𝑠𝑠 − 𝑧𝑧 2 𝑠𝑠 − 𝑝𝑝 1 𝑠𝑠 − 𝑝𝑝 2 If we change k, we need to work out the pole-zeros again and produce another pole-zero plot – time consuming!!! A root-locus plot (rlocus in Matlab) will show us ALL pole-zero locations from k=0 to k = infinity Gain <3.99 UNSTABLE Gain = 3.99 OCSILLATES Gain > 3.99 STABLE For the SAME system, we have three stability criteria depending upon the gain of the system!

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