# Open loop transfer function 50/(s^3+2s^2+s+4) ?

Posted 1 year ago
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 When I examine open loop transfer function L(s)= 50/(s^3+2s^2+s+4) using Routh-Hurowitz as well as the Roots[L(s)] function in Mathematica, both show two poles in right half plane. Yet, when I do a NyquistPlot[L(s) ], there is no encirclement of -1. Why is that? Answer
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Posted 1 year ago
 Perhaps your syntax is incorrect. The documentation is a reliable friend. ls = 50/(s^3 + 2 s^2 + s + 4) tf = TransferFunctionModel[{{ls}}, s] NyquistPlot[tf] Solve[0 == 1/ls, s] Answer
Posted 1 year ago
 You can explicitly see that the closed-loop system has 2 unstable poles In:= csys = SystemsModelFeedbackConnect@tf; TransferFunctionPoles[csys][[1, 1]] // N Out= {-4.47411, 1.23705 - 3.2464 I, 1.23705 + 3.2464 I} Or use the Routh–Hurwitz criterion which shows two sign changes in the first column. In:= ControlStabilityArray[csys][[1, 1]] // TableForm `  Answer
Posted 1 year ago
 Oops! Never mind, my mistake. There should be zero encirclement's since Z=N+P.=0+2=2. Answer
Posted 1 year ago
 Hi Craig, Thanks for the response. Attached is the code. Attachments: Answer