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

Don Patterson, Consultant
Posted 5 years ago

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?
POSTED BY: Don Patterson
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You can explicitly see that the closed-loop system has 2 unstable poles

In[25]:= csys = SystemsModelFeedbackConnect@tf;
TransferFunctionPoles[csys][[1, 1]] // N

Out[26]= {-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[27]:= Control`StabilityArray[csys][[1, 1]] // TableForm

enter image description here
POSTED BY: Suba Thomas

Oops! Never mind, my mistake. There should be zero encirclement's since Z=N+P.=0+2=2.

POSTED BY: Don Patterson

Hi Craig, Thanks for the response. Attached is the code.

Attachments:
NyquistPlot.pdf
POSTED BY: Don Patterson

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]
POSTED BY: W. Craig Carter
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