Wolfram Mathematica

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4:33 AM

BodePlot[TransferFunctionModel[Unevaluated[{{1/((s + 1) (s + 100))}}],
s, SamplingPeriod ->None, SystemsModelLabels -> None], {0.01, 10},
StabilityMargins -> True]
BodePlot[1/((s + 1) (s + 100)), {0.01, 10}, StabilityMargins -> True]

Both generate the same plot

so why do you need TransferFunctionModel?

Nasser

5:08 AM

@anhnha it is a short cut. From help it says

anhnha

5:27 AM

Thanks

1 hour later…

yode

6:47 AM

@Szabolcs There is any measure can judge two graph is similarity?

Such as this two graph:

9 hours later…

yode

3:25 PM

g1=Graph[{1\[UndirectedEdge]2,2\[UndirectedEdge]3,3\[UndirectedEdge]4,4\[UndirectedEdge]1,5\[UndirectedEdge]6,6\[UndirectedEdge]7,7\[UndirectedEdge]8,8\[UndirectedEdge]5,1\[UndirectedEdge]5,2\[UndirectedEdge]6}];
g2=Graph[{1\[UndirectedEdge]2,2\[UndirectedEdge]3,3\[UndirectedEdge]4,4\[UndirectedEdge]1,5\[UndirectedEdge]6,6\[UndirectedEdge]7,7\[UndirectedEdge]8,8\[UndirectedEdge]5,1\[UndirectedEdge]5,2\[UndirectedEdge]7}];

1 hour later…