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| (5) |
The elements of are all positive if and only if , by inspection. Thus, you can use the information returned even when the direct call to Hurwitz fails.
Separate calls to Hurwitz in the cases and give nontrivial gcds between and its paraconjugate. Thus, the stability criteria are satisfied only as above.
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| (9) |
Notice that the last term has coefficient . Thus, you can say unequivocally that is not Hurwitz, for any value of .
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By inspecting , notice that is Hurwitz only if , and , and . This can be simplified to the conditions
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evalc and the Hurwitz function assume that symbolic parameters have real values.
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| (15) |
The coefficients of can be inspected according to rules, but it is a tedious process.
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| (23) |
Examination of the above for real values of is a way to determine whether the polynomial is Hurwitz.
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| (24) |
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In the previous example, might be zero. Thus, Hurwitz cannot determine whether all the zeros are in the left half plane.