Let's say I have the following transfer function:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]G(s)=\frac { s-1 }{ { s }^{ 4 }+2{ s }^{ 3 }+{ 3 }s^{ 2 }+{ 4s }+5 }[/tex]

Which is run through MATLAB to obtain the pole-zero format:

[tex]G(s)=\frac { s-1 }{ ({ s }^{ 2 }+2.576s+2.394)({ s }^{ 2 }-0.5756s+2.088) } [/tex]

Using a quadratic solver such as this one, both poles are found to be complex.

I still can't tell the difference between a pole and a zero in terms of system stability. From my understanding of poles and zeroes, roots that are located on the left hand side make the system stable while those on the right hand side make it unstable. Therefore, am I correct to assume that since the zero is 1, the system is unstable? And since the poles are complex, the system oscillates forever without reaching a steady state?

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# Homework Help: Control system stability

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