Why is this transfer function unstable? The poles are \$0\$ and \$-4\$ , so they are in the negative half-plane, theoretically it was to be stable. Can anyone explain to me what makes it unstable? $$G(s)=\frac<10>$$
103 3 3 bronze badges asked Feb 26, 2018 at 10:59 Mathias Scroccaro Mathias Scroccaro 3 1 1 bronze badge \$\begingroup\$ 0 is not part of - ve half plane. \$\endgroup\$ Commented Feb 26, 2018 at 14:05\$\begingroup\$ This system is marginally stable due to having a pole at the imaginary axis. \$\endgroup\$
Commented Apr 8, 2018 at 23:15G(s) is equivalent to an integrator (\$\frac\$) followed by \$\frac\$ and, given that integrators are unstable, the whole TF is unstable. The pole at 0,0 means it is on the cusp of instability so, in any practical implementation that doesn't use it (say) within a feedback loop that might create stability, it is unstable.
answered Feb 26, 2018 at 11:18 466k 28 28 gold badges 380 380 silver badges 834 834 bronze badges \$\begingroup\$First of all you should note that the Laplace transform is strictly defined over some region of conversion (ROC) which should not include any of the poles for bounded output. In this case ROC doesn't include imaginary axis, so The Fourier transform will not converge. Also you can easily check step response of the system using final value theorem. It's unbounded for the given transfer function.
answered Feb 26, 2018 at 11:29 78 1 1 silver badge 5 5 bronze badges\$\begingroup\$ Thanks for answering. It is a good approach to check the final value of the step response, I did not remember this strategy \$\endgroup\$
Commented Feb 27, 2018 at 16:04 \$\begingroup\$\$\dfrac\$ is the equivalence of an integration, this means positive feedback.
In bodeplots you are measuring the 0 dB gain and the phase margin, if you have a 180 degree phase shift (=0 phase margin) and a gain above 0 dB at some frequency, then you will accumulate that frequency and oscillate => unstable.
The feedback is set to be negative, -1, 180 degrees is also -1 on the unit circle. (-1)(-1) = 1 = positive feedback = unstable.
Bottom line: A pole in the origin without a zero to counter it => unstable.
answered Feb 26, 2018 at 11:22 Harry Svensson Harry Svensson 8,239 3 3 gold badges 34 34 silver badges 53 53 bronze badgesTo subscribe to this RSS feed, copy and paste this URL into your RSS reader.
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