Stability of a BIBO transfer function

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?

$$G(s)=\frac{10}{s^2+4s}$$

• 0 is not part of - ve half plane... – Deep Feb 26 '18 at 14:05
• This system is marginally stable due to having a pole at the imaginary axis. – CroCo Apr 8 '18 at 23:15

G(s) is equivalent to an integrator ($\frac{1}{s}$) followed by $\frac{10}{s+4}$ 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.

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.

• Thanks for answering. It is a good approach to check the final value of the step response, I did not remember this strategy – Mathias Scroccaro Feb 27 '18 at 16:04

$G(s)=\dfrac{10}{s^2+4s}=\dfrac{10}{s(s+4)}=\dfrac{10}{s+4}\dfrac{1}{s}$

$\dfrac{1}{s}$ 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.