I have a rather simple question I would say.

I know that systems having real poles to the right side of the plane will inevitably be instable (for LTI systems at least).

But I'm confused with systems having poles at origin with poles at the left side of the plane.

For example say we have this kind of systems: $$ H_1(s) = \frac{K}{s(1 + \tau s)} $$ $$ H_2(s) = \frac{K}{s(1 + \tau_1 s)(1 + \tau_2 s)} $$ $$ H_3(s) = \frac{K}{s^{2} (1 + \tau s)} $$ $$ \tau_i > 0$$

Those are only examples among others. If I analyze the response to an impulse, it would make sense to me that H1 and H2 are stable, and not H3 as the response y3(t) would contain a ramp.

But... if I check the response to a step... H1 and H2 are not stable anymore (time response contains a ramp).

So finally my question boils down to : In reference to what kind of excitation do we call a system stable in the EE jargon? Because from what I see, H1 and H2 can be considered both stable and instable depending on whether the excitation U(s) is an impulse or a step (for example).

So do we only refer to the transfer function (a.k.a. impule response) "as-is" (for pole analysis) or... sorry I'm confused.

TL;DR Are H1 and H2 considered stable, and H3 is unstable?

EDIT: My question refers to open loop systems.


  • \$\begingroup\$ H1 and H2 have 1/s which is an integration therefore, for a unit step the output will keep rising. This is what an integrator does but it is stable. \$\endgroup\$
    – Andy aka
    Nov 18, 2015 at 8:46
  • \$\begingroup\$ And H3 has two of them. \$\endgroup\$ Nov 18, 2015 at 9:03
  • \$\begingroup\$ As system TFs, none of them are bounded input / bounded output stable. Apply a step to any and the response goes to infinity. If they are open-loop TFs then the first two would be stable in unity feedback closed-loop (assuming K is suitably chosen in H2), but the third would not. \$\endgroup\$
    – Chu
    Nov 18, 2015 at 9:26
  • \$\begingroup\$ Isn't a stable system a system that converges to a constant value though? So basically if I understand correctly, when determining system stability by pole analysis, we only consider the poles of the transfer function without any excitation? I'm sorry I feel my core question hasn't been responded to clearly. \$\endgroup\$
    – Yannick
    Nov 18, 2015 at 11:49
  • \$\begingroup\$ The pole is at zero, so neither left-plane nor right-plane. This qualifies as 'marginally stable', so you could say not stable, and not unstable. BIBO stability is a more strict definition and, as the DC gain is infinite, the systems are not BIBO stable. But I don't really know why it's necessary to hang a particular label on a system; the TF tells the complete story. \$\endgroup\$
    – Chu
    Nov 18, 2015 at 12:24

2 Answers 2


The transfer function of a stable (LTI) system needs to have all its poles in the left half-plane, i.e. any pole \$s_{\infty}\$ must satisfy


If this condition is satisfied, then any bounded input signal \$|x(t)|\le K\$ will result in a bounded output signal \$|y(t)|\le L\$ with some positive constants \$K\$ and \$L\$. This concept is called BIBO-stability. Poles on the imaginary axis, i.e. poles with \$\text{Re}(s_{\infty})=0\$ do not satisfy (1), and, consequently, systems with such poles are not stable in the BIBO sense.

In some contexts, systems with poles on the imaginary axis are called marginally stable, but such systems will generally produce unbounded outputs for bounded input signals.

  • \$\begingroup\$ So marginally stable systems can produce (for example H1/H2 with a step excitation?) for example a ramp to a bound signal (I suppose you refer to bound signal as a signal that converges to a constant over infinite), but still be considered stable technically? Correct? It's a bit "interesting" in the sense that I don't get what is stability in the marginal case then... isn't stable means "converges to a value", which we know a ramp wouldn't. To this point its vocabulary but still... \$\endgroup\$
    – Yannick
    Nov 18, 2015 at 13:55
  • \$\begingroup\$ @Yannick: From a practical point of view, marginally stable systems are unstable, because they generally produce unbounded outputs for bounded inputs. For specific input signals (such as an impulse), the output will not blow up, but it won't decay either. \$\endgroup\$
    – Matt L.
    Nov 18, 2015 at 14:43
  • \$\begingroup\$ @Yannick You are talking about steps, ramps,..etc. None of this has to do with stability criterion, frequency analysis is the point. If you excite with sweep generator you get a freq. response. All DC signals and amplitude rising with step or ramp excitation have no meaning. \$\endgroup\$ Nov 22, 2015 at 16:10

The instable point resides in -1,0j like an attractor, the curve has to pass at safe distance - stability margin, or alpha max. Each pure integration, rotates the plane for 90 degrees, H1 is 1st order + 1x integration. If you look 1st order Nyqvist diagram it is in 4-th quadrant, adding an integration the whole characteristics is rotated into 3-rd quadrant. A rule of thumb is: nr of poles determine the nr of x-y axis crossings in Nyqvist diagram.
With regard to K, all of systems can be stable or unstable in the closed loop control. In our jargon the system becomes unstable when it pass the point of non return, it is attracted into point -1 and will not escape, this means that it begins to oscillate at system's natural frequency.
What you are looking with step responses are actualy not the stability, because your thought is that if the response is a ramp that increases over all limits, this system is unstable, but you're wrong.


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