We define system to be stable if for all bounded inputs, it will give bounded output at all times.

But does unstable system means it will have unbounded output for any input?

This question is motivated from a different problem I came across which was to find dc gain of a system having closed loop poles on right - hand side of complex plane. The answer states that since poles are in RH s-plane, dc gain can't be defined. My question is it valid to deduce that unstable system would be always unstable or are there inputs to which 'unstable' system might be stable?

  • \$\begingroup\$ Even with no input at all there is probably a noise source in the system somewhere. \$\endgroup\$
    – The Photon
    Dec 18, 2019 at 5:42
  • 1
    \$\begingroup\$ An integrator is unstable but if fed with a pure sine wave the output will be bounded. \$\endgroup\$
    – rotom407
    Dec 18, 2019 at 5:58
  • \$\begingroup\$ Stability is a property of the system, not the input signal(s). \$\endgroup\$
    – Chu
    Dec 18, 2019 at 7:26
  • \$\begingroup\$ ....continuing Chus comment: .....however, the internal resistance/impedance of a connected source (i.e termination of an input node without any input signal) may have an influence on stability \$\endgroup\$
    – LvW
    Dec 18, 2019 at 8:39

1 Answer 1


Yes. There are inputs for which the output of an unstable system is bounded. For example, we can stabilise some unstable systems with feedback. In this case, someone who sees only the input and output of the unstable system (without seeing the feedback mechanism) would see bounded outputs for the unstable system.

However, you may not be able to find the DC gain if the DC input is not in the list of inputs for which the (unstable) system has bounded outputs.


Consider as an example, the first order system A $$ \frac{dx}{dt} = x+u\\ x(0) = 1 $$ The state is bounded for the input \$u(t) = -1\$. The output grows for all other constant inputs.

  • \$\begingroup\$ It took me a while to realise how this answer is true and excellent. Thank you! \$\endgroup\$
    – Deep
    Dec 23, 2019 at 11:01
  • \$\begingroup\$ followup question : Suppose I 'record' what input is being supplied to originally unstable system (while it is working in a closed loop), then disconnect the feedback loop BUT provide exact replica of the input to an unstable system (without connecting feedback), output would be stable right? \$\endgroup\$
    – Deep
    Dec 23, 2019 at 11:08
  • \$\begingroup\$ I mean 'bounded' instead of stable in my last comment in the last line. \$\endgroup\$
    – Deep
    Dec 23, 2019 at 11:09
  • \$\begingroup\$ Output would be bounded only for ideal case (exact same input, exact same initial conditions etc). For practical systems, output is not likely to remain bounded. Even if you simulate the system by solving the equations numerically in a computation, the output is still likely to grow unbounded. \$\endgroup\$
    – AJN
    Dec 23, 2019 at 12:33
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    \$\begingroup\$ @AJN the example you gave is not bounded for \$ u(t) = 1 \$, it is bounded for \$ u(t) = -1 \$. \$\endgroup\$
    – jDAQ
    Dec 23, 2019 at 17:38

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