I know that for a system to be BIBO stable its impulse response must be absolutely integrable and the impulse response $ h(t)= u(t)$ integrates to approach infinity (i guess)

I proceeded as$$ \int_{-\infty}^{\infty}|u(t)|dt $$ $$ =\int_{0}^{\infty}dt $$ $$ =t\Big|_0^{\infty} $$ $$ =\infty$$

So the system must be unstable right? I m unsure 😅 Can anyone explain in terms of poles of the $H(s)$?

  • $\begingroup$ Have you tried inverse Laplace? $\endgroup$ Commented May 17, 2021 at 12:28
  • $\begingroup$ @a concerned citizen : yes...I know about it but how will it help? $\endgroup$ Commented May 17, 2021 at 12:35
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    $\begingroup$ that system is what we call "marginally stable" or "critically stable". i would prefer to call it "unstable" but the output is bounded given a bounded input. so a single pole on the $j\omega$ axis (or the unit circle in the $z$-plane) will not make your system totally unstable. however a double pole on the $j\omega$ axis will. $\endgroup$ Commented May 17, 2021 at 13:06
  • $\begingroup$ @robert bristow-johnson : okay thanks👍 $\endgroup$ Commented May 17, 2021 at 13:13
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    $\begingroup$ @robertbristow-johnson: The output is not bounded for any bounded input. Try $x(t)=u(t)$, which is clearly bounded, yet the output grows without bounds. $\endgroup$
    – Matt L.
    Commented May 17, 2021 at 13:13

2 Answers 2


At first, the meaning of BIBO stability should be considered which really helps to understand what it actually means even though dealing with words rather than numbers seems abstract in terms of engineering.

BIBO stability or Bounded Input-Bounded Output stability states that for every bounded input $x(t)$ i.e. $|x(t)| < \infty$, output of the system of concern $y(t)$ is bounded i.e. $|y(t)| < \infty$. So, basically, according to that statement; if every input that is not ever-increasing function of time, so is resultant output of the system.

Let's consider the system with the impulse response (actually, the transfer function of the system in s-domain) $h(t) = u(t)$. The Laplace transform of this impulse response is;

$$H(s) = \frac{1}{s}$$

Before moving onto BIBO stability, let's give some information about the system. According to the transfer function of the system, there is just one trivial pole just at the origin i.e. at $s = 0$. That situation points out the marginal stability in which the system can be either stable or unstable depending upon the working environment, input signals, etc. which can result in the shifting of the pole of the system to the right-hand side of the s-domain which definitely causes unstability and possible runaway of the system output.

Now, let's put BIBO stability into play. Let's assume that input of the system is $x(t) = sin({\omega}t)$ which is a bounded signal as it oscillates at the angular frequency $\omega$ and in the value interval $[-1, 1]$. The signal's Laplace transform is;

$$X(s) = \frac{\omega}{s^2 + {\omega}^2}$$

Accordingly, output of the system becomes;

$$Y(s) = X(s)H(s) = \frac{\omega}{s^3 + s{\omega}^2}$$

If the last value theorem is applied to the output in order to find out whether the output is bounded, we obtain;

$$Y(\infty) = \lim_{s \to 0} sY(s) = \frac{1}{\omega}$$

So, this result means that as time approaches infinity (practically, a huge amount of time), the output of the system approaches the value $\frac{1}{\omega}$ and terminates at that level. This shows a BIBO stability situation and for this input, the system has BIBO stability.

However, let's think about another input $x(t) = u(t)$. Laplace transform of this input is;

$$X(s) = \frac{1}{s}$$

Now, let's apply this input into the system and get the resultant output. The output becomes;

$$Y(s) = X(s)H(s) = \frac{1}{s^2}$$

By applying the last value theorem, the output becomes;

$$Y(\infty) = \lim_{s \to 0} sY(s) = \infty$$

This result means that as time approaches infinity, the output of the system approaches infinity. Alternatively, in time domain, the output function is;

$$y(t) = t$$

It is obvious that amplitude of the output signal is proportional to time i.e. the output is an ever-increasing function of time. So, we can't talk about BIBO stability of the system with this input.

But, why do we get two different stability behaviours for the same system? It is because marginal stability of the system as the system's nature. According to marginal stability; if the input signal's angular frequency is equal to the natural frequency $\omega_{n}$ of the system's transfer function -in our case, it is $\omega_{n} = 0 \frac{rad}{s}$-, the output of the system ends up in resonance and runaway.

Consequently, the system of concern isn't absolutely stable. The system's stability is actually on a red line so that it can be either stable or unstable depending upon the operating conditions e.g. temperature, input signal's characteristics and so on.

Actually, the most important behaviour that is expected from a system is stability which ensures the system to be under our control which is a must and the nature of control theory. But, sometimes, we deliberately put a system into unstable operating modes e.g. when designing an oscillator as we want to see resonance at the output stage of an oscillator. This is made possible by Barkhausen criterion.

  • $\begingroup$ Thank you so much for the answer 👍 $\endgroup$ Commented May 17, 2021 at 13:52

Your derivation is correct. The impulse response is not absolutely integrable, hence the system is not BIBO stable. From the corresponding transfer function $H(s)=1/s$, you can see that there is a single pole at the origin. Systems with single poles on the imaginary axis, like the integrator in your example, are also called marginally stable. These systems are not BIBO stable, transients do not decay, and certain inputs can cause unbounded outputs.

  • $\begingroup$ okay...Thanks 👍 $\endgroup$ Commented May 17, 2021 at 13:18
  • $\begingroup$ In addition to "marginally stable" you'll also see such systems described as "metastable". Different word, means the same thing. $\endgroup$
    – TimWescott
    Commented May 17, 2021 at 15:39

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