The Dirchlet conditions state that if the signal is absolutely summable then it the DTFT of the signal definitely exists. This is a sufficient condition but not necessary condition.

There are systems like Ideal Low Pass Filter, which are not absolutely summable. The condition of absolutely summability is a necessary and sufficient condition for stability. Therefore, I can find a bounded input for which the system produces unbounded input.

For such systems, which are not absolutely summable, but are absolutely square summable, we define the DTFT in the same manner and argue that even though non-zero error exists, the energy of the error is 0, which essentially gives rise to Gibbs Phenomenon.

But the condition of $y[n] = x[n] * h[n]$ is independent of stability, and simply depends on Linear and shift invariance property. From this definition, I can find a bounded input which will produce unbounded output. But the relation, $Y(w) = X(w)H(w)$ is also independent of stability. Even if I take into account Gibbs Phenomenon, I am unable to deduce the existence of a bounded input which will produce and unbounded output, and this essentially suggests the fact that bounded input will give bounded output.

Where am I going wrong?

  • $\begingroup$ For clarity, you're using the specific BIBO definition of stability, right? $\endgroup$ Commented Feb 5, 2017 at 18:52
  • $\begingroup$ Yes. I don't know if there are any other definitions for stability? $\endgroup$
    – Arka Sadhu
    Commented Feb 5, 2017 at 19:01
  • $\begingroup$ Oh, you will love Lyapunov. He came up with three definitions alone mentioned in that link, as well as two methods for determining them. The reason why they matter is because BIBO stability is weaker than state-space stability. It doesn't guarantee your system will be stable in all situations. $\endgroup$ Commented Feb 5, 2017 at 19:33

1 Answer 1


It is generally not true that the relation $Y(\omega)=X(\omega)H(\omega)$ is independent of the system's stability. For systems with a rational transfer function (i.e., systems that can be described by differential or difference equations), the frequency response, which is the Fourier transform of the impulse response, only exists for stable systems. For marginally stable systems with poles on the imaginary axis (for continuous-time systems), or on the unit circle (for discrete-time systems), the Fourier transform exists if we allow Dirac delta impulses in the expression for the Fourier transform. So for marginally stable systems you can see that the output can become unbounded, even for bounded input signals, due to the Dirac delta impulses in their frequency response. For unstable causal systems with transfer function poles in the right half plane, the frequency response doesn't exist.

For systems for which there exists no transfer function, but which do have a frequency response, such as ideal frequency-selective filters (low pass, high pass, etc.), the relation $Y(\omega)=X(\omega)H(\omega)$ holds, regardless of stability. Ideal frequency-selective filters are unstable (and non-causal), but their frequency response $H(\omega)$ is finite. This means that the frequency response of the bounded input signal which generates an unbounded output (if it exists) must be unbounded for at least one value of $\omega$, and the inverse Fourier transform of $X(\omega)H(\omega)$ must result in an unbounded time-domain signal. Also note that there is no guarantee that the Fourier transform $X(\omega)$ of the bounded input signal that generates an unbounded output signal even exists. And if it exists, it's generally not immediately clear from the expression $X(\omega)H(\omega)$ that the corresponding inverse transform is unbounded.

I think it is a misunderstanding that the relation $Y(\omega)=X(\omega)H(\omega)$ somehow implies that a bounded input must result in a bounded output. It's just not so easy to see whether the output is unbounded because one would have to compute


and check for each value of $n$ if the expression becomes unbounded. Direct time domain analysis is more straightforward in this case.

  • $\begingroup$ Since the absolute summability is necessary and sufficient condition for stability, there always exists a bounded input which produces unbounded output. $x[n] = \frac{h*[-n]}{|h*[-n]|} , h[n] \neq 0$ and $0$ otherwise, always does this. The problem is that finding DTFT of this kind of signal is non-trivial. The problem is using $Y(w) = X(w)H(w)$ makes it seem like bounded input is giving bounded output, and I feel there is some kind of fallacy here, which I am not being able to spot. Would appreciate if you could throw some light on this aspect. $\endgroup$
    – Arka Sadhu
    Commented Feb 5, 2017 at 16:27
  • $\begingroup$ @ArkaSadhu: I think that in general there is no guarantee that $X(\omega)$ exists for the bounded signal that results in unbounded output. If it exists it must be such that the inverse transform of $X(\omega)H(\omega)$ is unbounded for at least one value of (the time index) $n$. The point is that this may not be immediately clear from looking at the expression for $X(\omega)H(\omega)$. So actually the relation $Y(\omega)=X(\omega)H(\omega)$ does not "make it seem like bounded input is giving bounded output". $\endgroup$
    – Matt L.
    Commented Feb 5, 2017 at 20:47

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