I am trying to find a general way to find system stability.I have applied these methods and struck in confusion
For eg:
$$y(t)= \int_{-\infty}^{t} x(\tau) \sin(4\tau) d\tau$$
then find whether the system is stable or unstable?

Process (1):
If $$y(t) = \int_{-\infty}^{t} x(\tau)\sin(4\tau) d\tau$$
then $$h(t) = \int_{-\infty}^{t} \delta(\tau) \sin(4\tau) d\tau$$
$$\implies h(t) = \int_{-\infty}^{\infty} \delta(\tau)\sin(4\tau) u(t-\tau) d\tau$$
$$\implies h(t) = [\sin(4\tau)u(t-\tau)]|_{\tau=0}$$ $$\implies h(t)=0$$.

so system is stable as $$\int_{-\infty}^{\infty} |h(t)| dt= 0 < \infty$$ ;

Process (2):
If $$y(t)= \int_{-\infty}^{t} x(\tau)\sin(4\tau) d\tau$$
then when $x(t)=\sin(4t)$
$$y(t)= \int_{-\infty}^{t} \sin(4\tau)\sin(4\tau) d\tau = \infty $$ so the system produces unbounded output $y(t)$ for bounded input $x(t)= \sin(4t)$
Hence the system is unstable

Thus actually what will be the system stability and which process is right and which is wrong and how it is wrong?? please explain....

  • $\begingroup$ This is totally unreadable, please use Latex formatting for the formulas. $\endgroup$
    – Matt L.
    Feb 7, 2018 at 13:32
  • $\begingroup$ i believe that you are trying to convey convolution, and if you are, it is not correct. $\endgroup$
    – user28715
    Feb 7, 2018 at 16:44
  • $\begingroup$ i have tried my best to edit this post so that it is clear to you ;please now remove the closed from my post @Peter K $\endgroup$
    – Suresh
    Feb 15, 2018 at 13:57
  • 1
    $\begingroup$ @user9198116 Re-opened, but please learn more MathJax! :-) $\endgroup$
    – Peter K.
    Feb 15, 2018 at 15:03

1 Answer 1


Your first method assumes that the system is entirely characterized by its response to the input $x(t)=\delta(t)$. This would only be the case if the system were linear and time-invariant (LTI). In that case, the input-output relation would have the form


where $h(t)$ is the system's impulse response. However, the given input-output relation is not of the form $(1)$, and hence the system is not LTI. Consequently, the system's response to $x(t)=\delta(t)$ does not say much about the system's behavior in general. That's why your first method of determining the stability is not valid.

The second method is valid. You found a bounded input signal for which the output is unbounded, and in this way you've shown that the system is not BIBO-stable.


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