# BIBO stability of $y(t)=\int_{-\infty}^{t}{x(\tau)d\tau}$

How can I prove that the LTI system with (output $y(t)$, input $x(t)$)

$$y(t)=\int_{-\infty}^{t}{x(\tau)d\tau}$$

is BIBO (bounded-input/bounded output) stable?

• Why do you think it is? (Hint: It isn't). – Matt L. Aug 29 '17 at 11:20
• Aw, Matt! You gave away the answer! :-) – Dilip Sarwate Aug 29 '17 at 11:24
• @DilipSarwate: Well, I guess the solution to the homework problem must include a good argument as to why the system isn't stable. This is still up to the OP, and with your answer this shouldn't be a big problem :) – Matt L. Aug 29 '17 at 11:26
• Let me suggest that you enhance your (LTI) system questions with definitions you have at hand, and clear definitions of the concepts your are using, possibly with references. If your questions are homework, please tag them as such, and provide us with your initial thoughts. – Laurent Duval Aug 29 '17 at 19:04
• @LaurentDuval my question is a my curiosity. I thought it while I was studying. – Gennaro Arguzzi Aug 30 '17 at 7:07

Let us try with another hints:

• could you imagine a bounded input signal which could result in a non bounded output?
• general suggestion whenever analyzing a system: try a few "simple to compute" input signals, looking a the outputs, this could help you guessing some properties: an impulse, a ramp, a step, etc.
• if $x(t)$ is a step function, the system will be not BIBO stable because when t is infinity also $y(t)$ is infinity? – Gennaro Arguzzi Aug 29 '17 at 14:20
• Yep, at least a bounded input yields an unbounded output. A constant $x$ would work too – Laurent Duval Aug 29 '17 at 14:32
• Is the system also invertible because $x(t)=\frac{dy(t)}{dt}$? – Gennaro Arguzzi Aug 29 '17 at 16:01
• Oh, come on, Laurent! Infinity is not a real number, and $y(\infty)=\infty$ when $x(t) = u(t)$ does not prove anything. The OP has little understanding of the concept of BIBO stability (cf. his comment on my answer) and your acquiescence is not helpful at all. – Dilip Sarwate Aug 29 '17 at 18:04
• I was in the same mood as you, providing hints. I believe you chose the sufficient/necessary setting, I played on intuition (after Matt's comment), hoping for question improvement. I disagree with your premise of infinity not being a real number: because of history, either personal (my beginnings in DSP with distributions) and that of mathematics, from [Alexandroff extension]( en.wikipedia.org/wiki/Alexandroff_extension) to non standard analysis. Interesting book by Graham & Kantor : Naming Infinity: A True Story of Religious Mysticism and Mathematical Creativity – Laurent Duval Aug 29 '17 at 19:00

Since this is most likely homework, here is a hint.

Write the integral you have displayed in the form $\int_{-\infty}^\infty x(\tau)h(t-\tau) d\tau$ where you get to choose what the function $h(\cdot)$ is to make it all work out. Then, $h(t)$ is the impulse response of the LTI system, Do you know the criterion for BIBO stability of an LTI system in terms of $h(t)$? Does your $h(t)$ satisfy the criterion?

Edit in response to OP's comment and definition of BIBO stability:

Your understanding of the definition of BIBO stability is faulty.

A system with input $x(t)$ and output $y(t)$ is said to be bounded-input bounded-output (BIBO) stable if it has the property that whenever $x(t)$ is bounded (that is, there exists a (finite) real number $M$ such that $|x(t)| \leq M$ for all real numbers $t$), then the output is also bounded (that is, there exists a (finite) real number $N$ (possibly dependent on $M$) such that $|y(t)| \leq N$ for all real numbers $t$).

Note that while $N$ might depend on $M$, it does not depend on $t$, the same $N$ must be an upper bound on $|y(t)|$ for all $t$.

In your case, with $y(t) = \int_{-\infty}^t x(\tau) d\tau$, what happens if $x(t) = u(t)$ for all real numbers $t$ (step input)? This is a bounded input. Does there exist a finite number $N$ such that $|y(t)| \leq N$ for all real numbers $t$?

• Hi @DilipSarwate. I know only that a system is BIBO stable when $\mid x(t) \mid < m<+\infty \implies \mid y(t) \mid < n<+\infty$. – Gennaro Arguzzi Aug 29 '17 at 14:17