# Is Ideal LPF BIBO unstable?

In one of other discussions : How to find frequency response, stability, and causality of a linear system?

I found a comment which was quite strong and definitely caught my attention.

An ideal low-pass filter is an example of a system that is not BIBO stable even though its frequency response is bounded for all $f$

I am following the definition of stability as per here in wiki http://en.wikipedia.org/wiki/BIBO_stability

Can anyone give me a proof that ideal LPF can indeed be BIBO unstable?

Of course, ideal LPF with infinite gain can produce unbounded output. The question is restricted to LPF when gain is finite.

• An ideal LPF has impulse response of the form $h(t) =\text{sinc}(t)$ which does not satisfy the condition $\int_{-\infty}^{\infty}|h(t)|dt < \infty$ needed for BIBO stability. Thus, the response at $t=0$ to the bounded signal $x(t) = \text{sgn}(\text{sinc}(t))$ (which switches back and forth between $+1$ and $-1$) is $$\int h(-t)x(t)dt = \int h(t)x(t)dt = \int |h(t)|dt = \infty$$ and so an ideal LPF is not a BIBO-stable system. – Dilip Sarwate Dec 26 '11 at 16:38

This answer is a response to a comment by the OP on on yoda's answer.

Suppose that $h(t)$, the impulse response of a continuous-time linear time-invariant system, has the property that $$\int_{-\infty}^{\infty} |h(t)| \mathrm dt = M$$ for some finite number $M$. Then, for each and every bounded input $x(t)$, the output $y(t)$ is bounded also. If $|x(t)| \leq \hat{M}$ for all $t$ where $\hat{M}$ is some finite number, then $|y(t)| \leq \hat{M}M$ for all $t$ where $\hat{M}M$ is also a finite number. The proof is straightforward. \begin{align*} |y(t)| &= \left |\int_{-\infty}^\infty h(\tau)x(t - \tau)\mathrm d\tau\right |\\ &\leq \int_{-\infty}^\infty |h(\tau)x(t - \tau)|\mathrm d\tau\\ &\leq \int_{-\infty}^\infty |h(\tau)|\cdot|x(t - \tau)|\mathrm d\tau\\ &\leq \hat{M}\int_{-\infty}^\infty |h(\tau)|\mathrm d\tau\\ &= \hat{M}M. \end{align*} In other words, $y(t)$ is bounded whenever $x(t)$ is bounded.

Thus, the condition $\displaystyle\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$ is sufficient for BIBO-stability.

The condition $\displaystyle\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$ is also necessary for BIBO-stability.

Assume that every bounded input produces a bounded output. Now consider the input $x(t) = \text{sgn}(h(-t)) ~\forall~ t$. This is clearly bounded, ($|x(t)| \leq 1$ for all $t$), and at $t=0$, it produces output \begin{align*} y(0) &= \int_{-\infty}^\infty h(0-\tau)x(-\tau)\mathrm d\tau\\ &= \int_{-\infty}^\infty h(-\tau)\text{sgn}(h(-\tau))\mathrm d\tau &= \int_{-\infty}^\infty |h(-\tau)|\mathrm d\tau\\ &= \int_{-\infty}^\infty |h(t)|\mathrm dt. \end{align*} Our assumption that the system is BIBO stable means that $y(0)$ is necessarily finite, that is, $$\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$$

The proof for discrete-time systems is similar with the obvious change that all the integrals are replaced by sums.

Ideal LPFs are not BIBO-stable systems because the impulse response is not absolutely integrable, as stated in the answer by yoda. But his answer does not really answer the question

Can anyone give me a proof that ideal LPF can indeed be BIBO unstable?

A specific example of a bounded input signal that produces an unbounded output from an ideal LPF (and thus proves that the system is not BIBO-stable) can be constructed as outlined above (see also my comment on the main question).

A necessary condition for BIBO stability is the existence of the $L^1$ norm (or $\ell^1$ norm for discrete systems) of the impulse response. From the wiki article you cited,

For a continuous time linear time invariant (LTI) system, the condition for BIBO stability is that the impulse response be absolutely integrable, i.e., its L1 norm exist.

$$\int_{-\infty}^\infty |h(t)|\ dt=\Vert h(t)\Vert_1<\infty$$

The impulse response of an ideal LPF is the $\text{sinc}$ function, which only has the $L^2$ norm and not the $L^1$ norm. In other words, $\text{sinc}(t)$ is not absolutely summable or

$$\int_{-\infty}^\infty |\text{sinc}(t)|\ dt=\infty$$

Hence, an ideal LPF is not BIBO stable despite its frequency response being bounded for all $f$.

• From what i thought impulse response be absolutely summable, i.e., its L1 norm exist. is a sufficient condition that a system is BIBO stable. However, is this a necessary condition that must hold? – Dipan Mehta Dec 26 '11 at 16:53

The fourier transform of ideal lpf is sinc function in time domain which exist from -infinite to +infinite so it is noncausal and area within it is infinite so unbounded. ..

• Welcome to DSP.SE! Thanks for your answer, but I do not think that it adds anything to the existing answers. Furthermore, it is not true that the area under the sinc function is unbounded, it is the area under the magnitude of the sinc function that is unbounded. The latter causes the system's instability. – Matt L. Jan 5 '15 at 8:11