# Effect of BIBO-Instability on the frequency response of a ideal LPF

I recently came across this post stating that the continuous ideal LPF is BIBO-unstable since the impulse response is not absolutely integrable, and this post stating some examples.

I have been trying to understand this using the frequency response. For example, take a system with bounded input $$x(t) = \sin(2\pi B t)u(t)$$ where $$u(t)$$ indicates the Heaviside unit step function and the impulse response of the ideal LPF $$h(t) = \frac{\sin(2\pi B t)}{2\pi B t}$$. If we find the convolution of these two functions $$y(t) = x(t) * h(t)$$ for $$t = 0$$, we would find that the integral diverges: the output is unbounded due to $$h(t)$$ not being absolutely summable. If we would use frequency-domain multiplication instead of time-domain convolution for this system, would we find the same result? And maybe more importantly: does the convolution property of the Fourier Transform always hold if the Fourier Transforms of the two signals exist?

Thanks in advance for any help, this has been bugging me for the last couple of days. I feel that there may be something larger to this problem which I don't quite grasp yet.

• "If we would use frequency-domain multiplication instead of time-domain convolution for this system, would we find the same result?" Yes, because if it works in one domain it works in the other. What is keeping you from trying it out? Jun 5, 2023 at 4:00

Note that BIBO-unstable systems can have well-defined output signals for a large class of input signals, i.e., there are many cases for which the convolution integral doesn't diverge. The example in your question is problematic because the frequency of the sinusoid at the input is exactly at the band edge of the ideal lowpass. That's why the result is ill-defined anyway, no matter if you're trying to solve it in the time domain or in the frequency domain. If instead you had a sinusoidal input signal $$x(t)=\sin(\omega_0t)$$ with $$\omega_0$$ not equal to the ideal lowpass filter's band edge, then the convolution integral would converge, and you would obtain the same result by multiplying the corresponding Fourier transforms.