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Let's assume a system $h(t)= e^{j2t}$. This system has no region of convergence. What will be the output if I provide any input to this system?

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  • $\begingroup$ ROC is a term used in Laplace transform and Z transform, not for Fourier transform. If a signal is absolutely integrable, then its Fourier transform exists. dsp.stackexchange.com/questions/53875/… $\endgroup$
    – ZR Han
    Dec 15, 2021 at 9:55

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It does have a Fourier transform. The Fourier transform of $h(t)=e^{j2t}$ is given by

$$H(j\omega)=2\pi\delta(\omega-2)\tag{1}$$

You are right that you can't treat that system with the Laplace transform.

A system with frequency response $(1)$ is not bounded-input-bounded-output (BIBO) stable, i.e., there are bounded input signal for which the output is unbounded.

You can compute the output for a given input using the convolution integral:

$$\begin{align}y(t)&=\int_{-\infty}^{\infty}x(\tau)e^{2j(t-\tau)}d\tau\\&=e^{2jt}\int_{-\infty}^{\infty}x(\tau)e^{-2j\tau}d\tau\\&=e^{2jt}X(j2)\tag{2}\end{align}$$

where $X(j\omega)$ is the Fourier transform of the input signal $x(t)$, assuming that the integral in $(2)$ exists.

From $(2)$ we see that all input signals with a finite Fourier transform at $\omega=2$ will result in an output signal that is just a scaled version of the system's impulse response.

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