# What will the output of a system which has no Fourier transform?

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?

• 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/… Dec 15, 2021 at 9:55

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.