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I am an electrical engineer (first post on the Math SE) but I am looking for a mathematical explanation of something that we, as EE, do often.

When dealing with LTI systems, we find the output in the time domain by the use of convolution:

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

Where $h(t)$ is the impulse response of the sytem and $x(t)$ is the input.

If we say that the input is of the form $x(t)=e^{st}$, with $s=\sigma+j\omega$ — a complex exponential — the convolution integral becomes:

$$y(t)=e^{st}\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau =e^{st}H(s)$$

where $H(s)$ is the Laplace transform (bilateral at this point but I've rarely seen this used in EE) of $h(t)$. So far so good.

In electrical engineering, we're usually interested in the response of the system to a non-decaying sinusoidal input, so we make $\sigma=0$ because by using Euler's formula we can express a sinusoidal input as a complex exponential.

Now, here's where my question comes up. Say, for example, that $h(t)=u(t)$, where $u(t)$ is the unit step function. The convolution integral results in:

$$y(t)=e^{st}\int_{-\infty}^{\infty}u(t)e^{-st}d\tau =e^{st}\int_{0}^{\infty}e^{-st}d\tau$$

$$y(t) =e^{st}\int_{0}^{\infty}e^{-(\sigma +j\omega)\tau}d\tau=e^{st}\int_0^\infty e^{-\sigma \tau}e^{-j\omega\tau}d\tau$$

So, we are essentially finding the Fourier transform of $e^{-\sigma t}$ and this only converges for $\sigma>0$ ( i.e. the Laplace transform exists). Therefore the region of convergence of the Laplace transform is for $\sigma>0$.

In EE if we wish to find the so called frequency response (by drawing a bode plot), we make the substitution $s=j\omega$ in the Laplace transform and plot the the frequency response. The problem here is that, the Laplace transform of $h(t)=u(t)$ is $H(s)=\dfrac{1}{s}$ but the region of convergence for the Laplace Transform is $\sigma>0$.

So for a purely sinusoidal input, we shouldn't set $\sigma=0$ because it isn't in the ROC and that is a necessary condition for the Laplace Transform to exist. But in EE, we still find the frequency response of $H(s)=\dfrac{1}{s}$ (an integrator) by evaluating $H(\omega)=\dfrac{1}{j\omega}$ ( setting $\sigma=0$), after that we draw the Bode plot.

Have we broken any rules here? It has been my understanding that if the ROC does not include the $j\omega$ axis, then we can't find the Fourier transform by just making $s=j\omega$ in the Laplace transform transfer function (which would make sense since the Fourier transform does not exist unless we used some generalized function, e.g Dirac delta function)

Appreciate any help!

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    $\begingroup$ Random thought: the region of converges does include the $j\omega$ axis (as a limit from the plane with $\sigma>0$), provided the response is causal, i.e., $h(\tau <0 ) =0$. This is the way the Fourier transform is "saved" in physics and EE. $\endgroup$ – Fabian Apr 9 '18 at 17:29

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