In engineering we are mainly interested in linear time-invariant (LTI) systems which are bounded input, bounded output (BIBO). It's easy to prove that BIBO condition is equivalent to $$\int_{-\infty}^{+\infty}|h(t)|dt < \infty,$$where $h(t)$ is the impulse response. It's often stated that $h(t)$ being absolutely integrable is equivalent to $H(f) = \mathcal{F}\{h\}$ exists where $\mathcal{F}\{.\}$ denotes Fourier transform, $$H(f) = \mathcal{F}\{h\} = \int_{-\infty}^{+\infty}h(t)\exp(-2\pi ift)dt.$$ It can be proved that $$\int_{-\infty}^{+\infty}|h(t)|dt < \infty \ \ \text{converges} \implies H(f) \ \ \text{exists}$$ but examples like $h(t) = \text{sinc}(t)$ and $h(t) = \sin(t^2)$ shows that the other direction doesn't hold in general. Now I'm looking for additional assumptions for $H(f)$ to make the claim true or a reliable source which mentions assumptions. Specifically, I'm interested in the case $H(f) = P(f)/Q(f)$ where $P(f)$ and $Q(f)$ are the arbitrary polynomials, since systems with rational transfer function are widely used in practice.
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2$\begingroup$ There is no reasonable condition on Fourier transform which implies that the original is integrable. Even in the case of Fourier series. $\endgroup$Alexandre Eremenko– Alexandre Eremenko2025-04-24 14:19:39 +00:00Commented Apr 24 at 14:19
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$\begingroup$ @AlexandreEremenko Thanks. What about the case $H(f)$ is rational? $\endgroup$S.H.W– S.H.W2025-04-24 14:35:11 +00:00Commented Apr 24 at 14:35
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$\begingroup$ For a rational function, one can use partial fraction decomposition to compute the inverse Fourier transform explicitly. It is integrable if and only if the rational function has no real poles, and no pole at infinity. In other words when the denominator has no real zeros, and its degree is greater than that of the numerator. $\endgroup$Alexandre Eremenko– Alexandre Eremenko2025-04-25 14:20:38 +00:00Commented Apr 25 at 14:20
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$\begingroup$ @AlexandreEremenko Would you mind writing a proof for this case so I can accept it? $\endgroup$S.H.W– S.H.W2025-04-25 14:30:58 +00:00Commented Apr 25 at 14:30
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2$\begingroup$ Speaking generally, the Fourier transform maps $L^1$ to $C_0$, but it is not onto. We do not have a full classification of what the range of the Fourier transform is, only that it is dense in $C_0$. $\endgroup$Cameron L. Williams– Cameron L. Williams2025-04-25 15:07:51 +00:00Commented Apr 25 at 15:07
1 Answer
For a rational function, one can write the partial fraction decomposition $$r(z)=\sum_{j,k}\frac{a_{k,j}}{(z-c_k)^j}+p(z),$$ where $p$ is a polynomial. Since the Fourier transform of an integrable function must be bounded on the real line and must tend to $0$ on the real line (this is the Riemann-Lebesgue Lemma), we conclude that $p=0$, and that there are no poles on the real line. This gives a necessary condition.
To prove that it is also sufficient use the explicit formula for the inverse Fourier transform of $$1/(z-c)$$ From this formula, we only need the fact that this Fourier transform is exponentially decreasing at $\infty$ when $c$ is not real. Therefore when we differentiate the transform to obtain $1/(z-c)^j$, the original is multiplied on $x$ and remains integrable. So the whole sum will be integrable when none of $c_k$ is real.
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$\begingroup$ Thanks for the explanation. In engineering literature, it's often stated that "An LTI system with a rational transfer function is stable if the region of convergence (ROC) of its transfer function includes the $j\omega$-axis." Assuming the degree of the denominator is greater than that of the numerator, I believe this statement is equivalent to the condition you've proven since the Fourier transform exists if and only if the Laplace transform has no poles on the $j\omega$-axis. Is that correct? $\endgroup$S.H.W– S.H.W2025-04-25 16:34:40 +00:00Commented Apr 25 at 16:34
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$\begingroup$ I am not familiar with engineering terminology. $\endgroup$Alexandre Eremenko– Alexandre Eremenko2025-04-26 13:15:54 +00:00Commented Apr 26 at 13:15