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Today I tried to get Fourier transform of step function ($u(t)$). But I got a result which seems is not correct. I want to know what is incorrect in my work?

With attention to this relation: \begin{align} \mathrm{u(t)=\frac{1}{2}[1+sgn(t)]} \end{align} we know that: \begin{align} \mathfrak{F}(u(t))=\left (\pi\delta(\omega) + \frac{1}{j\omega}\right ) \end{align}

Today I tried to experience another way. Suppose:

$$ \begin{align} g(t) = \begin{cases} e^{-at} & \text{for } t > 0\\ 0 & \text{otherwise } \end{cases} \end{align} $$

So:

$$ \begin{align*} \mathfrak{F}(g(t)) = &\int_{0}^{\infty} e^{-at} e^{-i\omega t}\, dt\\ &=\int_{0}^{\infty}e^{-(a+i\omega) t}\, dt\\ &=\big[ \frac{e^{-(a+i\omega) t}}{-(a+i\omega)} \big]_{0}^{\infty}\\ &=\frac{1}{a+i\omega} \end{align*} $$

It is obvious that $\lim\limits_{a\to 0^+}g(t)= u(t)$. Thus:

$$ \begin{align} \mathfrak{F}(u(t)) &= \mathfrak{F}(\lim\limits_{a\to 0^+}g(t))\\ &=\lim\limits_{a\to 0^+}(\mathfrak{F}(g(t))\\ &=\lim\limits_{a\to 0^+}(\frac{1}{a+i\omega})\\ &=\frac{1}{i\omega} \end{align} $$ What is incorrect in my work?

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  • $\begingroup$ The integral of the limit is not always the limit of the integral. $\endgroup$ Apr 12, 2023 at 22:15
  • $\begingroup$ @ThomasAndrews Well, but when it is correct? I saw that in getting Fourier transform of sgn function. $\endgroup$ Apr 12, 2023 at 22:18
  • $\begingroup$ You need to prove it is correct to switch the limit and integral. Here, of course, the only place where it is incorrect is when $\omega=0.$ $\endgroup$ Apr 12, 2023 at 22:32
  • $\begingroup$ But the real problem is that $1/(a+wi)$ does not converge to anything when $\omega=0.$ $\endgroup$ Apr 12, 2023 at 22:34
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    $\begingroup$ To add to the comment posted by @Gonçalo , the Fourier transform is a distribution on the Schwartz space $\mathbb{S}$ of functions. The limit $\lim_{a\to 0^+}\left(\frac1{a+i\omega}\right)$ is a distributional limit, such that for any $\phi\in \mathbb{S}$, we have $$\lim_{a\to 0^+} \int_{-\infty}^\infty \frac{\phi(\omega)}{a+i\omega}\,d\omega=\pi \phi(0)+\text{PV}\int_{-\infty}^\infty \frac{\phi(\omega)}{i\omega}\,d\omega$$ $\endgroup$
    – Mark Viola
    Apr 13, 2023 at 13:23

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