I want to start by the definition of Fourier transformation. It is $$\mathcal{F}[H(-t)]=\int_{-\infty}^0 e^{-j{\omega}t}\,dt =\left.\frac{1}{-j\omega}e^{-j{\omega}t}\right|_{-\infty}^0 $$ But when $t \rightarrow{-\infty}$, the result goes to the infinity, right?

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    $\begingroup$ No, it's undefined, not $\infty$. Similarly, the limit $\lim_{t \to \infty} e^{-i \omega t}$ is also undefined. The Fourier transform of the step function requires some sophistication to establish, and simply taking the limit in the Riemann integral does not work. See this, for instance, and also look up some answers on MSE, starting perhaps with this and... $\endgroup$ – stochasticboy321 Dec 28 '16 at 4:17
  • $\begingroup$ (contd.) this. Once you grok the idea for $H(t)$, $H(-t)$ should be easy. $\endgroup$ – stochasticboy321 Dec 28 '16 at 4:18

Finding the Fourier Transform of the unit step function, $H(-t)$, is as easy as $1,2,3$.

STEP $1$:

The Fourier Transform of $f(t)=1$ is$\int_{-\infty}^\infty (1)e^{i\omega t}\,dt =2\pi \delta(\omega)$, since the inverse Fourier Transform of $2\pi \delta(\omega)$ is $\frac{1}{2\pi}\int_{-\infty}^\infty(2\pi \delta(\omega))\,e^{-i\omega t}\,d\omega=1$.

STEP $2$:

The Fourier Transform of the signum function can be evaluated as

$$\begin{align} \mathscr{F}\{\text{sgn}(t)\}(\omega)&=\lim_{a\to 0}\int_{-\infty}^\infty \text{sgn}(t)e^{-a|t|}e^{i\omega t}\,dt\\\\ &=\lim_{a\to 0}\left(\frac{i2\omega}{\omega^2+a^2}\right)\\\\ &=\frac{i2}{\omega} \end{align}$$

STEP $3$:

We note the $H(-t)=\frac12-\frac12\text{sgn}(t)$ and hence

$$\mathscr{F}\{\text{sgn}(t)\}(\omega)=\pi \delta(\omega)-\frac{i}{\omega}$$

And we are done!

  • $\begingroup$ Actually I want to know the Fourier transform of H(-t) instead of H(t). $\endgroup$ – Alex Thomson Dec 28 '16 at 4:57
  • $\begingroup$ @AlexThomson That should be a snap now. Simply follow the $1,2,3$ recipe and replace $H(t)$ with $H(-t)=\frac12(1-\text{sgn}(t))$. $\endgroup$ – Mark Viola Dec 28 '16 at 4:58
  • $\begingroup$ You are allowed to mention "the Fourier transform of distributions" $\endgroup$ – reuns Dec 28 '16 at 5:00
  • $\begingroup$ @user1952009 yes, of course. But then we could have simply written the answer without showing an heuristic way forward. $\endgroup$ – Mark Viola Dec 28 '16 at 5:01
  • $\begingroup$ I see, I understand how it is proven now. Thank you for your thorough explain. $\endgroup$ – Alex Thomson Dec 28 '16 at 5:08

As stochasticboy321 mentioned, taking the "Fourier transform" of $H(t)$ requires some slightly more sophisticated machinery. Details are given in this post: Heaviside step function fourier transform and principal values.

The idea is that one considers generalized functions which are not defined by pointwise values, but by their "action" on rapidly decreasing functions (functions in the Schwartz class). One defines genuine functions to act by integration and then one defines derivatives and Fourier transforms of generalized functions by abstracting what occurs in the genuine function case. In this way you can compute a Fourier transform of $H(t)$.

To compute the Fourier transform then of $H(-t)$, just use the time reversal property of Fourier transforms.

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    $\begingroup$ Yes, this method works quickly and explicitly. $\endgroup$ – Alex Thomson Dec 28 '16 at 5:10

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