What is the Fourier transformation of H(-t)?

Question

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?

Answer 1

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.

Answer 2

EDIT: ALTERNATIVE AND RIGOROUS DEVELOPMENT

In THIS ANSWER, I evaluated the Fourier Transform of the Heaviside (i.e., unit step) function using a rigorous distributional approach.

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

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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 $\frac{1}{2\pi}\int_{-\infty}^\infty(2\pi \delta(\omega))\,e^{-i\omega t}\,d\omega=1$.

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STEP $2$:

The Fourier Transform of the signum function can be evaluated as

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

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STEP $3$:

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

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

And we are done!

When applying $\mathscr{F}\{H(-t)\}(\omega)$ to a test function $\phi(\omega)$, the integral of the term $\frac{i}{\omega}$ must interpreted as a Cauchy Principal Value.

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