I'm trying to calculate the Fourier transform of the unit step function,

$$\mathcal{F}[u(t)] \ = \int_{-\infty}^{\infty}u(t)e^{-i\omega t}dt \ = \int_{0}^{\infty}e^{-i\omega t} dt. \tag{1}$$

This simplifies to,

$$U(\omega) = (i\omega)^{-1},\ (\omega \not = 0). \tag{2}$$

However, my book claims that $(1)$ simplifies to $ \pi \delta(\omega) + (i\omega)^{-1}. \tag{3}$

Here, $\delta(\omega)$ is the unit impulse function. I don't have my book with me right now but I think they use the differentiation property to derive it by calculating the transform of the derivative of $u$ (which is $\delta$).

My question is, isn't the appearance of $\delta$ in the result they obtain irrelevant? Since, at $\omega = 0$, $(3) = \infty$ and elsewhere $(3) = (2).$ So, why would they write $(3)$ instead of $(2)$? I should mention this is not in a mathematics textbook, but an engineering textbook.

  • $\begingroup$ How do you get (2) from (1)? $\endgroup$ – Artem May 15 '13 at 23:07
  • $\begingroup$ @Artem: consider $$\lim_{\epsilon \to 0^+}\int_0^{\infty} dt\, e^{-\epsilon t} \, e^{-i \omega t}$$ $\endgroup$ – Ron Gordon May 15 '13 at 23:50

The reason why you get a delta function is because the step function is actually defined as

$$\theta(t) = \begin{cases} \\ 1 & t \gt 0\\1/2 & t=0\\0 & t \lt 0\end{cases}$$

That nonzero value at $t=0$ is a bit troublesome. Better to consider the signum funciton $\text{sgn}(t) = 2 \theta(t)-1$. The FT of $\text{sgn}(t)$ is

$$\int_0^{\infty} dt \, e^{-i \omega t} - \int_0^{\infty} dt \, e^{i \omega t} = \frac{2}{i \omega}$$

Note that the value at $t=0$, being zero, does not contribute to the FT. The FT of $\theta(t)$ follows from this, because

$$\theta(t) = \frac12 \text{sgn}(t) + \frac12$$

so that its FT is

$$\frac{1}{i \omega} + \frac12 \int_{-\infty}^{\infty} dt \, e^{-i \omega t} = \frac{1}{i \omega} + \pi \delta(\omega)$$

  • $\begingroup$ Could you explain how you evaluated the second integral for $\mathcal{F}[\text{sgn}(t)]$? $\endgroup$ – ThisIsNotAnId May 17 '13 at 5:25
  • $\begingroup$ Just substitution $t \rightarrow -t$, so that $$\int_{-\infty}^0 dt \, e^{-i\omega t}=\int_0^{\infty} dt\,e^{i \omega t}$$ $\endgroup$ – Ron Gordon May 17 '13 at 6:46
  • $\begingroup$ Yes, then $$\int_0^\infty dt e^{i\omega t} = \left .(i\omega)^{-1} e^{i\omega t} \right |_0^\infty = \infty - (i\omega)^{-1}, (\omega \not = 0).$$ Is that correct? I'm not sure what $\infty - (i\omega)^{-1}$ evaluates to, or even if I should get that since since $e^{i\omega \infty}$ seems to be undefined because of the $i$. $\endgroup$ – ThisIsNotAnId May 17 '13 at 16:25
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    $\begingroup$ The integral is defined in the following sense: $$\lim_{\epsilon \to 0} \int_0^{\infty} dt \, e^{-\epsilon t} e^{i \omega t}$$ $\endgroup$ – Ron Gordon May 17 '13 at 16:57
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    $\begingroup$ @ThisIsNotAnId: that's not right. The integral I posted is simply $$\frac{1}{\epsilon-i \omega}$$ $\endgroup$ – Ron Gordon May 19 '13 at 0:13

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