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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.

ThisIsNotAnId
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1 Answers1

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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

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

Ron Gordon
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  • Could you explain how you evaluated the second integral for $\mathcal{F}[\text{sgn}(t)]$? – ThisIsNotAnId May 17 '13 at 05:25
  • Just substitution $t \rightarrow -t$, so that $$\int_{-\infty}^0 dt , e^{-i\omega t}=\int_0^{\infty} dt,e^{i \omega t}$$ – Ron Gordon May 17 '13 at 06:46
  • 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 $i$. – ThisIsNotAnId May 17 '13 at 16:25
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    The integral is defined in the following sense: $$\lim_{\epsilon \to 0} \int_0^{\infty} dt , e^{-\epsilon t} e^{i \omega t}$$ – Ron Gordon May 17 '13 at 16:57
  • I'm not really sure how this stands from a rigorous perspective. To evaluate your latest integral, $$\lim_{\epsilon \to 0} \int_0^{\infty} dt , e^{-\epsilon t} e^{i \omega t} = \lim_{\epsilon \to 0}\left [(i\omega - \epsilon )^{-1} (e^{i\omega t}) \left .\right |0^\infty\right ] = (e^{i\omega t}) \left .\right |_0^\infty \lim{\epsilon \to 0} \left [(i\omega - \epsilon )^{-1} \right ] = (e^{i \omega \infty} - 1) \cdot (i\omega)^{-1}.$$

    How may I simplify the last expression? I'm not sure what expressions of the form "$\pm i\infty$" mean.

     – ThisIsNotAnId May 19 '13 at 00:09
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    @ThisIsNotAnId: that's not right. The integral I posted is simply $$\frac{1}{\epsilon-i \omega}$$ – Ron Gordon May 19 '13 at 00:13
  • I made a mistake. After doing it again, I get $$\lim_{\epsilon \to 0} \int_0^{\infty} dt , e^{-\epsilon t} e^{i \omega t} = \lim_{\epsilon \to 0}\left [-(\epsilon - i\omega)^{-1} (e^{-(\epsilon - i\omega) t}) \left .\right |0^\infty\right ] = (e^{-(\epsilon - i\omega) t}) \left .\right |_0^\infty \lim{\epsilon \to 0} \left [-(\epsilon - i\omega)^{-1} \right ] = (0 - 1) \cdot (i\omega)^{-1} = -(i\omega)^{-1}$$ – ThisIsNotAnId May 19 '13 at 00:20
  • @ThisIsNotAnId: that's more like it. And that's why the above analysis makes sense. – Ron Gordon May 19 '13 at 00:22
  • Ah, I see. The rest works out just the same. I must say, it was very clever of you to define the integral in that way. I just wish they did that in my engineering book too! Haha – ThisIsNotAnId May 19 '13 at 00:22
  • I'm still not sure about how the $\pi \delta (\omega)$ helps. Since at $\omega = 0$, from my OP, $(3) = \pi - i \infty$ (which I'm not even sure how to define); and, for $\omega \not = 0, (3) = (2).$ Since you're an engineer, do you know how the expression $(3)$ is useful over $(2)$ in light of the above? I am really grateful for all your time on this. – ThisIsNotAnId May 19 '13 at 00:27
  • @ThisIsNotAnId: it's not a question of being more useful - that's the answer. We typically do not evaluate delta functions - they end up being used for the sifting property. – Ron Gordon May 19 '13 at 00:36
  • Many thanks for everything. – ThisIsNotAnId May 19 '13 at 04:46
  • @RonGordon Hi Ron, my friend. I hope that you're doing well and staying safe and healthy. I've posted a solution HERE that provide a rigorous methodology to evaluate the Fourier Transform of the unit step function. As always, I'd appreciate your feedback. – Mark Viola Apr 22 '21 at 02:11