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author | Prefetch | 2022-10-14 23:25:28 +0200 |
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committer | Prefetch | 2022-10-14 23:25:28 +0200 |
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diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md new file mode 100644 index 0000000..30b5f5d --- /dev/null +++ b/source/know/concept/heaviside-step-function/index.md @@ -0,0 +1,96 @@ +--- +title: "Heaviside step function" +date: 2021-02-25 +categories: +- Mathematics +- Physics +layout: "concept" +--- + +The **Heaviside step function** $\Theta(t)$, +is a discontinuous function used for enforcing causality +or for representing a signal switched on at $t = 0$. +It is defined as: + +$$\begin{aligned} + \boxed{ + \Theta(t) = + \begin{cases} + 0 & \mathrm{if}\: t < 0 \\ + 1 & \mathrm{if}\: t > 1 + \end{cases} + } +\end{aligned}$$ + +The value of $\Theta(t \!=\! 0)$ varies between definitions; +common choices are $0$, $1$ and $1/2$. +In practice, this rarely matters, and some authors even +change their definition on the fly for convenience. +For physicists, $\Theta(0) = 1$ is generally best, such that: + +$$\begin{aligned} + \boxed{ + \forall n \in \mathbb{R}: \Theta^n(t) = \Theta(t) + } +\end{aligned}$$ + +Unsurprisingly, the first-order derivative of $\Theta(t)$ is +the [Dirac delta function](/know/concept/dirac-delta-function/): + +$$\begin{aligned} + \boxed{ + \Theta'(t) = \delta(t) + } +\end{aligned}$$ + +The [Fourier transform](/know/concept/fourier-transform/) +of $\Theta(t)$ is as follows, +where $\pv{}$ is the Cauchy principal value, +$A$ and $s$ are constants from the FT's definition, +and $\mathrm{sgn}$ is the signum function: + +$$\begin{aligned} + \boxed{ + \tilde{\Theta}(\omega) + = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-fourier"/> +<label for="proof-fourier">Proof</label> +<div class="hidden"> +<label for="proof-fourier">Proof.</label> +In this case, it is easiest to use $\Theta(0) = 1/2$, +such that the Heaviside step function can be expressed +using the signum function $\mathrm{sgn}(t)$: + +$$\begin{aligned} + \Theta(t) = \frac{1}{2} + \frac{\mathrm{sgn}(t)}{2} +\end{aligned}$$ + +We then take the Fourier transform, +where $A$ and $s$ are constants from its definition: + +$$\begin{aligned} + \tilde{\Theta}(\omega) + = \hat{\mathcal{F}}\{\Theta(t)\} + = \frac{A}{2} \Big( \int_{-\infty}^\infty \exp(i s \omega t) \dd{t} + \int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t} \Big) +\end{aligned}$$ + +The first term is proportional to the Dirac delta function. +The second integral is problematic, so we take the Cauchy principal value $\pv{}$ +and look up the integral: + +$$\begin{aligned} + \tilde{\Theta}(\omega) + &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}} + = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}} +\end{aligned}$$ +</div> +</div> + +The use of $\pv{}$ without an integral is an abuse of notation, +and means that this result only makes sense when wrapped in an integral. +Formally, $\pv{\{1 / \omega\}}$ is a [Schwartz distribution](/know/concept/schwartz-distribution/). + |