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Diffstat (limited to 'source/know/concept/heaviside-step-function')
| -rw-r--r-- | source/know/concept/heaviside-step-function/index.md | 33 |
1 files changed, 17 insertions, 16 deletions
diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md index 1933037..15d1729 100644 --- a/source/know/concept/heaviside-step-function/index.md +++ b/source/know/concept/heaviside-step-function/index.md @@ -8,9 +8,9 @@ categories: layout: "concept" --- -The **Heaviside step function** $\Theta(t)$, +The **Heaviside step function** $$\Theta(t)$$, is a discontinuous function used for enforcing causality -or for representing a signal switched on at $t = 0$. +or for representing a signal switched on at $$t = 0$$. It is defined as: $$\begin{aligned} @@ -23,11 +23,11 @@ $$\begin{aligned} } \end{aligned}$$ -The value of $\Theta(t \!=\! 0)$ varies between definitions; -common choices are $0$, $1$ and $1/2$. +The value of $$\Theta(t \!=\! 0)$$ varies between definitions; +common choices are $$0$$, $$1$$ and $$1/2$$. In practice, this rarely matters, and some authors even change their definition on the fly for convenience. -For physicists, $\Theta(0) = 1$ is generally best, such that: +For physicists, $$\Theta(0) = 1$$ is generally best, such that: $$\begin{aligned} \boxed{ @@ -35,7 +35,7 @@ $$\begin{aligned} } \end{aligned}$$ -Unsurprisingly, the first-order derivative of $\Theta(t)$ is +Unsurprisingly, the first-order derivative of $$\Theta(t)$$ is the [Dirac delta function](/know/concept/dirac-delta-function/): $$\begin{aligned} @@ -45,10 +45,10 @@ $$\begin{aligned} \end{aligned}$$ The [Fourier transform](/know/concept/fourier-transform/) -of $\Theta(t)$ is as follows, -where $\pv{}$ is the Cauchy principal value, -$A$ and $s$ are constants from the FT's definition, -and $\mathrm{sgn}$ is the signum function: +of $$\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{ @@ -62,16 +62,16 @@ $$\begin{aligned} <label for="proof-fourier">Proof</label> <div class="hidden" markdown="1"> <label for="proof-fourier">Proof.</label> -In this case, it is easiest to use $\Theta(0) = 1/2$, +In this case, it is easiest to use $$\Theta(0) = 1/2$$, such that the Heaviside step function can be expressed -using the signum function $\mathrm{sgn}(t)$: +using the signum function $$\mathrm{sgn}(t)$$: $$\begin{aligned} \Theta(t) = \frac{1}{2} + \frac{\mathrm{sgn}(t)}{2} \end{aligned}$$ We then take the Fourier transform, -where $A$ and $s$ are constants from its definition: +where $$A$$ and $$s$$ are constants from its definition: $$\begin{aligned} \tilde{\Theta}(\omega) @@ -80,7 +80,7 @@ $$\begin{aligned} \end{aligned}$$ The first term is proportional to the Dirac delta function. -The second integral is problematic, so we take the Cauchy principal value $\pv{}$ +The second integral is problematic, so we take the Cauchy principal value $$\pv{}$$ and look up the integral: $$\begin{aligned} @@ -88,10 +88,11 @@ $$\begin{aligned} &= 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}$$ + </div> </div> -The use of $\pv{}$ without an integral is an abuse of notation, +The use of $$\pv{}$$ without an integral is an abuse of notation, and means that this result only makes sense when wrapped in an integral. -Formally, $\pv{\{1 / \omega\}}$ is a [Schwartz distribution](/know/concept/schwartz-distribution/). +Formally, $$\pv{\{1 / \omega\}}$$ is a [Schwartz distribution](/know/concept/schwartz-distribution/). |