I am looking for a smooth staircase equation $f(h,w,x)$ that is a function of the step height $h$, step width $w$ in the range $x$.
I cannot use the unit step or other similar functions since they are just one step. I have been experimenting with various sigmoid curves and while I can get a single smooth step I cannot get to realize the staircase shape. The closest staircase function I have found is given in this paper in equation (18) and depicted in Fig. 4 and it is a close example of what I want (i.e generate a staircase in the range $x$ for arbitrary step heights and widths) but it is not smooth at all.
Regarding smooth steps, a likely starting point I found is here but it gives a smooth function of just a single step. I have been unable to modify the equation to make it into a staircase. I would like to specify arbitrary step heights and widths and generate a smooth staircase in the range $x$ specified.
Edit (Extra info): The smooth function I mention above has the problem that the upper, horizontal line is not equal in length to the lower, horizontal line which is why I have been unable to adapt it into a staircase function
Edit 2
Including some pictures
Edit 2
Plot of $s$ with a steep slope showing a different width on the first horizontal line
6 Answers
$\begingroup$We can start with a simple soft staircase function:
and then feed it into itself:
$$ y(x) = f(f(x)) $$
then again:
$$ y(x) = f(f(f(x))) $$
and again:
$$ y(x) = f^4(x) $$
As you can see, each iteration makes the "flat" part of the step longer, and the rise steeper.
The period and the height of each step is $ 2 \pi $, so multiply $x$ by $2 \pi / w$ and $y$ by $h / 2 \pi$ to reach your desired scale.
In reality, the curve is only truly flat (zero derivative) at the centre of each step — at every $ 2 \pi k $ — and only close to flat on either side of that point.
Configurability is limited: The softness of the step can only be specified in integer amounts (the number of times we reapply $f$ to itself), and it requires many/infinite applications to make the step really sharp.
$\endgroup$ 1 $\begingroup$Here is an example based on Math536's answer: Wolfram link
$$f(h,w,a,x) = h \left[\frac{\tanh \left( \frac{ax}{w}-a\left\lfloor \frac{x}{w} \right\rfloor-\frac{a}{2}\right)}{2\tanh\left(\frac{a}{2}\right) } + \frac{1}{2} + \left\lfloor \frac{x}{w} \right\rfloor\right]$$
Where h is the step height, w is the period, and a is the smoothness
$\endgroup$ $\begingroup$Let $s : [0,1] \to [0,1]$ be a smooth function representing a single step. Assume that there exists some $\epsilon > 0$ such that $s(x) = 0$ for all $x < \epsilon$ and $s(x) = 1$ for all $x > 1 - \epsilon$. Setting $$ f(x) = s(x - \lfloor x \rfloor) + \lfloor x \rfloor$$ then gives us a smooth staircase with steps of height and width $1$. By rescaling $f$, we can get steps of arbitrary width $w$ and height $h$: $$f(h,w,x) = h f(x/w) = h(s(x/w - \lfloor x/w \rfloor) + \lfloor x/w \rfloor).$$
$\endgroup$ 9 $\begingroup$I used the following function, where the width of the flat part of each step is b and the width of the curved part is c, and each step is 1 unit tall:
$$ f(x) = \lfloor \frac{x}{b+c} \rfloor + \begin{cases} 0, & \text{if } x-(b+c) \lfloor \frac{x}{b+c} \rfloor, \\ s(\frac{x - (b+c) \lfloor \frac{x}{b+c} \rfloor - b}{c}) , & \text{otherwise}. \end{cases} $$
I used $\cos$ for my stepping function:
$$s(u) = \frac{1}{2} - \frac{1}{2} \cos (\large{\pi} \small{u})$$
It might look a bit simpler in Javascript code, using some temporary variables and modulus (%):
function stair (x, b, c) { const width = b + c; const base = Math.floor(x / width); // base of this step const o = x % width; // offset, between 0 and width return base + (o < b ? 0 : step((o - b) / c));
}
function step (u) { return 0.5 - 0.5 * Math.cos(Math.PI * u);
}although Javascript's % restricts us: x must be non-negative, and b and c must be integers .
Those restrictions can be lifted if we change how o is calculated:
const o = x - width * base;In Gnuplot you can experiment using:
gnuplot> f=4; c=1; plot [x=0:20] floor(x/(f+c)) + ((floor(x)%(f+c))<f ? 0 : 0.5-0.5*cos(((x - (f+c)*floor(x/(f+c))) - f) * pi/c))You can create a smooth staircase function $f(h, w, x, t)$, which starts of as a straight line at time $t=0$, and gets progressively more pronounced steps as times passes, by solving the ordinary differential equation
$$\frac{\operatorname{d}f(h, w, x, t)}{\operatorname{d}t} = -\sin\left(2\pi \frac{f(h, w, x, t)}{h}\right)h$$
with the boundary condition
$$f(h, w, x, 0) = \frac{h\,x}{w}.$$
The solution to this ODE is
$$ f(h, w, x, t) = \begin{cases}%{ll} f(h, w, x, 0), && \text{if } \displaystyle\frac{x}{w} + 0.5 \in \mathbb{Z}\\ \displaystyle h\left(\frac{\displaystyle \tan^{-1}\left[s\,\tan\left(\frac{\pi\,x}{w}\right)\right]}{\pi} + \left\lfloor \frac{x}{w} + \frac{1}{2} \right\rfloor\right), && \text{otherwise} \end{cases}, $$
where $s=e^{-t}$, which can be interpreted as a smoothness parameter, so we can make $f$ a function of $s$ instead of a function of $t$, i.e., $f(h, w, x, s)$. $s=0$ corresponds to $t=\infty$ and means no smoothness at all (perfectly sharp steps) and $s=1$ corresponds to $t=0$ and means maximum smoothness (a straight line). Values of $s$ greater than one corresponds to negative values of $t$, and will create a basically the same type of staircase but shifted half a step. (The first case in the bracket has been added because the second case is undefined in the middle of a step since $\tan$ becomes singular at those points.)
Plots from Wolfram Alpha:
Stair cases of Height=H and Width =W may be represented by:
$$S\left(x\right)=H\left(\frac{x}{W}+\frac{1}{2}-\frac{\arctan\left(\tan\left(\pi\left(\frac{x}{W}+\frac{1}{2}\right)\right)\right)}{\pi}\right)$$
Here you can observe
$\endgroup$ 2
