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For a signal processing analysis, I need a function that behaves like the image below, but I cannot figure out which formula replicates this dynamics.

enter image description here

Features:

  • $y(x) = xk$ for some constant $k$ at each point $x\in\mathbb{Z}^{+}$

  • at each $x\in\mathbb{Z}^{+}$, $\frac{dy}{dt} = 0$

  • Given any range $[x, x+1]$ for $x\in\mathbb{Z}^{+}$, we should have $\frac{d^{2}y}{dt^{2}}>0$ within $[x, x+0.5]$ and $\frac{d^{2}y}{dt^{2}}<0$ in $[x+0.5, x+1]$.

Lemmon
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    Does $$y = k\left(x - \frac{\sin{(\pi x)} \cos{(\pi x)}}{\pi}\right)$$ work? – Lemmon Jan 05 '23 at 03:23
  • @Lemmon: That's exactly what I wanted! If you supply an answer, I will accept it. (Can you explain what intuition did lead you to find out this complicated function does the job?) –  Jan 05 '23 at 03:29
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    Lemmon's function can be further recast as $$ y = c f(2\pi x), \qquad\text{where}\qquad f(x) = x - \sin x. $$ The constant $c$ is related to $k$ in Lemmon's comment via $c=k/2\pi$. – Sangchul Lee Jan 05 '23 at 03:42

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We want a function $y(x)$ such that $\frac{\mathrm{d}y}{\mathrm{d}x}$ is everywhere-positive and periodic with period $1$, and $\frac{\mathrm{d}^2y}{{\mathrm{d}x}^2}$ is positive on $[n, n + \frac 1 2]$ and negative on $[n+\frac 1 2, n + 1]$.

We can define the derivative of our function to be $y'(x) := A \sin^2{(\pi x)}$ where $A$ is an arbitrary scaling constant. This fulfils our requirements on the first and second derivatives.

Integrating, we get $y(x) = \int{y'(x)\, \mathrm{d}x} = \int{A \sin^2{(\pi x)}\, \mathrm{d}x} = \frac A 2 \left(x - \frac{\sin{(\pi x)} \cos{(\pi x)}}{\pi}\right)$.

We can then scale this function to fulfil the requirement that $y(n) = kn$ for $n \in \mathbb{Z}^+$, by setting $A := 2k$, giving:

$$y = k\left(x - \frac{\sin{(\pi x)} \cos{(\pi x)}}{\pi}\right) = k\left(x- \frac{\sin{(2 \pi x)}}{2 \pi}\right)$$

Lemmon
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