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I was trying to figure out a way to represent frequency modulation (FM) with sinusoids, and thus graphed the function $f(x) = \sin(\sin(x) \cdot x)$. Needless to say, the graph was not what I wanted. I couldn't understand why the graph is shaped as such. Since $\sin(x)$ oscillates between $-1$ and $1$, I would've thought that the wavelength of $f(x)$ would not go below $2\pi$. Is there something I'm missing? I could always confirm it mathematically using derivatives and such, but I'm looking for intuition.

P.S. I would also appreciate it if anyone could also share and explain the equation representing FM with sinusoids, but I'll probably ask another question just for that.

The graph of f(x)

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    $\begingroup$ because your function oscillates around zero frequency, while most FM satisfy that the frequency is modulated around a much larger carrier frequency. In order to get that you would need to add a bias frequency $\omega_b$ much larger than one to your shape: $\sin((\sin(x) + \omega_b) x)$ $\endgroup$
    – lurscher
    Jun 28, 2020 at 5:27

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The zeroes of $\sin(x\sin x)$ occur at the solutions of $x\sin(x)=\pi k$:

enter image description here

this gives that $\sin(x\sin(x))$ is not periodic (the density of zeroes increases as we move away from the origin) and explains its non-negativity over a large neighbourhood of the origin.

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  • $\begingroup$ Thanks for the answer! Just to clarify, the graph looks that way due to the higher number of solutions as we move away from the origin. However, there is one anomaly that I have yet to understand. The frequent "plateaus" I assume is when $x \sin(x)$ peaks and the gradient reaches zero. However, there is a little divot in the middle of the plateau, which I do not understand. It corresponds with the x-coordinate of the peaks. If someone could explain it, I would be ever grateful, thanks! $\endgroup$
    – chematwork
    Jun 29, 2020 at 1:09
  • $\begingroup$ The divot is especially obvious at x=2.029. The divot gets less obvious as the function furthers from the origin. I assume it's because the peaks mark a small change in gradient. $\endgroup$
    – chematwork
    Jun 29, 2020 at 1:16
  • $\begingroup$ @chematwork: sometimes two consecutive solutions of $x\sin x\in\pi \mathbb{Z}$ are associated to the same multiple of $\pi$: in the picture, from left to right, consider the lower yellow, upper green, lower green, upper purple, lower blue, upper blue, lower purple lines. Approximately in the middle of these intervals $x\sin x$ has a stationary point, which translates into a small bulge in the graph of $\sin(x\sin x)$. $\endgroup$ Jun 29, 2020 at 1:16
  • $\begingroup$ Why is there a bulge, though? $\endgroup$
    – chematwork
    Jun 29, 2020 at 1:24
  • $\begingroup$ @chematwork: if $x_0$ is a stationary point of $x\sin x$ it is also a stationary point of $\sin(x \sin x)$, but there is a conflict of concavities. If we assume that $a,b$ are consecutive zeroes associated to the same value of $x\sin x$, and $\sin(x\sin x)$ is positive over $(a,b)$, then $\sin(x\sin x)$ is mostly concave on $(a,b)$, but it is convex in a small neighourhood of the stationary point of $x\sin x$. $\endgroup$ Jun 29, 2020 at 1:32
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Answering here so I will understand later on when I revisit this topic:

Imagine $f(x)=x\sin(x)$ running through the values from $-x$ to $x$ in one period (as $\sin(x)$ traverses through $-1$ and $1$, but here $x$ also acts as a variable amplitude). It will traverse many points where $x\sin(x)=k\pi$. As $x$ increases down the $x$-axis, the "faster" $f(x)$ traverses the $x$-values as it takes the same amount of time to run through all the values of $x$ (sharper gradient).

When plugging $f(x)$ into another sine function, i.e. $\sin(x\sin x)$, it could be thought as traversing through multiple wavelengths faster and faster as $x$ increases. This can be proven knowing that every time $f(x) = k\pi$, $\sin(f(x)) = 0$. As $x$ increases, the number of points where $f(x) = k\pi$ increases as well, so the frequency of $\sin(x\sin x)$ must increase with $x$.

To model FM, we need to instead set a variable that will represent the change in frequency and vary it in accordance to the signal wave. This takes away the modulated frequency's direct dependence on the $x$-values and solves the problem.

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  • $\begingroup$ The weird shape of the graph is due to the fact the frequency is changing. If we input a constant frequency and vary the constant through a range of values, the same effect does not appear (there is no "going through" of values, so there isn't the sudden rise and drop of frequency) $\endgroup$
    – chematwork
    Feb 28, 2021 at 13:49

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