Method for determining probe angle by analyzing skewed sine wave I've got a fun problem and would be curious to get feedback on how some of you would go about solving this.

Imagine I have a probe and am scanning the surface of some material. This material surface is described by a perfect cosine signal. If my probe is perfectly normal to the surface, my probe will output a "1 to 1" match of the surface signal (excluding some DC offset).

However, if my probe is not perfectly normal to the surface, the collected signal would appear to be a skewed sinusoid. I've put together a simple diagram of set up.



Currently, my method for determining the probe angle would be to determine the frequency content of the signal. In the case of the probe being perfectly normal to the surface, I would expect that frequency domain would show a single impulse corresponding to the surface frequency.

In the case of the angled probe, I would expect to a find a range of frequencies. Obviously the peak would still correspond to the dominant spatial frequency, however, there would now be a certain bandwidth associated the frequency domain. This bandwidth is associated with the fact that the collect signal is a skew sinusoid.

My question would then be, how would I correlate signal bandwidth to probe angle?It would be nice to get an analytical solution to signal bandwidth and probe angle.

Furthermore, if anyone else could see a way of determining probe angle that isn't frequency domain related, I would be interested in hearing it.

Edit: As I think about this problem more, I'm becoming less confident in whether or not probe angle would create a skewed sinusoid. Assuming the angle doesn't exceed the max slope found in the signal, I'm starting to think the resulting signal would simply be a phased offset sinusoid...
I will assume:


The probe is like a line, infinitely narrow.
We move the probe along the $x$ dimension, parallel to the large-scale surface, and from each position, probe the surface:
The probe is extended to the direction it is pointing to until it meets the surface. (An alternative assumption would be that probing is done to the direction $y$, that of the normal of the large-scale surface, in which case the waveform measured would be sinusoidal unless the probe stem meets the surface before the tip. I wonder if your edit describes this scenario.)


Equation of line starting at coordinates $(x_0, 0)$ and with a positive slope $k = \frac{\Delta y}{\Delta x} \ge 1$ (where $1$ is the maximum slope of the sinusoid of Eq. 2, to ensure continuity):

$$y = k (x - x_0)\tag1\\
\Rightarrow x = x_0 + \frac yk.$$

Equation of the first period of the constant-biased sinusoid, spanning $0 \le x \le 2\pi$:

$$y = 1 - \cos(x)\tag2\\
\Rightarrow x = \begin{cases}
\arccos(1 - y)&amp;\text{if }0 \le x \le \pi\\
2\pi - \arccos(1 - y)&amp;\text{if }\pi \le x \le 2\pi.
\end{cases}$$

By Eq. 2 the period starts at $(0, 0)$ and ends at $(2\pi, 0)$, which makes the corresponding period of the skewed sinusoid also start and end at these points. We obtain the waveform by solving $y$ as function of $x_0$ from the system of Eqs. 1 and 2. Unfortunately there is no closed form solution, but we can still make some plots:

Figure 1. The skewed waveform for different values of the constant $k$. At $k=\infty$, the waveform is equal to the constant-biased sinusoid. The waveform shown represents the probe tip $y$ position as function of the coordinate $x_0$ at which the probe shaft crosses the $y$ axis. If you desire the "probe length" waveform, you need to multiply this waveform by a factor $\tfrac{\sqrt{k^2 + 1}}{k}$.

I doubt that you will be able to find an analytical solution in frequency domain. What we can do is determine $k$ based on the waveform peak location $x_0 = x_p$ by setting $y = 2$, for which the system of Eqs. 1 and 2 gives:

$$k = \frac{2}{\pi - x_p}.\tag{3}$$

To get the angle $\phi$ by which the probe angle deviates (moderately) from the normal of the large-scale surface:

$$\phi = \operatorname{atan2}\left(1, k\right) = \operatorname{atan2}\left(\pi - x_p, 2\right).\tag{4}$$