Statistics of the distance traveled until successful connectivity for unmanned vehicles

Abstract

In this paper, we consider a scenario where a robot needs to establish connectivity with a remote operator or another robot, as it moves along a path. We are interested in answering the following question: what is the distance traveled by the robot along the path before it finds a connected spot? More specifically, we are interested in characterizing the statistics of the distance traveled along the path before it gets connected, in realistic channel environments experiencing path loss, shadowing and multipath effects. We develop an exact mathematical analysis of these statistics for straight-line paths and also mathematically characterize a more general space of loop-free paths (beyond straight paths) for which the analysis holds, based on the properties of the path such as its curvature. Finally, we confirm our theoretical analysis using extensive numerical results with real channel parameters from downtown San Francisco.

Appendix

1.1 Proof of Lemma 2

Proof

Let $r(s)=(x(s), y(s))$ be the equation of the path parameterized by arc length. Since the path is parameterized by arc length, we have

$$\begin{aligned} \Vert r'(s)\Vert ^2 = |x'(s)|^2 + |y'(s)|^2 = 1. \end{aligned}$$

(17)

Moreover, we have the curvature constraint

$$\begin{aligned} \Vert r''(s)\Vert ^2 = |x''(s)|^2 + |y''(s)|^2 \le \kappa ^2. \end{aligned}$$

(18)

Let $s_0$ denote the current point, i.e., the center of the ball. Without loss of generality, let $(x(s_0), y(s_0)) = (0,0)$ and let the tangent at $s_0$ be parallel to the x-axis, i.e., $x'(s_0)=-1$, $y'(s_0)=0$, as shown in Fig. 3c.

We first prove that no point of $r_{\text {ball}}$ can lie outside the shaded region of Fig. 3c. Note that the shaded region has a boundary on the left corresponding to $x=-d_{\text {th}}$, and the two other boundaries correspond to circular arcs with curvature $\kappa$. Let us consider traveling backward along the path. For a given distance $d_x$ traveled along the negative x-axis (i.e., $x(s)=-d_x$), the path which maximizes the distance traveled along the y-axis |y(s)|, is the one that minimizes the x-axis velocity $|x'(s)|$ and maximizes the y-axis velocity $|y'(s)|$ the most. This corresponds to the circular path $(R_c\cos (s/R_c), R_c\sin (s/R_c))$ with constant curvature $\kappa$. Thus, for any path satisfying (17) and (18), the y-axis coordinate is bounded above and below by the circular arc. This implies that the segment $r_{\text {ball}}$ lies within the shaded region.

We next show that if $\kappa < 1/d_{\text {th}}$, then $r_{\text {ball}}$ cannot loop within the ball. Note that, by definition, $r_{\text {ball}}$ loops within the ball if $x'(s) > 0$ for some point on the path within the shaded region. The circular path with curvature $\kappa$ is the path that maximizes $x'(s)$. From Fig. 3c, we can see that if $\kappa = 1/d_{\text {th}}$, then $x'(s)=0$ at $x(s)=-d_{\text {th}}$ for the circular path. Thus, if $\kappa < 1/ d_{\text {th}}$, we have $x'(s)>0$ for any point of the path within the shaded region.

Finally, we determine the bound on the length of $r_{\text {ball}}$. If we travel a distance of $d_{\text {th}}$ along the negative x-axis, then we are guaranteed to have exit the ball. The path that maximizes its length before covering $d_{\text {th}}$ along the negative x-axis, would be the one that reduces the x-axis velocity $|x'(s)|$ the most. This maximal length path corresponds to the circular path with constant curvature $\kappa$. Any other path satisfying (17) and (18) would exit the shaded region before this circular path, i.e., the length of the segment of any path would be less that the length of this circular arc. The length of this circular arc can be found from the geometry of the figure. The chord length can be seen to be $2R_{c}\sin (\phi /2)$ where $R_c = 1/\kappa$. Moreover, we have $\cos (\phi /2) = \frac{d_{\text {th}}}{2R_{c}\sin (\phi /2)}$ which implies that $\phi = \sin ^{-1}\left( \frac{d_{\text {th}}}{R_c}\right)$. This gives us the arc length as $2\pi R_c \times \frac{\phi }{2\pi } = R_c \sin ^{-1}\left( \frac{d_{\text {th}}}{R_c}\right) = \frac{1}{\kappa }\sin ^{-1}\left( \kappa d_{\text {th}}\right)$. $\square$

1.2 Proof of Lemma 5

Proof

Using (8), we can show that $m = \alpha _1\Gamma _{\text {SH},-1} + \alpha _r\Gamma _{\text {SH},r}$ where

$$\begin{aligned} \alpha _1&= \frac{e^{-d_1/\beta _{\text {SH}}}-e^{-(d_{1r}+d_{r})/\beta _{\text {SH}}}}{1-e^{-2d_{1r}/\beta _{\text {SH}}}},\\ \alpha _r&= \frac{e^{-d_r/\beta _{\text {SH}}}-e^{-(d_1+d_{1r})/\beta _{\text {SH}}}}{1-e^{-2d_{1r}/\beta _{\text {SH}}}}. \end{aligned}$$

Then, the difference in mean $\varDelta m = m - {\hat{m}}$ is distributed as ${\mathcal {N}}(0, \sigma _{\varDelta m}^2)$, where using (10) we have

$$\begin{aligned} \sigma _{\varDelta m}^2 = \sigma _{\text {SH}}^2 \frac{\left( e^{-d_r/\beta _{\text {SH}}}-e^{-(d_1+d_{1r})/\beta _{\text {SH}}}\right) ^2}{1-e^{-2d_{1r}/\beta _{\text {SH}}}}. \end{aligned}$$

Moreover, using (9) we can calculate

$$\begin{aligned} \frac{\sigma ^2}{\sigma ^2_{\text {SH}}}&= 1 - \frac{e^{-2d_{1}/\beta _{\text {SH}}} + e^{-2d_{r}/\beta _{\text {SH}}}-2e^{-(d_{1}+d_{r}+d_{1r})/\beta _{\text {SH}}}}{1-e^{-2d_{1r}/\beta _{\text {SH}}}}. \end{aligned}$$

The difference in variance $\varDelta \sigma ^2 = \sigma ^2 - {\hat{\sigma }}^2$ can be calculated as

$$\begin{aligned} \varDelta \sigma ^2&= -\sigma _{\text {SH}}^2\frac{\left( e^{-d_r/\beta _{\text {SH}}}-e^{-(d_1+d_{1r})/\beta _{\text {SH}}}\right) ^2}{1-e^{-2d_{1r}/\beta _{\text {SH}}}} \\&= - \sigma _{\varDelta m}^2. \end{aligned}$$

From (11), we then have

$$\begin{aligned} KL&= \frac{\sigma _{\varDelta m}^2}{2{\hat{\sigma }}^2}\chi _1^{2} + \frac{1}{2}\left( -\frac{|\varDelta \sigma ^2|}{{\hat{\sigma }}^2} - \log _{e}\left( 1-\frac{|\varDelta \sigma ^2|}{{\hat{\sigma }}^2}\right) \right) . \end{aligned}$$

Since ${\mathbb {E}}[\chi _{1}^{2}] = 1$ and $\mathrm {Var}[\chi _{1}^2] = 2$, we can calculate the mean $m_{KL}$ and the standard deviation $\sigma _{KL}$ to be as stated in the lemma. $\square$

1.3 Proof of Lemma 6

Proof

Consider all possible locations of the general point (see Fig. 4a) at a fixed distance $d_{r}$. From the geometry of Fig. 4a, we can see that $d_{1r} = \sqrt{d_1^2+d_r^2 -2d_1d_r\cos \theta }$. Varying $\theta$, results in varying $d_{1r}$ which can take values in $[d_{r}-d_1, d_r+d_1]$. From Lemma 5, we can see that the $\theta$ that has a maximum impact on the KL divergence is the one that would minimize $m_{KL}$ and $\sigma _{KL}$. This would occur when we maximize $\sigma _{\varDelta m}^{2} = \sigma _{\text {SH}}^2 e^{-d_r/\beta _{\text {SH}}}\frac{(1-e^{-(z-z_{l})})^2}{1-e^{-2z}}$ where $z = d_{1r}/\beta _{\text {SH}}$ and $z_{l} = (d_r-d_1)/\beta _{\text {SH}}$. We wish to maximize $h(z) = \frac{(1-e^{-(z-z_{l})})^2}{1-e^{-2z}}$. Taking it’s derivative gives us

$$\begin{aligned} \frac{{\mathrm {d}}}{{\mathrm {d}} z}h(z) = \frac{2(1-e^{-(z-z_l)})}{(1-e^{-2z})^2}(e^{-(z-z_l)}-e^{-2z}). \end{aligned}$$

Then $\frac{{\mathrm {d}}}{{\mathrm {d}} z}h(z)> 0$ if $z>-z_l$, which is true as long as $d_r > d_1$.

Thus, maximizing $\sigma _{\varDelta m}^{2}$ occurs at $\theta = \pi$ where $d_{1r}$ takes its maximum value of $d_1+d_r$. Setting $\theta = \pi$ gives us

$$\begin{aligned} \sigma _{\varDelta m}^{2} = \sigma _{\text {SH}}^2 \frac{\left( e^{-d_r/\beta _{\text {SH}}} - e^{-(2d_1 + d_r)/\beta _{\text {SH}}}\right) ^2}{1 - e^{-2(d_1+d_r)/\beta _{\text {SH}}}}. \end{aligned}$$

From Lemma 5, we can see that satisfying the KL divergence parameters implies that $\frac{\sigma _{\varDelta m}^2}{{\hat{\sigma }}^2} \le 1-e^{-2\epsilon _{m}}$, and $\frac{\sigma _{\varDelta m}^2}{{\hat{\sigma }}^2} \le \sqrt{2} \epsilon _{\sigma }$. Let $\epsilon _d = \min \left\{ 1-e^{-2\epsilon _{m}}, \sqrt{2}\epsilon _{\sigma }\right\}$. Thus, we obtain the constraint

$$\begin{aligned} \frac{e^{-2d_r/\beta _{\text {SH}}}(1-\rho ^2)^2}{(1 - \rho ^2e^{-2d_r/\beta _{\text {SH}}})(1-\rho ^2)} \le \epsilon _d, \end{aligned}$$

which in turn gives us the constraint

$$\begin{aligned} d_{r} \ge \frac{\beta _{\text {SH}}}{2} \log _e\left( \rho ^2 + \frac{1 - \rho ^{2}}{\epsilon _d}\right) . \end{aligned}$$

$\square$

1.4 Proof of Lemma 7

Proof

Consider the scenario of Fig. 4b where $d_{1} = \varDelta d$. We will choose the location of the general point ($\Gamma _{\text {SH},r}$), which lies within the shaded region, such that it maximizes the impact (in terms of the KL divergence) on the approximation. From Lemma 5, we can see that the point that has a maximum impact on the KL divergence is the one that would maximize $\sigma _{\varDelta m}^{2}$. From the proof of Lemma 6, we know that for a fixed $d_r$ and varying $\theta$, the maximum value of $\sigma _{\varDelta m}^2$ occurs at the maximum value of $d_{1r}$. This occurs at the boundary of the shaded region, i.e., at a point on the circular arc. Since this holds for all $d_1<d_r\le d_{\text {th}}$, we know that the point that maximizes $\sigma _{\varDelta m}^{2}$ lies on the circular path with constant curvature $\kappa$.

We thus consider the setting in Fig. 4c with a fixed curvature $\kappa$. From the geometry of the figure, we have the following relations: $d_1 = 2R_c \sin \left( \frac{\varDelta \phi }{2}\right)$, $d_{1r} = 2R_c \sin \left( \frac{\phi }{2}\right)$ and $d_{r} = 2R_c \sin \left( \frac{\phi + \varDelta \phi }{2}\right)$. Since $d_1 = \varDelta d$, we have $\varDelta \phi = 2\sin ^{-1}(\kappa \varDelta d/2)$. From Lemma 3, we have the constraint that $\kappa < 1/d_{\text {th}}$. This guarantees that the path will leave the ball. Moreover, from the geometry of the figure, we can see that this will occur at the angle $\phi$ such that $d_{r} = 2R_c \sin \left( \frac{\phi + \varDelta \phi }{2}\right) = d_{\text {th}}$. This occurs at $\phi = h_{\text {cons}}(\kappa ) = 2\sin ^{-1}(\frac{\kappa d_{\text {th}}}{2})-\varDelta \phi$.

From Lemma 5, we can see that satisfying the KL divergence parameters implies that $\frac{\sigma _{\varDelta m}^2}{{\hat{\sigma }}^2} \le 1-e^{-2\epsilon _{m}}$, and $\frac{\sigma _{\varDelta m}^2}{{\hat{\sigma }}^2} \le \sqrt{2} \epsilon _{\sigma }$. Let $\epsilon _d = \min \left\{ 1-e^{-2\epsilon _{m}}, \sqrt{2}\epsilon _{\sigma }\right\}$. Thus, the point on the path that maximizes the KL divergence occurs at the angle

$$\begin{aligned} \arg \max _{0<\phi \le h_{\text {cons}}(\kappa )} h_{\text {opt}}(\kappa , \phi ), \end{aligned}$$

where

$$\begin{aligned} h_{\text {opt}}(\kappa , \phi )&= \frac{\sigma _{\varDelta m}^2}{{\hat{\sigma }}^2}\\&= \frac{\left( e^{-\frac{2}{\kappa \beta _{\text {SH}}}\sin \left( \frac{\phi + \varDelta \phi }{2}\right) } - \rho e^{-\frac{2}{\kappa \beta _{\text {SH}}}\sin \left( \frac{\phi }{2}\right) } \right) ^2}{\left( 1 - e^{-\frac{4}{\kappa \beta _{\text {SH}}}\sin \left( \frac{\phi }{2}\right) }\right) \left( 1-\rho ^2\right) }. \end{aligned}$$

We wish to find the maximum curvature $\kappa$, such that this maximum impact still satisfies the KL divergence parameters, i.e.,

$$\begin{aligned} \max _{0<\phi \le h_{\text {cons}}(\kappa )} h_{\text {opt}}(\kappa , \phi ) \le \epsilon _d. \end{aligned}$$

This results in the optimization problem stated in the lemma. $\square$