Introduction to Ultrasonic Mid-Air Haptic Effects

Abstract

In this chapter, we discuss the basic physical principles of ultrasonic mid-air haptics. Our aim is to provide a holistic introduction to the technology, facilitate a better understanding of these principles, and help newcomers to join this exciting field of research in following the rest of this book. To that end, we have assumed a simplified and idealized situation and divide our discussion into four sub-topics: acoustic radiation pressure, phased array focusing, vibrotactile stimulation, and by-product audible sounds.

Appendices

Appendix

A. Derivation of Spatial Distribution of Ultrasound on Focal Plane

The process by which a phased array forms a single ultrasonic focal point can be expressed using the following mathematical formulas. We assume the coordinate system shown in Fig. 11. The XY coordinate is on the surface of the phased array $\left( {x,y} \right)$. The focal plane, $\left( {x_{\rm{f}},y_{\rm{f}} } \right)$, is separated by the distance $r$ [m] from the surface. Let the center coordinate of the $m$-th and $n$-th transducers in the X and Y directions, respectively, be $\left( {x_{m} ,y_{n} ,0} \right)$, and the center coordinate of the focal point be $\left( {x_{\rm{c}} ,y_{\rm{c}} ,r} \right)$.

Fig. 11

Coordinate system for the formulation of phased array focusing © IEEE. Reprinted, with permission, from Hoshi et al. (2010)

Here, we ignore the directivity of the transducer, and assume that a spherical wave is radiated from each transducer with an appropriate phase to generate a single focal point. The RMS ultrasonic sound pressure $p_{r}$ [Pa] (inversely proportional to $r$) is produced on the focal plane by each transducer and is a paraxial approximation. A square-shaped phased array is assumed; this is the shape of the majority of current ultrasonic mid-air haptic devices. The $N$ transducers are lined up in both the X and Y directions. At this time, the distribution of ultrasonic sound pressure generated on the focal plane is given as follows:

$$\begin{aligned} & p\left( {x_{\mathrm{f}} ,y_{\mathrm{f}} } \right) = \mathop \sum \limits_{m = 0}^{N - 1} \mathop \sum \limits_{n = 0}^{N - 1} \sqrt 2 ~p_r {\text{e}}^{ - \mathrm{j}kr^{\prime} } {\text{e}}^{\,\mathrm{j}\left( {kr^{\prime\prime} - \omega t} \right)} \\ & \quad \approx \sqrt 2 ~p_r {\text{e}}^{ - \mathrm{j}\omega t} {\text{e}}^{\,\mathrm{j}\left\{ {\frac{k}{{2r}}\left( {x_{\mathrm{f}} ^2 + y_{\mathrm{f}} ^2 - x_{\mathrm{c}} ^2 - y_{\mathrm{c}} ^2 } \right) - \xi \left( {\nu _x + \nu _y } \right)} \right\}} \mathop \sum \limits_{m = 0}^{N - 1} {\text{e}}^{ - \mathrm{j}dm\nu _x } \mathop \sum \limits_{n = 0}^{N - 1} {\text{e}}^{ - \mathrm{j}dn\nu _y } \\ & \quad = \sqrt 2 p_r {\text{e}}^{ - \mathrm{j}\omega t} {\text{e}}^{\,\mathrm{j}\left\{ {\frac{k}{{2r}}\left( {x_{\mathrm{f}} ^2 + y_{\mathrm{f}} ^2 - x_{\mathrm{c}} ^2 - y_{\mathrm{c}} ^2 } \right) - \xi \left( {\nu _x + \nu _y } \right)} \right\}} \left( {\frac{{1 - {\text{e}}^{ - \mathrm{j}Nd\nu _x } }}{{1 - {\text{e}}^{ - \mathrm{j}d\nu _x } }}} \right)\left( {\frac{{1 - {\text{e}}^{ - \mathrm{j}Nd\nu _y } }}{{1 - {\text{e}}^{ - \mathrm{j}d\nu _y } }}} \right) \\ & \quad = \sqrt 2 p_r {\text{e}}^{ - \mathrm{j}\omega t} {\text{e}}^{\,\mathrm{j}\left\{ {\frac{k}{{2r}}\left( {x_{\mathrm{f}} ^2 + y_{\mathrm{f}} ^2 - x_{\mathrm{c}} ^2 - y_{\mathrm{c}} ^2 } \right) - \xi \left( {\nu _x + \nu _y } \right)} \right\}} {\text{e}}^{ - \mathrm{j}\frac{{\left( {N - 1} \right)d}}{2}\left( {\nu _x + \nu _y } \right)} \\ & \quad \quad \quad \times \left( {\frac{{{\text{e}}^{\,\mathrm{j}\frac{{Nd\nu _x }}{2}} - {\text{e}}^{ - \mathrm{j}\frac{{Nd\nu _x }}{2}} }}{{{\text{e}}^{\,\mathrm{j}\frac{{d\nu _x }}{2}} - {\text{e}}^{ - \mathrm{j}\frac{{d\nu _x }}{2}} }}} \right)\left( {\frac{{{\text{e}}^{\,\mathrm{j}\frac{{Nd\nu _y }}{2}} - {\text{e}}^{ - \mathrm{j}\frac{{Nd\nu _y }}{2}} }}{{{\text{e}}^{\,\mathrm{j}\frac{{d\nu _y }}{2}} - {\text{e}}^{ - \mathrm{j}\frac{{d\nu _y }}{2}} }}} \right) \\ & \quad = \sqrt 2 ~p_r ~{\text{e}}^{\,\mathrm{j}\left\{ {\varphi \left( {x_{\mathrm{f}} ,~y_{\mathrm{f}} } \right) - \omega t} \right\}} ~\frac{{\sin \frac{{Nd\nu _x }}{2}}}{{\sin \frac{{d\nu _x }}{2}}}~\frac{{\sin \frac{{Nd\nu _y }}{2}}}{{\sin \frac{{d\nu _y }}{2}}} \\ & \quad = \sqrt 2 ~p_r ~{\text{e}}^{\,\mathrm{j}\left\{ {\varphi \left( {x_{\mathrm{f}} ,~y_{\mathrm{f}} } \right) - \omega t} \right\}} \left( {\frac{{\frac{{Nd\nu _x }}{2}~\frac{{Nd\nu _y }}{2}}}{{\frac{{d\nu _x }}{2}~\frac{{d\nu _y }}{2}}}} \right)\frac{{\sin \left( {\frac{{Nd\nu _x }}{2}} \right)\sin \left( {\frac{{Nd\nu _y }}{2}} \right)/\left( {\frac{{Nd\nu _x }}{2}~\frac{{Nd\nu _y }}{2}} \right)}}{{\sin \left( {\frac{{d\nu _x }}{2}} \right)\sin \left( {\frac{{d\nu _y }}{2}} \right)/\left( {\frac{{d\nu _x }}{2}~\frac{{d\nu _y }}{2}} \right)}} \\ & \quad = \sqrt 2 ~p_r ~N^2 ~\frac{{\mathrm{sinc}\left( {\frac{{Nd\nu _x }}{2},\frac{{Nd\nu _y }}{2}} \right)}}{{\mathrm{sinc}\left( {\frac{{d\nu _x }}{2},\frac{{d\nu _y }}{2}} \right)}}~{\text{e}}^{\,\mathrm{j}\left\{ {\varphi \left( {x_{\mathrm{f}} ,~y_{\mathrm{f}} } \right) - \omega t} \right\}} \\ \end{aligned}$$

In the above, $\rm{j}$ is an imaginary unit. $k$ [rad/m] and $\omega$ [rad/s] are the wavenumber and angular frequency, respectively. $t$ [s] is the time. The definition of the sinc function is $\rm{sinc} \mathit{\left( {x,y} \right) \equiv \sin \left( x \right)\sin \left( y \right)/xy}$.

The first line represents the operation of multiplying the spherical wave coming from a transducer $\sqrt 2 p_{r} {\text{e}}^{{\,\mathrm{j}\left( {kr^{\prime \prime } - \omega t} \right)}}$ by the phase control factor to focus the waves ${\text{e}}^{{ - \mathrm{j}kr^{\prime } }}$ and then adding them together. In the second line, the Fresnel approximation is applied to $r^{\prime}$ and $r^{\prime\prime}$, where $r^{\prime}$ [m] is the distance from the $m$-th row and $n$-th column transducer to the focal point, and $r^{\prime\prime}$ [m] is the distance from the transducer to the arbitrary position $\left( {x_{\mathrm{f}} ,y_{\mathrm{f}} } \right)$ on the focal plane.

$$\begin{aligned} r^{\prime} & \equiv \sqrt {\left( {x_{m} - x_{\mathrm{c}} } \right)^{2} + \left( {y_{n} - y_{\mathrm{c}} } \right)^{2} + r^{2} } \\ & \approx r + \frac{{\left( {x_{m} - x_{\mathrm{c}} } \right)^{2} + \left( {y_{n} - y_{\mathrm{c}} } \right)^{2} }}{2r} \\ r^{\prime\prime} & \equiv \sqrt {\left( {x_{m} - x_{\mathrm{f}} } \right)^{2} + \left( {y_{n} - y_{\mathrm{f}} } \right)^{2} + r^{2} } \\ & \approx r + \frac{{\left( {x_{m} - x_{\mathrm{f}} } \right)^{2} + \left( {y_{n} - y_{\mathrm{f}} } \right)^{2} }}{2r} \\ \end{aligned}$$

Furthermore, changes of the variables are applied as follows:

$$\begin{aligned} \nu_{x} & \equiv \frac{k}{r}\left( {x_{\mathrm{f}} - x_{\mathrm{c}} } \right) \\ \nu_{y} & \equiv \frac{k}{r}\left( {y_{\mathrm{f}} - y_{\mathrm{c}} } \right) \\ \end{aligned}$$

Then, the positions of the $m$-th and $n$-th transducers are written in another form, that is, $\left( {x_{m} ,y_{n} } \right) = \left( {md + \xi ,nd + \xi } \right)$, using the interval between the centers of neighboring transducers $d$ [m] and an offset $\xi = - \left( {N - 1} \right)d/2$. The third line is obtained by the summation formula of the geometric series, i.e., $\sum\nolimits_{n = 0}^{N - 1} {\alpha^{n} = \left( {1 - \alpha^{N} } \right)/\left( {1 - \alpha } \right)}$. The fourth line is the middle of the formula transformation to the fifth line. By extracting ${\text{e}}^{ - \mathrm{j}\alpha /2}$ from $\left( {1 - {\text{e}}^{ - \mathrm{j}\alpha } } \right)$, a form of $\left( {{\text{e}}^{\,\mathrm{j}\alpha /2} - {\text{e}}^{ - \mathrm{j}\alpha /2} } \right)$ is obtained. The fifth line is derived because $\left( {{\text{e}}^{\,\mathrm{j}\alpha /2} - {\text{e}}^{ - \mathrm{j}\alpha /2} } \right) = 2\mathrm{j}\sin \left( {\alpha /2} \right)$. $\varphi \left( {x_{\mathrm{f}} ,y_{\mathrm{f}} } \right)$ is the phase delay and depends on the position on the focal plane:

$$\varphi \left( {x_{\mathrm{f}} ,y_{\mathrm{f}} } \right) \equiv \frac{k}{2r}\left( {x_{\mathrm{f}}^{2} + y_{\mathrm{f}}^{2} - x_{\mathrm{c}}^{2} - y_{\mathrm{c}}^{2} } \right) - \left\{ {\xi + \frac{{\left( {N - 1} \right)d}}{2}} \right\}\left( {\nu_{x} + \nu_{y} } \right)$$

The sixth line is again the middle of the formula transformation to the seventh line. By extracting $\alpha /2$ from $\sin \left( {\alpha /2} \right)$, a form of $\sin \left( {\alpha /2} \right)/\left( {\alpha /2} \right)$ is obtained; thus, the sinc function finally appears in the seventh line.