**Nanomedicine,
Volume I: Basic Capabilities**

**©
1999 Robert A. Freitas Jr. All Rights
Reserved.**

Robert A. Freitas Jr., Nanomedicine, Volume I: Basic Capabilities, Landes Bioscience, Georgetown, TX, 1999

**7.2.5.3 Acoustic Cables
and Transmission Lines**

Consider an acoustic compression signal of frequency n_{acoust}
traveling down a fluid-filled transmission cable of length l_{cable}
and radius r_{cable}. In order to be detected at a signal/noise ratio
SNR = 2, each pulse cycle must transfer a minimum energy kT e^{SNR}
to a receiver of volume L^{3} at the cable terminus; hence from Eqn.
4.29, the maximum detectable pulse frequency n_{acoust}
= I_{rec} L^{2} / kT e^{SNR} (Hz), where I_{rec}
is the acoustic power intensity at the receiver. To obtain this intensity at
the receiver, the input signal power at the other end of the cable must be P_{0}
__>__ p r_{cable}^{2} I_{trans}
(watts) where I_{trans} is the power intensity of the signal transmitter.
Combining these relations with Eqn. 6.44 for
acoustic attenuation in cables gives

_{}
{Eqn. 7.18}

where a_{tube} is given
by Eqn. 6.45 with a ~n_{acoust}^{1/2}
dependency. To avoid cavitation in pure water (Section
6.4.3.3) and to ensure safety in the unlikely event of cable detachment
in vivo, maximum I_{trans} = 10^{4} watts/m^{2} of acoustic
energy (Fig.
6.8). For a cable of diameter 1 micron terminating on a receiver of volume
(680 nm)^{3} sensitive to 10^{-6 }atm displacements (Section
4.5.1), then n_{acoust} ~ 1 GHz (10^{9}
bits/sec) for a cable of length l_{cable} = 14 microns (~8000 zJ/bit).
A 1000-micron long cable can transmit up to ~1 MHz; n_{acoust}
~ 1 KHz at l_{cable} ~ 5 cm; and n_{acoust}
~ 1 Hz at l_{cable} ~ 2 meters. For intradevice communications, n_{acoust}
~ 1 GHz for L = 300 nm, l_{cable} = 1 micron, and r_{tube} =
50 nm, though pure diamond fiber may be more efficient in this case.

A solid diamond acoustic transmission line transfers signals
over great distances at GHz frequencies with almost no power losses, in part
because of the extreme stiffness of diamond. Consider a rod of volume V_{rod}
at temperature T_{rod}, made of a material with a thermal coefficient
of volume expansion b and heat capacity C_{V},
to which a pressure pulse DP is applied at one end
that travels at velocity v_{sound} (~17,300 m/sec for diamond) to the
other end. In the worst-case thermodynamic cycle, Drexler ^{10}
gives the total energy dissipation per pulse W_{max} as

_{}
{Eqn. 7.19}

For diamond at T_{rod} = 310 K, b
= 3.5 x 10^{-6} /K and C_{V} = 1.8 x 10^{6} joule/m^{3}-K.^{460,567}
Thus a 1-atm pulse that is applied to a transmission line consisting of a diamond
rod 1 micron in length and (10 nm)^{2} in cross sectional area (V_{rod}
= 10^{-22 }m^{3}) requires a pulse input energy of ~10,000 zJ,
but suffers an energy loss during transmission of at most W_{max} =
2 x 10^{-6} zJ. Under smooth mechanical cycling, nanomechanical systems
may approach the isothermal limit and significantly reduce dissipation still
further,^{10} to ~1% W_{max} at
~1 GHz and ~0.001% W_{max} at ~1 MHz. Thus even at n_{rod}
~ 1 GHz the losses using 1-atm pulses amount to only ~20 zJ and the transmission
of energy is still ~99.8% efficient.

Last updated on 19 February 2003