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# Interstellar Communications -Theory

` `
```Preface:  This document (RADOBS.9) is in support of the discussion document

Handy conversions:
1 Astronomical Unit (A.U.) = 1.496 X 10^11 m
1 Light Year (L.Y.) = 9.461 X 10^15 m
1 Light Year (L.Y.) = 63,242 A.U.
1 Parsec (psc) = 3.26 L.Y.

The relationship between Apparent Stellar Magnitude (m) and the brightness
or intensity of a star may be expressed in the form:

m = -[19 + (2.5).log(I)]                                                 (1)

where I = received intensity (W/m^2).  The threshold for naked-eye
visibility is m = +6.

The Sun's total output (EIRP) = 3.90 X 10^26 watts.  Here are several
intensities and corresponding magnitudes as a function of range R:

At R = 1 A.U. (1.496 X 10^11 m):

I =  1.39 kW/m^2
m = -26.8

At R = 10 L.Y. (9.461 X 10^16 m):

I =  3.48 X 10-^9 W/m^2
m = +2.2

At R = 100 L.Y. (9.461 X 10^17 m):

I =  3.48 X 10^-11 W/m^2
m = +7.2*

At R = 1000 L.Y. (9.461 X 10^18 m):

I =  3.48 X 10^-13 W/m^2
m = +12.2*

* Not visible to the naked eye.

----------------------------

With no allowance for the Fraunhofer dark line absorption, the Planckian
(black body) starlight continuum level (spectral energy density) is given
by:

2.PI.h.f^3r^2
Npl = -----------------------  W/m^2.Hz                                  (2)
c^2[e^(h.f/k.T) - 1]R^2

where  h  = Planck's constant (6.63 X 10^-34 J.s),
c  = velocity of light (3 X 10^8 m/s),
Wl = wavelength (656 nm),
f  = frequency (c/Wl = 4.57 X 10^14 Hz),
k  = Boltzmann's constant (1.38 X 10^-23 J/K),
T  = temperature (5778 K),
r  = radius of star (6.96 X 10^8 m),
R  = distance of receiver (10 L.Y. = 9.461 X 10^16 m).

At R = 1 A.U.:

Npl = 2.19 X 10^-12 W/m^2.Hz

At R = 10 L.Y.:

Npl = 5.47 X 10^-24 W/m^2.Hz

----------------------------

For the purposes of this analysis we shall assume a fully (uniformly)
illuminated aperture and not a beam with a Gaussian intensity profile, as
might be obtained from a TEMoo single-mode laser.

The diffraction limited half-power (-3 dB) beamwidth is given by:

(57.3).Wl
THETA = ---------  degrees                                               (3)
d

where d = diameter of telescope (10 m).

THETA = 0.0135 arc seconds

----------------------------

The half-power (-3 dB) beam diameter is given by:

(1.22).Wl.R
D = -----------  meters                                                  (4)
d

At R = 1 A.U.:

D = 12 km

At R = 10 L.Y.:

D = 7.57 X 10^9 m = 0.0506 A.U.

----------------------------

The gain of an antenna is given by:

4.PI.At
G = -------                                                              (5)
Wl^2

where At = area of transmitting telescope mirror (78.5 m^2).

G = 153.6 dB

----------------------------

For a mean transmitted power Pt, the Effective Isotropic Radiated
Power (EIRP) is given by:

EIRP =  G.Pt  Watts                                                      (6)

For Pt = 1 GW:

EIRP = 2.29 X 10^24 W

----------------------------

The intensity of the beam at the receiver is given by:

EIRP
I = --------  W/m^2                                                      (7)
4.PI.R^2

At R = 1 A.U.:

I = 8.1 W/m^2

At R = 10 L.Y.:

I = 2.04 X 10^-11 W/m^2

----------------------------

The signal power received by a telescope is given by:

Ps = I.Ar  Watts                                                         (8)

where Ar = area of receiving telescope mirror (78.5 m^2).

At R = 1 A.U.

Ps  = 0.64 kW  Phew!!

At R = 10 L.Y.:

Ps = 1.60 X 10^-9 W

----------------------------

The effective system noise temperature of an optical receiver may be
expressed in the form:

h.f
Teff = -----  K                                                          (9)
eta.k

where eta = photodetector quantum efficiency (0.5).

Teff = 43,900 K

----------------------------

The Carrier-To-Noise Ratio in a perfect shot (quantum) noise limited optical
heterodyne system is given by:

eta.Ps
CNR = ------                                                            (10)
hfB

where  Ps = received optical power (1.6 nW),
B  = Intermediate Frequency bandwidth (30 MHz).

At R = 10 L.Y.:
CNR = 19 dB

There was no need to do a calculation for R = 1 A.U. since the optical
receiver went up in smoke!  Hence, CNR effectively equal to zero!

December 23, 1990
BBOARD No. 277

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
* Dr. Stuart A. Kingsley                       Copyright (c), 1990        *
* AMIEE, SMIEEE                                                           *
* Consultant                            "Where No Photon Has Gone Before" *
*                                                   __________            *
* FIBERDYNE OPTOELECTRONICS                        /          \           *
* 545 Northview Drive                          ---   hf >> kT   ---       *
* Columbus, Ohio 43209                             \__________/           *
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* Tel. (614) 258-7402                     .  .  .  .  .  .  .  .  .  .  . *
* skingsle@magnus.ircc.ohio-state.edu         ..    ..    ..    ..    ..  *
* CompuServe: 72376,3545                                                  *
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
```
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