Publications Department
Summary
In theory it is possible to receive a planetary radar that might well be used by a society having a technology only slightly more advanced than our own over a range of 200 to 600 light years, using a 5 meter dish and current electronics and software operated at 10 GHz. This was demonstrated in Small Parabolic Dish Antennas and the Search for Extra-Terrestrial Intelligence, by M. Mallette, Radio Astronomy, the Journal of the Society of Amateur Radio Astronomers, November-December, 1996 ("Small Dish Antennas and SETI")
The question that ought be answered before a radio telescope having any size antenna is used for a search for extra-terrestrial intelligence ("SETI"), is whether or not that particular radio telescope could receive an ETI signal that could be encountered. That question may be answered by determining the flux of the smallest natural source that the radio telescope can observe. Knowing the flux of the smallest natural source that the radio telescope can observe, the radio telescope’s "practical efficiency" may be determined and expressed as a percentage of the flux receiving ability of a radio telescope having the same size antenna, with a 100K system temperature, with the same integration time and bandwidth and with 100% illumination. The "practical efficiency", so determined, may be used with John Kraus’s SETI range formula to determine the range at which the radio telescope under consideration can receive a possible planetary radar transmission.
Determining the practical efficiency of a radio telescope is also useful in non-SETI radio astronomy for comparison of a radio telescope to an ideal system.
Theoretical Range of a 5 meter SETI system.
In Small Dish Antennas and SETI, it was demonstrated that an amateur radio telescope could have a system temperature of 100K. Moreover, a band width of 2 Hertz, based on the use of available software is possible, and is consistent with the limitations imposed by the drift in the Doppler shift of a signal caused by the turning of the Earth and of the ETI planet. A signal to noise ratio of 1 to 3 ought not to prevent detection of a signal by current software. Finally, an ETI planetary radar the equivalent of 10 MegaWatts or 100 MegaWatts into the Arecibo dish would be required to give the planetary radar range desirable if the operator of the planetary radar wanted to determine the exact orbit of an asteroid in time to do something with the information.
Radio Astronomy , Kraus, 2nd edition("Kraus") is the standard textbook on Radio Astronomy and is used by most professionals and amateurs in that field. At page 12-3, Kraus sets forth the following SETI range formula ("Range Formula"):
R = ( (P*AT*AR)/((S/N)* k * Tsys * (L)^2 * BW))^.5
Note that the Range Formula is here stated as used by the Mercury Math Solver software. It is identical to the formula in Kraus.
The range, R, is in meters. To convert to Range in light years ("Rly") divide by 10^16:
Rly = R/10^16
The variables in the Range Formula are as follows:
P = Output power of the ETI transmitter in watts.
AT = .78 * .5 * DT^2 , or the effective area of the ETI dish antenna in square meters.
AR = .78 * .5 * DR^2 , or the effective area of receive dish antenna in square meters.
DT = effective diameter of transmitter dish in meters.
DR = effective diameter of receive dish in meters.
k = 1.38 * 10^-23 , which is Boltzman's constant in Watt seconds per degree Kelvin.
Tsys = Receiver system temperature in degrees Kelvin.
BW = Receiver bandwidth in Hz, assumed identical to ETI transmitter bandwidth.
S/N = Signal to Noise ratio with S and N in the same units.
L = 300000000/F which is the wavelength ( speed of light divided by the frequency) in meters.
F = Frequency in Hz.
Using 100 MegaWatts for the ETI transmitter and 300 meters for the antenna in the range formula, together with the variables of the receive system described above will permit the calculation of the range at any frequency between 1 GHz and 10 GHz. The use of higher frequencies in a radar system should result in greater resolution, therefore, in Small Dish Antennas and SETI the range for a 10 GHz system is calculated:
R = ( (P*AT*AR)/((S/N)* k * Tsys * (L)^2 * BW))^.5 ; Kraus formula
Rly = R/10^16 ; convert R to range in light years
P = 100000000 ; power of trans. in watts
AT = .78 * .5 * DT^2 ; effective area of trans. dish calculated in square meters.
AR = .78 * .5 * DR^2 ;effective area of receive dish calculated in square meters.
DT = 300 ; diameter of trans dish in meters.
DR = 5 ; diameter of receive dish in meters.
k = 1.38 * 10^-23 ; Boltzman's constant in Watt seconds per degree Kelvin.
Tsys = 100 ; system temp in degrees Kelvin.
BW = 2 ; bandwidth in Hz.
S = 1 ; signal
N = 3 ; noise (same units as signal )
L = 300000000/F ; wavelength in meters.
F = 10000000000 ; frequency in Hz.
BEW = L/DR * 57 ; width of antenna beam in deg.
T = BEW * 4 ; time object at celestial equator is in beamwidth in minutes.
Spo = (4/3)*3.1415*(Rly^3) ; Volume of sphere that can be searched in light years cubed.
Ns = Spo/350 ; number of stars in such sphere.
Solution
Variables:
R = +6.4289596156089E+18
BW = +2.0000000000000000
Rly = +642.89596156089
BEW = +0.342000000000000 { = +171 / 500 }
T = +1.368000000000000 { = +171 / 125 }
Spo = +1113006994.85269
Ns = +3180019.9852934
The range in light years is 642. The beamwidth of the receive antenna is .34 degree, and a source will be in the beamwidth for 1.3 minutes.
However, this is theory. The question at hand is whether or not a particular radio telescope, which is not an ideal system but operates in the real world, is suitable for use as a SETI receiving system.
The minimum flux a radio telescope can observe.
The formula to calculate the minimum flux that a radio telescope can detect is given in Kraus at page 3-45. The formula below is the formula from Kraus, stating the minimum flux in Janskies. (Minimum Flux Formula)
S = 2 * k * KS * Tsys / (Ae * SQRT(BW * tau )) * 1 * 10^26 ; Formula for min. detectable flux density from Radio Astronomy 2nd ed page 6-58 S is in Janskies, 10^-26 watts per meter squared per hertz
k = 1.38 * 10^-23 ; Boltzman's constant
KS = 1 ; receiver constant is 1 for total power receiver
Tsys = ? ; system temperature in degrees Kelvin
Ae = EF * .7854 * D^2 ; effective aperture in meters^2, dish area times * practical efficiency (EF)
BW = ? ; bandwidth of receiver in Hertz
tau = ? ; integration time in seconds
S = ? ; smallest observable flux in Janskies
D = ? ; diameter of dish in meters
Note that the effective area of the receiving antenna has the same effect on SETI range as it does on the maximum flux that can be detected by a radio telescope. The larger the effective area of the receiving antenna is the lower flux it can observe and the longer the SETI range becomes. The following is the formula is used to calculate the effective area of the receiving dish:
Ae = EF * .7854 * D^2 ; effective aperture in meters^2, dish area times * practical efficiency (EF)
For an ideal dish antenna, EF is determined by the ability of the feedhorn to illuminate the dish. The perfect dish antenna, in theory, would have EF = 1, that is, the dish would be 100% illuminated. However, the best professional dish antennas have less than 100 % illumination and the illumination is often assumed to be approximately 50% (.5 ) . That is the number plugged into the determination of AR, (the effective receive dish area) of the Range Formula above.
Comparison of actual radio telescope performance with an standard radio telescope.
The above formulas require knowledge of the noise temperature of the system and the illumination factor. The system noise temperature is difficult to determine. Although 50% or 55% are often used to determine the effective area of the dish, taking into account illumination, the actual determination of that factor is difficult, to say the least.
In determining the performance of a particular radio telescope, it is possible to avoid such matters by comparing the radio telescope under consideration with a standard radio telescope having a fixed system total noise temperature, perfect 100% illumination and the same physical dish size, band width and integration time as the radio telescope under consideration.
Suppose we define EF as the ability of a radio telescope to observe flux, stated as the percentage of a standard radio telescope having a system temperature of 100K, having the same dish area as the radio telescope being compared, having ideal 100% illumination, and having the same bandwidth and integration time as the radio telescope being compared.
An easy method of visualizing the matter is to consider the standard radio telescope with a system temperature of 100K, 100% illumination, a perfect dish, and the same bandwidth and integration time as the radio telescope in question. Now cover part of the standard radio telescope’s dish with microwave absorbing material. Keep covering more and more of the dish area of the standard radio telescope until the standard radio telescope can only observe objects having flux that can be observed by the radio telescope being considered. As the effective area of our standard, perfectly illuminated, perfect dish (and no interference) radio telescope is reduced, the performance of the standard radio telescope is reduced until it reaches the actual performance of the real radio telescope, which is not 100% illuminated, does not have a dish with a perfect surface and may have a system temperature higher than 100K, not to mention interference.
So EF, the practical efficiency of a radio telescope, is the percent of the dish area of an ideal radio telescope that would be needed to observe the smallest flux that the real radio telescope can observe. This is the concept needed to create a simple method to determine the suitability of a particular real radio telescope for SETI.
Determining the practical efficiency of a radio telescope.
It is rather simple to determine the object with the smallest flux that a radio telescope can observe. The radio telescope operator can observe objects with known flux at the frequency of the radio telescope and compare the results of observing an object of known flux (usually by graphing the data) with the baseline fluctuation to determine the smallest flux that will not be washed out by baseline fluctuation. The smallest flux that can be observed can also be directly determined by the ability to observe one source and not another source with slightly less flux.
When the smallest flux that can be received is determined, the practical efficiency of the radio telescope can be determined by using the modified Minimum Flux Formula and solving for EF. This is best explained by an example. A radio telescope operating at 3.78 GHZ will be used in the example.
Assume that a radio telescope being considered for SETI use has a 5 meter diameter and an IF bandwidth of 35 MHz. The operating frequency is 3.78 GHz, although that factor is not involved in the modified Minimum Flux Formula. The radio telescope being considered has observed calibration sources of 93 Janskies and, from the fluctuations in the baseline and rise above the baseline of the calibration source at 93 Janskies it appears that the minimum source observable at 10 seconds integration time is 25 Janskies. Using the modified Minimum Flux Formula and solving for EF:
S = 2 * k * KS * Tsys / (Ae * SQRT(BW * tau )) * 1 * 10^26
; Formula for min. detectable flux density from Radio Astronomy, 2nd ed page 6-58; S is in Janskies, 10^-26 watts per meter squared per hertz
k = 1.38 * 10^-23 ; Boltzman's constant
KS = 1 ; receiver constant is 1 for total power receiver
Tsys = 100 ; system temperature in Kelvin
Ae = EF * .7854 * D^2 ; effective aperture in meters^2; dish area times * practical efficiency (EF)
BW = 35000000 ; bandwidth of receiver in Hertz
tau = 10 ; integration time in seconds
S = 25 ; smallest observable source
D = 5 ; diameter of dish in meters
Solution
Variables:
S = +25.000000000000000
k = +1.380000000000000E-23
KS = +1.0000000000000000
Tsys = +100.00000000000000
Ae = 19.635*EF
= +0.59011282214263
BW = +35000000.000000000
tau = +10.000000000000000
EF = +0.030054128960664
D = +5.0000000000000000
So EF is only 3%!! In other words, this particular radio telescope, that can observe a minimum flux of 25 Janskies, is only 3% as efficient as a standard radio telescope that has no interference, a perfect surface 100% ( perfect ) illumination and a system temperature of 100K., with, of course, the same bandwidth, integration time and dish diameter.
Using the comparison with a standard radio telescope to determine the SETI range.
When the above calculation has determined the percentage of the surface of the standard dish necessary to give the same results as the radio telescope under consideration, it is rather simple to calculate the SETI range of the radio telescope under consideration. Note that the 3.78 GHz operating frequency is important in the SETI Range Formula as it determines the transmitting antenna gain.
A minor modification of the SETI Range Formula is used to adjust for the fact that the radio telescope in question does not perform as well as the standard 100 percent illuminated radio telescope. That modification is to use the calculated EF in place of the illumination percentage (often assumed at 55% or 50%). The calculation for the radio telescope discussed above that can observe sources down to 25 Janskies is as follows:
R = ( (P*AT*AR)/((S/N)* k * tsys * (L)^2 * BW))^.5 ; Krause formula
Rly = R/10^16 ; convert R to range in light years
P = 100000000 ; power of trans. in watts
AT = .78 * .5 * DT^2 ; effective area of trans. dish- efficiency assumed
AR = .78 * EF * DR^2 ; area of receive dish calculated
; * practical efficiency
DT = 300 ; diameter of trans dish in meters
DR = 5 ; diameter of receive dish in meters
k = 1.38 * 10^-23 ; Boltzman's constant
tsys = 100 ; system temp in Kelvin
BW = 2 ; bandwidth
S = 1 ; signal
N = 3 ; noise (same units as signal )
L = 300000000/F ; wavelength in meters
F = 3780000000 ; frequency in cycles per sec
EF = .03 ; practical efficiency of radio telescope
BEW = L/DR * 57 ; width of antenna beam in deg
T = BEW * 4 ; time object at celestial equator is in beam
Spo = (4/3)*3.1415*(Rly^3) ; Volume of sphere that can be searched in light years cubed
Ns = Spo/350 ; number of stars in such sphere
Nss = Ns*BEW/180 ; number of stars per adv. scan of one dec.
Solution
Variables:
R = +5.9526195001061E+17
P = +100000000.00000000
AT = +35100.0000000000
AR = +0.585000000000000 { = +117 / 200 }
S = +1.0000000000000000
N = +3.0000000000000000
k = +1.380000000000000E-23
tsys = +100.00000000000000
L = +0.0793650793650794 { = +5 / 63 }
BW = +2.0000000000000000
Rly = +59.526195001061
DT = +300.00000000000000
EF = +0.03000000000000000 { = +3 / 100 }
DR = +5.0000000000000000
F = +3780000000.0000000
BEW = +0.904761904761905 { = +19 / 21 }
T = +3.61904761904762 { = +76 / 21 }
Spo = +883487.01082012
Ns = +2524.2486023432
Nss = +12.688022075270
This is not very good. The range of 60 light years, which assumes 100 million watts into a dish the size of Arecibo, will only bring 13 stars into range in an average 24 hour scan of one 5 degree segment of the sky. The total number of stars that can be searched, given the ability to move the dish to the southern hemisphere to look at the entire sky, is only 2524 . That result does not to indicate that this radio telescope is a good candidate for a SETI operation.
Notice that the bandwidth has been adjusted to the 2 Hertz bandwidth that is available for the SETI operation through the use of FFT software.
Note also that the system temperature remains the same. This is based on the assumption that the same LNA (or LNB ), feedhorn and feedline will be used for the SETI search as for the observation of the natural source. If the LNA has over 20 dB gain, the noise temperature of the system will primarily be determined by the antenna noise temperature and the LNB noise temperature and the change of noise temperature between the total power radio astronomy receiver and the narrowband SETI receiver will be insignificant. The system temperature of a receiver consisting of an 20 dB gain LNA that has a noise temperature of 20K (20 degrees Kelvin) followed by a receiver having a noise temperature of 100K is 20K + 100K/100 = 21K. See Krause p. 7-26. Clearly the difference between a 100K receiver and a 300K receiver in such a system will not be significant.
Note also that if the assumed total system temperature of 100K is replaced with an assumed total system temperature of 200K in both the modified Minimum Flux Formula and the Range Formula, the result will be the same!
Now, suppose that the same radio telescope could observe flux from natural objects down to 2 Janskies. The variable EF would then be 38%, not bad considering that illumination will only equal 55% and there will be other factors, such as dish shape, to reduce performance.
The same radio telescope, if it can observe flux from natural objects down to 2 Janskies, will have a SETI range of 212 light years (observing a transmitted signal from a 100 million watt transmitter into a 300 meter dish). The number of stars observed in 24 hours of operation, at an average declination, would be 571. If the radio telescope could be moved to the southern hemisphere when necessary, the total number of stars that could be observed is 114,000. That radio telescope would be a reasonable choice as a SETI system.
Note that it is not a valid conclusion that a system must be able to observe a 2 Jansky source to be usable for SETI. Radio telescopes that have only a narrow band SETI receiver will not be able to observe natural sources with as low a flux as the same radio telescope would be able to observe were it equipped with a wideband receiver. However, by substituting the actual bandwidth of the narrow SETI receiver and its integration time in the modified Minimum Flux Formula, the EF of that system can be calculated and used in the modified Range Formula to determine SETI range.
In testing a radio telescope with a narrowband SETI receiver, it may be best to set the receiver to AM mode, select the widest bandwidth, and rectify the audio from the receiver with a diode, resistor and capacitor network having an integration time of 10 seconds or so and convert the output to digital data with an analog to digital converter. In addition, the FFT program you have may have the ability to measure total power. Note, however, that if you use a sideband bandwidth and an FFT program, the very narrow bandwidth may make it difficult to observe most natural sources.
Conclusion
Determining the smallest flux that a radio telescope can observe makes it possible to determine the overall practical efficiency of that radio telescope as compared to a perfect radio telescope with the same dish diameter and bandwidth, and with a system temperature of 100 degrees Kelvin and perfect 100 % dish illumination. Such comparison, expressed as a percentage of the dish area of the perfect radio telescope necessary to produce the same result as the real radiotelescope, is useful to compare radio telescope efficiency.
The practical SETI range of a radio telescope may be determined by the following procedure:
The performance of a radio telescope or the noise level of the location in which it is operated ( such as in a large city) may result in a determination that a particular radio telescope is not a suitable radio telescope for a SETI operation. In that event, the radio telescope performance must be improved or interference avoided by changing location or frequency.
If the radio telescope performance cannot be improved or interference avoided, then science emulates art. In the words of Ellie Arroway in the novel Contact by Carl Sagan, "I need a bigger antenna".
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