View previous topic :: View next topic 
Author 
Message 
btb4198 Antenna Theory Regular
Joined: 19 May 2011 Posts: 20

Posted: Sat Jun 25, 2011 4:59 pm Post subject: convert dbi to watt 


how do you convert a 9 dbi gain to watt?
and 3 dbm to db? 

Back to top 


Schubert Antenna Wizard
Joined: 08 Apr 2009 Posts: 161

Posted: Sun Jun 26, 2011 5:24 am Post subject: 


9 dBi gain to Watts doesn't make sense. dBi is the gain of an antenna relative to a dipole.... It is a measure of how efficient and directive an antenna is, which isn't a function of how much power you input to the antenna.
3 dBm = 3 mW. Assuming "dB" means Watts in dB, then 3 dBm = 3 dBw  30 = 27 dBW = 27 dB. Note, the 30 dB comes from:
1 W = 1000 mW
0 dB = 30 dBm 

Back to top 


R. Fry Antenna Theory Regular
Joined: 06 Jun 2011 Posts: 36 Location: Illinois USA

Posted: Sun Jun 26, 2011 10:42 am Post subject: Re: convert dbi to watt 


btb4198 wrote:  how do you convert a 9 dbi gain to watt? 
"dBi" is the unit of measure for the gain of an antenna relative to the gain of an isotropic radiator. The radiation pattern of an isotropic source is perfectly spherical. An isotropic radiator exists only in theory, but is used as a reference for the gain of real antennas.
The peak gain of a 1/2wave, centerfed dipole in free space is 2.15 dBi, because due to its directional radiation pattern that gain is 2.15 dB (1.64X) greater than from an isotropic radiator.
The gain of an antenna sometimes is given with respect to that of a 1/2wave, centerfed dipole in free space. In that case the unit of measure is "dBd." Gain in dBd can be converted to dBi by adding 2.15 dB to the gain expressed in dBd. So an antenna with a gain of 3 dBd has a gain of 5.15 dBi.
Some antenna manufacturers prefer to use dBi when specifying antenna gain because it results in a larger, more impressive number. But for the same antenna input power an antenna with 3 dBd gain in a given direction produces exactly the same radiated field as one with 5.15 dBi gain in that direction.
The power in watts that any antenna radiates is a function of the power flowing into its input terminals, and the gain of that antenna in the direction of interest.
The term "dB" has no specific meaning or value unless the unit of measure (watts, volts etc), and the reference value are known.
"dBm" as commonly used is the unit of measure for power expressed in decibels with respect to 1 milliwatt (mW). In this case the unit of measure (watt) is understood both by the writer and readers. More correctly the unit should be written dB(mW).
The equation for a power relationship in decibels is 10*log(P1/P2). Rearranging that equation to convert the power in 3 dBm to power to milliwatts...
10^(dBm/10) = 10^(3/10) = 1.995 mW (approx) 

Back to top 


btb4198 Antenna Theory Regular
Joined: 19 May 2011 Posts: 20

Posted: Sun Jun 26, 2011 12:07 pm Post subject: 


so went you use the Friis Transmission formula and you want to get Pr in watt and you have gr and the gt in dbi what do you have to convert it to ? dbi to what watt, right? 

Back to top 


R. Fry Antenna Theory Regular
Joined: 06 Jun 2011 Posts: 36 Location: Illinois USA

Posted: Sun Jun 26, 2011 3:20 pm Post subject: 


Convert the published gain for a freespace path in the direction of interest as given in dBi to a multiplier, using 10^(dBi/10).
As examples, a gain of zero dBi (an isotropic source) is a multiplier of 10^(0/10) = 1, and a gain of 2.15 dBi is a multiplier of 10^(2.15/10) = 1.64 (approx).
The gains of both the tx and rx antenna must be calculated for the Friis equation, and used for Gt and Gr, respectively.
Both the path distance R and the wavelength (lambda) must be entered in the same unit of measure.
Pt is the power in watts delivered to the transmit antenna, and Pr is the impedancematched power in watts available at the terminals of the receive antenna, assuming that the polarization of both antennas is the same.
Doing the math for a 1,000 meter, 100 MHz freespace path with 100 watts applied to a tx antenna with 2.15 dBi gain, and a rx antenna with 6 dBi gain:
Pr = Pt*Gt*Gr*(lambda/(4*pi*R))^2
Pr = 100*1.64*3.98*(3/(12.566*1000))^2 = 0.000 037 watts, approx, or about 37 microwatts.
Note that this equation is not valid for path lengths less than several freespace wavelengths. 

Back to top 


