Well, the question your asking is basically impossible. It's like asking to convert a gallon of water into cans of pop. It is possible to convert the gallon of water into cans of water not soda. So you can convert dBm to watts, not dBi.
Well, the question your asking is basically impossible. It's like asking to convert a gallon of water into cans of pop. It is possible to convert the gallon of water into cans of water not soda. So you can convert dBm to watts, not dBi.
20 dbm
To convert dB to dBi (decibels isotropic), you need to add 2.15 dB to the dB value. This adjustment accounts for the difference in reference points used in the two scales. So, if you have a value of x dB, the equivalent value in dBi would be x + 2.15 dBi.
the first convert the power in dBm to MW, the define of dBm=10 log (P MW) -10 log ( 1mw). example: let P=-2 dBm convert this to dB? answer: Pmw= inv log(-2/10)=0.630mw*1000 micw/mw=630 microw 10log(630)=28dB
There are a few terms that need to be understood. [dBm,dBd,dBi and dB] For antennas, a common reference unit is the dBi, which states the gain of an antenna as referenced to an ISOTROPIC source. An Isotropic source is the perfect omnidirectional radiator, a true "Point Source", and does not exist in nature. Consider it a source which is the center of the sphere and the energy is coming equally out of it as a sphere. Now in reality nothing like that exists. It's also 2.41 dB BIGGER than the next common unit of antenna gain, the dBd, When you convert that to a real antenna. So a simple dipole antenna has a gain of 2.41dBi, and a gain of 0dBd, since we're comparing it to itself. Now lets talk about dBm, dBm is not is reference to anything else but the used as an actual gain ( say amplifiers ) , P(dBm)=10*LOG(1000*Power in milliwatts,10), an amp with an output of 30dBm puts out 1 Watt. It is not in refernce to two power level but directly correlating the gain of a device. Now when you talk about a dB it is a relative measure of two different power levels. 10log (p1/p2, 10).
PdBm = 10*log10(1000*W)
The 'm' in dBm means the power is referenced to 1mW. So, the power in dBm equals 10 times the log of the power in mW, or P(dBm) = 10*log(P(mW)/1mW). For example, 1W = 1000mW, so 10*log(1000/1) = 30dBm.
dBm is defined as power ratio in decibel (dB) referenced to one milliwatt (mW). It is an abbreviation for dB with respect to 1 mW and the "m" in dBm stands for milliwatt. dBm is different from dB. dBm represents absolute power, whereas in audio engineering the decibel is usually a voltage ratio of two values and is used then to represent gain or attenuation of an audio amplifier, or an audio damping pad.
dBm is defined as power ratio in decibel (dB) referenced to one milliwatt (mW). It is an abbreviation for dB with respect to 1 mW and the "m" in dBm stands for milliwatt. dBm is different from dB. dBm represents absolute power, whereas in audio engineering the decibel is usually a voltage ratio of two values and is used then to represent gain or attenuation of an audio amplifier, or an audio damping pad.
Put the power in milliwatts in cell A2, and then use the following formula to get the power in dBm. =10 * LOG(A2)
dBm us almost exactly the same as dB. The only difference is that there is a reference of 1 Watt = 0 dB, and 1 mW = 0 dBm. dBm is defined as power ratio in decibel (dB) referenced to one milliwatt (mW). It is an abbreviation for dB with respect to 1 mW and the "m" in dBm stands for milliwatt. dBm is different from dB. dBm represents absolute power, whereas in audio engineering the decibel is usually a voltage ratio of two values and is used then to represent gain or attenuation of an audio amplifier, or an audio damping pad. PdBm = 10*log10(1000*10W) = 40dBm
Here's how to convert dB units (with usually a 1 Watt or whatever 1 value as reference) to dBm units (with a 1 miliWatt reference value):x= value to be convertedx [dB]= x + 30 [dBm]Proof:P= 1 Watt--> 10*log10(1)= 0 [dB] (this is 1 Watt in dB)--> 10*log10(1/(1*10^(-3)))= 10*log(1*10^3)= 30 dBm (this is 1 Watt to dBm)Now, if you do whatever number of examples you want to do, you'll end up in concluding the conversion dB to dBm is totally linear without of actually having to proof the linear properties. (i'm too lazy to write it here).Hope this helps....Regards,STMI