2006-12-04 12:25:08 · 5 answers · asked by dmb 1 in Science & Mathematics ➔ Engineering
Power (in dBm) is the Power of a signal with respect to 1mW signal.
Power (in dBm) equals to,
10 log (Power of test signal in watt / Power of the unit signal in watt)
Power of test signal is 1 Watt == 1000mW
Power of unit signal is 1mW
therefore, Power (dBm) = 10 iog ( 1000/1)
= 10 log ( 10 ^3)
= 10*3 log 10 (log is base 10)
= 30 dBm.
2006-12-04 16:14:13 · answer #1 · answered by Siby Mathew 2 · 0⤊ 0⤋
1 Watt Dbm
2016-12-14 14:16:43 · answer #2 · answered by carle 4 · 0⤊ 0⤋
Well, 1 milliwatt is 0 dBm. 1 watt is 1000 milliwatts. Log 1000 =3. You multiply this by 10 to get 30 dBm. Likewise 1 microwatt = 1/1000 milliwatt. Log 1/1000= -3. Multiply by 10 to get 1 microwatt = -30 dBm.
2006-12-04 14:32:52 · answer #3 · answered by zee_prime 6 · 0⤊ 0⤋
The term "dBm" implies that 1 mW (milliWatt) is taken as reference. Thus,
1 W/ 1 mW = 1 000.
Converting 1 000 to dB yields:
10 log 1 000 = 10 × 3 = 30 dB.
That's 30 dB over 1 mW.
2006-12-04 12:35:30 · answer #4 · answered by Jicotillo 6 · 1⤊ 0⤋
30 dBm according to http://en.wikipedia.org/wiki/DBm
2006-12-04 12:29:31 · answer #5 · answered by F.G. 5 · 0⤊ 0⤋