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If transmitter power is increased from 200 Watts to 400 Watts how do you convert to Db form?
Decibels (db) is relative power, log base 2, times 3. Increasing power from 200 watts to 400 watts is doubling power, so the decibel change is +3 db.800 watts would be +6 db, 1600 watts would be +9 db, 100 watts would be -3 db, 50 watts would be -6 db, and so on.
Calculate the power out put of an amplifier that has an input of 15 μw and a gain of 25dB?
If you want to work in watts, convert 25dB to a scalling factor: 3dB = 2 x input 10dB = 10 x input 20dB = 100 x input ...25dB = 10 ^ (25/10) = 316.2 x input So the output is 15 micro Watts x 316.2 = (4700)/(10^6) = 4.7 milli watts If you want to work in dB, then convert 15 micro watts to dB: 10 * log |P| = dB = 10*log |15 x 10^6| = -48.2dB ***When you have very small (ie negative) dB, it is often referred to in dBm, or 1/1000 of dB ( 30 dBm = 0 dB) so the output is -18.2dBm + 25 = 6.8dBm, or -23.2dB
If a 13 dB gain antenna has 2 watts of power at the input what is the effective radiated power?
40 watts
What is the formula to convert voltage levels to decibels?
Here you find the formula to convert voltage levels to decibels. Scroll down to related links and look at "How to calculate dB?"
What is a 1-dB compression point?
Devices, such as amplifiers can't be linear over all input values. At some point they just can't output the required output power. I.e. an amplifier that increases input power by a factor of 10, may not be able to amplify a signal that comes in that is, let's say 10 watts. The point where the device is outputing 1 dB less POWER (which is roughly running at 80%) than it should is the 1 dB compression point. So lets say a 10 watt signal is input, and that the signal should be amplified by a factor of 10, and should output 100 watts. Let's also say the system output power is actually 1 dB down from the expected value and outputting roughly 80 watts, 10 watts is the input 1 dB compression point. Also, look here: http://www.rfcafe.com/references/electrical/p1db.htm
If transmitter power is increased from 200 Watts to 400 Watts how do you convert to Db form?
Decibels (db) is relative power, log base 2, times 3. Increasing power from 200 watts to 400 watts is doubling power, so the decibel change is +3 db.800 watts would be +6 db, 1600 watts would be +9 db, 100 watts would be -3 db, 50 watts would be -6 db, and so on.
25dB equal to how many watts?
That depends on you. If you want, you can say 1 watt should be 0 dB.
How do you convert 40w to db?
To convert power in watts to decibels (dB), you can use the formula: dB = 10 * log10(P2/P1), where P1 is the reference power (usually 1 watt) and P2 is the power you are converting (40 watts in this case). Plugging in the values, you would calculate: dB = 10 * log10(40/1) = 16.02 dB.
What is -2 db in watts?
To convert decibels (dB) to watts, you can use the formula: ( P = P_0 \times 10^{(dB/10)} ), where ( P_0 ) is the reference power level (usually 1 watt). For -2 dB, the calculation would be ( P = 1 \times 10^{-2/10} ), which equals approximately 0.63095 watts. Thus, -2 dB corresponds to about 0.631 watts.
Convert 100W into decibels?
Given P = 100 watts. Reference sound intensity Po = 10^−12 W. Reference sound intensity level LPo = 0 dB. Get power level LP in dB when entering sound power P in watts. Power level LP = 10×log (P / Po) dB = 10×log (100 / 10^−12) = 140 decibels (dB).
How do you convert a .db file?
How do I convert a .db file?
32 db is equal to how many watts?
Converting between dB and watts requires additional information about the reference power level. If we assume a reference power level of 1 watt, 32 dB would be equal to 1,000 watts.
Calculate the power out put of an amplifier that has an input of 15 μw and a gain of 25dB?
If you want to work in watts, convert 25dB to a scalling factor: 3dB = 2 x input 10dB = 10 x input 20dB = 100 x input ...25dB = 10 ^ (25/10) = 316.2 x input So the output is 15 micro Watts x 316.2 = (4700)/(10^6) = 4.7 milli watts If you want to work in dB, then convert 15 micro watts to dB: 10 * log |P| = dB = 10*log |15 x 10^6| = -48.2dB ***When you have very small (ie negative) dB, it is often referred to in dBm, or 1/1000 of dB ( 30 dBm = 0 dB) so the output is -18.2dBm + 25 = 6.8dBm, or -23.2dB
Calculate the power out put of an amplifier that has an input of 15wand a gain of 25dB?
If you want to work in watts, convert 25dB to a scalling factor: 3dB = 2 x input 10dB = 10 x input 20dB = 100 x input ...25dB = 10 ^ (25/10) = 316.2 x input So the output is 15 Watts x 316.2 = 4.7kW If you want to work in dB, then convert 15 watts to dB: 10 * log |P| = dB = 10*log |15| = 11.76dB so the output is 11.76 + 25 = 36.76dB
How many watts is 98db?
It depends on the reference. If the reference electric power is P0 = 1 W (0 dB) then 98 dB equals 6309573444.8 Watts.
How do you convert 15W to dB?
"dB" expresses a ratio between two different power levels. 15W can't be expressed in dB. It can only be expressed as some number of dB more or less than some other power. If the reference level is 1 watt, then 15W is written as " +17 dBW ". That means " 17 dB more power than 1 watt ". In most of electronics and telecommunications, the reference level is usually 1 milliwatt. Using that common reference level, 15W is written as " +47 dBm ". That means " 47 dB more power than 1 milliwatt ". ========================================================= The question starts from 15 watts. Using 1 watt as the reference level ... (15 watts) divided by (1 watt) = 15. 10 log(15) = +11.76 dBW (11.76 dB more power than 1 watt) Using 1 milliwatt as the reference level ... (15 watts) divided by (0.001 watt) = 15,000 10log(15,000) = +41.76 dBm (41.76 dB more power than 1 milliwatt)
How do you get power output with a given power input of 20dbm and three power gain which is Ap1 equals 10 ap2 equals 4 ap3 equals 23?
Two ways to do it. In this particular problem, it's a matter of opinionwhich one is easier and which one is harder.Way #1:Convert dBm to watts, multiply by gains, convert output watts to dBm.+20 dBm = 0.1 watt.Output power = (0.1 watt) x (ap1) x (ap2) x (ap3) = 0.1 x 10 x 4 x 23 = 92 watts = +49.64 dBmWay #2:Convert each gain ratio to dB, then add all dB to input power.ap1 = 10 = 10 dBap2 = 4 = 6.02 dBap3 = 23 = 13.62 dB+20 dBm + 10dB + 6.02 dB + 13.62 dB = +49.64 dBm
