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What voltage ratio does the gain represent if an amplifier has a 30 dB voltage gain?
20*log(x) = 30dB X= voltage ratio output/input Therefore X = 31,7
Why you use 20 log db instead of 10 log db in amplifiers?
you use 20log for voltages and 10log for power 10log(vo^2/vi^2) = 2*10log(vo/vi) = 20log(vo/vi)
What is the implication of negative voltage gain?
If the input and output are expressed in volts, then negative gain means a 180° phase shift. If the ratio of output to input is expressed in dB, then negative gain means attenuation, i.e., less power out than power in.
If the input power is 375 milliwatts and the output power is 1.79 watts What would be the dB gain of this amplifier circuit?
If you're doing anything with amplifier circuits, you really need to understand dB and be able to calculate it on your own. Here is everything you need to know about dB: The definition. Please memorize this: dB gain = 10 log [ (final power) divided by (original power) ] In your example: Original power = 375 mW = 0.375 W Final power = 1.79 W (final) / (original) = ( 1.79 / 0.375 ) = 4.7733 log ( 4.7733 ) = 0.678 Gain = 10 times the log = 6.78 dB
A receiver requires 10nW as input power if all the system losses add up to 50dB then how much power is required from the source?
here, the power required by the receiver is the output power and that required from the source is input power. Gain in dB=10 log(output power/input power) we have, loss in dB = -gain in dB = 10 log(input power/output power) or, 50 = 10 log(input power/10nW) or, anti-log(5) = input power/10 nW so the power required from the source is antilog(5)*10nW = 1 mW
How much is a decibel?
A decibel (dB) has meaning only when compared a quantity (P1, V1, or I1) with a reference (P0, V0, or I0). Since it is a ratio of two like quantities, it is dimensionless. For a power ratio, power gain = 10 * log10(P1/Po) in [dB]. For a voltage ratio, voltage gain = 20 * log10(V1/Vo) in [dB]. For example, when P1 = 100 * P0, the power gain = 10 * log10(100) [dB] = 20 dB.
How do you determine the ratio of an amplitude?
Amplitude ratio to dB conversion:For amplitude of waves like voltage, current and sound pressure level:GdB = 20 log10(A2 / A1)A2 is the amplitude level.A1 is the referenced amplitude level.GdB is the amplitude ratio or gain in dB.dB to amplitude ratio conversion:A2 = A1 · 10(GdB / 20)A2 is the amplitude level.A1 is the referenced amplitude level.GdB is the amplitude ratio or gain in dB.
What is the signal to noise ratio formula in dB?
The signal-to-noise ratio (SNR) formula in decibels (dB) is calculated as 10 times the logarithm base 10 of the ratio of the signal power to the noise power. The formula is: SNR(dB) 10 log10(signal power / noise power).
Convert signals from voltage levels to their corresponding decibel equivalents and from decibel levels to their corresponding voltage or current levels?
This question contains a fundamental misunderstanding of a decibel.A decibel (always shortened to db or dB ) is not a level ; it is a ratio between two levels.The ratio is not voltages either - it is a power ratio. A power ratio can be converted to a voltage ratio if you understand Ohm's law and if the circuit impedance is the same for both measurements.To convert a power ratio to dB, take the log of the ratio (to base 10) then multiply by ten. In figures, if ratio is x, calculate 10 log10 (x)examples:if power ratio is 10, log(10) is 1, = 10 dBif power ratio is two, log(2) is .3010, = 3dB.power ratio 100, 20 dB, 1000 30 db and so on.It does sometimes seem that db figures are used to define levels. You need to look closely to see how this magic works.23 dbW is 200 Watts. Because dbW means db relative to 1Watt. More common in electronics is dbm - relative to a milliwatt.Warning. If you are interested in voltage levels, remember that a 6db increase in power is 4 times the power, but only twice the voltage.And why is it sometimes dB ? Well, the original scientific unit was the Bel. This was an inconveniently large unit, so engineers started using the deciBel, one tenth of a Bel. Nowadays db seems to be more common than dB.
What does dB equals to?
dB (decibel) is a logarithmic measure of the ratio of two power values, for example, two signal strengths. This is often used for power gain or power loss. For example, a loss of 10 dB means that the signal degrades by a factor of 10, a loss of 20 dB means that the signal degrades by a factor of 100, and a loss of 30 dB means that the signal degrades by a factor of 1000.
What is processing gain?
The process gain (or 'processing gain') is the ratio of the spread (or RF) bandwidth to the unspread (or baseband) bandwidth. It is usually expressed in decibels (dB).For example, if a 1 kHz signal is spread to 100 kHz, the process gain expressed as a numerical ratio would be 100,000/1,000 = 100. Or in decibels, 10log10(100) = 20 dB.
What voltage ratio does the gain represent if an amplifier has a 30 dB voltage gain?
20*log(x) = 30dB X= voltage ratio output/input Therefore X = 31,7
What is gained?
The process gain (or 'processing gain') is the ratio of the spread (or RF) bandwidth to the unspread (or baseband) bandwidth. It is usually expressed in decibels (dB).For example, if a 1 kHz signal is spread to 100 kHz, the process gain expressed as a numerical ratio would be 100,000/1,000 = 100. Or in decibels, 10log10(100) = 20 dB.
What is the dB value for a gain of 10?
10dB It is a logarithmic relationship, so a gain of 20 is 13dB, and a gain of 100 is 20dB
What it the math for change in decibels?
A decibel is a measure on the logarithmic scale so a change from d1 dB to d2 dB is a measure of the power ratio of 10(d2 - d1)/10 . Thus, an increase of 1 dB is equivalent to the power ratio increasing by a multiple of 100.1, that is to a multiple of 1.259
What is dB in Electronics Engineering?
Deci-bels (dB) are a logarithmic measure of system/network amplification or gain (G). The logarithms are taken to the base 10 and multiplied by 20:dB = 20logG (where G is the gain of the system)To give an example of what this means:G = 0.01, so dB gain = 20log0.01 = 20 * -2 = -40dBG = 0.1, so dB gain = 20log0.1 = 20 * -1 = -20dBG = 1, so dB gain = 20log1 = 20 * 0 = 0dBG = 10, so dB gain = 20log10 = 20 * 1 = 20dBG = 100, so dB gain = 20log100 = 20 * 2 = 40dBG = 10000, so dB gain = 20log10000 = 20 * 4 = 80dBThere are other uses for dB in electronics, where logarithms are useful they are preferred. The reasons logs can be useful is because the natural log of an exponential curve is a straight line, allowing for easier ways of understanding the behaviour of a system.
How many milliwatts in 36 dB?
To convert decibels to milliwatts, you can use the formula: milliwatts = 10^(dB/10). In this case, for 36 dB, the calculation would be: 10^(36/10) = 10^3.6 = 3981.07 milliwatts. Therefore, there are approximately 3981.07 milliwatts in 36 dB.
