DB (decibels) gain is the log based 2 times 3 relative power change measured from input to output of a circuit. For instance, +3db is twice the power, +6db is four times, +9db is eight times. Similarly, -3db is half the power, -6db is quarter the power, and -9 db is eighth the power.
Keep in mind that DB is relative power, not voltage, so if you are measuring voltage, then +3db is square root of 2 (1.414) times the voltage, etc.
The voltage gain of an amplifier is 200. The decibel voltage gain is? Answer Gain in dB = 20 * log 200 = 46 dB
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.
10dB It is a logarithmic relationship, so a gain of 20 is 13dB, and a gain of 100 is 20dB
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.
Gain in decibels is a logarithmic base 2 scale with a multiplier of 3, so 3 db is twice the power, 6 db is four times, etc. Gain of -3.5 db means a power loss of 21.17, or a ratio of about -2.25. Note, however, that this is power, not voltage. Since power is voltage times amperes, and since amperes is voltage divided by resistance, in order to achieve a power loss of 2.25, the voltage must change by the square root of that, so the voltage changed by -1.5.
10 dB gain means a voltage ratio of 3.16227766 to 1.
using bode-m plots , one can tranfer dB magnitude to gain and gain to frequency
The voltage gain of an amplifier is 200. The decibel voltage gain is? Answer Gain in dB = 20 * log 200 = 46 dB
db gain is defined as power gain, not voltage gain. Please restate you question in terms of power, or provide details of input and output impedance.
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.
3 db is double the power, so an input of 1 KW would yield an output of 2 KW for a 3 db gain.
10dB It is a logarithmic relationship, so a gain of 20 is 13dB, and a gain of 100 is 20dB
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.
databand-a unit to measure gain power
increases by 6 dB
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.
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