6 dB is not good nor bad as an attenuation, if the context and the signal that is attenuated is not specified.
As a rule of thumb, addition between dBs is multiplication between linear attenuation, thus since 3dB is a division of the power by two, 6 dB means a division of the power by 4.
Just to give an idea:
In free space, without obstacles and reflections, the power of a normal voice attenuates 6 dB in dry air in about 230 cm (assuming 180° directional sound emission of the mouth, that is reasonable). This is the combined action of intrinsic air attenuation (the sound ordered power is converted by traversing air in disordered oscillations of molecules: i.e. heath) that is very small (about 1 dB/km) and of the increase of the area where the power is distributed, that going farer and farer from the source is a sphere of greater and greater radius so that power decrease due to this last effect as the square power of distance from source.
The humidity further reduces air intrinsic attenuation, but since the dominant phenomenon is the other, we can say that practically nothing changes.
If music is considered, the situation is more complex. Music frequencies are much more extended than voice frequencies and attenuation of the material depends on frequency (while attenuation due to distance does not, as it is intuitive).
High frequencies attenuates in air much more than low frequencies, for example 50 kHz attenuates about 0.9 dB/m in dry air for 180° directional sound emission.
Thus attenuation also imply a small distortion of music due to its dependence on frequency. However this effect is not so relevant to be problematic when hearing music in concert halls, due both to the small distance and to the logarithmic sensitivity characteristic of out ears. Moreover in concert halls there are a lot of reflections that are exploited in the design of a good concert hall to concentrate the sound on the public, thus the real attenuation is much reduced by the fact that power diverging from the source is redirected towards the hall center by reflections.
A problem to avoid in such a design is to exploit reflections from objects too far from the source or the hall. In this case reflections recover energy, but are also delayed with respect to the original sound, creating a bad echo effect.
Light exhibit similar behavior, but much more evident due to the huge frequency extension of visible light.
100 percent means full voltage or 0 dB.When 75 percent of the voltage is lost you still have 25 percent of the voltage.25 percent means damped to (-)12 dB.
A: Clipping only occurs if the input surpasses a threshold like the Vbe of a transistor. The 20 Db is really a change3 in voltage of a 100 that is not a small change
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.
6 dB per octave. The slope outside the pass band is 6 times the order, in dB/octave.
The decibel scale is a logarithmic scale where each change in three dB represents a power factor change of two. (3 dB is power times two, 6 dB is power times four, 9 dB is power times 8, etc. Similarly, -3dB is power divided by two, -6 dB is power divided by four, etc.) Zero dB is assigned some arbitrary reference power. One example is 1 mV across 600 ohms. If you double the voltage into a constant resistance, the power quadruples, so 2 mV would be +6 dB, 4 mV would be +12 dB, etc. The letter after dB is the reference power. In the case of dBm, it means that 0 dB is 1 milliwatt, so 2 milliwatt is +3 dB, etc. There are many dB scales, such as dBa, used in sound measurements. Still, fundamentally, 3 dB is a doubling of power, -3 dB is a halving of power, so, for any arbitrary scale, say dBq, then saying +6dBq is saying a power four times higher than 0 dBq. In the end, dBm plus dBm is delta dB, with no scale.
6 dB is a "good" attenuation.
db=20log(V1/V2) or 10log(P1/P2) Example: db=20log (100/50) db=20log(2) db=20*.3010 db=6.02
20 dB
1 db
100 percent means full voltage or 0 dB.When 75 percent of the voltage is lost you still have 25 percent of the voltage.25 percent means damped to (-)12 dB.
dB per kilometerNote: Depends on frequency of the radio signal, and on the rate at which rain is falling.
A: Clipping only occurs if the input surpasses a threshold like the Vbe of a transistor. The 20 Db is really a change3 in voltage of a 100 that is not a small change
dB or dBW relativ to 1W : dBm for dB relativ to 1 milliwatt. 0 dBm= 1 mW = -30dBW . Its interesting,but these terms are used interchangeably at times, erroneously.The term dBm is used by communications engineers and it is absolute.Most Power meters commercially available have this scale. It is : dBm=10log(base 10)(P1/P2) at two different Power points.Power is read always across a 50 Ohm resistor.IEEE has made this a standard and 1mW=0dBm(-40dBm is 100nW accross 50 Ohm and +20dBm is 100mW). As far as I know dB(not DB)refers simly to gain(and loss/attenuation as -dB).It is different this time as the equation is not Power but a simple ratio: dB=20log(base 10)(Gain or attenuation).One can have a reference and above that reference he is talking in positive dBs(Gain) and below in negative dBs(loss or attenuation).Such scales can then be modified as we have done for Acoustic Emission where we talk in dBae and our refence is the "perfect" sensor giving us ONLY 1microVolt(!) Peak output noise.Anyway a good goal.So for 40dBae a sensor/Amplifier(40dB Gain)output across a 50 Ohm resistor is 10mV.
100 percent is 0 dB.50 percent is - 6 dB.45 percent is -6.935749724 dB.10 percent is - 20 dB.
I have seen this alot on patient reports. A low attenuation lesion on a certain area of your liver just basically means there is something there it could be a scar or shadow. The low attenuation is a measurement. How strong it is. Low attenuation is good and usually not a concern. If it showed high than it is cause for concern. Repeat your test in 4-6 months just to make sure it does not change.
When an optical signal of a given wavelength travels in the fiber it looses power. The amount of loss of power per Km length of fiber is called its attenuation. A=10*LOG10(POUT/PIN) dB/Km Where POuT is optical power after 1 Km PIN is th epower launched in the Fiber.
When an optical signal of a given wavelength travels in the fiber it looses power. The amount of loss of power per Km length of fiber is called its attenuation. A=10*LOG10(POUT/PIN) dB/Km Where POuT is optical power after 1 Km PIN is th epower launched in the Fiber.