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
If I say jetplane = 130 dB, most people are content.Laymen think that a specific decibel value belongs to a noise source. Like helicopter = 100 dB, jackhammer = 110 dB, rock band 120 dB. That is not correct.Alway is forgotten that the distance from the sound source to the measuring point plays an important role. Sound level decreases by (−)6 dB per doubling of distance from the source. Without any distance a dB value is really nonsense.
The weighting filter A takes off low frequencies. So the measured value in dBA must be less than in dB. Look up at "A-weighting - Wikipedia".
If I say tank = 130 dB, most people are content. Laymen think that a specific decibel value belongs to a noise source. Like helicopter = 100 dB, jackhammer = 110 dB, rock band 120 dB. That is not correct. Alway is forgotten that the distance from the sound source to the measuring point plays an important role. Sound level decreases by (−)6 dB per doubling of distance from the source. Without any distance a dB value is really nonsense.
The sound being measured is equal to the reference value.
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
You must find a resistance value for 0 dB as reference. If 1 Ohm = 0 dB then 10 ohms = 20 dB and 100 ohms = 40 dB.
Send me the reference value (0 dB) for frequency.
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.
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.
The equation for the attenuation of a gamma-ray flux passing through a path of length x in a sample with linear attenuation coefficient u can be expressed as I = I0 e ^-u*x. For most environments, the value of u is not known and must be estimated by measurements.
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.