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A signal starts at point X. As it travels to point Y, it loses 10db. At point Y, the signal

is boosted by 5db. As the signal travels to point Z, it loses 7db. What is the db

strength of the signal at point Z?

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Ovayo Yokwe

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3y ago

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What is the deicbel loss of a signal that starts at point A with a strength of 2000 watts and ends at point B with a strength of 400 watts?

What is the decibel loss of a signal that starts at point A with a strength of 2000 watts and ends at point B with a strength of 400 watts?That is impossiple. If you have a light bulb of 100 watts. This power stays fixed in the lamp and does not decrease with distance!If you mean a sound source, only the sound pressure in pascals or the sound pressure level in dB is decreasing with distance from the source. Or the sound intensity in watts per square meter or the sound intensity level in watts is decreasing with distance.Scroll down to related links and look at "Sound intensity I and the inverse square law 1/r2"This question is talking about the fundamentals between data and signals in data communication. It has nothing to do with a light bulb and is not impossible. Here is how it is done. The decibel is a relative measure of signal loss or gain and is expressed as dB = 10 × log10 (P2 / P1) in which P2 and P1 are the ending and beginning power levels of the signal expressed in watts.So in this case where a signal starts with a strength of 2000 watts and ends with a strength of 400 watts the calculation would be:dB = 10 × log10 (400 /2000)dB = 10 × log10 (.2)dB = 10 × (-.69)dB = -6.9dB


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What is the deicbel loss of a signal that starts at point A with a strength of 2000 watts and ends at point B with a strength of 400 watts?

What is the decibel loss of a signal that starts at point A with a strength of 2000 watts and ends at point B with a strength of 400 watts?That is impossiple. If you have a light bulb of 100 watts. This power stays fixed in the lamp and does not decrease with distance!If you mean a sound source, only the sound pressure in pascals or the sound pressure level in dB is decreasing with distance from the source. Or the sound intensity in watts per square meter or the sound intensity level in watts is decreasing with distance.Scroll down to related links and look at "Sound intensity I and the inverse square law 1/r2"This question is talking about the fundamentals between data and signals in data communication. It has nothing to do with a light bulb and is not impossible. Here is how it is done. The decibel is a relative measure of signal loss or gain and is expressed as dB = 10 × log10 (P2 / P1) in which P2 and P1 are the ending and beginning power levels of the signal expressed in watts.So in this case where a signal starts with a strength of 2000 watts and ends with a strength of 400 watts the calculation would be:dB = 10 × log10 (400 /2000)dB = 10 × log10 (.2)dB = 10 × (-.69)dB = -6.9dB


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