Sound & Music: dB and Watts

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Homework Statement



  1. A stereo amplifier is rated at 130W output at 1150Hz. The power output drops by 10.0dB at 15.3kHz. What is the power output in watts at 15.3kHz?

Homework Equations


R=I/Io
Io = 10^-12 w/m2
bel = log(r)

The Attempt at a Solution


I have no idea where to start
 

Answers and Replies

  • #2
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This problem has nothing to do with sound radiating through the air. It has to do with how much the amplifier's output power rolls off vs. frequency. So if you connected the output of the amplifier to some kind of power meter or spectrum analyzer, you would measure a lower power out of the amp at 15.3 kHz than at 1150 Hz. If you measured 130 W out at 1150 Hz, how many Watts would you measure at 15.3 kHz by the power meter or spectrum analyzer?
 
  • #3
gneill
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I have no idea where to start
You'll need to do a bit better than that for an attempt. Perhaps you should start by investigating "decibel power ratio" as a search term.
 
  • #4
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Well I initially tried:
1150/15300 = 0.075 to get the ratio. And multiplied that by 130. But I know that can't be it because I don't think the relationship is linear like that. I don't know where to start because I don't understand the relationship between dB, Intensity, Watts and R
 
  • #5
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I have tried to look it up online. I see this equation:
RatioB = log10(P1 / P0)
So I tried log10(1150/15300) = -1.12399 But I don't know what this number is. I am assuming bels. So I would get -11.24 dB, but I don't know how I would turn that into watts.
I tried 10^-11.24 = .000000000005754. I don't think that can mean anything
 
  • #6
gneill
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The frequency values are unimportant: They don't enter into the calculation at all. The only important numbers are the 130 W and the 10 dB drop. 10 dB represents a power ratio in decibels. The formula that you found,

RatioB = log10(P1 / P0)

shows how to calculate the dB value from the actual power ratio (ratio of Watt values). Remember that the given 10 dB is said to be a drop, so make sure you give it the appropriate sign.
 
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  • #7
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So then it is:
1B = log10(P1/130)?
And I am looking for P1?
 
  • #8
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I can't figure out how to isolate P1 to solve for it. I don't know how to get it out of the log.
 
  • #9
collinsmark
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I can't figure out how to isolate P1 to solve for it. I don't know how to get it out of the log.

Exponentiate both sides of the equation. (hint: [itex] 10^{\mathrm{log}_{10} x} = x [/itex])

Edit: Oh, and don't forget @gneill's advice about the sign. It's a 10 dB drop.
 
  • #10
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Sorry but I still have no idea where to go with this. Thanks for your help though.
 
  • #11
gneill
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Sorry but I still have no idea where to go with this. Thanks for your help though.
You need to review the properties of the log function. It will come in handy again and again in your math and science studies.

The equation that @collinsmark gave in post #9 is the key to this problem. It shows you how to "get at" the argument of the log.
 
  • #12
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Here are a few examples of converting power in Watts to power in dBW. dBW means dB relative to 1 Watt. It may be more common to work with dBm (dB relative to 1 milliwatt), but I think it will be easier to work in dBW.

Convert 1 Watt to dBW: Power in dBW = 10log(1) = 0.000 dBW
Convert 2 Watts to dBW: Power in dBW = 10log(2) = 3.010 dBW
Convert 4 Watts to dBW: Power in dBW = 10log(4) = 6.021 dBW
Convert 8 Watts to dBW: Power in dBW = 10log(8) = 9.031 dBW
Convert 16 Watts to dBW: Power in dBW = 10log(16) = 12.041 dBW

Convert 1 Watt to dBW: Power in dBW = 10log(1) = 0.000 dBW
Convert 10 Watts to dBW: Power in dBW = 10log(10) = 10.000 dBW
Convert 100 Watts to dBW: Power in dBW = 10log(100) = 20.000 dBW
Convert 1000 Watts to dBW: Power in dBW = 10log(1000) = 30.000 dBW
Convert 10000 Watts to dBW: Power in dBW = 10log(10000) = 40.000 dBW

Note that for the first set of 5, the difference between any 2 adjacent calculations is approximately 3.0 dB. It is fairly well known for folks used to working with dB that doubling the power (in Watts) is equivalent to about a 3 dB increase. As you can see, every time the power (in Watts) doubled, the power in dBW increased by approximately 3 dB. NOTE: When you talk about the relative difference between 2 power levels, you use dB, not dBW. dBW is an absolute power, whereas dB is a relative power.

And for the next set of 5, the difference between 2 adjacent calculations is exactly 10.000 dB. If you increase the power in Watts by a factor of 10 times, the power in dBW increases by 10 dB.

Here is an example problem:
Let's say we have an amplifier manufacturer who sells an amplifier that outputs 50.0 Watts. Let's say that they happen to find a new output filter that has an insertion loss that is 0.4 dB less than what they currently have. That means that, using the new filter, the amplifier output power will increase by 0.4 dB. So using the new filter, what will the output power be in Watts?

The current amplifier output power is 50 Watts. In dBW, that is 10log(50) = 17.0 dBW
Our new filter has 0.4 dB less loss, which means the output power of our amplifier will increase to 17.0 + 0.4 = 17.4 dBW
Next convert that back to Watts by the inverse process, so: 17.4 dBW = log-1(17.4/10) = 54.8 Watts

Edit: I wanted to take that example problem a little further.
Let's say we have a 0.4 dB increase on an amplifier that outputs 2 Watts. What will its new output power be?
Convert 2 Watts to dBW: Power in dBW = 10log(2) = 3.01 dBW
Now add 0.4 dB so that the new output power will be 3.01 dBW + 0.4 dB = 3.41 dBW
Next, convert dBW back to Watts: 3.41 dBW = log-1(3.41/10) = 2.193 Watts

So we see that a 0.4 dB increase caused the 50 Watt amplifier to increase by 4.82 Watts, whereas a 0.4 dB increase caused the 2 Watt output to only increase by 0.193 Watts. But in both cases, it is the same percentage increase (or fractional increase). 4.8/50 = 0.096 and 0.193/2 = 0.096.

So for any output power, a 0.4 dB increase will cause a fractional increase of 0.096 in the Wattage. So a 0.4 dB increase for a 100 Watt amplifier would result in an output power of (100)(1 + 0.096) = 109.6 Watts.
 
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