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Calculating power and energy based on frequency and amplitude (pascal) 
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#1
Jul414, 06:30 PM

P: 173

Hello,
I'm working on a project to absorb sound energy and I'd like to ask for hints, or methods, of calculating the energy received by the microphone, or energy per second (power). I took measurements using a very sensitive microphone which outputs a voltage to a DAQ card in a PC. Using labview I convert this voltage into Pascals using the sensitivity and pickup or response of the microphone. Now I have an FFT of the time trace showing me the amplitude in Pascals (I think...) and the frequency as domain. I'd like to now figure out how I can calculate how much sound energy the microphone is receiving at each frequency. (Then when I add in my acoustic damping I can say and measure the change in peak height at each frequency I can get a number for how much energy I'm asborbing). I know that energy is proportional to the square of the amplitude, so I can get a "Factor of X decrease" in energy or power. But, I'd like to get an actual number in microwatts or microjoules of what effect this absorption has. I also know I'll have to make an assumption that the sound field is diffuse (or evenly spread). One quick question: pressure recordings should not depend on microphone orientation with respect to the direction of sound waves travelling, since pressure is directionless  correct? I've tried some dimensional analysis starting with N/m^2, 1/s, and trying to "get" to units of energy, but it feels like there's a lot of different arbitrary paths to get there. Suggestions, hints, provoking thought experiments, and the like, are more than welcome!!! Best, H2bro 


#2
Jul414, 08:27 PM

P: 143

Can't you just convert to Decibels and see how a sound insulator effect the DB level? I would think you could just use a spectrum analyzer which reads out in DB. Does your project demand to readout microwatts?
I found this link that does it for you: http://www.sengpielaudio.com/calculatorsoundlevel.htm Any help? 


#3
Jul414, 11:03 PM

P: 173

I am doing this as a 1st year undergraduate working in a physics lab over the summer. I'd like to be able to quantify and prepare a report of what my project accomplished, and ideally, I can report something along the lines of "these insulators dissipate X microwatts of sound energy each across a frequency from 5  100hz" or something along those lines.
In other words, it is easy to measure the difference in sound pressure level. I can even say something like "I've reduced the ambient sound energy by a factor of X at Y frequency". But it is much more interesting, in my eyes, to give it an absolute reference scale. 


#4
Jul414, 11:31 PM

P: 316

Calculating power and energy based on frequency and amplitude (pascal)
If you can read voltage data in real time, then make a simple circuit with the microphone and a resistor. Divide voltage data by the resistance to get the current. multiply voltage and current together and you've got the power output.



#5
Jul514, 01:44 PM

P: 173

In other words, there's no good argument to believe that the actual changes in electric potential of the capacitors is capturing ALL of the sound energy that hits the diaphragm. Maybe I could go about making some kind of transfer function but the manufacturers already did that when they gave me a calibrated voltage/pascal rating. Also, between the microphone and my data acquisition module is a large amplifier. This is necessary since the microvolt fluctuations passed down the (low noise) cable won't get picked up by a computer DAQ card (they are likely puny compared to voltages within the card itself). I have a feeling there is a more straightforward mathematical approach. I have a signal with a Pascal Amplitude and domain of 20 seconds. I take an FFT of this to produce an Amplitude (still pascals? I'll have to investigate that) versus frequency domain. I'd now like to produce an "X energy in 20 seconds at this frequency (or range of freqs)" or "X energy/20 seconds (power) at this freq". 


#6
Jul514, 02:17 PM

P: 316




#7
Jul514, 04:45 PM

P: 173

I'm interested in measuring the amount of sound energy absorbed by some acoustic attenuation I'll be placing in the room. sorry if that wasn't clear.
Also, I mentioned that the microphone is connected to a large amplifier, which drastically increases the voltage. The real solution is a mathematical technique of FFT analysis of which I am unaware. I do appreciate your thoughtful suggestion and I enjoyed the clever workaround to find the power dissipated. 


#8
Jul514, 08:55 PM

P: 316

Oh, I misunderstood then. I thought you wanted to find how much energy the microphone absorbed. as for your question about pressure recordings and microphone orientation, the design of the enclosure around the microphone will effect the magnitude of the readings, and so I would expect orientation to have an effect too. This can easily be checked by recording the same source in different orientations.



#9
Jul514, 10:19 PM

P: 316

Since I've eaten up a lot of your thread, I figure I owed you one. I've spent the last hour or two thinking about this.
Given p (pressure, pascals) and f (frequency) E (rms energy density) [itex]= \frac{p^2}{c^2\rho}[/itex] from wikipedia (I know. but it's a starting point) but it's also on this nice acoustic equations cheat sheet. where c is the speed of sound and [itex]\rho[/itex] is the density of air. Assuming you aren't intentionally creating large pressures, c and [itex]\rho[/itex] can be assumed constant. (as you pointed out, there's a nice [itex]p^2[/itex] term.) If you're uncomfortable assuming [itex]\rho[/itex] is constant, you can use from the ideal gas law [itex]\rho = \frac{p}{RT}[/itex] and assume R and T are constant. (The specific gas constant for dry air is 287.058 J/(kg·K)) Since you have P(f), this will give you E(f), rms energy density at the point of measurement, as a function of frequency. Do this all around the room and you can create a 'map' of the energy density in the room. Create a model density function from those points to fit your 'map', and integrate it over the room's volume, and you'll have the total rms energy. Do this with and without your absorber in place, and then look at the differences (J absorbed vs f) or the ratios with some standard value (dB vs f). 


#10
Jul3014, 03:30 PM

P: 173

best 


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