Friday, February 22, 2008

Audio Engineering Session 6

This weeks theory focused on acoustics and it is going to be a bit more difficult to condense down into bite size chunks like I have in past blogs. It is also testing my understanding of the topic at hand to the maximum. So with that being said, let me go make a sandwhich cause I think we are in for a long read.
There was mention of a computer application that will calculate the ideal speaker locations for any given room. You simply input the room data, such as dimensions and furniture etc and let the program calculate the optimum speaker placement. However, due to the complexity and countless permutations it can take a week or two to generate the information. Yes, a week or two (you read that right). A specific application was not mentioned as we were talking about the application functionality in general. I did some hunting on the web and found this one. TOA Speaker System Design Software. I am downloading it for free from their website as I write this blog. I'll be checking it out and I'll post my review in a separate blog entry. http://www.toaelectronics.com/speaker_software.asp UPDATE: I only had a chance to briefly check out this software package and I'm not sure that it will do what I described earlier. I will dig into this a little further at a different time.

We moved on to discussing "Room Modes" and why they are important to an Audio Engineer. A Room Mode Calculator can be found here http://www.mcsquared.com/metricmodes.htm Now then, I will try and explain room modes. A "Room Mode" is a set of frequencies that resonate between parallel surfaces. Room modes are calculated using the simple formula; frequency equals velocity divided by wavelength. For every 1000 hertz there are 9 modes. If you are like me you are asking yourself, why do I need to know about room modes? It is important to understand room modes as it will affect the quality of sound as the frequencies build up. Certain frequencies will get louder. The best example is to use the worst type of room, and that would be a cube. In this example we will say that this cube room is 10' x 10' x 10' to make the math simple. Using the formula frequency=velocity divided by wavelength we can calculate the frequencies. We know that the wavelength is 10 feet and the velocity is 1100 so we can then tell that the frequency equals 110 hz. This figure increases by 110 hz so then you would have 110, 220, 330, 440, 550, 660 etc. However, in this example we can cut the 110 hz in half to 55 hz. 55 hz is equal to the A string on a bass guitar. So in that example room the A string would sound louder because the room will resonate the frequencies more and they pile up to make it louder. Now that the problem frequencies can be identified, we can move on to treating the room. More on treating rooms in a future blog.

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