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#1 |
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diyAudio Member
Join Date: Mar 2002
Location: Stockholm, Sweden
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Hi all,
I want to actively bi-amp my 2-way Hales System Two Signature speakers. Paul Hales advised me to mesure the electrical transfer function provided to each driver and duplicate that response with an active crossover. Measurement is easy to do with the outboard crossovers, he said. Pardon my ignorance, but how do I do this? Thomas B. |
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#2 |
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diyAudio Member
Join Date: Jun 2001
Location: UK
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The transfer function is simply a swept-frequency plot of a circuits phase / amplitude response.
It can be easily done with a PC, half decent soundcard, and SpectraLAB FFT analyser software, or similar. A 30day trial is available which should give you enough time to do the job. Andy. |
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#3 |
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diyAudio Member
Join Date: Mar 2002
Location: Stockholm, Sweden
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Hi Andy,
Thanks for the reply. Do I measure each driver independently? If so, how? Do you know of any MacIntosh software? |
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#4 |
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diyAudio Member
Join Date: Mar 2002
Location: Upstate NY
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#5 |
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diyAudio Member
Join Date: Dec 2001
Location: Bremerton, WA.
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Thomas,
You'll have to measure each crossover output individually. Use both channels of the line-input from your sound card. One channel is the reference and attaches to the crossover input, and the other channel attaches to the appropriate crossover output. You can use white/pink noise for the input to your power amp and then select the transfer function mode from your spectrum analyzer program. It should draw you a smooth curve in just a few seconds. Be careful with the voltage since the inputs of your sound card can be damaged. SpectraPlus/Lab will allow you to save the results in ascii format which will allow importation as a "target" into design programs like LspCAD or SoundEasy. You can then use the optimization feature of these programs to automatically duplicate the responses with an active equivalent. Keep in mind that you might be able to achieve a superior result to the original with the extra versatility of an active crossover. A MAC software application should work equivalently. Cheers, Davey. |
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