 Posted: Fri May 06, 2011 1:52 pm |
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| voetbal |
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| Joined: 06 May 2011 |
| Posts: 4 |
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| the results givin for RT schroeder is weird. Like in the tutorial p96 they say it is incorrect. But they dont give a proper answer. So if i see on the graphic p96 the RT601 is then 5,8s or am i wrong. How can i find the RT60 on this curve.
Can anyone help me please.
Thanks and best regards!!
a student in a need of help |
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| Posted: Mon May 09, 2011 2:11 pm |
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| Waldemar |
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| Joined: 05 Dec 2005 |
| Posts: 112 |
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| As you can read at page 95, the graph at page 96 shows the curves for "No Noise Compensation". To get better results for T20 and T30 you should calculate the RT with "Noise Calculation" ON in the Options then. |
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_________________
Best Regards
SDA, Waldemar Richert |
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| Posted: Mon May 16, 2011 10:11 am |
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| voetbal |
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| thanks waldemar,
i will now use it with noise compensation.
But I have a questtion: If i want to know the T60 from the curve on p95 is it then T30*2? so 6.13*2=12.26s I dont think this is realistic. Or i am doping it wrong, i dont know. How would you get the T60 from this grafic and diagram?
voetbal |
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| Posted: Mon May 16, 2011 10:27 am |
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| Waldemar |
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| Joined: 05 Dec 2005 |
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| The numbers for the T10 ... T30 are extrapolated already for 60 dB decay.
So the T30 = 6.13s is the value for T60.
As far as the curve is not linear you should take the T20 = 5.4s as Reverberation Time (T60). The inaccuracy is about 1/2 second! |
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_________________
Best Regards
SDA, Waldemar Richert |
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