# Clarity C-80 and Impulse response



## thaumas (Nov 6, 2012)

Hi to you all,

I've used REW for a few weeks now regarding an acoustic analysis of an amphitheater in my university. I arrive to some good result but I have some hard time to extract information from my graphs. My goal is to compute the clarity C-80 which is ten time the log(10) of the integration of energy on the 80 first milliseconds divided by the integration of energy on the rest of the time [80ms to infinity].

C-80 = 10 log( ∫[0,80] E(t) dt / ∫[80,∞] E(t) dt)

My guess was to take the impulse response of my amphitheater, then as my schroeder integral was quite linear I approximate my decreasing of intensity by

Lp(t) = - at,

with Lp in dB intensity in function of time. Now I remember my formula

Lp = 20 log (P(t)/P0)

with P0 = 2*10⁵, then I extract P(t)²

P(t)² = 10^( -at /10) - P0²

which is the energy E(t), so finally I integrate (I use wolfram Alpha) (b = a/10)

∫ [0,t]P(t)² dt = [-t^(-bt) / (b log 10) + P0²t] [0,t]

With a = 28,0 dB/S my final value for C-80 is 13,5dB which, first, is huge and then I have lots of difficulties to find the meaning of the values of C-80. I look up lots of stuff but it was really confusing and unclear. I know that it should be between +/- 3dB but I couldn't find what does this number tells us and does this value apply for a room made for speech. Finally I did the same with another amphitheater which is really good acoustically speaking and my value for C-80 was - 13dB. So I wonder...
One of my worried is that my approximation made the result to far away from reality :| and I don't now what to do otherwise.

I upload the two impulses responses that allow me those computation. The red one is the main amphitheater with a C-80 of 13,5 dB the green one correspond to the C-80 of -13 dB

Well thanks for any of your thought.

Thaumas


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## jamin (Jun 3, 2006)

I’ll go out on a limb and take a crack at this. Remember that the coffee may not be working and my limbs often crack!

For 20Log(P(t)/Po)=C*t

I get (P(t))^2 = (Po)^2*10 ^ (C*t /10)
And letting 
K=(Po)^2
E(t)=(P(t))^2
D=C/10

Yields
E(t)=K*10^(D*t)

With the integral of E(t) dt showing as
K*(10^(D*t)) / (D*Ln(10))

And since C_80 is the ratio of integrals the constants 
K
And
D*Ln(10)
Both drop out leaving only

[10^(D*(.08)) - 10^(D*(0))] / [10^(D*(infinity)) - 10^(D*(.08))] 
to be evaluated.

Realizing that D = C/10
And C would be (–a) in your example.

This may yield a better result.
Or, I could have missed completely !!

PS. FWIW Po is equal to 2E(-5) not 2E(5)


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## thaumas (Nov 6, 2012)

Thanks Jamin! I see that your limbs didn't crack!  I finallly get the mistake now.:gulp: And I find a much better result around C-80 = -1,7 for the red graph (the main study) even though the other one is still a mystery C-80 = 20. :| 
Now that the computation are fixed I would need advice about the meaning of the result if you anybody know something about it'll be great  by the way I still wonder about the aproximation :/

thanks for just thinking about 

Thaumas


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