Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

Frequency-weighting

927 views
Skip to first unread message

Jeff Stacey

unread,
Jul 26, 1996, 3:00:00 AM7/26/96
to

I am looking for the equations to calculate dB weighting factors,
particularly the one for the B-weight. (I have found equations for
A & C on the 'net, but I would like a reference that has them.)
If anyone can help me, I would greatly appreciate it.

Jeff Stacey
j...@altair.com

PS - tables/graphs are not what I would like

Neil Glenister

unread,
Aug 2, 1996, 3:00:00 AM8/2/96
to

In Article<31F8B1...@altair.com>, <j...@altair.com> write:

The s-domain transfer function for C-weighting is :-


Hc(s)= 4*pi^2*12200^2*s^2
_______________________________
(s+2*pi*20.6)^2(s+2*pi*12200)^2


Adding an extra real-axis pole to the C-weighting transfer function
gives us B-weighting :-


Hb(s)= 4*pi^2*12200^2*s^3
_____________________________________________
(s+2*pi*20.6)^2(s+2*pi*12200)^2(s+2*pi*158.5)

Adding two real-axis poles to the C-weighting transfer function gives
us A-weighting :-


Ha(s)= 4*pi^2*12200^2*s^4
______________________________________________________________
(s+2*pi*20.6)^2(s+2*pi*12200)^2(s+2*pi*107.7)(s+2*pi*737.9)

where pi=3.14159...etc and s is the complex variable.

If you are only interested in the steady-state response then the
weightings may be expressed in terms of frequency alone :-

Rc(f)= 12200^2*f^2
_________________________
(f^2+20.6^2)(f^2+12200^2)


Rb(f)= 12200^3*f^3
____________________________________________
(f^2+20.6^2)(f^2+12200^2)((f^2+158.5^2)^0.5)


Ra(f)= 12200^4*f^4
_____________________________________________________________
(f^2+20.6^2)(f^2+12200^2)((f^2+107.7^2)^0.5)((f^2+737.9^2)^0.5)

These filters show a loss at 1kHz of 2.0dB ,0.17dB , 0.06dB for A , B
and C weightings respectively and , since it is usual to normalise the
response of each filter to 1kHz , this loss must be added to the to the
modulus . In other words the responses may be expressed (in dB's) as follows
:-

C= 0.06 + 20*log(Rc(f))

B= 0.17 + 20*log(Rb(f))

A= 2.0 + 20*log(Ra(f))

I don't have the transfer function fo D-weighting easily to hand , but
I can tell you the position of the poles and zeroes (from IEC 537)

Poles Zeroes
-282.7 + j0 -519.8 + j876.2
-1160 + j0 -519.8 - j876.2
-1712 + j2628 0 + j0
-1712 - j2628


Regards ,

Neil R. K. Glenister

0 new messages