# Thread: Re: pole-zero for Milne-Shaw seismometer

Started: 2012-01-23 19:18:12
Last activity: 2012-01-25 00:17:47
Topics:
Sheila Peacock
2012-01-23 19:18:12
Caroline Holden wrote, on 20th January 2012:

Hello,

could you please let me know what would be the sac polezero file for a Milne-Shaw instrument with the following characteristics:
Seismometer period Ts=12
Damping ratio 20:1
Magnification 250

Thank you very much for your help!

Caroline

PS: my guess is:

Zero 2
poles 2
-0.3613 0.3790
-0.3613 0.3790
Constant 250*2*pi

I looked up in Scherbaum ("Of Poles and Zeros", Kluwer, 2001) and
found the following.

First of all, the second pole must be the complex conjugate of the
first, so the second 0.3790 should be negative.

Secondly, I'm not sure how the "damping ratio" above fits into the
equations on p. 57 of Scherbaum. Does it correspond to the amplitude
ratio of two consecutive maxima, or of two consecutive extreme values
i.e. maximum and the following minimum?
The logarithmic decrement, Lambda, is given by ln (ratio of two consecutive
maxima), or 2 x ln (ratio of two consecutive extremes).

On p. 64 of Scherbaum the poles for an underdamped system are given by

p1,2 = -(h +/- i sqrt (1-h^2)) x omega0

where h is the damping coefficient and omega0 is given by

2 x pi / T0,

and T0 is your 12 s.

On page 57 the logarithmic decrement Lambda is given as

Lambda = 2 x pi x h / sqrt(1-h^2).

I reproduced your two poles using these formulae and with the second
definition of Lambda, i.e. that your "damping ratio" 20:1 is the
ratio of two consecutive extremes of the response. I got a
damping coefficient h = 0.69 which is close to the value of
0.7 said by my senior colleague to be favoured by operators
of this type of seismometer.

The two zeros are necessary to convert the response in
acceleration to displacement.

If the magnification of 250 is the magnification of displacement
then the constant required by SAC is equal to the ratio of the
zeros polynomial to the poles polynomial at 12 s (Scherbaum p. 40),
multiplied by the magnification. Here it is 1.38x250 = 345.0
(I use a program for doing the polynomial ratio - it needs
the ability to handle complex numbers, which is in fortran or Matlab).

There should be a further factor introduced by the digitiser,
with units of digital counts per nanometre (or similar), which
will need to be incorporated into the constant.

I hope that's some help - and that it's correct, which is why I'm
sending it to the list, so that others can tell me if it isn't.
I find it very easy to get mixed up about whether to multiply or
divide the constant by the magnification.

Regards,
Sheila Peacock
AWE Blacknest.

• Caroline Holden
2012-01-25 00:17:47
Thank you very much Sheila, that's very clear now,

only one last point, regarding the constant,
I read somewhere that we need to multiply it by 2*pi to follow sac
conventions ?
A0 then becomes 1.38*250*2*pi

Kind regards,

Caroline

------------------------------------------
GNS Science - Te Pu Ao
1 Fairway Drive, PO Box 30368, Lower Hutt, New Zealand
Web: www.gns.cri.nz
Email: c.holden<at>gns.cri.nz

Sheila Peacock <sheila<at>blacknest.gov.uk>
Sent by: sac-help-bounces<at>iris.washington.edu
24/01/2012 00:34

To
"sac-help<at>iris.washington.edu" <sac-help<at>iris.washington.edu>
cc

Subject
Re: [SAC-HELP] pole-zero for Milne-Shaw seismometer

Caroline Holden wrote, on 20th January 2012:

Hello,

could you please let me know what would be the sac polezero file for a
Milne-Shaw instrument with the following characteristics:
Seismometer period Ts=12
Damping ratio 20:1
Magnification 250

Thank you very much for your help!

Caroline

PS: my guess is:

Zero 2
poles 2
-0.3613 0.3790
-0.3613 0.3790
Constant 250*2*pi

I looked up in Scherbaum ("Of Poles and Zeros", Kluwer, 2001) and
found the following.

First of all, the second pole must be the complex conjugate of the
first, so the second 0.3790 should be negative.

Secondly, I'm not sure how the "damping ratio" above fits into the
equations on p. 57 of Scherbaum. Does it correspond to the amplitude
ratio of two consecutive maxima, or of two consecutive extreme values
i.e. maximum and the following minimum?
The logarithmic decrement, Lambda, is given by ln (ratio of two
consecutive
maxima), or 2 x ln (ratio of two consecutive extremes).

On p. 64 of Scherbaum the poles for an underdamped system are given by

p1,2 = -(h +/- i sqrt (1-h^2)) x omega0

where h is the damping coefficient and omega0 is given by

2 x pi / T0,

and T0 is your 12 s.

On page 57 the logarithmic decrement Lambda is given as

Lambda = 2 x pi x h / sqrt(1-h^2).

I reproduced your two poles using these formulae and with the second
definition of Lambda, i.e. that your "damping ratio" 20:1 is the
ratio of two consecutive extremes of the response. I got a
damping coefficient h = 0.69 which is close to the value of
0.7 said by my senior colleague to be favoured by operators
of this type of seismometer.

The two zeros are necessary to convert the response in
acceleration to displacement.

If the magnification of 250 is the magnification of displacement
then the constant required by SAC is equal to the ratio of the
zeros polynomial to the poles polynomial at 12 s (Scherbaum p. 40),
multiplied by the magnification. Here it is 1.38x250 = 345.0
(I use a program for doing the polynomial ratio - it needs
the ability to handle complex numbers, which is in fortran or Matlab).

There should be a further factor introduced by the digitiser,
with units of digital counts per nanometre (or similar), which
will need to be incorporated into the constant.

I hope that's some help - and that it's correct, which is why I'm
sending it to the list, so that others can tell me if it isn't.
I find it very easy to get mixed up about whether to multiply or
divide the constant by the magnification.

Regards,
Sheila Peacock
AWE Blacknest.

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