ГОСТ 31191.5—2007
-0.00078 -0.00261 0.00771 0.00600 -0.00908 0.00504 0.00135;
-0.00405 -0.00210 0.00520 0.00176 -0.00465 -0.00198 0.00451.
0.00563 0.00218 -0.00105 0.00195 0.00296 -0.00190 0.00306;
-0.00372 0.00037 -0.00045 -0.00197 0.00289 -0.00448 0.00216;
-0.31088 -0.95883 -0.67105 0.14423 0.04063 0.07029 1.0330);
w = (57.96539 52.32773 49.78227 53.1688556.02619 -27.79550 72.34446. 21.51959).
%calculat»on of alz(t)
fort = (9 ; length (asz));
forJ=1 :7
x(t. j) = sum (alz (t- 1:-1:1 - 4). *4v(1 : 4.))) + sum <asz(t -1 :-1 : t- 8). *w(5:12. J))
♦w(13.J);
x(t. l) = lanh (x (t. J)):
end
alz(t) = 8um(W(1 ; 7). *x(t. 1 :7)) *W(8);
end
alz=atz(9:length (asz));
%call the function CountPeaks lo calculate Dz
Dz=CouhtPeaks (alz. ’z’);
%plot the result Infigure 3
plot (time, alz)
title (zfUe)
legend <(’Dz = ‘ , num2str(Dz)). 1)
%add the time column to the calculated response and the calculated value Dz to the last rowand second
%cotumn
alz=(time. aiz;0Dz);
((’save’. path, zfile,*_al. txt alz-ascii -tabs •))
function Dk=CountPeaks (alk. xyz)
%CountPeaks: Calculates Dk from the inputsignal alk.
%Dk:Calculated output value Dk.
%alk:Input vector: One column with the response alk.
%xyz:String expression: ‘x’, у or ’z’ depending on which direction the Input vector represents.
Dk=0;
1=1;
Id1=1;
id2=1;
%set thearray pointers idl and id2 each time signal changes sign (♦/-) and find the maximum between id1 and
%ld2
for 1=1 : length (alk)-1
if (((alk (i) > 0) & (alk (i ♦ 1)< 0)) | ((alk (i) < 0) & (alk (i ♦ 1)> 0)))
id2=l;
ifalk (Id2) < 0
(mx. md) =mm ((alk (id1 : id2))).
•f((xyz==’ z ’) |(xyz==’ 2 ’))
mx=0,
end
else
(mx. md)«max ((alk (Id1 : Id2)));
end
Dk=Dk»mxA6;
Id1=ld2;
end
end
Dk=Dk A (1/6)
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