Calculation of cosh, sinh big numbers

Question

john 2014-2-27

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/119422-calculation-of-cosh-sinh-big-numbers

评论： john 2014-3-9

采纳的回答： Roger Stafford

在 MATLAB Online 中打开

Hi,

let t is time. How can I calculate and plot:

for i=1:10
  y=0.912*sin(314.0*t) - 3.31*cos(314.0*t) + (3.31*(cosh(1399.0*t) + 0.484*sinh(1399.0*t)))/exp(1500.0*t)
current(:,i)=(eval(y))'
end

I got NaN from cosh and sinh for time 0.51....how can I block cosh, sinh if actual value is NaN?

Thank you

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Answer 1

Roger Stafford 2014-2-27

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https://ww2.mathworks.cn/matlabcentral/answers/119422-calculation-of-cosh-sinh-big-numbers#answer_126425

编辑：Roger Stafford 2014-2-27

All three quantities, cosh(1399.0*t), sinh(1399.0*t), and exp(1500.0*t), will fall far beyond the realmax limit. I would suggest you refer to the definitions of sinh and cosh and modify your expression as follows. Replace cosh(1399.0*t)/exp(1500.0*t) by (exp(-101*t)+exp(-2899*t))/2 and similarly with sinh. That way you will not get NaNs even though these are equivalent expressions.

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Roger Stafford 2014-3-9

在 MATLAB Online 中打开

I would replace that particular expression with this equivalent one:

 a=[ -5.1     0.794  -0.794  -0.794   4.06    4.06  ;
      5.1    -0.794   0.794   0.794   4.06    4.06  ;
      5.1    -0.794   0.794   0.794   4.06    4.06  ;
      0.912  -3.31    3.31    3.31    0.484   0.484 ;
      0.912  -3.31    3.31    3.31    0.484   0.484 ;
      4.19    2.52   -2.52   -2.52   -0.644  -0.644 ] * ...
      [ sin(314.0*t)*exp(-1500.0*t);
        cos(314.0*t)*exp(-1500.0*t);
        exp(-101*t)/2              ;
        exp(-1899*t)/2             ;
        exp(-101*t)/2              ;
       -exp(-1899*t)/2             ];

I don't know any way of doing this sort of thing "automatically", but maybe the method you see here will suggest some kind of simplification for whatever you are doing. You will notice that the terms in the above never grow large with t. In fact they will essentially disappear after t takes on appreciable sizes.

Note: I would suggest that you let t = 0.001*rand several times and for each value of t compare your original 'a' with the above 'a' to ensure that they are essentially equivalent and that I didn't make some algebraic blunder.

john 2014-3-9