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Asked by Lorenzo
on 19 Nov 2012

Dear all; I'm trying to work out a script which should give a ver small value as a result. The problem is that I keep getting 0 instead…

Any help would be really appreciated.

Here's the script:

rho=7860; B=-1.00000000011701; BL=23.562; Bj=46.9362549800797;

k=1;

int1(k)=(exp(-2*BL(k))*(2*exp(2*BL(k))*sin(2*BL(k))+(-4*exp(3*BL(k))-4*exp(BL(k)))*sin(BL(k))+(4*exp(BL(k))-4*exp(3*BL(k)))*cos(BL(k))+exp(4*BL(k))+8*BL(k)*exp(2*BL(k))-1))/(8*Bj(k));

int2(k)=-(exp(-2*BL(k))*(2*exp(2*BL(k))*sin(2*BL(k))+(4*exp(3*BL(k))+4*exp(BL(k)))*sin(BL(k))+(4*exp(BL(k))-4*exp(3*BL(k)))*cos(BL(k))-exp(4*BL(k))+1))/(8*Bj(k));

int3(k)=-(exp(-BL(k))*((exp(2*BL(k))-1)*sin(BL(k))-exp(BL(k))*((exp(BL(k))+exp(-BL(k)))/2)^2+exp(BL(k))*cos(BL(k))^2))/(2*Bj(k));

m_cap1=rho.*(int1+B.^2.*int2+2.*B.*int3);

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Answer by Matt J
on 19 Nov 2012

Edited by Matt J
on 19 Nov 2012

You haven't said what variable in your code is the problematic one.

Regardless, computers are inevitably discrete things. If you get too close to 0, then the number will have to get rounded to zero in order to be represented in floating point.

Matt J
on 21 Nov 2012

If Matt Fig's suggestion doesn't work for you (because you don't have the Symbolic Toolbox), then there is this and

http://www.mathworks.com/matlabcentral/fileexchange/22725-variable-precision-integer-arithmetic

and others like it on the FEX.

However, it would be wise to re-evaluate your approach. It looks very dubious to be working with numbers as huge as

>> exp(4*BL)

ans =

8.5386e+40

and still requiring precision<1.

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