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How does RF laminate size changes vs temperature? Practical calculations question.

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Terminator3

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For example, laminate have Thermal Coefficient of Expansion (CTE) of 18 PPM/'C ( 18 × 10−6) in X-Y plane.
I want to calculate thermal expansion in mm for temperature swept from 0 to 100'[C].

Is that right:
18 x (10^-6) * 100[C] = 18 x (10^-4) = 0.0018
so 10.0 x 10.0 mm board becomes 10+10x0.0018 = 10.018 x 10.018 mm?

Thanks
 

On high frequencies it is a huge problem, cant use cheap board for cheap oscillator, frequency shift is too big for transceivers even in room temp.
 

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