Since 10 is midpoint between 5 and 25, linear interpolation would simply be the average. But, in general, interp1 is your friend...
>> interp1([5 25]',y,15)
ans =
6.40 7.50 6.30 8.05 9.45 8.60 6.20 6.15 6.10 6.55 6.65 ...
>> mean(y)
ans =
6.40 7.50 6.30 8.05 9.45 8.60 6.20 6.15 6.10 6.55 6.65 ...
Logarithmically, ...
>> 10.^(interp1([5 25]',log10(y),15))
ans =
Columns 1 through 9
6.3718 7.3485 6.2929 7.4900 9.1531 7.9825 6.0828 6.0216 6.0000
Columns 10 through 18
6.4653 6.5666 6.5177 6.9282 5.9127 5.4083 6.7454 7.2360 5.5678
Columns 19 through 24
5.6569 5.3479 5.0299 6.2992 7.8975 7.9875
>>
Best Answer