# displaying indexed values correctly

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harley on 7 Aug 2013
hi, trying to display some values but am not sure how to go about it. I want to do is; find Vb at half of Vb_ev and then find the corresponding pH value at that point. see last 2 lines. any help will be appreciated.
format short
Kw = 1e-14;
Ka = 1.755388e-5;
Ca = 0.5;
Cb = 0.1;
Va = 100;
Vb = 0.05:0.05:2000;
Ma = (Ca * Va) / 1000;
Mb = (Cb .* Vb) ./ 1000;
for i = 1:length(Mb)
M_excess = Ma - Mb(i);
if abs(M_excess)<eps
Hplus = Ka * ((Ma_final * 0.999999) ./ Mb_final);
Vb_ev = Vb(i);
pH_ev = pH(end);
elseif M_excess > 0
Ma_final = (M_excess * 1000) ./ (Va + Vb(i));
Mb_final = (Mb(i) * 1000) ./ (Va + Vb(i));
Hplus = Ka * (Ma_final ./ Mb_final);
elseif M_excess < 0
OH = (M_excess * 1000 * (-1)) ./ (Va + Vb(i));
Hplus = Kw ./ OH;
end
pH(i) = -log10(Hplus);
end
%
%
Vb_Mid_Pt1 = uint16(Vb_ev)*0.5;
pH_Mid_Pt1 = pH(Vb_Mid_Pt1)
i tried the above but the pH_Mid_Pt1 gives me the value in the pH array that is equal to the actual numerical value of that from (Vb_ev)*0.5.
harley on 7 Aug 2013
its gives me the pH value in column 250 of the pH array, (using above code), i want it to give me the value of pH when Vb = 250 (half of Vb_ev). thanks again.

Richard Brown on 7 Aug 2013
Similar to what you asked before, I think what you want is to find the Vb element that is closest to half of Vb_ev, and then pluck out the corresponding pH element. The answer is very similar to what I gave you last time:
[~, idx] = min(abs(Vb - Vb_ev/2));
Vb_Mid_Pt1 = Vb(idx);
pH_Mid_Pt1 = pH(idx);
I think what's throwing you off is the distinction between values and indices. You are wanting to find certain indices in your array that have values near other specific ones. Hence the use of min with abs, which provides the index of the closest point.
Richard Brown on 7 Aug 2013
Indeed. And for this problem it is likely that you'd end up halfway between two sample values. Using interpolants would be the next step in sophistication