How to normalize the baseline of multiple Differential Pulse Voltammetry Curves?

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Samiha 2024-5-8

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https://ww2.mathworks.cn/matlabcentral/answers/2116786-how-to-normalize-the-baseline-of-multiple-differential-pulse-voltammetry-curves

评论： Star Strider 2024-5-9

DPV.xlsx

I have a few DPV curves for different concentration that I plotted in the same graph. But the baselines of the DPV curves do not match. So I want to normalize the baseline using Matlab. (So that the baseline is 0) My DPV curves looks like the following:

Can someone please help me with this? I also added the excel file with the data here.

Thank you!

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Answer 1

Star Strider 2024-5-8

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Without actually having the data to work with, my first approach would be to use the detrend function on each curve.

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Star Strider 2024-5-9

在 MATLAB Online 中打开

DPV.xlsx

My pleasure!

Using detrend did not work the way I wanted it to, so I created my own detrending approach. It calculates the slope and intercept (columns of the ‘B’ matrix) of the baseline (assumed to be linear) from the first and last elements of the vectors (and experimented with the 10th and last elements), and then creates a linear fit to the entire ‘Voltage’ variable and subtracts that from the original variable for each the ‘Current’ variables. That works better than detrend for these data, however it may be necessary to experiment further, depending on the result you want.

I am not certain that it is possible to perfectly detrend the baselline without distorting the curves themselves.

Try this —

T1 = readtable('DPV.xlsx','VariableNamingRule','preserve')

T1 = 61x8 table

Voltage Current (0.1)Current (0.3) Var3 Current (0.5) Current (0.7) Current (1) Current (2) Current (3) _______ __________________________ _______ _____________ _____________ ___________ ___________ ___________ -300 0.88631 0.14617 0.26339 0.52428 0.61456 0.51742 0.4885 -290 1.2188 0.14004 0.4203 0.6438 0.66459 0.55903 0.52387 -280 1.4655 0.17436 0.55381 0.62123 0.63863 0.55029 0.51182 -270 1.7461 0.35175 0.67868 0.62086 0.65448 0.54676 0.50542 -260 1.655 0.547 0.69532 0.62473 0.61814 0.53223 0.47636 -250 1.6687 0.78629 0.71277 0.63484 0.64034 0.54321 0.48141 -240 1.6236 0.97004 0.71882 0.63707 0.65091 0.55595 0.48179 -230 1.5832 1.115 0.72875 0.6572 0.66627 0.58214 0.49861 -220 1.6755 1.137 0.7347 0.65705 0.68499 0.58991 0.50568 -210 1.7379 1.1966 0.75801 0.69857 0.70984 0.63047 0.52962 -200 1.715 1.2094 0.77933 0.72121 0.74034 0.6553 0.55002 -190 1.7591 1.2586 0.85322 0.76694 0.79021 0.69411 0.58604 -180 1.8078 1.2954 0.84447 0.79708 0.81889 0.72845 0.59764 -170 1.8551 1.3269 0.88539 0.8395 0.8516 0.77739 0.63474 -160 1.9023 1.3562 0.91233 0.86336 0.87275 0.80403 0.67134 -150 1.9921 1.4461 0.97145 0.9101 0.91852 0.85444 0.72822

VN = T1.Properties.VariableNames;

figure

plot(T1.Voltage, T1{:,2:end})

grid

xlabel('Voltage (mV)')

ylabel('Current (\muA)')

title('Original')

legend(VN{2:end}, 'Location','best')

for k = 1:size(T1,2)-1

B(:,k) = [T1{[1 end],1} ones(2,1)] \ T1{[1 end],k+1};

% Q = [T1{:,1} ones(size(T1{:,1}))] * B(:,k)

Idt(:,k) = T1{:,k+1} - [T1{:,1} ones(size(T1{:,1}))] * B(:,k);

end

figure

plot(T1.Voltage, Idt)

grid

xlabel('Voltage (mV)')

ylabel('Current (\muA)')

title('Linear Detrending: [1 end]')

legend(VN{2:end}, 'Location','best')

for k = 1:size(T1,2)-1

B(:,k) = [T1{[10 end],1} ones(2,1)] \ T1{[10 end],k+1};

% Q = [T1{:,1} ones(size(T1{:,1}))] * B(:,k)

Idt(:,k) = T1{:,k+1} - [T1{:,1} ones(size(T1{:,1}))] * B(:,k);

end

figure

plot(T1.Voltage, Idt)

grid

xlabel('Voltage (mV)')

ylabel('Current (\muA)')

title('Linear Detrending: [10 end]')

legend(VN{2:end}, 'Location','best')

Most of the problems are in the region of the lowest voltages. It might be possible to simply use:

Idt(:,k) = max([zeros(size(Idt(:,k))) Idt(:,k)], [], 2);

to even out the baseline (delete the negative values) if the current response at the lowest voltages are not requred to be retained. If so, put this after the initial ‘Idt(:,k)’ calculation in any loop that begins with an index larger than 1 (for example, in the second loop here that begins with 10). That provides an even baselilne at the expense of eliminating any negative current values that may result from the detrending process.

.

Star Strider 2024-5-9

在 MATLAB Online 中打开

DPV_old.xlsx

This approach works for all except the first set (‘I1’), and I cannot devise an appropriate integer-order polynomial that works correctly with it. The others detrend well with a 5th-order polynomial.

Try this —

T1 = readtable('DPV_old.xlsx','VariableNamingRule','preserve')

T1 = 81x4 table

Voltage I1 I2 I3 _______ _________ _________ ________ -300 1.2998 2.6418 0.92055 -290 0.17632 0.20264 3.5631 -280 0.14736 0.15839 3.4154 -270 0.1227 0.14206 0.82172 -260 0.12101 0.12943 0.23768 -250 0.063406 0.094923 0.14846 -240 0.066088 0.075111 0.14889 -230 0.076057 0.064936 0.11877 -220 0.041663 0.049986 0.073133 -210 0.023613 0.029505 0.08501 -200 0.017722 0.028476 0.049472 -190 0.02553 0.025718 0.04199 -180 0.032825 0.028149 0.053867 -170 0.030347 0.0060319 0.038857 -160 0.0075581 0.02 0.013637 -150 0.032135 0.013491 0.027272

VN = T1.Properties.VariableNames;

figure

plot(T1.Voltage, T1{:,2:end})

% plot(T1{:,1}, T1{:,2})

grid

xlabel('Voltage (mV)')

ylabel('Current (\muA)')

title('Original')

legend(VN{2:end}, 'Location','best')

% hold on

% for k = 1:size(T1,2)-1

% % plot(T1{Lv,1}, T1{Lv,k+1}, 's');

% Lv1 = (T1{:,2} <= mean(T1{:,2})+3);

% Lv2 = ischange(T1{:,2}, 'linear', 'Threshold',1E-9);

% Lv = Lv1 & Lv2;

% Nv = find(Lv);

% Nv = Nv(3:end);

% plot(T1{Nv,1}, T1{Nv,k+1}, 's');

% Bv = polyfit(T1{Nv,1}, T1{Nv,k+1}, 5);

% rline = polyval(Bv, T1{:,1});

% plot(T1{:,1}, rline, '-r')

% end

% hold off

for k = 1:size(T1,2)-1

Lv1 = (T1{:,k+1} <= mean(T1{:,k})+3);

Lv2 = ischange(T1{:,k+1}, 'mean', 'Threshold',1E-11);

Lv = Lv1 & Lv2;

Nv = find(Lv);

Nv = Nv(3:end);

op = 5;

if k == 1

op = 2;

end

[B,S,mu] = polyfit(T1{Nv,1}, T1{Nv,k+1}, op);

dtv = polyval(B, T1{:,1}, S, mu);

Idt(:,k) = T1{:,k+1} - dtv;

end

figure

plot(T1.Voltage, Idt)

grid

xlabel('Voltage (mV)')

ylabel('Current (\muA)')

title('Polynimial Detrending')

legend(VN{2:end}, 'Location','best')

.

Samiha 2024-5-9

Thank you so much! This was really helpful.

Star Strider 2024-5-9

As always, my pleasure!

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