Comparing curves to an original signal

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Amanda Botelho Amaral 2021-12-14

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评论： Jon 2021-12-17

I'm trying to compare two curves. They represent the same situation. One curve is the original system and a second this system is undergoing some changes. I would like to compare them. What changes made to the second curve affect the first. And knowing these variations, is it possible for me to go back to the original curve? My curves are experimental data.

I thought about choosing an equation or model that best describes my curve and fitting it using curve fitting or optimization. But I still can't think of something to compare them

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Mathieu NOE 2021-12-15

sorry for my last sentence it was just making the whole think more confusing - forget it

just now I don't really see what you mean by "I would like to get the curve in yellow, starting from the purple one."

do you mean reduce the initial vertical gap so that both curves are overlaid ?

Amanda Botelho Amaral 2021-12-15

I think I confused you, sorry!

Something similar.... I would like to know a variable, or some element that I can compare.

For example, I have the equation of a curve (Linear), if I change the angular coefficient of this line, I have a new curve. So if I compare the two, can I know what the angular coefficient of each one is? I thought of something like that....

Sorry for my English, I'm learning!

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Answer 1

Jon 2021-12-15

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In terms of frequency response of a transfer function it looks like you could make the orange curve move towards the purple curve by adjusting the gain of the transfer function to change the low frequency value. You could provide a pole in the denominator whose value would adjust the high frequency roll off.

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Jon 2021-12-16

Each real pole provides a roll off (decreasing amplitude as a function of frequency) starting at a break frequency of approximately the radian locaton of the pole.

Each real zero provide a roll up (increasing amplitude as a function of frequency) starting a a break frequency of approximately the radina location of the zero.

So I placed on pole at 2*pi*1e-8 corresponding to where your function starts to roll off. I also initially place one pole at 2*pi*1e-5 where your function again rolls off. The decrease didn't look rapid enough so I added another pole close by at 2*pi*1.1e-5. To make the flat region, I placed a zero at 2*pi*1e-7 where the flat region begins. The increasing amplitude of this zero cancels the decrease from the low frequency pole, so the net is no increase with frequency in this region, giving the flat spot.

You can adjust those to get a better approximation, I just did it very roughly to show the idea. You can also add more to get steeper slopes.

The gain adjusts the low frequency amplitude

Usually transfer function frequency responses are plotted on log-log scales (Bode Plots) this gives a linear decrease and increase from each real pole, or zero respectively. Might be good to plot it that way.

Jon 2021-12-16

在 MATLAB Online 中打开

You can also do this without the Control System Toolbox, just evaluating the complex freqency response directly as the ratio of two polynomial evaluated for a range of purely imaginary frequencies (along the jw axis) Here's code to do that. Some is a repeat of earlier approach using the toolbox, but I thought I'd put it all here so you have it all in one place:

% define frequencies where curve will break
fz = 1e-7; % zeros, bring magnitude up
fp = [1e-8;1e-5;1.1e-5]; % poles bring magnitude down
% define steady state gain
ssgain = 40
% convert to radians/sec
zeroFit = 2*pi*fz;
poleFit = 2*pi*fp;
% calculate gain
gain = ssgain*prod(poleFit)/prod(zeroFit);
% define transfer function
num = poly(zeroFit); % polynomial with roots at zero locaions
den = poly(poleFit); % polynomial with roots at pole locations
% define anonymous function giving the frequency response
fresp = @(s) gain*polyval(num,s)./polyval(den,s);
% calculate the frequency response
w = 2*pi*logspace(-9,-4,1000)
jw = complex(0,w);
H = fresp(jw);
% plot the frequency response magnitude using semilog plot
f = w/2*pi;
semilogx(f,abs(H))

Amanda Botelho Amaral 2021-12-17

Thank you so much for the tip!

helped me a lot

Helped me a lot! I ended up commenting in the comment above about another question. Would you help me?

Jon 2021-12-17

Hi,

It looks like you were able to solve your second problem.

For the future, it would be good to start a new question if you have an new unrelated, or only distant related issue. That way the answer threads stay clean, and later people can find answers if they have the same question.

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Answer 2

Mathieu NOE 2021-12-15

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https://ww2.mathworks.cn/matlabcentral/answers/1610700-comparing-curves-to-an-original-signal#answer_855815

在 MATLAB Online 中打开

hello again

so the yellow curve can be shifted to match the purple one by using a linear equation like

y2_shifted = = y2*a+b;

the code provide some initail values for a and b but this could be refined by optimisation

a = ratio2*0.963

b = 1.4

the values 0.963 and 1.4 are my manual fine tuning , but again we could implement a more elegnat and automatic tuning.

see the results now : the yellow has been shifted as close as possible to the purple curve

I just considered that the last portion of the purple curve where the signal sinks would not be taken into account in this procedure...

code :

% extract_data_from_figures
data = extract_data_from_figures('ATE.fig');
x = data.Y(:,1);
y1 = data.Y(:,2);
y2 = data.Y(:,3);
% save amanda.mat x y1 y2 % save in case 
% y ratio between the two curves - measured at first x position
ratio2 = y1(1)./y2(1);
y2_shifted = y2*ratio2*0.963+1.4; 
figure(1),semilogx(x,y1,x,y2,x,y2_shifted,'r');
legend('y1','y2','y2*ratio2*0.963+1.4');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function data = extract_data_from_figures(filename)
%%
% 
% Input  : File name <filename.fig> 
%          with multiple plots in a single file
% Output : struct data
%        : data.names  contains names of the display object Yvalues
%        : data.Y      contains the actual plot values withthe first column
%                      containing the x-values
%
% Written by Chetanya Puri, 2019
% Last Modified: Nov 6, 2019
%
fig = openfig(filename); % Open figure and assign it to fig object
dataObjs = findobj(fig,'-property','YData'); % Find all graphic objects with YData, in our case line values
xval        = dataObjs(1).XData; % Find the X-axis value
Ymat = [xval(:)]; % Create a matrix with first column of x values
for i=1:length(dataObjs)
    legend_name{i,1} = dataObjs(i).DisplayName;
    yval        = dataObjs(i).YData;
    Ymat = [Ymat yval(:)]; % Keep appending column vectors
end
close(fig); % close the figure
data.names = ['X';legend_name]; 
data.Y = Ymat;
end

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Mathieu NOE 2021-12-17

在 MATLAB Online 中打开

hello

I simply executed this very slightly changed code and I got no such effect

I plot in log x scale so the signal trend is easier in my opinion

load('freq.mat');
load('Z.mat');
n = length(freq);
d_vec = (1:(n))./0.2;
t_vec = (1./freq).*d_vec;
ytime = ifft(ZZ, [], 2);
semilogx(t_vec,ytime);

Amanda Botelho Amaral 2021-12-17

I found the problem.

I forgot to put the real part of the curve to apply ifft.

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Comparing curves to an original signal

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Comparing curves to an original signal

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