Determine intercept of two Lines

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Daniel Murphy 2022-2-25

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1658715-determine-intercept-of-two-lines

评论： Voss 2022-2-26

Good afternoon,

I've developed the following data set, interpolating a line in a data set and looking to determine the points in which the interpolated data intercepts with a horizontal value. Relativly new to Mathlab, so this is most likely not the most concise approach. Thanks in advance for any advice or word of wisdom.

f = [1.337 1.378 1.4 1.417 1.438 1.453 1.462 1.477 1.487 1.493 1.497 1.5 1.513 1.52 1.53 1.54 1.55 1.567 1.605 1.628 1.658];

a = [.68 .9 1.15 1.5 2.2 3.05 4 7 8.6 8.15 7.6 7.1 5.4 4.7 3.8 3.4 3.1 2.6 1.95 1.7 1.5];

xq = 0:.01:1.7

hp = (1/sqrt(2))*a_max

figure

vq1 = interp1(f,a,xq);

plot(f,a,'ob',xq,vq1,':.')

hold on;

yline(hp,'b--')

xlabel('Frequency, Hz')

ylabel('Acceleration, 10^-3 g')

%Natural Frequency

[a_max, index] = max(a);

f_n = f(index)

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Daniel Murphy 2022-2-25

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Answer 1

Voss 2022-2-25

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1658715-determine-intercept-of-two-lines#answer_904760

编辑：Voss 2022-2-26

在 MATLAB Online 中打开

You can do another interpolation (actually two interpolations), splitting the curve into two parts: before the peak and after the peak. Treat acceleration as x (independent variable) and frequency as y (dependent variable) in the interpolations, in order to find which frequency corresponds to acceleration hp in each part of the curve.

f = [1.337 1.378 1.4 1.417 1.438 1.453 1.462 1.477 1.487 1.493 1.497 1.5 1.513 1.52 1.53 1.54 1.55 1.567 1.605 1.628 1.658];

a = [.68 .9 1.15 1.5 2.2 3.05 4 7 8.6 8.15 7.6 7.1 5.4 4.7 3.8 3.4 3.1 2.6 1.95 1.7 1.5];

xq = 0:.01:1.7

xq = 1×171

0 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 0.1000 0.1100 0.1200 0.1300 0.1400 0.1500 0.1600 0.1700 0.1800 0.1900 0.2000 0.2100 0.2200 0.2300 0.2400 0.2500 0.2600 0.2700 0.2800 0.2900

[a_max, index] = max(a);

hp = (1/sqrt(2))*a_max

hp = 6.0811

vq1 = interp1(f,a,xq);

% finding the frequencies where hp occurs, using the original data:

% f_hp = [0 0];

% f_hp(1) = interp1(a(1:index),f(1:index),hp);

% f_hp(2) = interp1(a(index:end),f(index:end),hp);

% finding the frequencies where hp occurs, using the interpolated data:

nan_vq = isnan(vq1); % first, need to remove extrapolated data (NaNs)

vq1_temp = vq1(~nan_vq);

xq_temp = xq(~nan_vq);

[~,index_vq] = max(vq1_temp); % the rest is the same

f_hp = [0 0];

f_hp(1) = interp1(vq1_temp(1:index_vq),xq_temp(1:index_vq),hp);

f_hp(2) = interp1(vq1_temp(index_vq:end),xq_temp(index_vq:end),hp);

figure();

plot(f,a,'ob',xq,vq1,':.')

hold on;

yline(hp,'b--')

plot(f_hp,[hp hp],'ok','LineWidth',2,'MarkerSize',9)

xlabel('Frequency, Hz')

ylabel('Acceleration, 10^-3 g')

%Natural Frequency

f_n = f(index)

f_n = 1.4870

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Daniel Murphy 2022-2-26

Thanks, I'm stepping through your code now to understand it for next time. Thanks for taking the time out of your day to help.

Voss 2022-2-26

No problem.

By the way, I would do it the way that's commented out in my answer, interpolating from the original data (f and a), not interpolating from the interpolated data (xq and vq1). (I left the xq/vq1 version uncommented because that was the phrasing of the question.)

I only mention it because the f/a version is a simpler and easier to understand, but the answers from one method vs the other should be very close. Let me know if you have any questions about it.

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