fitlm and exponential model help

2022 10 4
1 个回答
回答已采纳
更新时间：2022 10 5
8 次查看（30 天）

0 个投票

Hi
I have a set of data, which is voltage from a sensor aganist temperture.
As temperture increase the voltage output from the sensor will drop, even though the sensing enviroment in the same.
Therefore i need to be able to calulate the correction factor so the output is very close to when sensor is expose to 20 degrees.
I am using fitlm, but when i use exponential i am getting errors, which is not then when i use Quadratic
This is what i am using
no2Model = fitlm(iot_temperature, voltage_differnce_no2, "exponential");
Is this the correct command to use?

请先登录，再回答此问题。

 采纳的回答

Star Strider
Star Strider 2022-10-5
Ran in:

0 个投票

Ran in:
The "exponential" option for modelspec does not appear to exist.
The fitnlm function is likely more appropriate. That is more likely to produce the result you want.
Example —
t = linspace(0, 10, 25);
y = 3*exp(-0.25*t) + 0.5*randn(size(t)) + 1;
fcn = @(b,x) b(1).*exp(b(2).*x) + b(3);
B0 = randn(3,1);
mdl = fitnlm(t,y,fcn,B0)
mdl =
Nonlinear regression model: y ~ b1*exp(b2*x) + b3 Estimated Coefficients: Estimate SE tStat pValue ________ _______ _______ __________ b1 2.7569 0.44251 6.2301 2.8587e-06 b2 -0.2526 0.12197 -2.0711 0.050281 b3 1.0658 0.48102 2.2157 0.037369 Number of observations: 25, Error degrees of freedom: 22 Root Mean Squared Error: 0.519 R-Squared: 0.696, Adjusted R-Squared 0.669 F-statistic vs. constant model: 25.2, p-value = 2.01e-06
[yfit,yci] = predict(mdl,t(:));
figure
hp{1} = plot(t, y, '.', 'DisplayName','Data');
hold on
hp{2} = plot(t, yfit, '-r', 'DisplayName','Model Fit');
hp{3} = plot(t, yci, ':r', 'DisplayName','±95% CI');
hold off
grid
xlabel('t')
ylabel('y')
legend([hp{1},hp{2},hp{3}(1)], 'Location','best')
.

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Control System Toolbox 的更多信息

产品

版本

R2022b

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by