0
$\begingroup$

I have to calibrate power output of a diode to its input voltage. The voltage source has negligible error but the power meter has significant uncertainty (a couple %). I take a number of power readings at successive voltage inputs, plot a power vs voltage graph, and do a curve fit using least squares. I have the variances in the fit parameters. Now I put away the power meter and want to use the curve fit for calculating the power output as a function of input voltage. How do I estimate the uncertainty in the power determined this way?

$\endgroup$
  • 1
    $\begingroup$ What theoretical model are you fitting your data to? $\endgroup$ – Bunji Feb 1 at 0:34
  • $\begingroup$ power propto voltage^2 $\endgroup$ – cryonole Feb 1 at 2:38
0
$\begingroup$

I am a little unclear from your comment, but if you have not done so already, you should linearize your data. In other words, if you have a theoretical equation from the literature that says $P = \alpha V^2 + \beta$, where $\alpha$ and $\beta$ are constants, then you should plot $P$ on your vertical axis, and $V^2$ on your horizontal axis so that the slope of the line is $\alpha$, and the y-intercept is $\beta$. If your theoretical equation is more complicated than that, you may have to work a little harder -- regardless, the first step is to put your equation in $y=mx +b $ form.

Perform your weighted linear least-squares fit (so that the fit takes into account the uncertainties in your measurements), and obtain an $R^2$ value for the fit.

The uncertainty in the slope can than be found using $$ \sigma_m = m\sqrt{\frac{(1/R^2)-1}{n-2}}, $$ where $m$ and $\sigma_m$ are the slope and uncertainty in the slope respectively, and $n$ is the number of data points used to make the fit. Careful, the $R^2$ is already squared! Don't square it again!

The uncertainty in y-intercept can be found using the above result: $$ \sigma_b = \sigma_m \sqrt{\frac{\sum x^2}{n}} $$

So, now you have an equation such as $P = \alpha V^2 + \beta$, and you have uncertainties in the constants; namely, $\sigma_\alpha$, and $\sigma_\beta$. Use standard propagation of uncertainty in quadrature to propagate uncertainty from $\alpha V^2 + \beta$ to $P$ for each value of $V$.

$\endgroup$
  • $\begingroup$ Your last sentence is the answer I was looking for. Much thanks! $\endgroup$ – cryonole Feb 1 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.