Derivation of the uncertainties in the amplitude, phase, and frequency

As stated above, we wish to assume the ideal case for our error computations. We therefore assume that we have N measurements of the magnitudes, m_i, which are taken at times t_i, each of which are evenly separated by a time $\Delta t$. We assume that the times of the observations are error free, but that the brightness measurements m_i are subject to random errors, $\Delta m_i$. These errors are assumed to have an average of zero ( $\langle \Delta m_i \rangle = 0$), to have a root-mean-square amplitude which is constant in time ( $\langle \Delta m_i^2 \rangle = \sigma^2(m)$), and to be uncorrelated in time ( $\langle \Delta m_i \Delta m_j \rangle = 0$ for $i \neq j$). We now wish to analyze our time series data by fitting a sinusoid to it. Specifically, we fit the function

$$f(t) = a \sin(\omega t_i + \phi),$$

where the amplitude a, phase $\phi$, and (angular) frequency $\omega$ are yet to be determined. We define

$$\chi^2 \equiv \sum_{i=1}^{N} \left[m_i- f(t_i) \right]^2$$

$$\sum_{i=1}^{N} \left[m_i - a \sin(\omega t_i + \phi) \right]^2,$$ =

where the minimum in $\chi^2$ corresponds to the best fit solution of the model parameters. Minimizing $\chi^2$ with respect to a, $\phi$, and $\omega$, we obtain the following three relations:

$$\frac{\partial \chi^2}{\partial a} = 0 \Rightarrow a = \frac{2}{N} \sum_{i=1}^{N} m_i \sin(\omega t_i+\phi)$$ (1)

$$\frac{\partial \chi^2}{\partial \phi} = 0 \Rightarrow 0 = \sum_{i=1}^{N} m_i \cos(\omega t_i+\phi),$$ (2)

$$\frac{\partial \chi^2}{\partial \omega} = 0 \Rightarrow 0 = \sum_{i=1}^{N} m_i t_i \cos(\omega t_i+\phi),$$ (3)

where we have assumed that the time distribution of the data is such that the orthogonality relations $\sum_{i=1}^{N} \sin^2(\omega t_i+\phi) = N/2$ and $\sum_{i=1}^{N} \sin(\omega t_i+\phi) \cos(\omega t_i+\phi) = 0$ represent valid approximations. The above three equations must be satisfied simultaneously by the best fit solution. In general, the random errors in magnitude, $\Delta m_i$, produce small variations in the fit parameters $(\Delta a, \Delta \phi, \Delta \omega)$ from their ``true'' values. If we take a total differential of equation 1 with respect to $(m_i, a, \phi, \omega)$, we obtain

$$\Delta a \frac{2}{N} \sum_{i=1}^{N} \left[ \Delta m_i \sin(\omega t_i+\phi) \right.$$ =

$$\left. \hspace{1em} + \hspace{0.5em} m_i \cos(\omega t_i+\phi) \Delta \phi + m_i t_i \cos(\omega t_i+\phi) \Delta \omega \right]$$

$$\frac{2}{N} \sum_{i=1}^{N} \Delta m_i \sin(\omega t_i+\phi),$$ =

where the second and third terms have vanished through the application of equations 2 and 3, respectively. If we square this expression and then take a statistical average, we find

$$\langle (\Delta a)^2 \rangle \frac{4}{N^2} \sum_{i=1}^{N} \sum_{j=1}^{N} \langle \Delta m_i \Delta m_j \rangle \sin(\omega t_i+\phi) \sin(\omega t_j+\phi)$$ =

$$\frac{4}{N^2} \sum_{i=1}^{N} \langle (\Delta m_i)^2 \rangle \sin^2(\omega t_i+\phi)$$ =

$$\frac{4}{N^2} \sigma^2(m) \sum_{i=1}^{N} \sin^2(\omega t_i+\phi)$$ =

$$\frac{2}{N} \sigma^2(m),$$ =

where we have made use of the assumed statistical properties of $\Delta m_i$. Writing $\sigma^2(a) = \langle (\Delta a)^2 \rangle$, we find

$$\sigma(a) = \sqrt\frac{2}{N} \cdot \sigma(m).$$ (4)

We repeat this analysis in order to find the errors in $\phi$ and $\omega$. From equation 2, we have

$$\sum_{i=1}^{N} \left[ \Delta m_i \cos(\omega t_i+\phi) - \right.$$ 0 =

$$\hspace{0.1em} \left. m_i \Delta \phi\sin(\omega t_i+\phi) - m_i t_i \Delta \omega\sin(\omega t_i+\phi) \right].$$ (5)

Now we must make use of the fact that the signal without noise is just the sinusoidal solution which we are seeking, i.e., $m_i = a \sin(\omega t_i + \phi)$. Using this, we find that equation 5 reduces to

$$\sum_{i=1}^{N} \Delta m_i \cos(\omega t_i+\phi) = \Delta... ...c{a}{2} N + \Delta \omega \frac{a}{2} \sum_{i=1}^{N} t_i.$$ (6)

Applying this same set of steps to equation 3, we obtain

$$\sum_{i=1}^{N} \Delta m_i t_i \cos(\omega t_i+\phi) = \D... ...{N} t_i + \Delta \omega \frac{a}{2} \sum_{i=1}^{N} t_i^2.$$ (7)

Equations 6 and 7 form a system of linear equations for the errors $\Delta \phi$ and $\Delta \omega$. These equations can be greatly simplified by a specific choice of the zero point in time. We choose the zero point to be the ``average time'', so we require

$$\sum_{i=1}^{N} t_i = 0.$$

This choice has the advantage of decoupling equations 6 and 7. Using this zero point, the sum in the coefficient of $\Delta \omega$ in equation 7 becomes

$$\sum_{i=1}^{N} t_i^2 = {\Delta t}^2 \sum_{i=1}^{N} (i-N/2)^2 = \frac{{\Delta t}^2}{12} N^3 + O(N^2). \nonumber$$

Assuming that $N \gg 1$, we retain only the leading term in N. It is now straightforward to solve equations 6 and 7 for $\Delta \phi$ and $\Delta \omega$:

$$\Delta \phi \frac{2}{a N} \sum_{i=1}^{N} \Delta m_i \cos(\omega t_i+\phi)$$ =

$$\Delta \omega \frac{24}{a N^3 {\Delta t}^2} \sum_{i=1}^{N} \Delta m_i t_i \cos(\omega t_i+\phi)$$ =

If we now square both sides of the above equations, perform a statistical average, and then a summation over i, we find that

$$\sigma^2(\phi) \frac{2}{N} \, \frac{\sigma^2(m)}{a^2}$$ = (8)

$$\sigma^2(\omega) \frac{24}{N^3 {\Delta t}^2 } \, \frac{\sigma^2(m)}{a^2},$$ = (9)

where we have written $\sigma^2(\phi) \equiv \langle (\Delta \phi)^2 \rangle$ and $\sigma^2(\omega) \equiv \langle (\Delta \omega)^2 \rangle$. Rewriting these relations in terms of the actual frequency ( $\omega = 2 \pi f$) and taking a square root, we find that

$$\sigma(\phi) \sqrt{\frac{2}{N}} \, \frac{\sigma(m)}{a}$$ = (10)

$$\sigma(f) \sqrt{\frac{6}{N}} \frac{1}{\pi T} \, \frac{\sigma(m)}{a},$$ = (11)

where $T \equiv N \Delta t$ is the total time length of the data set. We note that if a different zero point for the time t_i is chosen, then the errors in $\phi$ and $\omega$ are no longer uncorrelated, which has the effect of changing the derived errors in $\phi$. For instance, if we choose the beginning of the run to be the zero point (i.e., t₁ = 0), then the derived error for the phase, $\sigma(\phi)$, is exactly twice the value given by equation 10. However, the relations for the errors in the amplitude and frequency are unchanged, as must be the case.