How do you solve a simple simultaneous equation to get unkowns for a sinusoid (with no phase shift) given a couple of points?

Question

This is probably a very simple question, but I am struggling to find the answer.

Given:

\begin{equation} y_1 = A \cos(\omega t_1)\\ y_2 = A \cos(\omega t_2)\\ y_3 = A \cos(\omega t_3) \end{equation}

(I can go up to as much as $y_n = A \cos(\omega t_n)$. Using only $y_1$ and $y_2$ is also, probably, fine.)

So I know the $y's$ and $t's$. How do I find $A$ and $\omega$?

I, essentially, want to use that to compute the peak of a discrete signal I'm assuming is a cosine for a very short interval which includes that peak. So I thought if I can obtain the equation of the sinuoid then I can differentiate it, set it to $0$ and then solve for the $t$. That way I can obtain $t$ and $y$ at the maximum/peak.

Answer 1

$$ \begin{cases} y_1 = A\cos(\omega t_1)\\ y_2 = A\cos(\omega t_2)\\ \end{cases}\tag{1} $$ If you have two unknowns, $A, \omega$ you need only two equations.

If there is another equation, the risk is that the third is not compatible with the other two, like in this example $$ \begin{cases} a+b=10\\ a-b=4\\ 2a+3b=1\\ \end{cases} $$ first two equations have $a=7,b=3$ as result, but this doesn't satisfy the third equation. The system is inconsistent.

Back to $(1)$.

You get $A$ from the first equation $A=\frac{y_1}{\cos\omega t_1}$ and plug it in the second:

$y_2=\frac{y_1}{\cos\omega t_1}\cos \omega t_2$

$y_2\cos\omega t_1-y_1\cos\omega t_2=0$

which can be solved with numerical methods once you have the actual values.

There is no formula to solve something like

$1.2 \cos (3.6 \omega)-0.75 \cos (2.7 \omega)=0$

Once you get $\omega$ you substitute and find $A$.

Answer 2

Considering that you have $n$ data points $(t_i,y_i)$, you want to adust $A$ and $\omega$ in order to minimize $$\text{SSQ}=\frac12\sum_{i=1}^n \big[A \cos(\omega t_i)-y_i\big]^2$$ Computing the partial derivatives and setting them equal to $0$ $$\frac{\partial \text{ SSQ}}{\partial A}=\sum_{i=1}^n \cos(\omega t_i)\big[A \cos(\omega t_i)-y_i\big]=0\tag1$$ $$\frac{\partial \text{ SSQ}}{\partial \omega}=-A\sum_{i=1}^n t_i \sin (\omega t_i)\big[A \cos(\omega t_i)-y_i\big]=0\tag2$$

From $(1)$ $$A=\frac {\sum_{i=1}^n y_i \cos(\omega t_i) } {\sum_{i=1}^n \cos^2(\omega t_i) }\tag 3$$ and plugging in $(2)$ you are left with an equation in $\omega$ which can write $$\Big[\sum_{i=1}^n y_i \cos(\omega t_i)\Big]\Big[\sum_{i=1}^n t_i \sin(2\omega t_i)\Big]-2\Big[\sum_{i=1}^n \cos^2(\omega t_i)\Big] \Big[\sum_{i=1}^n t_iy_i \sin(\omega t_i)\Big]=0$$

Example

Let us use the following data $$\left( \begin{array}{cc} t & y\\ 13 & 49 \\ 14 & 90 \\ 15 & 114 \\ 16 & 124 \\ 17 & 112 \\ 18 & 85 \\ 19 & 42 \end{array} \right)$$

I "know" that $0.3 \leq \omega \leq 0.5$. Just plotting the equation fot this range, the solution is close to $0.4$.

Starting Newton iterations, the path to solution is $$\left( \begin{array}{cc} n & \omega_n \\ 0 & 0.4000000000 \\ 1 & 0.3946667642 \\ 2 & 0.3945546651 \\ 3 & 0.3945546065 \end{array} \right)$$

Plug the solution in $(3)$ to get $A=123.17165$. The recomputed values of the $y$'s are

$$\{49.87,89.33,115.1,123.1,112.3,84.14,43.10\}$$ and the peak is at $t=15.9248$