Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is my first post on math.stackexchange (sorry if meta people remove the Hello (sometimes we do that over on stackoverflow ;P)!

I have a system wherein I know that the output is a sine wave, with a known frequency. My objective is to find the approximate (x,y) of the first peak (i.e., find the phase shift of the signal). An important point is that I do not need to know y, or the amplitude, of this peak. Essentially, I can poll the system at a given angular shift (represented by x), and receive a y value in return. I start with zero points, and want to poll the minimum number of x points in order to be able to know where to poll to receive a max y value.

I believe that I can describe the sine wave with only two points, yet I do not know how to calculate this (it's on a motion controller, so I have quite limited functionality). My thoughts so far: phase = -sin^-1(y) - wt + 2*pi*n, but I don't know how to easily fit this with two points.

Once I know the fitted sine wave, I will be able to determine which x should yield a max amplitude peak y, and then subsequently poll the x location.

If this can be done, the final solution would account for noise in the system (i.e. each y point polled will be within a given tolerance... thus, the two or more points polled to fit the sine wave would cause additive errors...), but I'll cross that bridge when I come to it.

Thanks! I think it's a pretty interesting problem :) Let me know if you need any further clarification!


share|cite|improve this question
Is there a DC offset to the signal or is it purely $f(t)=Y \sin (\omega t + \varphi)$ ? – ja72 May 6 '13 at 23:21
@ja72, I don't think there's $\omega$ since the frequency is known. – vadim123 May 6 '13 at 23:24
Ah, good point ja72. I am looking for a solution in which there can be a DC offset. – Kadaj Nakamura May 6 '13 at 23:24
@vadim123, yes, I suppose my notation isn't the best >_<;; So, in a system in which I did not know $\omega$, I think I would require at least 3 points. Since I know $\omega$, I believe it can be done in 2, I just cannot seem to arrive at a solution :'( – Kadaj Nakamura May 6 '13 at 23:26
@KadajNakamura - frequency is $f=\frac{\omega}{2 \pi}$. So $\omega$ is just frequency in rad/s. – ja72 May 7 '13 at 0:15
up vote 2 down vote accepted

Given the general equation $f(t) = Y \sin (\omega t + \varphi)$ where $\omega$ is known and two points, $y_1 = f(t_1)$ and $y_2=f(t_2)$ the solution is

$$ Y = \frac{ \sqrt{ y_1^2 + y_2^2 - 2 y_1 y_2 \cos (\omega(t_2-t_1))}}{\sin ( \omega(t_2-t_1))} $$

$$ \varphi = 2\pi - \tan^{-1} \left( \frac{y_2 \sin \omega t_1 - y_1 \sin \omega t_2}{y_2 \cos \omega t_1 - y_1 \cos \omega t_2} \right) $$


I expanded the sine function into two components

$$ f(t) = A \sin \omega t + B \cos \omega t $$

where $Y=\sqrt{A^2+B^2}$ and $\tan(\varphi) = \frac{B}{A}$. The two points are

$$ y_1 = A \sin \omega t_1 + B \cos \omega t_1 $$ $$ y_2 = A \sin \omega t_2 + B \cos \omega t_2 $$

or in matrix form

$$ \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{pmatrix} \sin \omega t_1 & \cos \omega t_1 \\ \sin \omega t_2 & \cos \omega t_2 \end{pmatrix} \begin{bmatrix} A \\ B \end{bmatrix} $$

with the inverse

$$\begin{pmatrix} \sin \omega t_1 & \cos \omega t_1 \\ \sin \omega t_2 & \cos \omega t_2 \end{pmatrix}^{-1} = \frac{1}{\sin( \omega (t_2-t_1))} \begin{pmatrix} \mbox{-}\cos \omega t_2 & \cos \omega t_1 \\ \sin \omega t_2 & \mbox{-}\sin \omega t_1 \end{pmatrix}$$


$$ \begin{bmatrix} A \\ B \end{bmatrix} = \frac{1}{\sin( \omega (t_2-t_1))} \begin{bmatrix} y_2 \cos \omega t_1 - y_1 \cos \omega t_2 \\ y_1 \sin \omega t_2 - y_2 \sin \omega t_1 \end{bmatrix} $$


$$ Y = \sqrt{A^2+B^2} = \sqrt{ \left( \frac{y_2 \cos \omega t_1 - y_1 \cos \omega t_2}{\sin( \omega (t_2-t_1))} \right)^2 + \left( \frac{y_1 \sin \omega t_2 - y_2 \sin \omega t_1}{\sin( \omega (t_2-t_1))} \right)^2 } $$


$$ \varphi = n \pi + \tan^{-1}\left( \frac{B}{A} \right) = n \pi + \tan^{-1}\left( \frac{y_1 \sin \omega t_2 - y_2 \sin \omega t_1}{y_2 \cos \omega t_1 - y_1 \cos \omega t_2} \right) $$

share|cite|improve this answer
This definitely looks promising! Thank you, @ja72 (especially considering how quickly you replied)! Have to run now, but will try it out later today/tomorrow =) Tried to bump it, but I don't have enough rep on here yet T^T – Kadaj Nakamura May 6 '13 at 23:59
Oh, I have a quick question -- can you point me in the right direction for this derivation? Is there a name for this type of problem or field that I can read up on it more? – Kadaj Nakamura May 7 '13 at 2:22
I used a 2x2 system to fit the function into the data points. Essentially this is a least squares fit, but with zero error. See edited reply for details. – ja72 May 7 '13 at 18:56
Awesome !! Thank you! – Kadaj Nakamura May 8 '13 at 2:59
Shouldn't the matrix inversion line read to the right of the determinate? --> \begin{pmatrix} \cos \omega t_2 & \mbox{-}\cos \omega t_1 \\ \mbox{-}\sin \omega t_2 & \sin \omega t_1 \end{pmatrix} – Kadaj Nakamura May 8 '13 at 3:28

According to the Nyquist limit, two samples are not enough to have it, considering that your signal is just a period of a sinus, corresponding to a sinc in frequency.

Imaging the two points are just the two zeros of your sinusoidal signal, how can you recover anything from that?

A good question would be how many points are then needed. I guess three, but does not have demonstration.

This is for total recovery of the signal. For just phase, two points are right.

share|cite|improve this answer
Welcome to our site! – kjetil b halvorsen Jan 20 '15 at 10:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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