Reconstructing a sine wave from an interval shorter than half its wavelength

Question

I know that the function that I search is a sine wave of the form $$A \cdot \sin(x) $$

where my $A$ is unknown. I have samples of the sine wave at discrete points within an interval that's shorter than $\lambda/2$ represented by the red line in the plot.

It must be possible to somehow determine $A=1$ through reconstruction or so from my samples given by the red line. How can i do that?

Answer 1

import numpy as np

#========================================================================
def main():

        omega = 0.1

        X = np.array( [ 0.5, 0.6, 0.65, 0.7 ] )
        
        C = np.zeros( 4 );  S = np.zeros( 4 );  U = np.zeros( 4 )
        
        for n in range( 4 ):
          C[n] = np.cos( omega * n )
          S[n] = np.sin( omega * n )
          U[n] = 1.0
          
        V = np.zeros( 3 )          
        M = np.zeros( ( 3, 3 ) ) 
         
        M[0,0] = C.dot( C );  M[0,1] = S.dot( C ); M[0,2] = U.dot( C )
        M[1,0] = M[0,1];      M[1,1] = S.dot( S ); M[1,2] = U.dot( S )
        M[2,0] = M[0,2];      M[2,1] = M[1,2];     M[2,2] = U.dot( U )
        
        V[0] = X.dot( C );    V[1] = X.dot( S );   V[2] = X.dot( U )
        
        R = np.linalg.solve( M, V )
        
        for n in range( 4 ):
          y = R[0] * C[n] + R[1] * S[n] + R[2] * U[n]
          print ( n, X[n], y )

#========================================================================
main()

0 0.5 0.502540182211
1 0.6 0.592404834029
2 0.65 0.657595165971
3 0.7 0.697459817789

Answer 2

If your signal is really as simple as

$$x(t)=A\sin(\omega_0t)\tag{1}$$

with known $\omega_0$, and you have observations $y(t_i)$, which are noisy samples of $x(t)$ at known time instances $t_i$, then a simple solution would be the least squares estimate

$$\hat{A}=\frac{\displaystyle\sum_iy(t_i)\sin(\omega_0t_i)}{\displaystyle\sum_i\sin^2(\omega_0t_i)}\tag{2}$$

Of course, this simple solution won't work if your signal actually has the form

$$x(t)=A\sin(\omega_0t+\phi)+c \tag{3}$$

with unknown phase $\phi$ and DC-offset $c$. However, you can compute optimum least squares estimates for that problem too. This is discussed in Cedron's answer.

Answer 3

Build a basis set with your frequency and match your signal. It is straightforward linear algebra:

$C$ is portion of cosine

$S$ is portion of the sine

$U$ is a vector of ones (DC)

$$ X = a C + b S + c U $$

$$ X \cdot C = a (C \cdot C) + b (S \cdot C) + c (U \cdot C) $$ $$ X \cdot S = a (C \cdot S) + b (S \cdot S) + c (U \cdot S) $$ $$ X \cdot U = a (C \cdot U) + b (S \cdot U) + c (U \cdot U) $$

Now you have three equations three unknowns, $a$, $b$, and $c$.

Best fit interpolation/extrapolation function:

$$ x[n] = a \cos[wn] + b \sin[wn] + c $$

$$ A = \sqrt{a^2+b^2} $$

Now, wouldn't it be handy if $C\cdot S=0$?

[Overengineered solution to account for any vertical or horizontal shifts, use Matt's if you know it is a simple multiple.]

Answer 4

If samples are results of precise measurements, the amplitude is any of ratios $sample_i/sin(ω·t_i)$.

If the noise is present, some kind of averaging is needed. Because a single parameter is calculated, and no information is given on noise distribution, a simple weighted average is the only available option: $A_{avg} = {Σ(sample_i·sin(ωt_i))\over{Σsin^2(ωt_i)}}$. A weighted sample variance is $(σ_w)^2 = {Σ\{sin(ωt_i)·(sample_i-A_{avg}·sin(ωt_i))^2\}\over{Σsin(ωt_i)}}$