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We want to calculate the amplitude $A$ and the phase angle $\phi$ of the oscillation $b(t)=A\sin(2t+\phi)$.

We have $t_k=(0,\pi/4, \pi/2, 3\pi/4)$ and $b_k=(1.6,1.1,-1.8,0.9)$

Use $\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$ and $\alpha=A\cos(\phi), \beta=A\sin(\phi)$ to get a linear problem.

We get $b(t)=\alpha\sin(2t)+\beta\cos(2t)$

Using the above, we get $b^T=A (\alpha, \beta)^T$

Using QR and/or normal equation [ code:https://hastebin.com/otezejobaj.pl ] we get $\alpha=0.1, \beta=1.7$

Now, I should write down the residual vectors for QR and for the normal equation

Question 1: Are the residual vectors here: $Ax_1 - b$ and $Ax_2-b$ with $x_1=(\alpha,\beta)$ from QR Method and $x_2=(\alpha,\beta)$ from the normal equation. (It's the same result here)?

Now, I should calculate $A$ and $\phi$. How should I do that numerically? Also, I noted the following:

(1)$\beta = A\sin(\phi) \Rightarrow 1.7=a\sin(\phi)$ and $b(0)=A\sin(\phi)=1.6$ which can't be.

Question 2: Is there a reason that (1) isn't legit?

Edit: Since I'm in a least square problem, I can't actually expect (1) to work, right? Anyway, Question 1 is the important question here.

2 Answers 2

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The four residuals are $$b(t_k)-b_k$$ evaluated with the computed $\alpha,\beta$, which you can group as a sum of squares

$$\sum_{k=1}^4(b(t_k)-b_k)^2.$$

Also,

$$A=\sqrt{\alpha^2+\beta^2},\\\tan\phi=\frac\beta\alpha.$$

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    To what is the sum of equal? What's the motivation behind it? And what's the step you did to get $\tan(\hi)=\beta\over\alpha$? And I should get the residuals to the QR method and normal equation each. – 2017-02-01
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    The $QR$ and normal equation methods give the same solution. – 2017-02-01
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    I know that - they are probably asking for both since that's the whole point. Numerically, they are also different, one more stable than the other iirc. But anyway, why did you do that sum and where comes that tangens from? – 2017-02-01
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    @xotix: is it so hard for you to take the quotient of two expressions ? – 2017-02-01
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    I didn't meant I wasn't able to get tangens I only asked what you did. I just like to get the details here. But anyway, thanks for your input. – 2017-02-01
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@Yves Daoust answers the issues succinctly. It is worthwhile to make the linear algebra more explicit.

The trial function is $y(x) = \alpha \sin (2t) + \beta \cos (2t)$, and the linear system is $$ \left[ \begin{array}{ll} \sin 2t_{1} & \cos 2t_{1} \\ \sin 2t_{2} & \cos 2t_{2} \\ \sin 2t_{3} & \cos 2t_{3} \\ \sin 2t_{4} & \cos 2t_{4} \end{array} \right] % \left[ \begin{array}{l} \alpha \\ \beta \end{array} \right] = \left[ \begin{array}{ll} b_{1} \\ b_{2} \\ b_{3} \\ b_{4} \end{array} \right]. $$ How to solve this system? Normal equations, $\mathbf{Q}\mathbf{R}$, SVD? A place to start is the product matrix: $$ \mathbf{A}^{\mathrm{T}}\mathbf{A} = \left[ \begin{array}{rrrr} 0 & 1 & 0 & -1 \\ 1 & 0 & -1 & 0 \end{array} \right] % \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \\ 0 & -1 \\ -1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right]. $$ The condition number of this matrix is unity - the best possible case. Without the need to resolve an ill-conditioned system, we don't need to us the $\mathbf{Q}\mathbf{R}$ or singular value decompositions.

The rank of the system matrix $\mathbf{A}$ is $\rho=2$. One way to see this is to note the two columns are linearly independent. The number of rows $m=4>\rho$, so the null space $\mathcal{N}(\mathbf{A})$ is trivial so $\mathbf{A}^{\dagger}\mathbf{A} = \mathbf{I}_{2}.$ The least squares solution $$ \left[ \begin{array}{c} \alpha \\ \beta \end{array} \right]_{LS} = \mathbf{A}^{\dagger}b + \left(\mathbf{I}_{2} - \mathbf{A}^{\dagger}\mathbf{A} \right)y, \quad y\in\mathbb{C}^{2} $$ is a unique point and $$ \mathbf{A}^{\dagger} = \left( \mathbf{A}^{\mathrm{T}}\mathbf{A} \right)^{-1} \mathbf{A}^{\mathrm{T}}. $$

Postscript: see the more general problem at Trigonometrical Least Squares (Linear Algebra)