Sinusoidal Amplitude Estimation

If the sinusoidal frequency $\omega_0$ and phase $\phi$ happen to be known, we obtain a simple linear least squares problem for the amplitude $A$ . That is, the error signal

$$x(n)-\hat{{\cal A}}e^{j\hat{\omega}_0n} = x(n)-{\hat A}e^{j(\omega_0 n+\phi)}$$ (6.36)

becomes linear in the unknown parameter ${\hat A}$ . As a result, the sum of squared errors

$$J({\hat A}) \isdef \sum_{n=0}^{N-1}\left\vert x(n)-{\hat A}e^{j(\omega_0 n+\phi)}\right\vert^2 \protect$$ (6.37)

becomes a simple quadratic (parabola) over the real line.^6.11 Quadratic forms in any number of dimensions are easy to minimize. For example, the ``bottom of the bowl'' can be reached in one step of Newton's method. From another point of view, the optimal parameter ${\hat A}$ can be obtained as the coefficient of orthogonal projection of the data $x(n)$ onto the space spanned by all values of ${\hat A}$ in the linear model ${\hat A} e^{j(\omega_0 n+\phi)}$ .

Yet a third way to minimize (5.37) is the method taught in elementary calculus: differentiate $J({\hat A})$ with respect to ${\hat A}$ , equate it to zero, and solve for ${\hat A}$ . In preparation for this, it is helpful to write (5.37) as

$$\begin{eqnarray*} J({\hat A}) &\isdef & \sum_{n=0}^{N-1} \left[x(n)-{\hat A}e^{j(\omega_0 n+\phi)}\right] \left[\overline{x(n)}-\overline{{\hat A}} e^{-j(\omega_0 n+\phi)}\right]\\ &=& \sum_{n=0}^{N-1} \left[ \left\vert x(n)\right\vert^2 - x(n)\overline{{\hat A}} e^{-j(\omega_0 n+\phi)} - \overline{x(n)}{\hat A}e^{j(\omega_0 n+\phi)} + {\hat A}^2 \right] \\ &=& \left\Vert\,x\,\right\Vert _2^2 - 2\mbox{re}\left\{\sum_{n=0}^{N-1} x(n)\overline{{\hat A}} e^{-j(\omega_0 n+\phi)}\right\} + N {\hat A}^2. \end{eqnarray*}$$

Differentiating with respect to ${\hat A}$ and equating to zero yields

$$0 = \frac{d J({\hat A})}{d{\hat A}} = - 2 \left\{\sum_{n=0}^{N-1} x(n) e^{-j(\omega_0 n+\phi)}\right\} + 2N{\hat A}.$$ (6.38)

Solving this for ${\hat A}$ gives the optimal least-squares amplitude estimate

$${\hat A}= \frac{1}{N} \left\{\sum_{n=0}^{N-1} x(n) e^{-j(\omega_0 n+\phi)}\right\} = \frac{1}{N} \left\{\hbox{\sc DTFT}_{\omega_0 }\left[e^{-j\phi} x\right]\right\}. \protect$$ (6.39)

That is, the optimal least-squares amplitude estimate may be found by the following steps:

1. Multiply the data $x(n)$ by $e^{-j\phi}$ to zero the known phase $\phi$ .
2. Take the DFT of the $N$ samples of $x$ , suitably zero padded to approximate the DTFT, and evaluate it at the known frequency $\omega_0$ .
3. Discard any imaginary part since it can only contain noise, by (5.39).
4. Divide by $N$ to obtain a properly normalized coefficient of projection [264] onto the sinusoid

   $$s_{\omega_0 }(n)\isdef e^{j\omega_0 n}.$$ (6.40)