Sinusoidal Peak Finding

For each sinusoidal component of a signal, we need to determine its frequency, amplitude, and phase (when needed). As a starting point, consider the windowed complex sinusoid with complex amplitude ${\cal A}_x$ and frequency $\omega _x$:

$$x_w(n) = w(n){\cal A}_xe^{j\omega_x nT}$$

As discussed in Chapter 1, the transform (DTFT) of this windowed signal is the convolution of a frequency domain delta function at $\omega _x$ [ $\delta(\omega - \omega_x)$], and the transform of the window function, $W(\omega)$, resulting in a shifted version of the window transform ${\cal A}_xW(\omega-\omega_x)$. Assuming $M$ is odd, we can show this as follows:

$$\begin{eqnarray*} X_w(\omega) &=& \sum_{n=-\infty}^{\infty}[w(n)x(n)]e^{ -j\omeg... ...mega-\omega_x)nT} \\ &=& \zbox {{\cal A}_xW(\omega-\omega_x)} \end{eqnarray*}$$

Hence,

$$\begin{eqnarray*} \vert X_w(\omega) \vert &=& \vert{\cal A}_x\vert \cdot \vert W... ...le X_w(\omega) &=& \angle {\cal A}_x+ \angle W(\omega-\omega_x). \end{eqnarray*}$$

At $\omega _x$, we have

$$\begin{eqnarray*} \vert X_w(\omega_x)\vert &=& \vert{\cal A}_x\vert\cdot \vert W... ...\\ \angle X_w(\omega_x)\vert &=& \angle {\cal A}_x+ \angle W(0) \end{eqnarray*}$$

If we scale the window to have a dc gain of 1, then the peak magnitude equals the amplitude of the sinusoid, i.e., $\vert X_w(\omega_x)\vert=\vert{\cal A}_x\vert\isdef a$, as shown in Fig.7.2.

Figure: Schematic diagram of a window transform amplitude-scaled by $a$ and frequency-shifted by $\omega _x$.

If we use a zero-phase (even) window, the phase at the peak equals the phase of the sinusoid, i.e., $\angle X_w(\omega_x) = \angle {\cal A}_x$.