Fundamental Frequency Model and Its Generalization

Abstract

In this chapter we have discussed the fundamental frequency model (FFM) and the generalized fundamental frequency model (GFFM). Both these models are special cases of the sinusoidal frequency model. But many real-life phenomena can be analyzed using such special models. In estimating unknown parameters of multiple sinusoidal model, there are several algorithms available, but the computational loads of these algorithms are usually quite high. Therefore, the FFM and the GFFM are very convenient approximations where inherent frequencies are harmonics of a fundamental frequency. We have discussed different developments of these models both from the classical and Bayesian points of view.

Appendix A

In order to prove Theorem 6.5, we need the following lemmas.

Lemma 6.1

(An, Chen, and Hannan [1]) Define,

$$ I_X(\lambda ) = \left| \frac{1}{n} \sum _{t=1}^n X(t) e^{i t \lambda } \right| ^2. $$

If $\{X(t)\}$ satisfies Assumption 6.1, then

$$\begin{aligned} \lim \sup _{n \rightarrow \infty } \max _{\lambda } \frac{n I_X(\lambda )}{\sigma ^2 \log n} \le 1 \hbox {a.s.} \end{aligned}$$

(6.31)

Lemma 6.2

(Kundu [11]) If $\{X(t)\}$ satisfies Assumption 6.1, then

$$ \lim _{n \rightarrow \infty } \sup _{\lambda } \frac{1}{n} \left| \sum _{t=1}^n X(t) e^{i \lambda t} \right| = 0 \hbox {a.s.} $$

6.1.1 Proof of Theorem 6.5

Observe that, we need to show

$$ IC(0)> IC(1)> \cdots> IC(p^0-1) > IC(p^0) < IC(p^0+1). $$

Consider two cases separately.

Case I: $L < p^0$

$$\begin{aligned}\lim _{n \rightarrow \infty } R(L)= & {} \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{t=1}^n \left( y(t) - \sum _{j=1}^L \widehat{\rho }_j \cos (j \widehat{\lambda } t - \widehat{\phi }_j) \right) ^2 \\= & {} \lim _{n \rightarrow \infty } \left[ \frac{1}{n} \mathbf{Y}^T \mathbf{Y} - 2 \sum _{j=1}^L \left| \frac{1}{n} \sum _{t=1}^n y(t) e^{i j t \widehat{\lambda }} \right| ^2 \right] = \sigma ^2 + \sum _{j=L+1}^{p^0} \rho _j^{0^2} \hbox {a.s.} \end{aligned}$$

Therefore, for $0 \le L < p^0-1$,

$$\begin{aligned}&\lim _{n \rightarrow \infty } \frac{1}{n} \left[ IC(L) - IC(L+1) \right] = \nonumber \\&\lim _{n \rightarrow \infty } \left[ \log \left( \sigma ^2 + \sum _{j=L+1}^{p^0} \rho _j^{0^2} \right) - \log \left( \sigma ^2 + \sum _{j=L+2}^{p^0} \rho _j^{0^2} \right) - \frac{2 C_n}{n} \right] \hbox {a.s.} \nonumber \\ \end{aligned}$$

(6.32)

and for $L = p^0 -1$,

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{1}{n} \left[ IC(p^0-1) - IC(p^0) \right] = \lim _{n \rightarrow \infty } \left[ \log \left( \sigma ^2 + \rho _{p^0}^{0^2} \right) - \log \sigma ^2 - \frac{2 C_n}{n} \right] \hbox {a.s.} \end{aligned}$$

(6.33)

Since $\frac{C_n}{n} \rightarrow 0$, therefore as $n \rightarrow \infty$ for $0 \le L \le p^0-1$,

$$ \lim _{n \rightarrow \infty } \frac{1}{n} \left[ IC(L) - IC(L+1) \right] > 0. $$

It implies that for large n, $IC(L) > IC(L+1)$, when $0 \le L \le p^0-1$.

Case II: $L = p^0+1$.

Now consider

$$\begin{aligned} R(p^0 + 1) = \frac{1}{n} \mathbf{Y}^T \mathbf{Y} - 2 \sum _{j=1}^{p^0} \left| \frac{1}{n} \sum _{t=1}^n y(t) e^{i \widehat{\lambda } j t} \right| ^2 - 2 \left| \frac{1}{n} \sum _{t=1}^n y(t) e^{i \widehat{\lambda } (p^0+1) t} \right| ^2. \end{aligned}$$

(6.34)

Note that $\widehat{\lambda } \rightarrow \lambda ^0$ a.s. as $n \rightarrow \infty$ (Nandi [14]). Therefore, for large n

$$\begin{aligned}&IC(p^0+1) - IC(p^0) \\= & {} n \left( \log R(p^0+1) - \log R(p^0) \right) + 2 C_n = n \left[ \log \frac{R(p^0+1)}{R(p^0)} \right] + 2 C_n \\\approx & {} n \left[ \log \left( 1 - \frac{2 \left| \frac{1}{n} \sum _{t=1}^n y(t) e^{i \lambda ^0 (p^0+1) t} \right| ^2}{\sigma ^2} \right) \right] + 2 C_n \hbox { (using lemma 6.2)} \\\approx & {} 2 \log n \left[ \frac{C_n}{\log n} - \frac{n \left| \frac{1}{n} \sum _{t=1}^n X(t) e^{i \lambda ^0 (p^0+1) t} \right| ^2}{\sigma ^2 \log n} \right] \\= & {} 2 \log n \left[ \frac{C_n}{\log n} - \frac{n I_X(\lambda ^0(p^0+1))}{\sigma ^2 \log n} \right] > 0 \quad \hbox {a.s.} \end{aligned}$$

Note that the last inequality follows because of the properties of $C_n$ and due to Lemma 6.1.    $\blacksquare$