Taylor Series Based Model

We must also consider the case where the above model does not apply: when $\alpha$ is too small for the large limits approximations to be valid. In that case, we may use the Taylor series approximation for an exponential signal with a small argument, namely:

$$y(n) = \exp(j\alpha n^2) \approx 1 + j \alpha n^2,$$ (10)

where the Taylor series error introduced is $O(\alpha^2 ((N-1)/2)^4)$, as the argument of the exponential will be at most $\alpha(\ensuremath{\frac{N-1}{2}})^2$. Hence, the model loses accuracy as the argument value strays from zero, though we can ensure that our model introduces less than 2.1% error, for example, if we require that $\alpha(\ensuremath{\frac{N-1}{2}})^2 \leq \pi/16$. The accuracy of the model versus $\alpha$ will be apparent in section 4. Applying the Taylor series approximation allows us to write the DFT of the windowed chirp signal as

$$Y(k) = \sum_{n={-(N-1)/2}}^{(N-1)/2} (1+j\alpha n^2) w(n) e^{-j2\pi k n/ K}.$$ (11)

where $w(n)$ is an arbitrary zero-phase symmetric windowing function whose DFT is $W(k)$. Applying Fourier theory and approximating frequency domain differentiation with DFT domain differencing (with large $K$), we may write

$$Y(k) = W(k) + i\alpha W''(k)$$ (12)

$$Y''(k) = W''(k) + i\alpha W''''(k).$$ (13)

where the number of $'$ superscripts indicates the order of differencing applied. Though the result in equation 12 may suggest an estimator in itself, it is not tolerant of phase offset in $y(n)$. An estimator robust to this may be found by analyzing the phase curvature of $Y(k)$ at $k=0$. Specifically, it may be shown that, for zero-phase symmetric $w(n)$

$$\left. \ensuremath{\frac{\Delta^2 \angle Y(k)}{\Delta k^2}}\right\vert _{k=0} = \ensuremath{\frac{ \Im Y_0 \Re Y''_0 - \Re Y_0 \Im Y''_0}{ (\Re Y_0 )^2 + (\Im Y_0 )^2 }}$$ (14)

where the subscript 0 indicates $k=0$. By substituting equations 12 and 13 into the above expression, we obtain that

$$\left. \ensuremath{\frac{\Delta^2 \angle Y(k)}{\Delta k^2}}\right... ...suremath{\frac{\alpha ((W''_0)^2 - W_0 W''''_0) }{W_0^2 + \alpha^2 (W''_0)^2}}.$$ (15)

For future convenience, we will denote the observed value of phase curvature at zero as $p$. The expression in equation 15 may be solved for $\alpha$ to obtain an estimator for small $\alpha$.