Fresnel Analysis Based Model

It may be shown [6] that for sufficiently large $\alpha$ and $N$,

$$Y(k) \approx Y(\omega) = \int_{-T}^T \exp(j( \alpha t^2- \omega t)) dt,$$ (3)

where we have applied the midpoint approximation to the definite integral in the Fourier transform of the analogous continuous time chirp, $y(t) = e^{j\alpha t^2}.$ Applying this approximation, it may be shown [6] that the real and imaginary parts of $Y(k)$ are given by:

$$\Re{Y(k)} \approx \pm\sqrt{\frac{\pi}{2\alpha}} \Big( \cos\left(\phi\right) \int_{l_1}^{l_2}\cos\left(\frac{\pi}{2}u^2\right) du$$

$$+ \sin\left(\phi\right) \int_{l_1}^{l_2}\sin\left(\frac{\pi}{2}u^2\right) du \Big)$$ (4)

$$\Im{Y(k)} \approx \pm\sqrt{\frac{\pi}{2\alpha}} \Big( \cos\left(\phi \right) \int_{l_1}^{l_2}\sin\left(\frac{\pi}{2}u^2\right) du$$

$$- \sin\left(\phi\right) \int_{l_1}^{l_2} \cos\left(\frac{\pi}{2}u^2\right) du \Big),$$ (5)

$\phi = \frac{\pi^2 k^2}{K^2\alpha}$; $l_1=\sqrt{\frac{2\alpha}{\pi}}\left(-\frac{N}{2}-\frac{\pi k}{\alpha K}\right)$; $l_2 = \sqrt{\frac{2\alpha}{\pi}}\left(\frac{N}{2}-\frac{\pi k}{\alpha K}\right)$. We recognize the integrals in the above expressions as Fresnel integrals. When $\alpha$, $N$, $k$,and $K$ are such that $l_1\ll-1$ and $l_2\gg1$, we may apply what we call the large limits approximations,

$$\int_{0}^{l_2}\sin\left(\frac{\pi}{2}u^2\right) du \approx \frac{1}{2} - \frac{1}{\pi l_2} \cos\left(\frac{\pi}{2} l_2^2\right)$$ (6)

$$\int_{0}^{l_2} \cos\left(\frac{\pi}{2}u^2\right) du \approx \frac{1}{2} + \frac{1}{\pi l_2} \sin\left(\frac{\pi}{2} l_2^2\right)$$ (7)

and where the odd symmetry of the Fresnel integrals leads to analogous negative results when the limits are $[l_1,0]$ rather than $[0,l_2]$ as above. These approximations will allow us to create an invertible model of the signal's DFT, and incur a modeling error of less than 1% when $\{-l_1,l_2\}\geq 3$. (When $\{-l_1,l_2\}\geq 2$ or $\{-l_1,l_2\}\geq 1$ the error bounds are 2.3% or 14% respectively.) The condition on the limits merits discussion. In an estimation context, $N$ and $K$ are in practice fixed and only certain small magnitude values of $k$ will be of interest. In our estimator, we will implicitly use $\vert k\vert\leq\ensuremath{\frac{KN\hat{\alpha}}{4\pi}}$ where $\hat{\alpha}$ is at the same time estimated. We see then that as $\alpha$ becomes very small, no range of $k$ will be valid. Requiring $\{-l_1,l_2\}\geq 3$ and solving for $\alpha$ we find a constraint of $\alpha N^2 \geq 72\pi$. (For $\{-l_1,l_2\}\geq 2.5$, we require $\alpha N^2 \geq 50\pi$.) Figure 1 should help clarify the relationship of the limits to $\alpha$ and $k$. In the top subplot, we see $l_1$ and $l_2$ plotted versus $\alpha$, for each of $k = -1, 0,$ and $1$, with $N$ fixed at 641 and $K$ fixed at 2048. The middle subplot is similar, but considers values of $k$ from -10 to 10. We see that the larger magnitude values of $k$ cause one of the limits to be too small for a given value of $\alpha$. We also see that in general, the model is valid for smaller $\alpha$ when smaller $k$ values are used. The bottom subplot is similar to the top two, but uses $k$ values ``dynamically'' selected by $\vert k\vert\leq\ensuremath{\frac{KN\hat{\alpha}}{4\pi}}$ (see Fresnel estimator discussion below).

Figure 1: The choice of $k$ and the value of $\alpha$ affect the limits $l_1$ and $l_2$, which in turn dictate Fresnel model validity. The values of $l_1$ and $l_2$ giving 1% model error bounds ($\pm 3$) are shown with dash-dot lines.

Returning to the model, it may be shown that substituting the approximations above in the present case, and applying a time domain Hann window via a frequency domain convolution [6] leads to the following expressions for the magnitude and phase of $Y(k)$, valid for $k$ as noted above:

$$\angle Y(k) \approx \pm \frac{\pi}{4} - \frac{\pi^2 k^2}{K^2\alpha}$$ (8)

$$\vert Y(k)\vert \approx \sqrt{\frac{\pi}{\vert\alpha}\vert} \left(\ensuremath{\frac{1}{2}... ...math{\frac{1}{2}}\cos\left(\ensuremath{\frac{2\pi^2 k}{KN\alpha}}\right)\right)$$ (9)

where the $\ensuremath{\frac{\pi}{4}}$ term in the phase expression is positive for $\alpha>0$ and negative for $\alpha<0$. As we will see in the next section, these models are invertible.