JOS

The JOS model assumes a time domain signal

$$f_J(t) = e^{-p t^2} = e^{-(\alpha-j \beta )t^2}$$ (1)

$$= e^{-\alpha t^2}e^{j\beta t^2}$$ (2)

where $\alpha$ is the Gaussian window parameter and $\beta$ is the chirp rate in radians per second squared. It may be shown that the Fourier transform is:

$$F_J(\omega) = \sqrt{\frac{\pi}{\alpha-j\beta}} \exp\left(\frac{-\omega^2}{4(\alpha-j \beta)}\right)$$ (3)

$$= \sqrt{\frac{\pi}{\alpha-j\beta}} \exp\left(\frac{-\alpha\omega^2}... ...\beta^2)})\right) \exp\left(\frac{-j\beta\omega^2}{4(\alpha^2+\beta^2)})\right)$$ (4)

and it has been observed that

$$\ln\vert F_J(\omega)\vert = \ln\left(\sqrt{\frac{\pi}{\alpha-j\beta}}\right) - \frac{\alpha\omega^2}{4(\alpha^2+\beta^2)}.$$ (5)

Though it is not the original design of the JOS estimator, we may solve for $\beta$ as follows:

$$x \equiv \frac{d^2}{d\omega^2}\ln\vert F_J(\omega)\vert = \frac{\alpha}{2(\alpha^2+\beta^2)}$$ (6)

$$\alpha^2 + \beta^2 = \frac{\alpha}{2 x}$$ (7)

$$\beta = \pm\sqrt{-\frac{\alpha}{2 x} - \alpha^2}$$ (8)

$$\beta = \pm\sqrt{-\frac{\alpha}{2 x} - \alpha^2}$$ (9)

Making the appropriate substitution $\alpha = \ensuremath{\frac{1}{2\sigma^2}}$ for the Gaussian parameter in terms of the variance, we have

$$\beta = \pm\sqrt{-\frac{1}{2 x}\frac{1}{2\sigma^2} - \left(\frac{1}{2\sigma^2}\right)^2}$$ (10)

$$\beta = \pm\frac{1}{2\sigma^2}\sqrt{-\frac{\sigma^2}{x} - 1 }$$ (11)