A Brief Justification of the Amplitude and Phase Estimator

To use the above expression, the authors must make use of Parseval's relation, as well as the assumption that the vast majority of energy for a chirp's FFT lies within some limited peak range. Specifically, Parseval's relation states that

$$\int_{-\infty}^\infty \vert z(t)\vert^2 dt = \frac{1}{2\pi} \int_{-\infty}^\infty \vert Z(\omega)\vert^2 d\omega.$$ (22)

Also, because inner products are preserved under the Fourier transform, we have that

$$\int_{-\infty}^\infty z^*(t) f_M(t) dt = \frac{1}{2\pi} \int_{-\infty}^\infty Z^*(\omega) F_M(\omega) d\omega.$$ (23)

Combining the two, we have that

$$\frac{\int_{-\infty}^\infty z^*(t) f_M(t) dt} {\int_{-\infty}^\infty \vert z(t)\vert^2 dt} = \frac{\frac{1}{2\pi} \int_{-\infty}^\infty Z^*(\omega) F_M(\omega) d\omega} {\frac{1}{2\pi}\int_{-\infty}^\infty \vert Z(\omega)\vert^2 d\omega}.$$ (24)

Substituting known expressions and considering windowing and the concentration of energy in the main lobe (between $L_1$ and $L_2$) of the FFT, we get:

$$\frac{\frac{1}{2\pi} \int_{L_1}^{L_2} Z^*(\omega) F_M(\omega) d\omega} {\frac{1}{2\pi}\int_{L_1}^{L_2} \vert Z(\omega)\vert^2 d\omega} = \frac{\int_{-N/2}^{N/2} 1 e^{j0} e^{-j(\beta t^2 + \omega_0 t)} A... ...\phi_0} e^{j(\beta t^2 + \omega_0 t)} dt} {\int_{-N/2}^{N/2} \vert 1\vert^2 dt}$$ (25)

$$= \frac{A e^{j\phi_0} \int_{-N/2}^{N/2} 1 dt} {\int_{-N/2}^{N/2} \vert 1\vert^2 dt}$$ (26)

$$= A e^{j\phi_0}$$ (27)

as we sought to prove.