Liu's Proof of the Hann case Phase

We now formally prove 15. Considering $Y_W^{\mathrm{Hann}}(\omega)$ as a trajectory on the complex plane, where $\omega$ can be thought of as the dummy variable, the following proof involves using the polar coordinates and expressing all derivatives in terms of the basis vectors of Frenet frames. Define

$$Y = Y_W^{\mathrm{Hann}}(\omega) = R (\cos \Theta + j \sin \Theta) = R \widehat{N}$$

and

$$\widehat{T} = -\sin \Theta + j \cos \Theta$$

Here, all the capitalized variables are functions of $\omega$. Also, for simplicity, we will use the Leibniz dot operator to notate $\frac{d}{d\omega}$. By inspection, we have

$$\dot{\widehat{N}} = \dot{\Theta}\widehat{T}$$ (18)

$$\dot{\widehat{T}} = -\dot{\Theta}\widehat{N}$$ (19)

which is the usual relation between the derivatives of the normal vector and the tangential vector of a trajectory on the complex plane. Now, with the aid of equations 18 and 19, the first and second derivatives can be concisely written as the following,

$$\dot{Y} = \dot{R}\widehat{N} + R\dot{\Theta}\widehat{T}$$ (20)

$$\ddot{Y} = [\ddot{R}-R\dot{\Theta}^2]\widehat{N} + [2\dot{R}\dot{\Theta} + R\ddot{\Theta}]\widehat{T}$$ (21)

Recall that the values of the derivates at $\omega = 0$ are

$$\dot{Y}(0) = 0$$ (22)

and

$$\ddot{Y}(0)= -j\frac{1}{2\alpha}Y(0) = -j\frac{1}{2\alpha}R(0)\widehat{N}(0) = -\frac{1}{2\alpha}R(0)\widehat{T}(0)$$ (23)

Combining equations 20 and 22, we have $\dot{R}(0)=0$ and $\dot{\Theta}(0)=0$. Plugging these into equation 21, we obtain

$$\ddot{Y}(0)= \ddot{R}(0)\widehat{N} + R(0)\ddot{\Theta}(0)\widehat{T}$$ (24)

Finally, equating equations 23 and 24, and noticing that $\widehat{N}$ and $\widehat{T}$ are perpendicular to each other, we prove that

$$\ddot{\Theta}(0)= -\frac{1}{2\alpha}$$