Hann Windowed Signal

We note that the above derivation assumed a rectangular window. When we use a Hann window, however, the math will eventually reveal a simpler second derivative at zero. We begin by considering the Hann windowed signal in the time domain.

$$y_W^{\mathrm{Hann}}(t) = w(t) y(t) = (1+\frac{1}{2}e^{j\omega_T t}+\frac{1}{2}e^{-j\omega_T t}) y(t)$$ (10)

where $\omega_T = \frac{\pi}{T}$. Therefore, the Fourier transform becomes:

$$Y_W^{\mathrm{Hann}}(\omega) = \int_{-T}^T w(t) y(t) e^{-j \omega t} dt$$ (11)

$$= Y_W^{\mathrm{Rec}}(\omega) + \frac{1}{2}Y_W^{\mathrm{Rec}}(\omega-\omega_T) + \frac{1}{2}Y_W^{\mathrm{Rec}}(\omega + \omega_T)$$ (12)

the first derivative becomes

$$\begin{eqnarray*} \frac{d Y_W^{\mathrm{Hann}}(\omega)}{d \omega} &=& \frac{d... ... &=& \frac{-1}{2\alpha}j\omega Y_W^{\mathrm{Hann}}(\omega)\\ \end{eqnarray*}$$

and the second derivative becomes

$$\frac{d^2Y_W^{\mathrm{Hann}}(\omega)}{d\omega^2} = \frac{-1}{2\alpha}{j\omega \frac{dY_W^{\mathrm{Hann}}(\omega)}{d\omega} + j Y_W^{\mathrm{Hann}}(\omega)}$$ (13)

We now evaluate this at $\omega = 0$:

$$\frac{d^2Y_W^{\mathrm{Hann}}(\omega)}{d\omega^2} \Big\vert _{\omega=0} = \frac{-j}{2\alpha}(Y_W^{\mathrm{Hann}}(0))$$ (14)

Due to the $-j$ factor, the orientation of $Y''(0)$, as a vector, is 90 degrees either ahead or behind $Y(0)$ on the complex plane, depending on if $\alpha$ is positive or negative. Now, we recall that $Y'(0) = 0$ still holds ( $Y(-w) = Y(w)$ is valid for any complex signal $y(-t) = y(t)$ ). Thus, we can rotate the coordinates by $-\angle(Y(0))$, and the resulting trajectory will touch the new x-axis at $\omega = 0$ and then trace backward. It touches the x-axis perpendicularly, which, with some vector calculus manipulations, proves phase concavity, namely

$$\frac{d^2 \Theta(\omega)}{d\omega ^2} = \frac{-1}{2\alpha}$$ (15)

where $\Theta(\omega)$ is the usual definition of phase spectrum. The proof is given in appendix A.