Rectangle Windowed Signal

Now, we consider the Fourier Transform of the rectangle-windowed time domain signal.

$$Y_W^{\mathrm{Rec}}(\omega) = \int_{-T}^T e^{j \alpha t^2} e^{-j \omega t} dt$$

We now can differentiate $Y_W^{\mathrm{Rec}}(\omega)$ with respect to frequency. This introduces a multiplicative factor of $(-jt)$ in the time domain. Thus, we may write:

$$\frac{d Y_W^{\mathrm{Rec}}(\omega)}{d \omega} = \int_{-T}^T (-jt) e^{j \alpha t^2} e^{-j \omega t} dt$$ (1)

$$= \frac{1}{2\alpha} \int_{-T}^T (-j2\alpha t) e^{j \alpha t^2 -j \omega t}dt$$ (2)

$$= \frac{1}{2\alpha} \int_{-T}^T -(j2\alpha t - j\omega) e^{j (\alpha t^2-\omega t)} - j\omega e^{j (\alpha t^2-\omega t)} dt$$ (3)

$$= \frac{-1}{2\alpha} \left. \big( \big[ e^{j(\alpha t^2-\omega t)} ... ... \right\vert _{-T}^T - \int_{-T}^T j\omega e^{j (\alpha t^2-\omega t)} \big) dt$$ (4)

$$= \frac{-1}{2\alpha} \big(-e^{j\alpha T^2}2j\sin{\omega T}+ j \omega Y_W^{\mathrm{Rec}}(\omega) \big)$$ (5)

Given this closed form iterative expression, we consider the derivative of the frequency response at 0:

$$\frac{d Y_W^{\mathrm{Rec}}(\omega)}{d \omega}\Big\vert _{\... ... T^2}2j\cdot 0 + j \cdot 0 \cdot Y_W^{\mathrm{Rec}}(0) ) = 0$$

Thus we show that there is a stationary point of $Y_W^{\mathrm{Rec}}(\omega)$ at $\omega = 0$. Because $Y_W^{\mathrm{Rec}(\omega)}$ is a complex variable, this is a more appropriate term than ``min'' or ``max.'' We proceed to calculate the second derivative of $Y_W^{\mathrm{Rec}}(\omega)$ with respect to $\omega$:

$$\frac{d^2 Y_W^{\mathrm{Rec}}(\omega)}{d \omega^2} = \frac{-1}{2\alpha}\big[ -j T\big(e^{j(\alpha T^2-\omega T)} + e^{... ...frac{dY^{\mathrm{Rec}}_W(\omega)}{d \omega} + jY_W^{\mathrm{Rec}}(\omega) \big]$$ (6)

$$= \frac{-1}{2\alpha}\big[ -j T e^{j\alpha T^2} 2 \cos(\omega T) + j... ...\frac{dY^{\mathrm{Rec}}_W(\omega)}{d\omega} + jY_W^{\mathrm{Rec}}(\omega) \big]$$ (7)

Now, to determine phase concavity, we examine the second derivative with respect to frequency at DC:

$$\left. \frac{d^2 Y_W^{\mathrm{Rec}}(\omega)}{d \omega^2} \right\vert _{\omega=0} = \frac{-1}{2\alpha}\big[ - j T e^{j\alpha T^2} 2 \cos(0) + j0 \cdo... ...thrm{Rec}}_W(0)}{d\omega}\right\vert _{\omega=0} + jY_W^{\mathrm{Rec}}(0) \big]$$ (8)

$$= \frac{-j}{2\alpha}Y_W^{\mathrm{Rec}}(0) +\frac{j}{\alpha}e^{j \alpha T^2}T$$ (9)

Thus, the first term suggests a clear phase relationship, even though the second term is less clear.