Definition of Approximation

Something not covered above was the effect of using the infinite limits approximation directly, while still using the rectangular window. Herein, we rigorously show why this is not a valid approach. We begin our pursuit of this doomed idea by stating the assumption:

$$\pm\sqrt{\frac{2\alpha}{\pi}}\left(-\frac{N}{2}-\frac{\pi k}{\alp... ...\alpha}{\pi}}\left(\frac{N}{2}-\frac{\pi k}{\alpha K}\right) \approx \pm\infty.$$ (147)

This approximation is a much stronger one than the large limits approximation, and allows a greatly simplified expression for the real and imaginary parts of $Y_a(k)$. Here, we consider why this is not valid for the rectangle window case. Though our result is discouraging, we recall that the infinite limits approximation may in fact be simulated by using a different windowing function, as has been covered earlier.

To explore the validity of the infinite limits approximation for the rectangle window case, we will compare this approximation to that made above (the large limits approximation). Since we know the large limits approximation is very accurate (to within 2.3% for $c = 2$) we will consider the error introduced by the infinite limits approximation relative to the large limits approximation, and judge its validity accordingly.

Presently, we compare the large limits approximation

$$\int_{\pm l_1}^{\pm l_2} \cos\left(\frac{\pi}{2}u^2\right) du \approx \pm\left(1 + \ensuremath{\frac{1}{-l_1\pi}}\sin\left(\ensuremath{... ...ath{\frac{1}{l_2\pi}}\sin\left(\ensuremath{\frac{\pi}{2}}(l_2)^2\right) \right)$$ (148)

$$% \int_{\pm l_1}^{\pm l_2} \sin\left(\frac{\pi}{2}u^2\right) du \approx \pm\left(1 - \ensuremath{\frac{1}{-l_1\pi}}\cos\left(\ensuremath{... ...th{\frac{1}{l_2\pi}}\cos\left(\ensuremath{\frac{\pi}{2}}(l_2)^2\right) \right),$$ (149)

and the infinite limits approximation

$$\int_{\pm l_1}^{\pm l_2} \cos\left(\frac{\pi}{2}u^2\right) du \approx \pm \int_{-\infty}^{\infty} \cos\left(\frac{\pi}{2}u^2\right) du = \pm1$$ (150)

$$\int_{\pm l_1}^{\pm l_2} \sin\left(\frac{\pi}{2}u^2\right) du \approx \pm \int_{-\infty}^{\infty} \sin\left(\frac{\pi}{2}u^2\right) du = \pm1.$$ (151)

and see that the infinite limits approximation allows us to drop the sinusoidal terms we included in the large limits approximation.