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> $X(w)=\int_{-T}^{T}1*e^{-iwt}dt=\frac{e^{-iwT}-e^{iwT}}{-jw}=\frac{e^{iwT}-e^{-iwT}}{jw}=\frac{2sin(wT)}{w}$ > $= 2 T\frac{sin(wT)}{wT}=2 Tsinc(\frac{wT}{\pi})$ Now let's take $T$ to infinity: > $lim_{T\rightarrow \infty} 2 Tsinc(\frac{wT}{\pi}) $ The sinc function gets zero'd at $\omega=\pm\pi/T$. So we have a function that can be approximated as an infinitely narrow isosceles triangle with a base of: >$\pi/T-(-\pi/T)=\frac{2\pi}{T}$ The height is obviously $2T$ so we can calculate the area easily: > $AreaUnderCurve=\frac{2\pi}{T}\cdot 2T\cdot \frac{1}{2}=2\pi$ We've found a function $lim_{T\rightarrow \infty} 2 Tsinc(\frac{wT}{\pi}) $ that is zero for all values except $x=0$. The limit at $x=0$ is $\infty$. So we can just write: > $lim_{T\rightarrow \infty} 2 Tsinc(\frac{wT}{\pi}) = 2\pi \delta(\omega)$
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Sat, 30 Jan 2021 00:48 GMT