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Answer by Matt L. for Fourier transform of an impulse-train sampled signal
Both results are correct. Hence, the equality$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=0}^{\infty}e^{-nT(1+j\omega)}\tag{1}$$holds true.One way...
View ArticleFourier transform of an impulse-train sampled signal
I'm trying to calculate the Fourier transform of an impulse-train sampled signal in two differnt ways but I end up with different results.Impulse-train sampling of a continous signal $x(t)$ with...
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