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To obtain fourier transform of u[n], u[n] - u[n-1] = delta[n] , taking fourier transform of both sides of the equation results in : U(w) - exp(-jw) U(w) = 1 , hence : U(w) = 1/(1-exp(-jw)) which is wrong and the right answer has an extra term. Which step is wrong in this possible solution? I know the right proof of fourier transform of u[n], my question is regarding the wrong part of this solution.

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  • $\begingroup$ Yes it is similar but not the same. I had seen that question but I could not find flaw of this solution. Now I got that this is wrong because we can only divide both sides by (1-exp(-jw)) when w is not zero. Hence, this solution is wrong. $\endgroup$ – Mopu Nov 13 at 18:55
  • $\begingroup$ Yes, sorry, I meant this question. $\endgroup$ – Matt L. Nov 13 at 19:00
  • $\begingroup$ Yes, that is the same. Thank you. $\endgroup$ – Mopu Nov 13 at 19:12
  • $\begingroup$ The way I would explain the flaw in that proof is shown here. $\endgroup$ – Matt L. Nov 14 at 11:48
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The DFT of a unit step response is $$U(\omega) = \frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega)$$ Applying the shift property as you did will give: $$\mathcal{F}(u[n] - u[n-1]) = U(\omega) - U(\omega)e^{-j \omega} = \frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega) - [\frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega)]e^{-j \omega}$$ that is $$\mathcal{F}(u[n] - u[n-1]) = \frac{1- e^{-j \omega}}{1 - e^{-j \omega}} +\pi\delta(\omega)( 1 - e^{-j\omega}) = 1+\pi\delta(\omega)( 1 - e^{-j\omega})$$ The second term is always zero because for $\omega = 0$, $1 - e^{-j\omega} = 0 $ and it's zero on any other point, so you get, $$\mathcal{F}(u[n] - u[n-1]) = 1 = \mathcal{F}(\delta[n])$$

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