Problem on  Discrete Time Fourier Transform (DTFT) – Discrete Time Signals Processing
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Problem on Discrete Time Fourier Transform (DTFT) – Discrete Time Signals Processing


Hi friends let us see a problem on DTFT on how to solve a problem using DTFT tool now a given sequence towards this X of N as A for N varying from zero less than equal to n less than equal to L and is zero otherwise and they’re asking us to find X of Omega so we wanted to find X of Omega from this given sequence X of N as A for 0 to L and is 0 otherwise so let us first draw and see how the signal looks like so if I draw the signal X of n with the n as the x axis and X of n as my function y axis then it will go from let us say this is my length L then for 0 1 2 3 and so on till L I will get the amplitude as a so there will be amplitude a everywhere it also will be a till L I will have an amplitude of L a and after that there are all zeros and therefore this also there are all zeros so 0 otherwise means except at L so there are many such things after L everything goes 0 and before 0 everything is 0 so this is how my signal X of n looks like and I have to find X of Omega so the solution is by definition of DTS D X of Omega is equal to summation n goes from 0 to infinity now I can write minus infinity to plus infinity but because the signal starts from 0 and below zero everything is zero it will be useless in writing from minus infinity to zero and then from zero to infinity because minus infinity to minus one every value will be zero so I am writing directly from n is equal to 0 to infinity X of n e raised to minus J Omega n so after putting the value of x of n my X of Omega will look like X of Omega is equal to summation now n is not going to infinity because after L the signals are going to 0 so after L the summation will all return to 0 so that is why I will put the limits on the summation as 0 to L and X of n e raised to minus J Omega n here we can put summation n is equal to 0 to L this I put a I raised to minus J Omega n which is X of Omega ok now you put a outside a summation n goes from 0 to L raised to minus J Omega n is what we got as a sequence now this is a finite series so the finite series output we will be getting as 1 minus ok so just to recall just to recall summation n goes from 0 to say K and a raise to n goes as 1 minus a raise to K minus 1 divided by sorry K plus 1 divided by 1 minus a ok so this is the series so I will get a into 1 minus instead of a over here we have erased – J Omega so we get e raise to K over here instead of n I will put a raise to minus J Omega L plus 1 divided by 1 minus e raised to J Omega so this is my X of Omega thank you

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