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When calculating DTFT of $\left( 1/2 \right)^{n} u \left[ n \right]$. We evaluate the sum as follows: DTFT

Please correct statements and answer questions below:

  1. So to go from STEP 1 to STEP 2, the limits of the series are changed from $-\infty$, $+\infty$ to 0, $+\infty$ because $u \left[ n \right]$ is $1$ for $n \geq 0$ and $0$ otherwise. Correct?

  2. In STEP 2 we substitute the value of $x \left[ n \right]$ so the term $u \left[ n \right]$ is not found in STEP 2 because it is evaluated as $1$. Correct?

  3. If term $u \left[ n \right]$ is no longer in STEP 2 equation how can we use the transform equation (indicated by arrow) in table when only $ \left( 1/2 \right)^{2}$ remains?

If you have a different explanation on how to transition from STEP 1 to STEP 3. Please feel free to comment.

My solution is below. I want to comment but I cannot attach images in comments.

Solution_DTFT

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  • $\begingroup$ This is not homework since it is a solution from a book. I am practicing these questions. See My notes attached below. $\endgroup$
    – Leo
    Commented Apr 3, 2019 at 21:54
  • $\begingroup$ $\displaystyle\left(\frac 12\right)^n e^{-j\omega n}= \left(\frac 12e^{-j\omega}\right)^n$ where $\displaystyle\left|\frac 12e^{-j\omega}\right| = \frac 12 < 1$ and so $$\sum_{n=0}^\infty \left(\frac 12\right)^n e^{-j\omega n} = \frac{1}{1-\frac 12e^{-j\omega}}$$ $\endgroup$
    – Dilip Sarwate
    Commented Apr 5, 2019 at 21:34
  • $\begingroup$ OK, thanks @DilipSarwate. This shows why the series formula was picked for the solution because of the magnitude of (1/2) being less than 1. But could you please answer the 3 questions posted above. Especially question 3, as we are solving the problem using formula that doesn't have u[n] in it. $\endgroup$
    – Leo
    Commented Apr 7, 2019 at 4:03
  • $\begingroup$ The value of $u[n]$ was included when in the transition from Step 1 to Step 2, the range of the summation was changed from all integers to all nonnegative integers. Since this bothers you, keep the range to all integers and put in $u[n]$ explicitly so that everything matches to the way you like to think. $\endgroup$
    – Dilip Sarwate
    Commented Apr 7, 2019 at 12:06
  • $\begingroup$ what is the book you are referring to $\endgroup$
    – vishal ranaweera
    Commented Apr 16, 2022 at 17:32

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