# Calculating DTFT

When calculating DTFT of (1/2)^n u[n]. We evaluate the sum as follows:

1) So to go from STEP 1 to STEP 2, the limits of the series are changed from -infinity, + infinity to 0,+ infinity because u[n] is 1 for n>=0 and 0 otherwise. Correct?

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

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

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

• So is this a homework? why don't you show us your progress as well? – Fat32 Apr 3 at 20:50
• This is not homework since it is a solution from a book. I am practicing these questions. See My notes attached below. – Leo Apr 3 at 21:54
• $\displaystyle\left(\frac 12\right)^n e^{-j\omega n}= \left(\frac 12e^{-j\omega}\right)^n$ where $\displaystyle\left|\frac 12e^{-j\omega}\right| = \frac 12 < 1$ and so $$\sum_{n=0}^\infty \left(\frac 12\right)^n e^{-j\omega n} = \frac{1}{1-\frac 12e^{-j\omega}}$$ – Dilip Sarwate Apr 5 at 21:34
• 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. – Leo Apr 7 at 4:03
• 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. – Dilip Sarwate Apr 7 at 12:06