0
$\begingroup$

How did we get this final answer in example in fig attached. I understand that delta(n) is a DT unit impulse where it is 1 at n = 0 and zero elsewhere. I understand that u[n] is a unit-step where it is 1 for n>=0 and 0 elsewhere.

How do these two functions equate in final answer below and how did we get each of the final form of the three terms in the final answer? and why is the 3rd term still in u[n-2] form? Convolution between two DT signals

$\endgroup$
1
$\begingroup$

Draw the three discrete time signals (shown on the left side) on top of one another. You will see the summation leading to $\delta[n]$ and $\delta[n-1]$ term as follows. For $n=0$, only the first term survives and hence $3\delta[n]$. For the second term, put $n=1$ that makes $3/2+2=7/2$ at index $1$.

Then, obviously third term extends in $u[n-2]$ form as a summation from all three terms at each subsequent index. This is because $n=2$ brings $3/4$ from the first term, $1$ from the second term and $1$ from the third term, totaling to $11/4$ onwards.

$\endgroup$
  • $\begingroup$ this answer deserves a check (IMHO) but I don't have enough points so someone should check it. $\endgroup$ – mark leeds Apr 13 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.