00:01
Okay, so first we'll do question a, we compute the discrete time signal, y of n squared, 3 times negative 0 .2 to n.
00:28
And this can be simplified to 9 times sum of 0 .04 to n.
00:45
Okay.
00:49
And this is a geometric series with the first term of 0 .04.
00:54
Excuse me, actually n starts from 3.
01:07
So again, this is a geometric series with a first term of a equal 0 .04 cube, and the common ratio of 0 .04.
01:18
So ey has a solution 9 times 0 .04 cube over 1 minus 0 .04, which is equal to 6 times 0 .04, which is equal to 6 times 10 to negative 4.
01:41
So for signal energy, the average power is 0, and the type of signal is, so signal, the average power of the energy signal is equal to 0.
02:00
And since the energy is finite and the power is 0, y n is an energy signal.
02:14
Okay, b.
02:24
Okay, so first, first of all, for a, we take the sum from n equals 3 because the unit step function u of n minus 3, shift the signal to start at n equals 3.
02:37
So that's why we sum from 3...