00:01
So here we have to find out the energies of the signal which is illustrated here.
00:05
So we are considering about the signals.
00:08
So first signal is given.
00:09
So in the x -axis, it is representing the time and in the y -axis, it is representing the ft.
00:15
So let's say this is the first signal.
00:17
This is zero point.
00:18
This is one point from here.
00:20
So energies of the given signal are calculated here.
00:23
That is from limit that is from minus infinity to plus infinity x of n raised to the power 2 of n.
00:31
So equation of the given line in the graph is represented as y minus y2 divided by x minus x2 become equals to y2 minus y1 that is divided by x2 minus x1.
00:41
So here we can say that x1, y1 and x2, y2 is the point through which the line passes.
00:48
So we can say that f of t minus 1 which is divided by t minus 1 is equals to 1 minus 0 that is divided by 1 minus 0.
00:56
So the value of f of t from here is equals to t.
00:59
So energy from here is equals to integral that is from 0 to 1 of f of t to its whole square which is multiplied by the dt.
01:09
So this from here is equals to integral that is limit from 0 to 1 of t raised to the power 2 that is multiplied by the dt.
01:15
So e from here is equals to 1 divided by 3 units.
01:22
Similarly, if we have to find the energy of the another signal, let's say we are considering about the another signal.
01:30
So this is another signal which we are having.
01:33
So this is a signal here that is f1 of t, t this is minus 1, this from here is 1.
01:39
So equation of the signal is f1 of t minus 1 which is divided by t minus of minus of 1 that is equals to 1 minus 0 which is divided by 1 minus 0.
01:48
So the f1 of t from here is equals to minus of t.
01:52
And if we are considering about the energy, e1 that is equals to integral that is from minus 1 to 1 of f1 of t raised to the power 2 which is multiplied by the dt that is equals to integral that is from minus 1 to 1 of t square which is multiplied by the dt.
02:07
So the value of e1 from here is equals to 1 divided by 3 units.
02:11
So this is the value of the e1 from here.
02:14
Now if we are considering about the this term that is t1, this from here is minus of 1.
02:20
So this is the system here we are considering let's say this is f2 of t...