The signal is f(t) = sin(t) and t1=0 and t2=2pi
Energy of sin(t)
The energy expended from t1 to t2 is:
$ E=\int_{t1}^{t2}{|f(t)|^2dt} $
Therefore for our signal:
$ E=\int_0^{2\pi}{|sin(t)|^2dt} $
$ =\frac{1}{2}\int_0^{2\pi}(1-cos(2t))dt $
$ =\frac{1}{2}(t-\frac{1}{2}sin(2t))|_{t=0}^{t=2\pi} $
$ =\frac{1}{2}(2\pi) $
Therefore the total energy for the signal sin(t) from t1 to t2 is: $ E=\pi $
Average Power in time interval [t1, t2]
The average power for a signal is given by:
$ P=\frac{1}{t2-t1}\int_{t1}^{t2}{|f(t)|^2dt} $
Therefore for our signal sin(t) from t1=0 to t2=2pi:
$ P=\frac{1}{2\pi-0}\int_{0}^{2\pi}{|sin(t)|^2dt} $
$ =\frac{1}{2\pi}\frac{1}{2}\int_0^{2\pi}(1-cos(2t))dt $
$ =\frac{1}{4\pi}(t-\frac{1}{2}sin(2t))|_{t=0}^{t=2\pi} $
$ =\frac{1}{4\pi}(2\pi) $
Therefore the average power for the signal sin(t) from t1 to t2 is: $ P=\frac{1}{2} $
