I would like to know how $\frac T2$ came for time period in the image below.Because I think the frequncy is 1 so time period should be 1.Is that correct? or can anyone help me.
.
The article from where I got this image is here
$\begingroup$
$\endgroup$
$\begingroup$
$\endgroup$
Note that the second half of the period of the sinusoid is simply equal to the first half times $-1$. Since the same is true for the square wave, the multiplication of the two is the same in both intervals $[0,T/2]$ and $[T/2,T]$. That's why the integral over the whole period equals twice the integral over the first half of the period.
-
$\begingroup$ :Thanks for your answer.But could you say how it becomes "part times -1". $\endgroup$ – justin Nov 21 '14 at 10:55
-
$\begingroup$ @justin: What I meant to say was that the second half of the period of the sine wave equals the inverted first half. $\endgroup$ – Matt L. Nov 21 '14 at 11:20
-
$\begingroup$ :okay.Then whether time period should become $T$.Whether we are only taking the integral over the first half of the signal(that means upto 0.5 second)?In that case it's right it should be $\frac T2$.I couldn't get why we should express timeperiod in $T$ when the timeperiod is given in numbers in the image. $\endgroup$ – justin Nov 21 '14 at 11:59
-
$\begingroup$ @justin: Draw $x(t)y(t)$ in the interval $[0,T]$, and you'll understand. $\endgroup$ – Matt L. Nov 21 '14 at 14:40
-
$\begingroup$ @justin: in case it helps: the first line in the correlation calculation is generic: it works for any value of $T$. In the second line, they substitute $T$ with 1 to get the value of the correlation in the particular case when $T=1$. $\endgroup$ – MBaz Nov 21 '14 at 16:43
