GATE 2015 EE – SET 1 – Question 9

y(t) = $$\frac{1}{T}\int_{t-T}^{T}U(\tau )d\tau $$

y(t)=$$\frac{1}{T}\int_{t-T}^{T}sin(\omega \tau )d\tau $$

U$$(\tau )$$= Sin$$(\omega \tau )$$
$$\omega$$ = 2$$ \pi (\frac{1}{2T})=\frac{\pi }{T}$$
$$\omega$$ T=$$\pi$$

$$\therefore $$ , Y(t) =$$\frac{1}{t}\left [ \frac{-C0S(\omega \tau )}{\omega } \right ]_{t-T}^{t}$$

Y(t) =$$\frac{1}{t}\left [ \frac{-C0S(\omega \tau )}{\omega } \right ]_{t}^{t-T}$$

Y(t)=$$ \frac{1}{\pi }\left [ cos\omega (t-T) -cos\omega t\right ] $$

y(t)= $$\frac{1}{\pi }\left [ -2cos\omega t \right ]$$ = -$$\frac{2}{\pi }cos\omega T$$

y(t)=- $$\frac{2}{\pi }\left [ -sin(90+\omega T) \right ]$$ =  $$\frac{2}{\pi }sin(90 +\omega T)$$

So u(t) = sin$$ (\omega T)$$

$$ \phi $$ =90