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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

 

 

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