0(t) Note that the same conclusion can be reached by noticing that i(t) is the same signal in Example 3.5(O& W, p 193)with Ti=and T=4 To find y(t), which has bo=0, we first calculate bo and then subtract it from g(t) 1/2 bo T 0()dt (1)dt <t< →()=()-→y(O)= 号<|<2 x(t)=y(t+1) 1.5<t<-0.5 0.5<t<2.5 ketches of y(t)and r t) are shown below y(t) 3/4 3/4 1/4� � yˆ(t) −3 −2 −1 0 1 2 3 t 1 −T 2 T 2 · · · · · · Note that the same conclusion can be reached by noticing that yˆ(t) is the same signal in Example 3.5 (O& W, p.193) with T 1 1 = 2 and T = 4. To find y(t), which has b0 = 0, we first calculate ˆb0 and then subtract it from yˆ(t) : ⎪ 1 1 ⎪ 1/2 1 ˆb0 = yˆ(t)dt = (1)dt = T T 4 −1/2 4 1 2 2 3 1 1 < t < ⇒ y(t) = yˆ(t) − 4 � y(t) = 4 , − 1 1 4 , < |t| < 2. − 2 3 −1.5 < t < −0.5 x(t) = y(t + 1) = 4 , ⇒ 1 4 − , −0.5 < t < 2.5 Sketches of y(t) and x(t) are shown below: y(t) −3 −2 −1 1 2 3 t 3/4 −1/4 · · · · · · x(t) −3 −2 1 2 3 t 3/4 −1/4 −1 · · · · · · 10