Y1t ht f1t ht t 15t 1 ht 15ht 1

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Unformatted text preview: e function is shifted. y1(t) = h(t) ∗ f1(t) =h(t) ∗ (δ(t)-1.5δ(t − 1)) = h(t) – 1.5h(t − 1) = [2u(t − 1) + u(t − 2) − 2u(t − 3) − u(t − 5)]- 1.5[2u(t − 2) + u(t − 3) − 2u(t − 4) − u(t − 6)] y1(t) =2u(t − 1) -2u(t − 2) – 3.5u(t - 3)+ 3u(t − 4)− u(t − 5) +1.5u(t − 6) 2 1 0 y1(t) -1 -2 -3 -4 0 1 2 3 t 4 5 6 7 Page 5 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo (ii) f2(t) = u(t)-1.5u(t − 1) y2(t) = h(t) ∗ f2(t) =h(t)*[u(t)-1.5u(t − 1)] =[2u(t − 1) + u(t − 2) − 2u(t − 3) − u(t − 5)] *[u(t)-1.5u(t − 1)] = 2r(t − 1) + r (t − 2) − 2 r (t − 3) − r (t − 5)- 3r(t − 2) - 1.5r (t − 3) + 3r (t − 4) + 1.5r (t − 6) y2(t)= 2r(t − 1) -2r (t − 2) – 3.5 r (t − 3) + 3r (t − 4) -r (t − 5) +1.5 r (t − 6) 2 1 0 y2(t) -1 -2 -3 -4 0 1 2 3 t 4 5 6 7 (iii) f3(t) = h(-t)=u(t + 5 ) + 2u(t+3) − u(t + 2) −2 u(t +1) y3(t) = h(t) ∗ f3(t) =[2u(t − 1) + u(t − 2) − 2u(t − 3) − u(t − 5)] * [u(t + 5 ) + 2u(t+3) − u(t + 2) −2 u(t +1)] =2r(t + 4) + r(t + 3) – 2r(t + 2) − r(t) + 4r(t + 2) + 2r...
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