(b)For this part, we plot the individual signals involved and take the sum of products 2 r(-2)6(t-)+(3-t) Figure 2.b: a(t-2)[8(t-3)+u(3
(b) For this part, we plot the individual signals involved and take the sum of products: 1 2 −1 1 2 3 4 5 6 x(t − 2) t −1 1 2 −1 1 2 3 4 5 6 �(t − 1 2 ) 1 2 (1) −1 1 2 3 4 5 6 u(3 − t) 2 −1 −1 1 2 3 4 5 6 x(t − 2)[�(t − 1 2 u(3 − t)] 1 2 ( 1 2 ) 1 2 -1 ) + t t t Figure 2.b: x(t − 2)[�(t − 1 2 ) + u(3 − t)] 6
Problem 3 (a) This problem can be solved in two stages. First we Hip the signal and then we shift by 345 1 234 (b) From the expression, all we have to do is take every odd sample. When you plug in n=0, n=l, n=2... into the expression 2n+l, you end up with the odd samples of the original signal as follows x2n+1
Problem 3 (a) This problem can be solved in two stages. First we flip the signal and then we shift by 2: 1 −1 −4 −3 −2 −1 1 2 3 4 5 x[−n] n 1 2 1 2 −1 2 x[2 − n] 1 −1 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 n 1 2 1 2 −1 2 (b) From the expression, all we have to do is take every odd sample. When you plug in n = 0, n = 1, n = 2 · · · into the expression 2n + 1, you end up with the odd samples of the original signal as follows: x[2n + 1] 1 −3 −2 −1 1 2 3 4 n 1 2 −1 2 7