ECE 101 problem set solution1 - ECE 101 Linear Systems...

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ECE 101 – Linear Systems, Winter 2008 Problem Set #1 Solutions January 17, 2008 (send comments/questions to [email protected]) Problem 1 [OW2, 1.21 (d,e) and 1.22 (f,h)] 1.21 (d),(e) Plots: -2 -1 0 1 2 3 4 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 [x(t) + x(-t)]u(t) -2 -1 0 1 2 3 4 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 x(4-t/2) 1
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1.22 (f),(h) Plots: -2 -1 0 1 2 -4 -3 -2 -1 0 1 2 3 4 x[n-2] ° [n-2] x[(n-1) 2 ] -2 -1 0 1 2 -4 -3 -2 -1 0 1 2 3 4 Problem 2 [OW2, 1.34] 1.34 (a) If x [ n ] is an odd signal, then x [ - n ] = - x [ n ] for all n . In particular, x [0] = x [ - 0] = - x [0] , implying that x [0] = 0. Therefore, X n = -∞ x [ n ] = x [0] + X n =1 ( x [ n ] + x [ - n ]) = 0 + X n =1 ( x [ n ] - x [ n ]) = 0 . (b) Define y [ n ] = x 1 [ n ] x 2 [ n ] . Then y [ - n ] = x 1 [ - n ] x 2 [ - n ] = ( - x 1 [ n ]) · x 2 [ n ] = - x 1 [ n ] x 2 [ n ] = - y [ n ] . This means that y [ n ] is an odd signal. (c) Since x [ n ] = x e [ n ] + x o [ n ], we have X n = -∞ x 2 [ n ] = X n = -∞ ( x e [ n ] + x o [ n ]) 2 = X n = -∞ x 2 e [ n ] + X n = -∞ x 2 o [ n ] + 2 X n = -∞ x e [ n ] x o [ n ] (1) From part (b), we know that x e [ n ] x o [ n ] is an odd signal. It then follows from part (a) that X n = -∞ x e [ n ] x o [ n ] = 0 , which implies that the last summation on the right of equation (1) equals zero. The desired result follows. 2
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(d) We can use the same reasoning as in part (c). All we need to show is that if y ( t ) is an odd signal, then Z -∞ y ( t ) dt = 0 . Since y ( t ) is odd, we know that it equals its odd part y o ( t ). Therefore y ( t ) = y o ( t ) = 1 2 ( y ( t ) - y ( - t )) .
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