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3 - ECE 483 Homework#3 Solution Problem 1 Given a...

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2/18/10 ECE 483 Homework #3 Solution Problem 1. Given a continuous-time signal e ( t ) = e - 3 t , t 0. (a) Find the signal e * ( t ) obtained by passing e ( t ) through an ideal sampler with sampling period T = 0 . 5 s, and its Laplace transform E * ( s ). (b) Suppose e * ( t ) is subsequently passed through a zero-order data hold to produce the signal ¯ e ( t ). Find ¯ e ( t ) and its Laplace transform ¯ E ( s ). (c) Repeat (a)(b) for the signal f ( t ) = e ( t 1) = e - 3( t - 1) u ( t 1), t 0. (d) Repeat (a)(b) for the signal g ( t ) = e ( t 0 . 1) = e - 3( t - 0 . 1) u ( t 0 . 1), t 0. Solution: (a) e * ( t ) = k =0 e ( kT ) δ ( t kT ) = k =0 e - 1 . 5 k δ ( t 0 . 5 k ), which is a sequence of impulses; E * ( s ) = k =0 e - 1 . 5 k e - 0 . 5 ks = 1 1 - e - 1 . 5 - 0 . 5 s . (b) ¯ E ( s ) = E * ( s ) 1 - e - sT s = 1 - e - 0 . 5 s s (1 - e - 1 . 5 - 0 . 5 s ) , and ¯ e ( t ) = e ( kT ) = e - 1 . 5 k , for kT t < ( k + 1) T, k = 0 , 1 , . . . , which is a stair-like signal. (c) For the given f ( t ), we have f * ( t ) = summationdisplay k =0 f ( kT ) δ ( t kT ) = summationdisplay k =2 e - 3(0 . 5 k - 1) δ ( t kT ) = summationdisplay k =2 e - 1 . 5 k +3 δ ( t 0 . 5 k ) F * ( s ) = summationdisplay k =2 e - 1 . 5 k +3 e - 0 . 5 ks = e - s 1 e - 1 . 5 - 0 . 5 s ¯ f ( t ) = braceleftBigg 0 if 0 t < 2 T e - 3(0 . 5 k - 1) if kT t < ( k + 1) T, k = 2 , 3 , . . . ¯ F ( s ) = F * ( s ) 1 e - T s s = e - s (1 e - 0 . 5 s ) s (1 e - 1 . 5 - 0 . 5 s ) . Note that since the time delay 1 = 2 T is an integer multiple of the sampling period T = 0 . 5, the time delay property of star transform applies, i.e., F * ( s ) = e - 2 sT E * ( s ). 1
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(d) For the given g ( t ), we have g * ( t ) = summationdisplay k =0 g ( kT ) δ ( t kT ) = summationdisplay k =1 e - 3(0 . 5 k - 0 . 1) δ ( t kT ) = summationdisplay k =1 e - 1 . 5 k +0 . 3 δ ( t 0 . 5 k ) G * ( s ) = summationdisplay k =1 e - 1 . 5 k +0 . 3 e - 0 . 5 ks = e - 0 . 5 s - 1 . 2 1 e - 1 . 5 - 0 . 5 s ¯ g ( t ) = braceleftBigg 0 if 0 t < T e - 3(0 . 5 k - 0 . 1) if kT t < ( k + 1) T, k = 1 , 2 , . . .
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