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Quiz_3_sol

# Quiz_3_sol - ECE 483 Quiz 3 Solution Problem 1(10 points...

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Unformatted text preview: ECE 483 Quiz # 3 Solution Problem 1. (10 points) Given the sample period T = 0.5 s, ﬁnd the star transform of s−1 . E (s) = (s + 2)2 Solution: First decompose E (s) as E (s) = s+2−3 1 3 = − . 2 (s + 2) s + 2 (s + 2)2 Then by looking up the z -transform table, we have E (z ) = 3T ze−2T z 1.5ze−1 z (z − 2.5e−1 ) z − = − = . z − e−2T (z − e−2T )2 z − e−1 (z − e−1 )2 (z − e−1 )2 λ−1 1 d λ−1 = (λ + 2)2 1 − z −1 eT λ dλ 1 − z −1 eT λ Alternatively, one can use the formula E (z ) = poles of E (λ) Residue of , λ=−2 which can be veriﬁed to give the same result as above. Therefore, the star transform is E ∗ (s) = E (z ) z =esT = e0.5s (e0.5s − 2.5e−1 ) 1 − 2.5e−1−0.5s = . (e0.5s − e−1 )2 (1 − e−1−0.5s )2 e(t) 1 s f (t) T f (kT ) Data hold ¯ f (kT ) Problem 2. (10 points) In the above system, the signal e(t) = e−2t · 1(t), the sample period is T = 0.5 s, and the data hold is the zero order data hold. Find (a) (2 points) The Laplace transform of e(t); Solution: E (s) = 1 . s+2 (b) (2 points) The Laplace transform of f (t); Solution: 1 1 1 F (s) = ·= . s+2 s s(s + 2) (c) (3 points) The z -transform of f (kT ); Solution: F (z ) = Z [F (s)] = Z 1 0.5z (1 − e−1 ) = . s(s + 2) (z − 1)(z − e−1 ) ¯ (d) (3 points) The Laplace transform of f (t); Solution: 1 − e−sT ¯ F (s) = F ∗ (s) · = F (z ) s 1 z =esT · 1 − e−sT 0.5(1 − e−1 ) = . s s(e0.5s − e−1 ) ...
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