pset14_sol_04

pset14_sol_04 - Unified Engineering II Spring 2004 Problem...

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Unformatted text preview: Unified Engineering II Spring 2004 Problem S21 (Signals and Systems) Solution: 1. The signal is plotted below: 1.2 1 g(t) 0.8 0.6 0.4 0.2 0 -10 -8 -6 -4 -2 0 Time, t 2 4 6 8 10 The signal is very smooth, almost like a Gaussian. Therefore, I expect that the duration bandwidth product will be close to the theoretical lower bound. 2. � Δt 2 �2 � 22 t g (t) dt =�2 g (t) dt The two integrals are easily evaluated for the given g (t). The result is � 7 t2 g 2 (t) dt = 2 � 5 g 2 (t) dt = 2 Therefore, � 7 Δt = 2 5 3. The time domain formula for the bandwidth is � �2 � 2 g (t) dt ˙ Δω =� 2 2 g (t) dt The numerator integral is � g 2 (t) dt = ˙ 1 2 Therefore, 2 Δω = √ 5 4. The duration­bandwidth product is √ 47 Δt Δω = ≈ 2.1166 5 which is very close to the theoretical lower limit of 2. This is not surprising, since the shape of g (t) is close to a gaussian. Unified Engineering II Spring 2004 Problem S22 (Signals and Systems) Solution: I used Mathematica to find some of the integrals, although you could use tables or integrate by parts. (a) ¯ t= � ∞ � 2 t7 e−2t/tau dt = 315 8 τ 16 t6 e−2t/tau dt = tg (t) dt = 45 7 τ 8 0 ¯ t= � � 2 ∞ g (t) dt = 0 Therefore, ¯7 t= τ 2 (b) � ¯ (t − t)2 g 2 (t) dt = Therefore, Δt = √ 315 9 τ 32 7 τ (c) � 9 g 2 (t) dt = τ 5 ˙ 8 Therefore, 2 Δω = √ 5τ (d) The duration­bandwidth product is � Δt Δω = 2 7 ≈ 2.366 5 which compares favotably with the theoretical lower bound Δt Δω ≥ 2 ...
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pset14_sol_04 - Unified Engineering II Spring 2004 Problem...

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