pset14_sol_04

# pset14_sol_04 - Uniﬁed Engineering II Spring 2004 Problem...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

pset14_sol_04 - Uniﬁed Engineering II Spring 2004 Problem...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online