102_1_final_2010_fall_solution

# 102_1_final_2010_fal - Systems and Signals EE102 Lee Fall 2010-11 Final Exam NAME You have 3 hours for 6 questions Show enough(neat work in the

This preview shows pages 1–5. Sign up to view the full content.

Systems and Signals Lee, Fall 2010-11 EE102 Final Exam NAME: You have 3 hours for 6 questions. Show enough (neat) work in the clear spaces on this exam to convince us that you derived, not guessed, your answers. Put your ﬁnal answers in the boxes at the bottom of the page. Closed notes, closed book, 1 letter sized handwritten sheets allowed. Problem Score 1 2 3 4 5 6 Total 1

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

View Full Document
Problem 1. Fourier Transforms (15 Points) Find the Fourier transform of the following signal f ( t ) t 0 1 2 3 4 5 - 5 - 4 - 3 - 2 - 1 sinc 2 ( t / 3 ) rect ( t 2 ) This is an inﬁnite sequence of rectangles, weighted by a sinc 2 (). Eliminate convolutions in your answer. Solution: We can write this signal as f ( t ) = rect (2 t ) * h sinc 2 ( t/ 3) δ 1 ( t ) i . The Fourier transform is then F ( ) = 1 2 π " 1 2 sinc ω/ 2 2 π ! [3Δ(3 ω/ 2 π ) * 2 πδ 2 π ( ω )] # = 3 2 sinc ± ω 4 π ² " X n = -∞ Δ(3 ω/ 2 π ) * δ ( ω - 2 πn ) # = 3 2 sinc ± ω 4 π ² " X n = -∞ Δ(3( ω - 2 πn ) / 2 π ) # = 3 2 X n = -∞ sinc ± ω 4 π ² Δ 3( ω - 2 πn ) 2 π ! F ( ) = 3 2 n = -∞ sinc ³ ω 4 π ´ Δ ³ 3( ω - 2 πn ) 2 π ´ 2
Problem 2. Laplace Transforms (15 Points) Find the Laplace transform of the following signal f ( t ) t p ( t ) p ( t - 2) p ( t - 4) 2 1 0 4 f ( t ) This is the sum of an inﬁnite sequence of delayed causal subpulses f ( t ) = X n =0 p ( t - 2 n ) Assume p ( t ) = e - 7 t u ( t ) and indicate region of convergence. Hint: L{ p ( t - n ) } = e - sn P ( s ) . Also, recall that X n =0 a n = 1 1 - a when simplyﬁng your answer. Solution: f ( t ) = X n =0 p ( t - 2 n ) so F ( s ) = X n =0 L{ p ( t - 2 n ) } = X n =0 e - 2 sn P ( s ) = P ( s ) X n =0 ( e - 2 s ) n = P ( s ) 1 1 - e - 2 s Then P ( s ) is L n e - 7 t u ( t ) o = 1 s + 7 Substituting this into the expression for F ( s ), F ( s ) = 1 ( s + 7)(1 - e - 2 s ) . ROC: σ > 0 F ( s ) = 1 ( s +7)(1 - e - 2 s ) , ROC: σ > 0 3

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

View Full Document
Problem 3. Discrete Time Fourier Transform (30 Points) An engineer purchased an expensive, ﬁnely-tuned discrete-time low-pass ﬁlter with an im- pulse response h [ n ] and frequency response H ( e ). Sadly, the ﬁlter did not exactly suit the needs of the application and need to be modiﬁed. Note: In this problem we adopt an intuitive notion of non-ideal low-pass ﬁlters. To be ”low-pass” H ( e ) merely should have larger magnitude at ω 0 , ± 2 π, ± 4 π ,..., than at ω ≈ ± π, ± 3 π, ± 5 π , etc Parts a) and b) are independent: you may solve them in any order. a) Consider a ﬁlter f [ n ] with time-domain response: f [ n ] = ( 1 - h [0] if n = 0 - h [ n ] if n 6 = 0 . Provide an expression for F ( e ) in terms of H ( e ). Solution. Rewrite the expression for f [ n ] in closed form: f [ n ] = δ [ n ] - h [ n ]. F
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/13/2011 for the course EE 102 taught by Professor Levan during the Fall '08 term at UCLA.

### Page1 / 22

102_1_final_2010_fal - Systems and Signals EE102 Lee Fall 2010-11 Final Exam NAME You have 3 hours for 6 questions Show enough(neat work in the

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online