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102_1_final_2010_fall_solution

102_1_final_2010_fall_solution - Systems and Signals EE102...

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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
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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 " X n = -∞ Δ(3 ω/ 2 π ) * δ ( ω - 2 πn ) # = 3 2 sinc ω 4 π " X n = -∞ Δ(3( ω - 2 πn ) / 2 π ) # = 3 2 X n = -∞ sinc ω 4 π Δ 3( ω - 2 πn ) 2 π ! F ( ) = 3 2 n = -∞ sinc ω 4 π Δ 3( ω - 2 πn ) 2 π 2
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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
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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 ( e ) = 1 - H ( e ) This part of the problem tested your ability to decompose signals into simpler elements. In particular, rewriting f [ n ] in closed form, as a sum of two signals whose Fourier Transform is known, was the most difficult part. F ( e ) = 1 - H ( e ) 4
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If H ( e ) is a low-pass filter, is F ( e ) necessarily a high-pass filter? If so argue convincingly. Otherwise, provide a counter-example.
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