midterm_2010_sol - EE380 – Communication Systems Spring...

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Unformatted text preview: EE380 – Communication Systems Spring 2010-2011 Midterm Examination – I (Solutions) Open Book, Open Notes; Calculators and Laptops without Network connection allowed Time Allowed = 180 minutes (March 5, 2011 from 3:00pm-6:00pm) “I certify that I have neither received nor given unpermitted aid on this examination and that I have reported all such incidents observed by me in which unpermitted aid is given.” Signature Name Student ID RETURN THIS QUESTION SHEET WITH YOUR ANSWER BOOK! There is usually an easier way to solve problems – try to recognize that! Problem 1 [15] Problem 2 [15] Problem 3 [15] Problem 4 [15] Problem 5 [15] Problem 6 [20] Problem 7 [15] Problem 8 [20] TOTAL [130] 1 Problem 1: We first define two signals over a limited time as follows: h ( t ) = 6 − t ; t ∈ [0 , 6) g ( t ) = cos( πt/ 2); t ∈ [0 , 4) Next, we define the following infinite-duration signals: h ( t ) =    6 − t t ∈ [0 , 6) else ˜ h ( t ) = periodic repetitions of h ( t ) without overlap ˜ g ( t ) = periodic repetitions of g ( t ) without overlap ˜ r ( t ) = ˜ g ( t ) + ˜ h ( t ) (a) What is the period of h ( t )? Solution: ∞ because h ( t ) is aperiodic. (b) What is the period of ˜ h ( t )? Solution: 6 ( h ( t ) is confined to 6, repetitions without overlap are done at 6) (c) What is the period of ˜ g ( t )? Solution: 4 (d) What is the period of ˜ r ( t )? Solution: 12 (e) Find H ( f ), i.e., the Fourier transform of h ( t ). Solution: By direct integration, several correct answers can be found given below: 3 jπf + 1 4 π 2 f 2 ( 1 − e − j 12 πf ) 3 jπf + 3 je − j 6 πf sinc (6 πf ) Other forms of solution are also possible and acceptable (if correct). However, H (0) must explicitly be evaluated as H ( f ) = integraltext 6 (6 − t ) dt = 18. (f) What is the Fourier transform of h ( t ) evaluated at f = 0? Solution: H ( f ) = integraltext 6 (6 − t ) dt = 18 (g) Suppose we evenly sample H ( f ) and would like to be able to reconstruct H ( f ) from those samples. What is the maximum separation allowed between two consecutive samples? Solution: Since h ( t ) is non-zero only from 0 to 6, the samples in frequency domain must be separated by no more than 1 6 units. Note that the time domain signal is NOT symmetric, therefore, the maximum separation is NOT 1 12 . (h) Continuing from part (g), how will you reconstruct H ( f ) from its samples? [Note: There is a precise answer which can be given in one sentence. We are allowing no more than two sentences. Longer answers will receive no credit.] 2 Solution: Multiple correct ways of answering this: 1) multiply the time domain signal with a rect function extending from 0 to 6 (i.e., A ⊓ ( t − 3 6 ) where A is a constant, or 2) convolve with a sinc function given by B sinc (6 f ) e − j 2 π 3 f where B is a constant....
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This note was uploaded on 03/11/2012 for the course EE EE380 taught by Professor Z.u during the Spring '11 term at Lahore University of Management Sciences.

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midterm_2010_sol - EE380 – Communication Systems Spring...

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