midterm_II_2010_sol - EE380 – Communication Systems...

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Unformatted text preview: EE380 – Communication Systems Spring 2010-2011 Midterm Examination – II – Solutions Open Book, Open Notes; Calculators allowed but laptops not allowed Time Allowed = 4 hours (April 17, 2011 from 10:00am-2:00pm) “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 INSTRUCTIONS: • MAKE SURE TO RETURN THIS QUESTION SHEET WITH YOUR ANSWER BOOK! • Perform your working and write your answers in the blue books; we will only look at those . • Compute final answers correct up to at least two decimal places. If numbers are too small, use exponent notation. • It is okay to leave the trivial steps and write the impor- tant ones. However, writing just the final answer that is incorrect will not earn you any partial credit. Use your good judgement in deciding what you want to write. • There is usually an easier way to solve problems – try to recognize that! • Read all the questions before you start working on the exam. The questions towards the end may be easier. Problem 1 [24] Problem 2 [30] Problem 3 [24] Problem 4 [12] Problem 5 [20] Problem 6 [20] Problem 7 [20] TOTAL [150] 1 Problem 1: [24 points] Let x 2 ( t ) = ∞ ∑ i =-∞ x 1 ( t − iT ) where T = 2 seconds. The Fourier transform of x 1 ( t ), X 1 ( f ), is as shown in Figure 1. 4 4-4 X 1 (f) f Figure 1: Fourier transform of x 1 ( t ) (a) [3 points] Sketch x 1 ( t ) labeling the significant points. Solution: Noting X 1 ( f ) = 4 ∧ ( f 8 ), we get x 1 ( t ) = 16 sinc 2 (4 πt ). Plot is shown in Figure 2(a).-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 2 4 6 8 10 12 14 16 t x 1 (t) (a) Plot of x 1 ( t )-3-2-1 1 2 3 2 4 6 8 10 12 14 16 t x 2 (t) (b) Plot of x 2 ( t ) Figure 2: Plots for parts (a) and (b) of the problem. (b) [4 points] Sketch x 2 ( t ). Again label the significant points. Solution: Noting x 2 ( t ) is simply repeated versions of x 1 ( t ), we get the plot shown in Figure 2(b). (c) [6 points] Sketch X 2 ( f ) and label the significant points. Solution: When x 2 ( t ) is the same as x 1 ( t ) repeated, then X 2 ( f ) is the sampled (and scaled down by T ) version of X 1 ( f ), as shown in Figure 3. f- 4 X 2 (f) 4 2 Figure 3: Plot of X 2 ( f ) 2 (d) [5 points] Find the Fourier series coefficients for the periodic signal x 2 ( t ). Solution: There are at least two ways to answer this. The easy way is to see that x 2 ( t ) is periodic extensions of x 1 ( t ), thus the Fourier series coefficients of x 2 ( t ) are, scaled down by T = 2, samples of the Fourier transform of x 1 ( t ). i.e. D i = 1 2 X 1 ( i 2 ). Same result is obtained by analysis as follows: x 2 ( t ) = ∞ summationdisplay i =-∞ x 1 ( t − iT ) = x 1 ( t ) ∗ ∞ summationdisplay i =-∞ δ ( t − 2 i ) (using T = 2) = x 1 ( t ) ∗ δ 2 ( t ) Then, X 2 ( f ) = X 1 ( f ) · 1 2 δ 1 2 ( f ) = X 1 ( f ) · 1 2 ∞ summationdisplay i =-∞ δ parenleftbigg...
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midterm_II_2010_sol - EE380 – Communication Systems...

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