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Unformatted text preview: EE 341 Spring 2011 Clark Page 1 of 11 Final Exam June 8, 2011 Your Name: Solutions Exam Instructions: 1. Open book and notes. But as listed in our course syllabus: “No turnedon electronic devices (calculators, laptops, ipods, cell phones, beepers, etc.) are allowed for exams.” Please turn all devices off now. 2. Do not open this exam until 8:30. 3. Put all your answers in the appropriate space. (2 points for not putting your solutions in their correct spaces on these exam pages.) But put your name at the top of extra worksheets as necessary. 4. Turn in your work and put your name at the top of all loose worksheets. No need to staple them. This work will be looked at for possible partial credit. 5. Justify all of your answers. 6. The exam will be collected promptly at 10:20. Continuing to work after the bell will cause you to lose points. 7. This exam has a total of 9 pages (including this page). 8. The weight (out of 100) of each section of each problem is specified in parentheses. 9. The total weight for this exam is 100 points. EE 341 Spring 2011 Clark Page 2 of 11 Problem 1 (20 points) For the input signal [ ] f n and the output signal [ ] g n , suppose we already know that [ ] 2 [ 1] [ ] g n g n f n Assume that [ ] 0 g n for n < 0 (meaning the system is initially at rest). In the following questions, show your steps and justify your answers for full credit. a) (6 points) What is the system impulse response? Does its Fourier transform exist? Applying the ztransform to both sides, we have 1 ( ) 2 ( ) ( ) G z z G z F z . The transfer function is the ratio 1 ( ) 1 ( ) ( ) 1 2 2 G z z H z F z z z with a ROC defined by 2 z since the system is causal. From a ztransform table we obtain the impulse response in the timedomain: [ ] 2 [ ] n h n u n . * Note: Another method is to define [ ] [ ] f n n and solve for [ ] [ ] h n g n for n = 0, 1, 2, 3, … until a pattern develops, and then conclude it is an exponential with base (–2). Since the ROC does not cover the unit circle, the Fourier transform of [ ] h n does not exist. b) (7 points) If [ ] [ ] f n u n , then what is [ ] g n ? With the impulse response in (a), we can compute the convolution [ ] [ ] [ ] g n h n u n . [ ] [ ] [ ] [ ] 2 [ ] [ ] 2 n n k n k k k k k g n h n k u k h n k u n k u n where the step functions [ ] [ ] u k u n k limit k between 0 and n , and n between 0 and . Breaking part the exponential and applying the finitesum formula yields 1 1 1/ 2 1 [ ] 2 [ ] 2 [ ] 2 1 1/ 2 n k n n n k g n u n u n and finally 2 1 [ ] 2 [ ] 3 3 n g n u n...
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This document was uploaded on 07/08/2011.
 Spring '09

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