finalSolutions - UNIVERSITY OF CALIFORNIA College of...

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UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences Professor Tse Spring 2005 EECS 20n — Final Exam Solutions [5 pts.] Problem 1 This is a linear constant coefficient system that is initially at rest . [15 pts.] Question 2 (a) (10 pts.) Overall period = 6 (b) (5 pts.) Overall period where are integers. No such exists because 2 is rational and is irrational, . y n () 1 4 -- yn 1 xn 1 2 1 + + = LTI q n 1 2 e i 2 π 3 ------ n 1 2 e i 2 π 3 n e i π n ++ = period 3 period 2 ω 0 2 π 6 = X 2 X 2 1 2 == X 3 1 = X k 0 otherwise = for 2 k 3 ≤≤ vt π t t cos + cos = period 2 period 2 π p 2 m 2 π n mn , , 2 π not periodic
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2 of 9 [10 pts.] Question 3 (a) (5 pts.) Yes, the system F can be linear. (b) (5 pts.) The system F cannot be time-invariant, because new frequencies have been created that did not exist in . [15 pts.] Question 4 (a) (7 pts.) Reading the signals off the diagram, we have: Written alternatively, the LCCDE describing the system above is: (b) (8 pts.) We note that if is the output of a delay block, then the input to the delay block must be , as shown below: Accordingly, we can label the original delay-adder-gain block diagram with and . We can now read off the diagram the expressions for , , and . Hence, the state-space equations are: ω 0 t () cos yt xt X ω yn 5 1 1 xn 1 + + = 5 1 1 ++ 1 + = s i n s i n 1 + s i n 1 + i 12 , = D s i n s 1 n 1 + s 2 n 1 + s 1 n 1 + s 2 n 1 + s 1 n 1 + s 2 n 5 } s 1 n 1 + + –5 s 1 n s 2 n 4 + == s 1 n + = s 2 n 1 + = s 2 n 1 + s 1 n = s 1 n 1 + s 2 n 1 + 5 –1 1 –0 s 1 n s 2 n 4 1 + = A b 1 0 s 1 n s 2 n 1
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This note was uploaded on 04/01/2008 for the course EECS 20n taught by Professor Babakayazifar during the Spring '08 term at University of California, Berkeley.

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finalSolutions - UNIVERSITY OF CALIFORNIA College of...

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