ee20-hw04-f09-sol

# ee20-hw04-f09-sol - EECS 20N: Structure and Interpretation...

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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences UNIVERSITY OF CALIFORNIA BERKELEY Problem Set 4 SOLUTIONS HW 4.1 Solution (a) x ( n ) = sin( 2 π 5 n ) + cos( 4 π 5 n ) = 1 2 i ( e i 2 π 5 n − e − i 2 π 5 n ) + 1 2 ( e i 4 π 5 n + e − i 4 π 5 n ) Therefore, p = 5 , ω = 2 π 5 . (b) x ( n ) = cos( 3 π 5 n ) + 1 3 cos( 4 π 5 n ) = 1 2 ( e i 3 π 5 n + e − i 3 π 5 n ) + 1 6 ( e i 4 π 5 n + e − i 4 π 5 n ) Therefore, p = 10 , ω = π 5 . (c) x ( n ) = cos( √ 2 π 5 n ) √ 2 π 5 p = 2 kπ ⇐⇒ p = 5 √ 2 k Since √ 2 is irrational, p cannot be an integer, and hence x is not periodic. (d) x ( n ) = cos( 2 π 3 n ) + ( − 1) n = cos( 2 π 3 n ) + cos( πn ) = 1 2 ( e i 2 π 3 n + e − i 2 π 3 n ) + 1 2 ( e iπn + e − iπn ) Therefore, p = 6 , ω = π 3 . (e) x ( n ) = ∞ summationdisplay l = −∞ δ ( n − lp ) A graphical representation of x ( n ) is shown below: (Note p is the period and ω = 2 π p is the fundamental frequency) 1 The Discrete Fourier Series (DFS) equations are : x ( n ) = ∑ k = ( p ) X k e ikw n Synthesis Equation X k = 1 p ∑ n = ( p ) x ( n ) e − ikw n Analysis Equation Note: ( p ) denotes an appropriately-chosen interval of p adjacent sam- ples, e.g., { , 1 , . . ., p − 1 } or { 1 , 2 , . . ., p } ,etc. Let ( p ) = { , 1 , . . ., p − 1 } . Note that x ( n ) = δ ( n ) in the interval ( p ) that we have chosen. Therefore, only one term in the right-hand side of the analysis equation is nonzero (the term corresponding to n = 0 ): X k = 1 p p − 1 summationdisplay n =0 x ( n ) e − ikw n = 1 p x (0) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =1 + x (1) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =0 e − ikw + . . . + x ( p − 1) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =0 e − ik ( p − 1) w bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =0 ⇒ X k = 1 p , ∀ k ∈ Z ⇒ x ( n ) = 1 p p − 1 summationdisplay k =0 e ikw n This shows that all frequency components { , w , αw , . . ., ( p − 1) w } are equally present in the impulse train signal x ( n ) ....
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## This note was uploaded on 12/24/2011 for the course ELECTRICAL 20 taught by Professor Babak during the Spring '11 term at University of California, Berkeley.

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ee20-hw04-f09-sol - EECS 20N: Structure and Interpretation...

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