hw05-f06-soln

hw05-f06-soln - EECS 20N Structure and Interpretation of...

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EECS 20N: Structure and Interpretation of Signals and Systems Department of EECS U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 5 Issued: 10 November 2006 Due: 17 November 2006, 5pm HW5.1 Parseval’s Identity for Periodic, Discrete-Time Signals: We can use con- cepts from inner-product spaces to derive Parseval’s identity for periodic discrete- time signals. Consider a DT signal x : Z R that is periodic with period p (and frequency ω 0 = 2 π p ). Parseval’s identity is given by: 1 p s n = a p A | x ( n ) | 2 = s k = a p A | X k | 2 . The left-hand side of Parseval’s identity, 1 p s n = a p A | x ( n ) | 2 , denotes the average power in one period of x . Each term | X k | 2 in the right-hand side of the identity represents the average power of the k th harmonic component of x , x k ( n ) = X k e ikω 0 n , where X k denotes the k th coef±cient in the complex exponential Fourier series ex- pansion of x : x ( n ) = s a p A X k e ikω 0 n = s a p A x k ( n ) . (a) The total energy of a discrete-time signal r (not necessarily periodic) over an interval a p A is E a p A ( r ) = s n = a p A | r ( n ) | 2 . Note that a p A can be, for example, [0 ,p 1] Z , or any other suitably-chosen length- p interval. The average power of r over the same time interval is P a p A ( r ) = 1 p s n = a p A | r ( n ) | 2 . It should now be clear why the left-hand side of Parseval’s identity denotes the average power of the periodic signal x over one period. Show that the average power of x k ( n ) , the k th harmonic component of x , is indeed | X k | 2 , as claimed in the problem statement. 1
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Solution x k ( n ) = X k e ikω 0 n average power of x k ( n ) = 1 p p 1 s n =0 | x k ( n ) | 2 = 1 p p 1 s n =0 | X k e ikω 0 n | 2 = 1 p p 1 s n =0 X k e ikω 0 n ( X k e ikω 0 n ) = 1 p p 1 s n =0 X k X k 1 = 1 p p | X k | 2 = | X k | 2 (b) The inner product of two discrete-time signals v,w : Z C over the interval a p A is deFned as follows: a v,w A : [ Z C ] 2 C v,w [ Z C ] , a v,w A = s n = a p A v ( n ) w ( n ) , where w ( n ) denotes the complex conjugate of w . Two signals v and w are orthogonal , written v w , if a v,w A = 0 . Prove the following orthogonality relationship among the harmonic complex exponentials φ k and φ l in the ±ourier series expansion of the p -periodic signal x : a φ k l A = ( k l ) , where δ ( · ) denotes the Kronecker delta function and φ k ( n ) = e ikω 0 n and φ l ( n ) = e ilω 0 n , k,l ∈ a p A . Note that the complex exponential ±ourier series expansion of the signal x can be written as x ( n ) = k = a p A X k φ k ( n ) . We have now shown that the func- tions φ k are mutually orthogonal, i.e., φ k φ l if k n = l .
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This note was uploaded on 04/02/2008 for the course EE 20N taught by Professor Ayazifar during the Fall '08 term at Berkeley.

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hw05-f06-soln - EECS 20N Structure and Interpretation of...

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