20102ee113_1_hw6_sol_spring10

20102ee113_1_hw6_sol_spring10 - EE113: Digital Signal...

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Unformatted text preview: EE113: Digital Signal Processing Prof. Mihaela van der Schaar Prepared by Nick Mastronarde, Yu Zhang, and Shaolei Ren Spring 2010 Homework #6 Solutions Problems from Prof. Sayed’s lecture notes: 13.7 Let 1 ( ) be the output of the lowpass filter, 2 ( ) the output of the first modulator, and ), 2 ( ), and 3 ( ) the output of the second modulator. The frequency responses 1( ) are shown below: 3( ( 4 ) −2 −4 1( 4 2 ) 4 2 −2 −8 8 2 2( ) 2 1 −2 −4 3( 4 2 ) 4 2 − − 34 −2 3 4 2 1 The energy of 1 2 3( ) is 3( ∣ − )∣2 =4 ( 1 2 /8 0 2 /8 2 = 2 16 2 1 3 3 8 1 =. 3 13.8 We see from the figure that ( Notice that the magnitude of odd function of . Hence, ( )= = 1 2 1 2 + ( ) can be written as ⎧ ( −2 ) ⎨ 4 ≤∣ ∣< 2 3 ( −2 ) )= 2 2 ≤∣ ∣≤ 4 ⎩ 0 otherwise. ) is an even function of and the phase of ( ) is an ( − −2 − 34 ) ( −2 ) 2 2 + 1 2 + 1 2 −4 −2 3 4 ( −2 ) 1 2 ( −2 ) 2 2 ( −2 ) . 4 These integral are of the same form, ( −2 ) = =− = ( +( −2) ) =− ( −2) 1 ( − 2) ( −2) ( −2) = ( −2) − ( − 2) . Using this result we get 1 ( )= ( − 2) + = 2( ( −2) 4 − 2 ( −2) 2 + ) + 2( ( −2) 2 − 2 ( −2) 34 ) − ( −2) 34 − 2 − ( −2) 2 − ( −2) 2 − 2 − ( −2) 4 1 3 sin ( − 2) − 2 sin ( − 2) ( − 2) 4 4 −2 3 3( − 2) 1 1 − sinc + sinc = sinc 4 4 2 4 2 ( + sin ( − 2) −2 2 2 where sinc( ) = sin( ) . Since is simply as shown below. ) is periodic, the DTFT of ( ) over the interval [0, 2 ] 2 ∣( 2 1 )∣ 2 3 2 2 ∠( 2 ) 2 3 2 2 −2 ∙ Let sinc( ) = sin( ) . Thus ( ) = ric,using Parseval’s relation we get ∞ sin( /3) Problem F = 1 sinc( /3). Then, since ( ) is symmet3 1 (0)2 − (1)2 2 )∣2 − 1 (0)2 − (1)2 . 2 ( )2 = =2 1 2 ∞ ( )2 − =−∞ = 11 ⋅ 22 ∣( − The Fourier transform of ( ) is given by ( where rect( ) = Thus, 1 2 and ∞ =2 ) = rect 2 1 3 1 ∣∣≤ 1 2 0 ∣ ∣ > 1. 2 = 1 2 /3 ∣( − )∣2 1 − /3 = 1 3 11 11 1 sin( /3) ()= − (0)2 − (1)2 = − − 23 23 9 2 2 ≈ 0.0351. 3 ∙ Using the identity cos 2 = cos2 − sin2 part, we get ∞ =1 = 1 − 2 sin2 , and the result from the first 2 6 2 8 2 6 cos ∞ 3 2 = =1 ∞ 1 2 −2 ∞ sin = =1 2 1 2 −2 =1 ∞ 2 =1 sin sin = = Problem G Multiplying by − (2 + /4) 6 2 −2 − 2 6 1 1 − ≈ 0.0492. 6 36 yields ⎧ − (4 −3 /4) ⎨ ) = 2 − (4 −3 /4) ⎩ 0 ) ≤∣ ∣< 2 3 2 ≤∣ ∣≤ 4 otherwise. 4 ′ ( Thus, the magnitude response is the same (see above), but the phase response changes (see below). ∠ 3 4 2 4 ′( −4 −2 − 34 − The inverse transform of (3 /4) in ′( ). ′( 2 3 2 2 ) is not a real sequence because of the complex exponential term ∞ =−∞ ∣ Problem H Suppose ( ) is absolutely summable (i.e. < ∞ such that ∣ ( )∣ < 1 for all ∣ ∣ ≥ ∣ ∣ ≥ . Then, ∞ ( )∣ < ∞). Then there exists a positive , which implies that ∣ ( )∣2 < ∣ ( )∣ for all ℰ= =−∞ ∣ ( )∣2 = ∣ ∣< ∣ ( )∣2 + ∣ ∣≥ ∣ ( )∣2 < ∣ ∣< ∣ ∣≥ ∣ ( )∣2 + ∣ ∣≥ ∣ ( )∣. But ∣ ∣< ∣ ( )∣2 is finite because absolutely summable. Hence, ( ) has finite energy. is finite, and ∣ ( )∣ is finite because ( ) is 4 Problem I The causality of the system cannot be determined from the figure. A causal system ℎ1 ( ) that has a linear phase response ∠ ( ) = − , is: ℎ1 ( ) = ( ) + 2 ( − 1) + ( − 2) 1( )=1+2 = = − − − + −2 − ( +2+ () ) (2 + 2 ( )) ∣ ∠ 1( 1( )∣ = 2 + 2 )=− . An example of a non-causal system with the same phase response is: ℎ2 ( ) = ( + 1) + ( ) + 4 ( − 1) + ( − 2) + ( − 3) 2( )= = = − − +1+4 ( 2 − + −2 + − −3 + ( )+2 +4+ ( )+2 (2 ) + −2 ) (4 + 2 (2 )) ∣ ∠ 2( 2( )∣ = 4 + 2 )=− . Thus, both the causal sequence ℎ1 ( ) and the non-causal sequence ℎ2 ( ) have a linear phase response ∠ ( ) = − , where = 1. 5 ...
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