10 pi x1 x g x2 1 5 4 x k0 2k

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Unformatted text preview: j 2ω )/3 1 1 − e −j 3 ω · 3 1 − e−jω ω sin( 32 ) 1 = e−jω 3 sin( ω ) 2 = |H (ω )| = ω ω 1 −jω sin( 32 ) 1 sin( 32 ) |e || |= | | 3 sin( ω ) 3 sin( ω ) 2 2 |H(omega)| 1 b) 1/3 pi/2 Figure 19: c) Figure 20: System diagram of problem 4. 10 pi X1 (ω ) = X (ω )G(ω ) X2 (ω ) = 1 5 4 X( k=0 ω − 2πk ω − 2πk )G( ) 5 5 X3 (ω ) = X2 (ω )H (ω ) = e−jω X4 (ω ) = X3 (5ω ) = ω sin( 32 ) 1 · 3 sin( ω ) 5 2 sin( 15ω ) 2 e −j 5 ω ω 3 sin( 52 ) · 1 5 4 X( k=0 ω − 2πk ω − 2πk )G( ) 5 5 4 X (ω − k=0 2πk 2πk )G(ω − ) 5 5 sin( 15ω ) 2 Y (ω ) = G(ω )X4 (ω ) = e−j 5ω ω X (ω )G(ω ) 15 sin( 52 ) (Because LPF filters out terms associate with k = 1, ..., 4) F (ω ) = sin( 15ω ) Y (ω ) 2 = e −j 5 ω ω G(ω ) X (ω ) 15 sin( 52 ) d G(omega) d) Figure 21: s e) The decimation by 5 before applying H (ω ) reduces the data size by a factor of 5 which reduces the computation of the filter. Besides, the first low-pass filter help cut off high frequency noise, which help H (ω ) work more efficiently. Problem 5 a) Since the highest...
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This note was uploaded on 01/16/2014 for the course ECE 438 taught by Professor Staff during the Fall '08 term at Purdue.

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