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# sol2 - ECE 544 MD Wavelets in Signal Processing Solution to...

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Unformatted text preview: ECE 544 - MD: Wavelets in Signal Processing Solution to Homework #2 1. Multirate identities 1. Using the fact that filtering followed by upsampling is equivalent to upsampling followed by upsampled filtering, the system can be redrawn as follows: 2 X(z) 2 Y(z) 2 2 H ( z ) 1 2 H ( z ) We also know that upsampling followed by downsampling is identity. Therefore the transfer function of this system is: Y ( z ) X ( z ) = H 1 ( z ) H 2 ( z ) 2. Using again the interchange of filtering and upsampling, we can redraw the system as follows: 2 2-1 Z E (z) 2 2 E (z) 1 X(z) y(z) The lower branch contains an upsampler followed by a delay and a downsampler. The output of such a system is 0. Therefore only the upper branch remains and the final transfer function of the system is: Y ( z ) X ( z ) = E ( z ) . 3. Call P ( z ) = H ( z ) G ( z ), we have thus that P ( z ) + P ( − z ) = 2 . In the original domain this corresponds to p n + ( − 1) n p n = 2 δ n , 2 p + 2 p 2 + ... + 2 p 2 n = 2 δ n . Therefore, p = 1, and p 2 n = 0 for n ∈ Z . 1 We can also write P ( z ) = P even ( z 2 )+ z- 1 P odd ( z 2 ) with P even ( z 2 ) = 1. Applying the result of the previous part, we obtain that the transfer function of this system is P even ( z ) = 1. The first system is thus unity. Similarly, call Q ( z ) = H ( z ) F ( z ). It can be shown that Q ( z ) = Q even ( z 2 ) + z- 1 Q odd ( z 2 ) with Q even ( z 2 ) = 0, therefore, the second system is zero. 2. Use of Multirate Identities 1. The z-transform of the output is Y ( z ) = D 8 { A ( z ) A ( z 2 ) A ( z 4 ) X ( z ) } = 1 8 7 summationdisplay k =0 A ( W k 8 z 1 / 8 ) A ( W k 8 z 1 / 4 ) A ( W k 8 z 1 / 2 ) X ( W k 8 z 1 / 8 ) , where D N is a downsample-by-N operator. In the frequency domain this is Y ( e jω ) = 1 8 7 summationdisplay k =0 A ( W k 8 e jω/ 8 ) A ( W k 8 e jω/ 4 ) A ( W k 8 e jω/ 2 ) X ( W k 8 e jω/ 8 ) ....
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sol2 - ECE 544 MD Wavelets in Signal Processing Solution to...

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