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lab 5

# lab 5 - ECE 101 Linear Systems Winter 2009 Lab 5...

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ECE 101 - Linear Systems, Winter 2009 Lab # 5 Solutions (send comments/questions to [email protected]) Problem 5.4(a), BDS The transform H id ( e ) is periodic with period 2 π . Thus, since the desired low-pass filter will satisfy H id = 1 for | ω | < π/ 2 over the range - π < ω < π , it follows that over the range 0 < ω < 2 π , it will be non-zero for 0 < ω < π/ 2 and 3 π/ 2 < ω < 2 π . 1 0 0 π π 2 Η ( ) e ω j id 1

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Problem 5.4(b), BDS From the 9 samples taken at spacings of (2 / 9) π in frequency, an interpola- tion of fl fl H ( e ) fl fl can be constructed and plotted. As can be seen, it contains the general features of the ideal filter. 0.5 1 1.5 2 0.2 0.4 0.6 0.8 1 1.2 1.4 Η( ) e ω j ω/π 2
Problem 5.4(c), BDS We are constrained to design an FIR filter that is both causal and of finite duration in n . In order to obtain a unique solution based on the samples specified in part ( b ), we set the number of non-zero n values to N = 9 (i.e., set N to the number of frequency samples). It follows that H ( e ) = N - 1 X n =0 h [ n ] e - jωn

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