lab 5

# Lab 5 - ECE 101 Linear Systems Winter 2009 Lab 5 Solutions(send comments/questions to [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ Problem 5.4(a BDS The transform Hid ej is

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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 H ( e ) 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

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## This note was uploaded on 02/24/2010 for the course ECE ECE25 taught by Professor Bill lin during the Spring '09 term at UCSD.

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Lab 5 - ECE 101 Linear Systems Winter 2009 Lab 5 Solutions(send comments/questions to [email protected] Problem 5.4(a BDS The transform Hid ej is

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