(New page: a) This problem is transformation of the independent variable. The transformation consists of a shift and time scaling. The resulting signal is shifted to the left by 5 and time scaled so...)
 
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[[Image:xoft_OldKiwi.doc]]
 
[[Image:xoft_OldKiwi.doc]]
  
b) This problem is finding the even and odd parts of a signal x[n].  x1[n] = (x[n] + x[-n])/2 is the even signal.   
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b) This problem is finding the even and odd parts of a signal x[n].  x1[n] = (x[n] + x[-n])/2 is the even signal.  It can be found by plotting x[n]/2 and x[-n]/2 then summing the two signals.  This is shown below.  Note that x1[n] is symmetric about the verticle axis.
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[[Image:x1ofn_OldKiwi.doc]]
  
 
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x2[n]= (x[n]-x[-n])/2 is the odd signal.  It can be found by plotting x[n]/2 and -x[n]/2 and summing the two signals.  This is shown below.  Note that x2[n] satisfies the condition that x[n]=-x[-n].
  x2[n] = (x[n]-x[-n])/2 is the odd signal.
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Revision as of 12:31, 30 June 2008

a) This problem is transformation of the independent variable. The transformation consists of a shift and time scaling. The resulting signal is shifted to the left by 5 and time scaled so the new times are divided by 2.

File:Xoft OldKiwi.doc

b) This problem is finding the even and odd parts of a signal x[n]. x1[n] = (x[n] + x[-n])/2 is the even signal. It can be found by plotting x[n]/2 and x[-n]/2 then summing the two signals. This is shown below. Note that x1[n] is symmetric about the verticle axis. File:X1ofn OldKiwi.doc

x2[n]= (x[n]-x[-n])/2 is the odd signal. It can be found by plotting x[n]/2 and -x[n]/2 and summing the two signals. This is shown below. Note that x2[n] satisfies the condition that x[n]=-x[-n].

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood