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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.   
 
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.   
  
[[Image:xoft_OldKiwi.doc]]
+
[[Image:xofttrans_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.
 
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.

Revision as of 12:49, 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:Xofttrans 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]. File:X2ofn OldKiwi.doc

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva