hw6ans_1 - to the left by m w and the neg. freq. portion to...

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Homework 6: Answers 1. (a) x ( t ) = 2 " ( t # 0.4) # 0.6 e # 0.3( t # 0.4) u ( t # 0.4) (b) x ( t ) = 3 ( t ) # ( t # 1) # ( t + 1) (c) x ( t ) = e " 2 t " e " 5 t ( ) u ( t ) (d) " 1 # sin(17 t ) 2. X ( j ) = j e # j 2 u + 6 $ ( ) # u # 6 ( ) [ ] 3. x ( t ) = 4 te " 2 t " 2 e t u ( " t ) 4. (a) (b)
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5. (a) X ( j " ) = 2 #$ ( % n 2 # ) n = (b) H ( j ) = 4 u + co ( ) # u # co ( ) [ ] (c) y ( t ) = 8cos(4 t ) + 8cos(2 t ) + 4 (d) Choose co such that 0 < co < 2 . This will filter out all frequencies except the DC term and the result will be y ( t ) = 4 . 6. (P 12.4) b)
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c) d) The output signal consists of only one copy of each half of the original spectrum. 7. (P-12.6) a) b) c) The system places the positive frequency portion on the negative frequency portion and the negative frequency portion on the positive frequency portion by translating the pos. freq. portion
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Unformatted text preview: to the left by m w and the neg. freq. portion to the right by m w . d) 8. (P-12.12) a) ! 160 &quot; s w For 160 = s w , 80 1 = s T , b) For 100 = s w , 50 1 = s T , For one period 50 50 &quot; &quot; # w , 50 20 , 20 , 20 , 20 50 , ) ( 4 3 80 80 4 3 &quot; &quot; &quot; &quot; &quot; &quot; # # &quot; &quot; # $ $ % $ $ &amp; + # + = w w w w w w jw X s s s s s s T A T A T A T A T A T A s c) 9. (P-12.14ab) a) ! 2000 2 &quot; s T b) ) 10 ( ) ( s T t x t y ! =...
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This homework help was uploaded on 10/10/2007 for the course ECE 2200 taught by Professor Johnson during the Spring '05 term at Cornell University (Engineering School).

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hw6ans_1 - to the left by m w and the neg. freq. portion to...

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