Assign2_sol - EE 3210 Assignment 2 Solution Question 1 Using the chain rule we have dx(t de 2(t 1 2 t 1 du(t 1 =e u(t 1 dt dt dt 2 t 1 2 t 1 =e(t 1

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EE 3210 Assignment 2: Solution Question 1 Using the chain rule, we have ) 1 ( 2 ) 1 ( ) 1 ( 2 ) 1 ( ) 1 ( ) 1 ( ) ( ) 1 ( 2 ) 1 ( 2 ) 1 ( 2 ) 1 ( 2 ) 1 ( 2 = = + = t u e t t u e t e dt de t u dt t du e dt t dx t t t t t δ Question 2 It is easy to see that T = 3, ω 0 = 2 π / 3, and a 0 = 1. For other values of k , we have [ 3 / 4 3 / 2 2 1 1 0 2 2 1 2 3 1 0 0 π jk jk t jk t jk k e e jk dt e dt e a = + = ] (1) It can also be written as [] + = + = 3 2 sin 3 sin 1 ) 1 ( ) 1 ( 2 1 3 / 2 3 / 3 / 4 3 / 2 k e k e k e e jk a jk jk jk jk k (2) Alternatively, one can regard the signal as the superposition of two periodic rectangular waveforms. The Fourier series of a periodic rectangular waveform has been found in the lecture notes. Using the time-shifting and linearity properties, one can show that jk jk k e k k e k k a + = 3 sin 1 3 sin 2 3 / (3)
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This note was uploaded on 11/22/2009 for the course ELECTRONIC EE3210 taught by Professor Sungchiwan during the Spring '09 term at École Normale Supérieure.

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Assign2_sol - EE 3210 Assignment 2 Solution Question 1 Using the chain rule we have dx(t de 2(t 1 2 t 1 du(t 1 =e u(t 1 dt dt dt 2 t 1 2 t 1 =e(t 1

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