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# Assign2_sol - EE 3210 Assignment 2 Solution Question 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 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 π ω ω 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 π π π π π π π π 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 π π π π π π jk jk k e k k e k k a

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