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Lesson_5 - Lesson5 Challenge4 Lesson5: Challenge5 Lesson 5...

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Lesson 5 Lesson 5 Challenge 4 Lesson 5:  Fourier series of signals      Properties of the Fourier Series Challenge 5
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  Lesson 5 Challenge 4 What is the 1 st  harmonic Fourier coefficient of the pulse process shown  below? T 0 =2 -2.0 | -1.0 | 0.0 | 1.0 | 2.0 1.0 0.0 ( 29 ( 29 π π π π π π π n n jn e dt e dt e t x dt e t x T a t jn t jn t jn t jn T T n ) 2 / sin 2 1 1 2 1 ) ( 2 1 1 2 / 1 2 / 1 2 / 1 1 2 / 2 / 0 2 / 1 1 0 0 = - = = = = - - - - - - - -
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  Lesson 5 Challenge 4 For n=1, a 1 =sin( π /2)/ π  ~ 0.318 Notice for n even harmonics (except n=0), that sin(n π /2)=0 and the  corresponding Fourier coefficients are likewise zero.  The signal spectrum  therefore contains only odd harmonics with a DC bias of a 0 =1/2.  {a 0 =1/2, a ± 1 =0.318, a ± 2 =0, a ± 3 =-0.106, a ± 4 =0, … }  Machine calculation
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  Lesson 5 Lesson 5 Fourier Series
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  Lesson 5 Motivation …-f 0  0 f 0  … …-f 0  0 f 0  … T 0 T 0 f 0 =1/T d=-50% Machine calculation Machine calculation What does this say about  bandwidth requirements? d=12.5%
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  Lesson 5 Motivation …-f 0  0 f 0  … …-f 0  0 f 0  … 12.5% Machine calculation Machine calculation Machine calculation Noise added
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  Lesson 5 Lesson 5 The Fourier series maps a periodic signal x(t) x(t); x(t)=x(t+T 0 )  (period T 0 ) into an equivalent frequency domain representation.   Fourier Series Periodic Real/Complex Spectrum Complex x(t) |X(f)|
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  Lesson 5 Lesson 5  Complex exponential form Synthesis equation Analysis equation or ( 29 ( 29 -∞ = = k kt T j k e a t x ) / 2 ( 0 π ( 29 ( 29 ( 29 (DC) harmonic k ; 1 (DC) harmonic 0 ; 1 th / 2 0 0 th 0 0 0 0 0 0
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