Lesson+28 - Berational i 1 Lesson 28 Getreal Lesson 28...

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Lesson 28 1 i  Be rational Get real
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2 Lesson 28 Challenge 27 Exponential Fourier Series Ch. 16-8 Challenge 28 Lesson 28
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Challenge 27 What is the Fourier approximation error squared (error energy) for a 4-term series representation of x(t)? n B A A nt B nt A A t x n n n n n n n π π π 1 0 1 1 0 1 0 2 1 2 sin 2 cos ; ; / ) ( ) ( ) ( 3 0 1 1 x(t) Determine the Fourier series representation of x(t) (using tables or previous lesson) Lesson 28
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4 ; 12 1 4 1 3 1 2 1 2 1 0 2 2 0 dt t 0 1 1 x(t) n B A A n n n π 1 0 1 0 2 1 : series rc Trignometi ; ; / DC (energy A 0 2 ) Lesson 28 Power x(t) = 1/3W
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5 ; 032 . 0 2 1 4 1 3 1 2 2 1 0 1 1 x(t) n B A A n n n π 1 0 1 0 2 1 : series rc Trignometi ; ; / 1 st harmonic (energy B 1 2 /2) Lesson 28 Power x(t) = 1/3W
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6 ; 02 . 0 8 1 2 1 4 1 3 1 2 2 2 2 0 1 1 x(t) n B A A n n n π 1 0 1 0 2 1 : series rc Trignometi ; ; / 2 nd harmonic (energy B 2 2 /2) Lesson 28 Power x(t) = 1/3W
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7 014 . 0 18 1 8 1 2 1 4 1 3 1 2 2 2 2 3 0 1 1 x(t) n B A A n n n π 1 0 1 0 2 1 : series rc Trignometi ; ; / 3 rd harmonic (energy B 3 2 /2) Lesson 28 Power x(t) = 1/3W
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8 Review-RMS/Power Lesson 28 Average power Assume T d i v T P 0 ) ( ) ( 1 ) cos( ) ( ) cos( ) ( 0 1 0 0 1 0 n n n n n n t n I I t i t n V V t v ) cos( 2 1 1 0 0 n n n n n I V I V P Peak values factor power ) cos( n n ) cos( 2 2 1 0 0 n n n n n I V I V P RMS values factor power ) cos( n n
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The Fourier series (CTFS) is used to analyze periodic continuous-time signals in the frequency domain. The CTFS synthesis equation can appear in three common forms: Exponential form: Trigonometric form: Compact form:  n t jn n 0 e c x(t) 9 ) sin( B ) cos( A x(t) 0 n 1 n 0 n 0 t n t n A 1 n 0 n 0 ) cos( C x(t) n t n C CTFS Review 0 =2 f 0 =2 /T 0 Lesson 28
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10 ( Exponential series ) Coefficient production rules: (Analysis equation) (Synthesis equation):  n t jn n 0 e c x(t) 0 0 0 ) ( 1 T t jn n dt e t x T c (2-sided spectrum) Preferred by some because of its similarity to Laplace transforms Lesson 28
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11 MATALB Symbolic tool – the Slacker’s friend >> syms Ck ker t >> for k=1:3 w0=2*pi; ker=exp(-j*k*w0*t); Ck=2* int (t*ker,0,0.5)+2*int((1-t)*ker,0.5,1); simplify(Ck) end   ans = -2/pi^2 ans = 0 ans = -2/9/pi^2 Computed C i  using MATLAB  integration and not MATLAB’s 
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