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chapter 15

# chapter 15 - Chapter 15 Fourier Series Exercises Ex 15.3-1...

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Chapter 15: – Fourier Series Exercises Ex. 15.3-1 Notice that ( ) 1 2 1 2 ( ) ( ) ( ) ( ) ( ) f t T f t T f t T f t f t f t = + = + = Therefore, f ( t ) is a periodic function having the same period, T . Next ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) 1 1 2 2 1 10 1 0 1 0 1 2 20 2 0 2 0 1 1 10 2 20 1 1 2 2 0 1 1 2 2 0 1 cos sin cos sin cos sin n n n n n n n n n n n f t k f t k f t k a a n t b n t k a a n t b n t k a k a k a k a n t k b k b n t ω ω ω ω ω ω = = = = + = + + + + + = + + + + + ) Ex. 15.3-1 f(t) = K is a Fourier Series. The coefficients are a 0 = K; a n = b n = 0 for n 1. Ex. 15.3-2 f(t) = Acos w 0 t is a Fourier Series. a 1 = A and all other coefficients are zero. 15-1

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Ex. 15.4-1 4 8 2 T π π = = , 0 2 rad 4 s T π ω = = Set origin at t = 0, so have an odd function; then a n = 0 for n = 0,1, . . . Also, f(t) has half wave symmetry, so b n = 0 for n = even. For odd n , we have ( ) ( ) ( ) ( ) ( ) 0 2 2 0 0 0 0 2 2 2 0 0 0 0 2 2 2 ( ) sin sin sin 4 sin 4 4 (1 cos ) 1, 3, 5, . . . 2 T T n T T T b f t n t dt n t dt n t dt T T T n t dt T n T n nf T n ω ω ω ω ω π π = = − + = = = = Finally, 0 0 4 1 ( ) sin ; odd and 4 rad s N n f t n t n n ω ω π = = Ex. 15.4-2 0 0 4 0 0 2 , 2 0 , 0 for all odd function with quarter wave symmety 0 = even 2 0 6 8 ( ) sin where ( ) = 6 2 6 4 Thus n n n n T T a a b n t t b f t n t dt f t t b π n π π ω π π ω π π π = = = = = = < < = < = 2 2 2 2 1 24 1 sin 3 24 1 so ( ) sin sin (2 ) 3 N n odd n n n n f t nt n π π π π = = 15-2
Ex. 15.4-3 a) is neither even nor odd. f ( t ) will contain both sine and cosine terms b) 1 wave symmetry no even harmonics 4 c) average value of f ( t ) = 0 a 0 = 0 Ex. 15.5-1 0 2 2 s, rad/s T T π ω π = = = 2 1 2 0 0 1 2 1 1 1 ( ) 2 2 2 1 1 1 (1 2 2 0 jn t jn t jn t n jn j n jn jn n C f t e dt e dt e dt e e e e jn jn n odd jn C n even π π π ) π π π π π π = = = + + = = π Finally, 3 5 3 5 2 1 1 1 1 ( ) ... ... 3 5 3 5 j t j t j t j t j t j t f t e e e e e e j π π π π π π π = + + + Ex. 15.5-3 4 2 / 4 0 4 4 1 1 1 2 2 T T j nt T j n j n j nt n T T T C e dt e e e T T j n j n π π ω π π 2 / 2 π = = = ( 1) ( 1) 2 odd 0 even , 0 1 2 0 n n n n C n n π = = n 15-3

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Ex. 15.6-1 Use the “stem plot” in Matlab to plot the required Fourier spectra: % Fourier Spectrum of a Pulse Train A = 8; % pulse amplitude T = 4; % period d = T/8; % pulse width pi = 3.14159; w0 = 2*pi/T; %fundamental frequency N = 49; n = linspace(-N,N,2*N+1); x = n*w0*d/2; % Eqn.15.6-3. Division by zero when n=0 causes Cn(N+1) to be NaN. Cn = (A*d/T)*sin(x)./x; Cn(N+1)=A*d/T; % Fix Cn(N+1); sin(0)/0 = 1 % Plot the spectrum using a stem plot stem(n,Cn,'filled'); xlabel('n'); ylabel('|Cn|'); title('Fourier Spectrum of Pules Train with d = T/8'); 15-4
Ex. 15.8-1 0 = 4 rad/s ω From Example 15.4-1: 0 2 1 odd 1 1 1 ( ) 3.24 sin sin 3.24 sin 4 sin12 sin 20 sin 28 2 9 25 49 N s n n n v t n t t t t t n π ω = = = + " 1 The network function of the circuit is ( ) ( ) ( ) o s 1 1 1 1 1 1 j C 4 j C R j R j

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chapter 15 - Chapter 15 Fourier Series Exercises Ex 15.3-1...

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