chap17

chap17 - Chapter 17 Solution 1(a This is periodic with =...

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Chapter 17, Solution 1. (a) This is periodic with ω = π which leads to T = 2 π / ω = 2 . (b) y(t) is not periodic although sin t and 4 cos 2 π t are independently periodic. (c) Since sin A cos B = 0.5[sin(A + B) + sin(A – B)], g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t which is harmonic or periodic with the fundamental frequency ω = 1 or T = 2 π / ω = 2 π . (d) h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and a constant is also periodic , h(t) is periodic. ω = 2 or T = 2 π / ω = π . (e) The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic . ω = 0.2 π or T = 2 π / ω = 10 . (f) p(t) = 10 is not periodic . (g) g(t) is not periodic . Chapter 17, Solution 2. (a) The frequency ratio is 6/5 = 1.2. The highest common factor is 1. ω = 1 = 2 π / T or T = 2 π . (b) ω = 2 or T = 2 π / ω = π . (c) f 3 ( t ) = 4 sin 2 600 π t = (4/2)(1 – cos 1200 π t ) ω = 1200 π or T = 2 π / ω = 2 π /(1200 π ) = 1/600 . (d) f 4 ( t ) = e j 10 t = cos 10 t + j sin 10 t . ω = 10 or T = 2 π / ω = 0.2 π .
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Chapter 17, Solution 3. T = 4, ω o = 2 π /T = π /2 g(t) = 5, 0 < t < 1 10, 1 < t < 2 0, 2 < t < 4 a o = (1/T) = 0.25[ + ] = T 0 dt ) t ( g 1 0 dt 5 2 1 dt 10 3.75 a n = (2/T) = (2/4)[ ω T 0 o dt ) t n cos( ) t ( g π 1 0 dt ) t 2 n cos( 5 + π 2 1 dt ) t 2 n cos( 10 ] = 0.5[ 1 0 t 2 n sin n 2 5 π π + 2 1 t 2 n sin n 2 π π 10 ] = (–1/(n π ))5 sin(n π /2) a n = (5/(n π ))(–1) (n+1)/2 , n = odd 0, n = even b n = (2/T) = (2/4)[ ω T 0 o dt ) t n sin( ) t ( g π 1 0 dt ) t 2 n sin( 5 + π 2 1 dt ) t 2 n sin( 10 ] = 0.5[ 1 0 t 2 n cos n 5 x 2 π π 2 1 t 2 n cos n 10 x 2 π π ] = (5/(n π ))[3 – 2 cos n π + cos(n π /2)] Chapter 17, Solution 4. f(t) = 10 – 5t, 0 < t < 2, T = 2, ω o = 2 π /T = π a o = (1/T) = (1/2) = T 0 dt ) t ( f 2 0 dt ) t 5 10 ( 2 0 2 )] 2 / t 5 ( t 10 [ 5 . 0 = 5 a n = (2/T) = (2/2) ω T 0 o dt ) t n cos( ) t ( f π 2 0 dt ) t n cos( ) t 5 10 ( = π 2 0 dt ) t n cos( ) 10 ( π 2 0 dt ) t n cos( ) t 5 ( = 2 0 2 2 t n cos n 5 π π + 2 0 t n sin n t 5 π π = [–5/(n 2 π 2 )](cos 2n π – 1) = 0
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b n = (2/2) π 2 0 dt ) t n sin( ) t 5 10 ( = π 2 0 dt ) t n sin( ) 10 ( π 2 0 dt ) t n sin( ) t 5 ( = 2 0 2 2 t n sin n 5 π π + 2 0 t n cos n t 5 π π = 0 + [10/(n π )](cos 2n π ) = 10/(n π ) Hence f(t) = ) t n sin( n 1 10 1 n π π + = 5 Chapter 17, Solution 5. 1 T / 2 , 2 T = π = ω π = 5 . 0 ] x 2 x 1 [ 2 1 dt ) t ( z T 1 a T 0 o = π π π = = π π π π π π = π π = π π = ω = 2 2 0 0 T 0 o n 0 nt sin n 2 nt .. sin n 1 ntdt cos 2 1 ntdt cos 1 1 dt n cos ) t ( z T 2 a π π π π π π = = π = π + π = π π = ω = 2 2 0 0 T 0 o n even n 0, odd n , n 6 nt cos n 2 nt cos n 1 ntdt sin 2 1 ntdt sin 1 1 dt n cos ) t ( z T 2 b Thus, nt sin n 6 5 . 0 ) t ( z odd n 1 n = = π + =
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Chapter 17, Solution 6. . 0 a , function odd an is this Since 3 2 6 ) 1 x 2 1 x 4 ( 2 1 dt ) t ( y 2 1 a 2 2 , 2 T n 2 0 o o = = = + = = π = π = ω = = = = = π π π + = = π π = π π π π = π π π π π = π π π π = π + π = ω = odd n 1 n even n , 0 odd n , n 4 2 1 1 0 2 1 1 0 2 0 o n ) t n sin( n 1 4 3 ) t ( y )) n cos( 1 ( n 2 )) n cos( 1 ( n 2 )) n cos( 1 ( n 4 )) n cos( ) n 2 (cos( n 2 ) 1 ) n (cos( n 4 ) t n cos( n 2 ) t n cos( n 4 dt ) t n sin( 2 dt ) t n sin( 4 dt ) t n sin( ) t ( y 2 2 b Chapter 17, Solution 7. 0
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