# Bài giải phần giải mạch P17

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## Bài giải phần giải mạch P17

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Chapter 17, Solution 1. (a) This is periodic with ω = π which leads to T = 2π/ω = 2. (b) y(t) is not periodic although sin t and 4 cos 2πt 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 = 2π/ω = 2π. (d) h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic...

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1. Chapter 17, Solution 1. (a) This is periodic with ω = π which leads to T = 2π/ω = 2. (b) y(t) is not periodic although sin t and 4 cos 2πt 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 = 2π/ω = 2π. (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 = 2π/ω = π. (e) The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic. ω = 0.2π or T = 2π/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 = 2π/T or T = 2π. (b) ω = 2 or T = 2π/ω = π. (c) f3(t) = 4 sin2 600π t = (4/2)(1 – cos 1200π t) ω = 1200π or T = 2π/ω = 2π/(1200π) = 1/600. (d) f4(t) = ej10t = cos 10t + jsin 10t. ω = 10 or T = 2π/ω = 0.2π.
2. Chapter 17, Solution 3. T = 4, ωo = 2π/T = π/2 g(t) = 5, 0
3. 2 bn = (2/2) ∫ (10 − 5t ) sin(nπt )dt 0 2 2 = ∫ (10) sin(nπt )dt – ∫ (5t ) sin(nπt )dt 0 0 2 2 −5 5t = 2 2 sin nπt + cos nπt = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ) n π 0 nπ 0 10 ∞ 1 Hence f(t) = 5 + ∑ sin(nπt ) π n =1 n Chapter 17, Solution 5. T = 2π, ω = 2π / T = 1 T 1 1 a o = ∫ z( t )dt = [1xπ − 2 xπ] = −0.5 T 2π 0 T π 2π 2 1 1 1 π 2 2π an = T ∫ z(t ) cos nωo dt = π ∫ 1cos ntdt − π ∫ 2 cos ntdt = nπ sin ..nt − 0 nπ sin nt π = 0 0 0 π T π 2π  6 2 1 1 1 π 2 2π  , n = odd b n = ∫ z( t ) cos nωo dt = ∫ 1sin ntdt − ∫ 2 sin ntdt = − nπ cos nt 0 + nπ cos nt π =  nπ T π π 0, n = even 0 0 π  Thus, ∞ 6 z( t ) = − 0.5 + ∑ sin nt n =1 nπ n =odd
4. Chapter 17, Solution 6. 2π T = 2, ωo = =π 2 1 2 1 6 ao = 2 ∫0 y(t )dt = 2 (4x1 + 2x1) = 2 = 3 Since this is an odd function, a n = 0. 2 2 1 2 bn = 2 ∫0 y(t ) sin(nωo t )dt = ∫0 4 sin(nπt )dt + ∫1 2 sin(nπt )dt −4 1 2 2 −4 2 = cos(nπt ) 0 − cos(nπt ) 1 = (cos(nπ) − 1) − (cos(2nπ) − cos(nπ)) nπ nπ nπ nπ 4 2 2 0, n = even = (1 − cos(nπ)) − (1 − cos(nπ)) = (1 − cos(nπ)) = 4 nπ nπ nπ , n = odd nπ 4 ∞ 1 y( t ) = 3 + ∑ sin(nπt ) π n =1 n n = odd Chapter 17, Solution 7. π T = 12, ω = 2π / T = , a0 = 0 6 T 4 10 2 1 a n = ∫ f ( t ) cos nωo dt = [ ∫ 10 cos nπt / 6dt + ∫ (−10) cos nπt / 6dt ] T 6 0 −2 4 10 10 10 10 = 4 sin nπt / 6 − 2 − sin nπt / 6 4 = [2 sin 2nπ / 3 + sin nπ / 3 − sin 5nπ / 3] nπ nπ nπ T 4 10 2 1 b n = ∫ f ( t ) sin nωo dt = [ ∫ 10 sin nπt / 6dt + ∫ (−10) sin nπt / 6dt ] T 6 0 −2 4
5. 10 10 10 10 =− 4 cos nπt / 6 − 2 + cos nπnt / 6 4 = [cos 5nπ / 3 + cos nπ / 3 − 2 sin 2nπ / 3] nπ nπ nπ ∞ f (t) = ∑ (a n cos nπt / 6 + b n sin nπt / 6) n =1 where an and bn are defined above. Chapter 17, Solution 8. f ( t ) = 2(1 + t ), - 1 < t < 1, T = 2, ωo = 2π / T = π T 1 1 1 1 a o = ∫ f ( t )dt = ∫ 2( t + 1)dt = t 2 + t =2 T 2 0 −1 −1 T 1 1 2 2  1 t 1  an = T ∫ f (t ) cos nωo dt = 2 ∫ 2(t + 1) cos nπtdt = 2 n 2 π 2 cos nπt + nπ sin nπt + nπ sin nπt  = 0     −1 0 −1 T 1 1 2 2  1 t 1  4 bn = T ∫ f (t ) sin nωodt = 2 ∫ 2(t + 1) sin nπtdt = 2 − n 2 π2 sin nπt − nπ cos nπt − nπ cos nπt  = − nπ cos nπ     −1 0 −1 4 ∞ (−1) n f (t) = 2 − ∑ π n =1 n cos nπt Chapter 17, Solution 9. f(t) is an even function, bn=0. T = 8, ω = 2π / T = π / 4 2  10 4 T 2 1 10 ∫ f (t )dt =  ∫ 10 cos πt / 4dt + 0 = ( ) sin πt / 4 2 ao = = = 3.183  4 π π 0 T 0 8 0
6. T /2 2 2 4 40 ∫ [ 10 cos πt / 4 cos nπt / 4dt +0] = 5∫ [cos πt (n + 1) / 4 + cos πt (n − 1) / 4]dt 8 ∫ an = f (t ) cos nω o dt = T 0 0 0 For n = 1, 2 2 2  a1 = 5∫ [cos πt / 2 + 1]dt = 5 sin πt / 2dt + t  = 10 0 π 0 For n>1, 2 20 π (n + 1)t 20 π (n − 1) 20 π (n + 1) 20 π (n − 1) an = sin + sin = sin + sin π (n + 1) 4 π (n − 1) 4 0 π (n + 1) 2 π (n − 1) 2 10 20 20 10 a2 = sin π + sin π / 2 = 6.3662, a3 = sin 2π + sin π = 0 π π 4π π Thus, a 0 = 3.183, a1 = 10, a 2 = 6.362, a3 = 0, b1 = 0 = b2 = b3 Chapter 17, Solution 10. T = 2, ωo = 2π / T = π T 1 1 1 1  4e − jnπt 1 2e − jnπt 2  h ( t )e − jnωo t dt =  ∫ 4e − jnπt dt + ∫ (−2)e − jnπt dt  =  2 T∫ cn =  2 − jnπ 0 − − jnπ 1  2 0  1      0 [ ]  6j j j − , n = odd cn = 4e − jπn − 4 − 2e − j2nπ + 2e − jnπ = [6 cos nπ − 6] =  nπ , 2nπ 2nπ  0, n = even  Thus, ∞  − j6  jnπt f (t ) = ∑  e n =−∞  nπ  n =odd
7. Chapter 17, Solution 11. T = 4, ω o = 2π / T = π / 2 T 1 1 0 c n = ∫ y( t )e − jnωo t dt =  ∫ ( t + 1)e − jnπt / 2 dt + ∫ (1)e − jnπt / 2 dt  1 T  −1 4 0   0 1  e − jnπt / 2 2 − jnπt / 2 0 2 − jnπt / 2 1  cn =  2 2 (− jnπt / 2 − 1) − e − −1 jnπ e 0 4 − n π / 4  jnπ   1 4 2 4 jnπ / 2 2 jnπ / 2 2 − jnπ / 2 2  =  2 2 − jnπ + 2 2 e ( jnπ / 2 − 1) + e − e + 4 n π n π jnπ jnπ jnπ   But e jnπ / 2 = cos nπ / 2 + j sin nπ / 2 = j sin nπ / 2, e − jnπ / 2 = cos nπ / 2 − j sin nπ / 2 = − j sin nπ / 2 1 cn = [1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2] n 2π2 ∞ 1 y( t ) = ∑ 2 2 [1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]e jnπt / 2 n = −∞ n π Chapter 17, Solution 12. A voltage source has a periodic waveform defined over its period as v(t) = t(2π - t) V, for all 0 < t < 2π Find the Fourier series for this voltage. v(t) = 2π t – t2, 0 < t < 2π, T = 2π, ωo = 2π/T = 1 ao = T 1 2π 1 2π 4π 3 2π 2 (1/T) ∫ f ( t )dt = ∫0 (2πt − t )dt = 2π (πt − t / 3) 2 2 3 = (1 − 2 / 3) = 0 2π 0 2π 3
8. 2π 2 T 1  2π 2πt  an = ∫ (2πt − t 2 ) cos(nt )dt =  2 cos(nt ) + sin(nt ) T 0 π n n 0 − 1 πn 3 [ 2nt cos(nt ) − 2 sin(nt ) + n 2 t 2 sin( nt ) ] 2π 0 2 1 −4 = (1 − 1) − 3 4nπ cos(2πn ) = 2 n 2 πn n 2 T 1 ∫0 (2nt − t ) sin(nt )dt = π ∫ (2nt − t ) sin(nt )dt 2 2 bn = T 2n 1 π 1 2π = (sin(nt ) − nt cos(nt )) 0 − 3 (2nt sin(nt ) + 2 cos(nt ) − n 2 t 2 cos(nt )) π n 2 πn 0 − 4 π 4π = + =0 n n 2π 2 ∞ 4 Hence, f(t) = − ∑ 2 cos(nt ) 3 n =1 n Chapter 17, Solution 13. T = 2π, ωo = 1 T 1 π 2π ao = (1/T) ∫ h( t )dt = [ ∫ 10 sin t dt + ∫ 20 sin( t − π) dt ] 0 2π 0 π = 1 2π [ π 2π − 10 cos t 0 − 20 cos( t − π) π = 30 π ] T an = (2/T) ∫ h( t ) cos(nω t )dt 0 o = [2/(2π)]  ∫ 10 sin t cos( nt )dt + 20 sin( t − π) cos( nt )dt  π 2π 0  ∫π   Since sin A cos B = 0.5[sin(A + B) + sin(A – B)] sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t] sin(t – π) = sin t cos π – cost sin π = –sin t sin(t – π)cos(nt) = –sin(t)cos(nt)
9. 1  π 10∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt − 20∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt  2π an = 2π  0  π   5  cos([1 + n ]t ) cos([1 − n ]t )  π  2 cos([1 + n ]t ) 2 cos([1 − n ]t )  2 π  =  − −  + +   π   1+ n 1− n 0  1+ n 1− n π   5  3 3 3 cos([1 + n ]π) 3 cos([1 − n ]π)  an = π 1 + n + 1 − n −  1+ n − 1− n   But, [1/(1+n)] + [1/(1-n)] = 1/(1–n2) cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ an = (5/π)[(6/(1–n2)) + (6 cos(nπ)/(1–n2))] = [30/(π(1–n2))](1 + cos nπ) = [–60/(π(n–1))], n = even = 0, n = odd T bn = (2/T) ∫ h ( t ) sin nωo t dt 0 π 2π = [2/(2π)][ ∫ 10 sin t sin nt dt + ∫ 20( − sin t ) sin nt dt 0 π But, sin A sin B = 0.5[cos(A–B) – cos(A+B)] sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)] π bn = (5/π){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ (1 + n )] 0 2π + [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ (1 + n )] π } 5  sin([1 − n ]π) sin([1 + n ]π)  = π −  1− n + 1+ n  = 0  30 60 ∞ cos( 2kt ) Thus, h(t) = π − ∑ π k = 1 ( 4k 2 − 1)
10. Chapter 17, Solution 14. Since cos(A + B) = cos A cos B – sin A sin B. ∞  10 10  f(t) = 2 + ∑  3 cos(nπ / 4) cos( 2nt ) − 3 sin(nπ / 4) sin( 2nt )  n =1  n + 1 n +1  Chapter 17, Solution 15. (a) Dcos ωt + Esin ωt = A cos(ωt - θ) where A = D 2 + E 2 , θ = tan-1(E/D) 16 1 A = + 6 , θ = tan-1((n2+1)/(4n3)) ( n + 1) 2 2 n ∞ 16 1  −1 n + 1  2 f(t) = 10 + ∑ + ( n 2 + 1) 2 n 6 cos 10nt − tan   4n 3  n =1   (b) Dcos ωt + Esin ωt = A sin(ωt + θ) where A = D 2 + E 2 , θ = tan-1(D/E) ∞ 16 1  4n 3  f(t) = 10 + ∑ + ( n 2 + 1) 2 n 6 sin 10nt + tan  −1  n2 + 1  n =1   Chapter 17, Solution 16. If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e. v2*(t) = v2(t) + 1. v2*(t) 2 1 t -2 -1 0 1 2 3 4 5
11. Comparing v2*(t) with v1(t) shows that v2*(t) = 2v1((t + to)/2) where (t + to)/2 = 0 at t = -1 or to = 1 Hence v2*(t) = 2v1((t + 1)/2) But v2*(t) = v2(t) + 1 v2(t) + 1 = 2v1((t+1)/2) v2(t) = -1 + 2v1((t+1)/2) 8   t + 1 1  t + 1 1  t + 1  = -1 + 1 − cos π 2  + 9 cos 3π 2  + 25 cos 5π 2  +  π2         8   πt π  1  3πt 3π  1  5πt 5π   v2(t) = − cos 2 + 2  + 9 cos 2 + 2  + 25 cos 2 + 2  +  π2         8   π t  1  3 πt  1  5 πt   v2(t) = − sin 2  + 9 sin 2  + 25 sin 2  +  π2         Chapter 17, Solution 17. We replace t by –t in each case and see if the function remains unchanged. (a) 1 – t, neither odd nor even. (b) t2 – 1, even (c) cos nπ(-t) sin nπ(-t) = - cos nπt sin nπt, odd (d) sin2 n(-t) = (-sin πt)2 = sin2 πt, even (e) e t, neither odd nor even.
12. Chapter 17, Solution 18. (a) T = 2 leads to ωo = 2π/T = π f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric. (b) T = 3 leads to ωo = 2π/3 f2(t) = f2(-t), showing that f2(t) is even. (c) T = 4 leads to ωo = π/2 f3(t) is even and half-wave symmetric. Chapter 17, Solution 19. This is a half-wave even symmetric function. ao = 0 = bn, ωo = 2π/T π/2 4 T/2  4t  an = T ∫0 1 − T  cos(nωo t )dt   = [4/(nπ)2](1 − cos nπ) = 8/(n2π2), n = odd = 0, n = even 8 ∞ 1  nπt  f (t) = π2 ∑ n = odd n 2 cos  2   Chapter 17, Solution 20. This is an even function. bn = 0, T = 6, ω = 2π/6 = π/3 2 2 2 ( 4 t − 4)dt ∫ 4 dt  T/2 3 ao = T ∫ 0 f ( t )dt = 6  ∫1  2  
13. 1 2 ( 2 t − 4 t ) + 4(3 − 2) = 2 2 = 3  1   4 T/4 an = T ∫ 0 f ( t ) cos( nπt / 3)dt 2 3 = (4/6)[ ∫ ( 4 t − 4) cos( nπt / 3)dt + ∫ 4 cos( nπt / 3)dt ] 1 2 2 3 16  9  nπt  3t  nπt  3  nπt  16  3  nπt  =  n 2 π 2 cos 3  + nπ sin 3  − nπ sin 3  + 6  nπ sin 3  6       1    2 = [24/(n2π2)][cos(2nπ/3) − cos(nπ/3)] 24 ∞ 1   2πn   πn    nπt  Thus f(t) = 2 + ∑ π 2 n =1 n2 cos 3  − cos 3   cos 3         At t = 2, f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3) + (1/4)(cos(4π/3) − cos(2π/3))cos(4π/3) + (1/9)(cos(2π) − cos(π))cos(2π) + -----] = 2 + 2.432(0.5 + 0 + 0.2222 + -----) f(2) = 3.756 Chapter 17, Solution 21. This is an even function. bn = 0, T = 4, ωo = 2π/T = π/2. f(t) = 2 − 2t, 0
14. = [8/(π2n2)][1 − cos(nπ/2)] 1 ∞ 8   nπ    nπt  f(t) = 2 + ∑n π 2 2 1 − cos 2   cos 2      n=1  Chapter 17, Solution 22. Calculate the Fourier coefficients for the function in Fig. 16.54. f(t) 4 t -5 -4 -3 -2 -1 0 1 2 3 4 5 Figure 17.61 For Prob. 17.22 This is an even function, therefore bn = 0. In addition, T=4 and ωo = π/2. 2 T2 2 1 2 1 ao = T ∫ 0 f ( t )dt = 4 ∫0 4tdt = t 0 = 1 4 T2 4 1 an = T ∫ 0 f ( t ) cos(ωo nt )dt = 4 ∫0 4 t cos( nπt / 2)dt 1  4 2t  = 4  2 2 cos( nπt / 2) + sin( nπt / 2) n π nπ 0 16 8 an = (cos( nπ / 2) − 1) + sin( nπ / 2) n π 2 2 nπ Chapter 17, Solution 23. f(t) is an odd function. f(t) = t, −1< t < 1 ao = 0 = an, T = 2, ωo = 2π/T = π
15. 4 T/2 4 1 bn = T ∫ 0 f ( t ) sin( nωo t )dt = 2 ∫0 t sin( nπt )dt 2 = [sin(nπt ) − nπt cos(nπt )] 10 n π 2 2 = −[2/(nπ)]cos(nπ) = 2(−1)n+1/(nπ) 2 ∞ ( −1) n + 1 f(t) = π ∑ n =1 n sin( nπt ) Chapter 17, Solution 24. (a) This is an odd function. ao = 0 = an, T = 2π, ωo = 2π/T = 1 4 T/2 bn = T ∫0 f ( t ) sin(ωo nt )dt f(t) = 1 + t/π, 0
16. ∞ 2 (c) f(t) = ∑ nπ [1 − 2 cos(nπ)] sin(nt ) π n =1 ∞ 2 f(π/2) = ∑ nπ [1 − 2 cos(nπ)] sin(nπ / 2) π n =1 For n = 1, f1 = (2/π)(1 + 2) = 6/π For n = 2, f2 = 0 For n = 3, f3 = [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π) For n = 4, f4 = 0 For n = 5, f5 = 6/(5π), ---- Thus, f(π/2) = 6/π − 6/(3π) + 6/(5π) − 6/(7π) --------- = (6/π)[1 − 1/3 + 1/5 − 1/7 + --------] f(π/2) ≅ 1.3824 which is within 8% of the exact value of 1.5. (d) From part (c) f(π/2) = 1.5 = (6/π)[1 − 1/3 + 1/5 − 1/7 + - - -] (3/2)(π/6) = [1 − 1/3 + 1/5 − 1/7 + - - -] or π/4 = 1 − 1/3 + 1/5 − 1/7 + - - - Chapter 17, Solution 25. This is an odd function since f(−t) = −f(t). ao = 0 = an, T = 3, ωo = 2π/3. 4 T/2 4 1 bn = T ∫ 0 f ( t ) sin( nωo t )dt = 3 ∫0 t sin(2πnt / 3)dt
17. 1 4 9  2πnt  3t  2πnt  =  4π 2 n 2 sin 3  − 2nπ cos 3  3     0 4 9  2 πn  3  2 πn   =  4π 2 n 2 sin 3  − 2nπ cos 3  3     ∞  3  2 πn  2  2π n    2π t  f(t) = ∑ π n 2 2 sin −  3  nπ cos   sin  3   3   n =1  Chapter 17, Solution 26. T = 4, ωo = 2π/T = π/2 1 T 1 1 f ( t )dt =  ∫ 1 dt + 2 dt + ∫ 1 dt  = 1 3 4 T ∫0 ∫ ao = 40  1 3   2 T T ∫0 an = f ( t ) cos( nωo t )dt 2 2 2 cos( nπt / 2)dt + ∫ 1 cos( nπt / 2)dt  3 4 4  ∫1 ∫ an = 1 cos( nπt / 2)dt +  2 3   2 nπt 2 4 nπt 3 2 nπt  4 = 2  sin + sin + sin   nπ  2 1 nπ 2 2 nπ 2 3 4  3nπ nπ  = nπ sin 2 − sin 2    2 T T ∫0 bn = f ( t ) sin( nωo t )dt 2 2 nπt 3 nπt 4 nπt  =  ∫1 1 sin 2 dt + 4 ∫ 2 2 sin 2 dt + ∫ 3 1 sin 2 dt    2 nπt 2 4 nπt 3 2 nπt  4 = 2− cos − cos − cos   nπ  2 1 nπ 2 2 nπ 2 3
18. 4 = [cos(nπ) − 1] nπ Hence f(t) = ∞ 4 1+ ∑ nπ [(sin( 3nπ / 2) − sin(nπ / 2)) cos( nπt / 2) + (cos( nπ) − 1) sin(nπt / 2)] n =1 Chapter 17, Solution 27. (a) odd symmetry. (b) ao = 0 = an, T = 4, ωo = 2π/T = π/2 f(t) = t, 0 < t < 1 = 0, 1
19. Chapter 17, Solution 28. This is half-wave symmetric since f(t − T/2) = −f(t). ao = 0, T = 2, ωo = 2π/2 = π 4 T/2 4 1 an = T ∫ 0 f ( t ) cos( nωo t )dt = 2 ∫0 ( 2 − 2 t ) cos( nπt )dt 1 1 1 t  = 4  sin( nπt ) − 2 2 cos( nπt ) − sin( nπt )  nπ n π nπ 0 = [4/(n2π2)][1 − cos(nπ)] = 8/(n2π2), n = odd 0, n = even 1 bn = 4 ∫ (1 − t ) sin( nπt )dt 0 1  1 1 t  = 4 − cos( nπt ) − 2 2 sin( nπt ) + cos( nπt )  nπ n π nπ 0 = 4/(nπ), n = odd ∞  8 4  f(t) = ∑n  k =1 π 2 2 cos( nπt ) + nπ sin(nπt )  , n = 2k − 1  Chapter 17, Solution 29. This function is half-wave symmetric. T = 2π, ωo = 2π/T = 1, f(t) = −t, 0 < t < π 2 2 [cos(nt ) + nt sin(nt )] 0 = 4/(n2π) π π For odd n, an = T ∫ 0 ( − t ) cos( nt )dt = − n π 2 2 2 [sin(nt ) − nt cos(nt )] 0 = −2/n π π bn = π ∫ 0 ( − t ) sin( nt )dt = − n π 2
20. Thus, ∞  2 1  f(t) = 2∑  2 cos( nt ) − sin(nt ) , n = 2k − 1 k =1  n π n  Chapter 17, Solution 30. T/2 1 1  T/2 T/2  cn = ∫ f ( t )e − jnωo t dt =  ∫−T / 2 f ( t ) cos nω o tdt − j∫−T / 2 f ( t ) sin nω o tdt  (1) T T  −T / 2 (a) The second term on the right hand side vanishes if f(t) is even. Hence T/2 2 cn = T ∫ f (t ) cos nωo tdt 0 (b) The first term on the right hand side of (1) vanishes if f(t) is odd. Hence, T/2 j2 cn = − T ∫ f (t ) sin nωo tdt 0 Chapter 17, Solution 31. 2π 2π If h ( t ) = f (αt ), T' = T / α  → ωo ' = = = αωo T' T / α T' T' 2 2 an '= T' ∫ h (t ) cos nωo ' tdt = T' ∫ f (αt ) cos nωo ' tdt 0 0 Let αt = λ, , d t = dλ / α , αT ' = T T 2α T ∫ an '= f (λ) cos nωo λdλ / α = a n 0 Similarly, bn ' = bn