Thông tin tài liệu
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 = 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 j10t = cos 10t + jsin 10t. ω = 10 or T = 2 π / ω = 0.2 π . Chapter 17, Solution 3. T = 4, ω o = 2π/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 dt5 ∫ 2 1 dt10 3.75 a n = (2/T) = (2/4)[ ∫ ω T 0 o dt)tncos()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/(nπ))5 sin(nπ/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)tnsin()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 5x2 π π − – 2 1 t 2 n cos n 10x2 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 = 2π/T = π a o = (1/T) = (1/2) = ∫ T 0 dt)t(f ∫ − 2 0 dt)t510( 2 0 2 )]2/t5(t10[5.0 − = 5 a n = (2/T) = (2/2) ∫ ω T 0 o dt)tncos()t(f ∫ π− 2 0 dt)tncos()t510( = – ∫ π 2 0 dt)tncos()10( ∫ π 2 0 dt)tncos()t5( = 2 0 22 tncos n 5 π π − + 2 0 tnsin n t5 π π = [–5/(n 2 π 2 )](cos 2nπ – 1) = 0 b n = (2/2) ∫ π− 2 0 dt)tnsin()t510( = – ∫ π 2 0 dt)tnsin()10( ∫ π 2 0 dt)tnsin()t5( = 2 0 22 tnsin n 5 π π − + 2 0 tncos n t5 π π = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ) Hence f(t) = )tnsin( n 110 1n π π + ∑ ∞ = 5 Chapter 17, Solution 5. 1T/2,2T =π=ωπ= 5.0]x2x1[ 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 on 0ntsin n 2 nt sin n 1 ntdtcos2 1 ntdtcos1 1 dtncos)t(z T 2 a ∫∫∫ π π π π π π = = π = π + π −= π − π =ω= 2 2 0 0 T 0 on evenn 0, oddn, n 6 ntcos n 2 ntcos n 1 ntdtsin2 1 ntdtsin1 1 dtncos)t(z T 2 b Thus, ntsin n 6 5.0)t(z oddn 1n ∑ ∞ = = π +−= Chapter 17, Solution 6. .0a,functionoddanisthisSince 3 2 6 )1x21x4( 2 1 dt)t(y 2 1 a 2 2 ,2T 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= oddn 1n evenn,0 oddn, n 4 2 1 1 0 2 1 1 0 2 0 on )tnsin( n 14 3)t(y ))ncos(1( n 2 ))ncos(1( n 2 ))ncos(1( n 4 ))ncos()n2(cos( n 2 )1)n(cos( n 4 )tncos( n 2 )tncos( n 4 dt)tnsin(2dt)tnsin(4dt)tnsin()t(y 2 2 b Chapter 17, Solution 7. 0a, 6 T/2,12T 0 = π =π=ω= ∫∫∫ π−+π=ω= − 10 4 4 2 T 0 on ]dt6/tncos)10(dt6/tncos10[ 6 1 dtncos)t(f T 2 a [] 3/n5sin3/nsin3/n2sin2 n 10 6/tnsin n 10 6/tnsin n 10 10 4 4 2 π−π+π π =π π −π π = − ∫∫∫ π−+π=ω= − 10 4 4 2 T 0 on ]dt6/tnsin)10(dt6/tnsin10[ 6 1 dtnsin)t(f T 2 b [] 3/n2sin23/ncos3/n5cos n 10 6/ntncos n 10 6/tncos n 10 10 4 4 2 π−π+π π =π π +π π −= − () ∑ ∞ = π+π= 1n nn 6/tnsinb6/tncosa)t(f where a n and b n are defined above. Chapter 17, Solution 8. π=π=ω=<<+= T/22,T1, t 1- ),t1(2)t(f o 2ttdt)1t(2 2 1 dt)t(f T 1 a 1 1 1 1 2 T 0 o =+=+== − − ∫∫ 0tnsin n 1 tnsin n t tncos n 1 2tdtncos)1t(2 2 2 dtncos)t(f T 2 a 1 1 22 1 1 T 0 on = π π +π π +π π =π+=ω= − − ∫∫ π π −= π π −π π −π π −=π+=ω= − − ∫∫ ncos n 4 tncos n 1 tncos n t tnsin n 1 2tdtnsin)1t(2 2 2 dtnsin)t(f T 2 b 1 1 22 1 1 T 0 on ∑ ∞ = π − π −= 1n n tncos n )1(4 2)t(f Chapter 17, Solution 9. f(t) is an even function, b n =0. 4//2,8 ππω === TT 183.3 10 4/sin) 4 ( 4 10 04/cos10 8 2 )( 1 2 0 2 00 === +== ∫∫ π π π π tdttdttf T a T o [] dtntntdttntdtntf T a T on ∫∫∫ −++=+== 2 0 2 0 2/ 0 4/)1(cos4/)1(cos5]04/cos4/cos10[ 8 40 cos)( 4 ππππω For n = 1, 102/sin 2 5]12/[cos5 2 0 2 0 1 = +=+= ∫ tdttdtta π π π For n>1, 2 )1( sin )1( 20 2 )1( sin )1( 20 4 )1( sin )1( 20 4 )1( sin )1( 20 2 0 − − + + + = − − + + + = n n n n n n tn n a n π π π π π π π π 0sin 10 2sin 4 20 ,3662.62/sin 20 sin 10 32 =+==+= π π π π π π π π aa Thus, 3213210 0,0,362.6,10,183.3 bbbaaaa ======= Chapter 17, Solution 10. π=π=ω= T/2,2T o π− − π− = −+== π−π− π−π− ω− ∫∫∫ 2 1 tjn 1 0 tjn T 0 1 0 2 1 tjntjn t o jn n jn e2 jn e4 2 1 dte)2(dte4 2 1 dte)t(h T 1 c [] , even n ,0 oddn , n j6 ]6ncos6[ n2 j e2e24e4 n2 j c jnn2jnj n = = π − =−π π =+−− π = π−π−π− Thus, tjn oddn n e n 6j )t(f π ∞ = −∞= ∑ π − = Chapter 17, Solution 11. 2/T/2,4T o π=π=ω= ∫∫ ∫ ++== − π−π− ω− T 0 0 1 1 0 2/tjn2/tjn t o jn n dte)1(dte)1t( 4 1 dte)t(y T 1 c π − π −−π− π− = π− − π− π− 1 0 2/tjn 0 1 2/tjn 22 2/tjn n e jn 2 e jn 2 )12/tjn( 4/n e 4 1 c π + π − π +−π π + π − π π−ππ jn 2 e jn 2 e jn 2 )12/jn(e n 4 jn 2 n 4 4 1 2/jn2/jn2/jn 2222 = But 2/nsinj2/nsinj2/ncose,2/nsinj2/nsinj2/ncose 2/jn2/jn π−=π−π=π=π+π= π−π [] 2/nsinn2/nsin)12/jn(j1 n 1 c 22 n ππ+π−π+ π = [] 2/tjn 22 n e2/nsinn2/nsin)12/jn(j1 n 1 )t(y π ∞ −∞= ππ+π−π+ π = ∑ Chapter 17, Solution 12. A voltage source has a periodic waveform defined over its period as v ( t ) = t (2 π - t ) V, for all 0 < t < 2 π Find the Fourier series for this voltage. v ( t ) = 2 π t – t 2 , 0 < t < 2 π , T = 2 π , ω o = 2 π / T = 1 a o = (1/T) dt)tt2( 2 1 dt)t(f 2 0 2 T 0 ∫∫ π −π π = 3 2 )3/21( 2 4 )3/tt( 2 1 23 2 0 32 π =− π π =−π π = π a n = π π + π π =−π ∫ 2 0 2 T 0 2 )ntsin( n t2 )ntcos( n 21 dt)ntcos()tt2( T 2 [] π +− π − 2 0 22 3 )ntsin(tn)ntsin(2)ntcos(nt2 n 1 232 n 4 )n2cos(n4 n 1 )11( n 2 − =ππ π −−= b n = ∫∫ − π =− dt)ntsin()tnt2( 1 dt)ntsin()tnt2( T 2 2 T 0 2 π π −+ π −− π = 2 0 22 3 0 2 ))ntcos(tn)ntcos(2)ntsin(nt2( n 1 ))ntcos(nt)nt(sin( n 1n2 0 n 4 n 4 = π + π− = Hence, f(t) = ∑ ∞ = − π 1n 2 2 )ntcos( n 4 3 2 Chapter 17, Solution 13. T = 2π, ω o = 1 a o = (1/T) dttsin10[ 2 1 dt)t(h 0 T 0 ∫∫ π π = + ]dt)tsin(20 2 ∫ π π π− [] π =π−−− π = π π π 30 )tcos(20tcos10 2 1 2 0 a n = (2/T) ∫ ω T 0 o dt)tncos()t(h = [2/(2π)] π−+ ∫∫ ππ π0 2 dt)ntcos()tsin(20dt)ntcos(tsin10 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) a n = −++−−++ π ∫∫ ππ π0 2 dt)]t]n1sin([)t]n1[sin([20dt)]t]n1sin([)t]n1[sin([10 2 1 = − − + + + + − − − + + − π π π π 2 0 n1 )t]n1cos([2 n1 )t]n1cos([2 n1 )t]n1cos([ n1 )t]n1cos([5 a n = − π− − + π+ − − + +π n1 )]n1cos([3 n1 )]n1cos([3 n1 3 n1 35 But, [1/(1+n)] + [1/(1-n)] = 1/(1–n 2 ) cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ a n = (5/π)[(6/(1–n 2 )) + (6 cos(nπ)/(1–n 2 ))] = [30/(π(1–n 2 ))](1 + cos nπ) = [–60/(π(n–1))], n = even = 0, n = odd b n = (2/T) ∫ ω T 0 o dttnsin)t(h = [2/(2π)][ + ∫ π 0 dtntsintsin10 ∫ π π − 2 dtntsin)tsin(20 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)] b n = (5/π){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ π + 0 )]n1( + [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ π π + 2 )]n1(} = + π+ + − π− − π n1 )]n1sin([ n1 )]n1sin([5 = 0 Thus, h(t) = ∑ ∞ = − π − π 1k 2 )1k4( )kt2cos(6030 Chapter 17, Solution 14. Since cos(A + B) = cos A cos B – sin A sin B. f(t) = ∑ ∞ = π + −π + + 1n 33 )nt2sin()4/nsin( 1n 10 )nt2cos()4/ncos( 1n 10 2 Chapter 17, Solution 15. (a) Dcos ωt + Esin ωt = A cos(ωt - θ) where A = 22 ED + , θ = tan -1 (E/D) A = 622 n 1 )1n( + + 16 , θ = tan -1 ((n 2 +1)/(4n 3 )) f(t) = ∑ + 10 ∞ = − + −+ + 1n 3 2 1 622 n4 1n tannt10cos n 1 )1n( 16 (b) Dcos ωt + Esin ωt = A sin(ωt + θ) where A = 22 ED + , θ = tan -1 (D/E) f(t) = ∑ + 10 ∞ = − + ++ + 1n 2 3 1 622 1n n4 tannt10sin n 1 )1n( 16 Chapter 17, Solution 16. If v 2 (t) is shifted by 1 along the vertical axis, we obtain v 2 * (t) shown below, i.e. v 2 * (t) = v 2 (t) + 1. 1 v 2 * (t) 2 t -2 -1 0 1 2 3 4 5
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