Bài giải mạch P18

Chapter 18, Solution 1. )2t()1t()1t()2t()t('f −δ+−δ−+δ−+δ= 2jjj2j eeee)(Fj ω−ω−ωω +−−=ωω ω−ω= cos22cos2 F(ω) = ω ω−ω j ]cos2[cos2 Chapter 18, Solution 2.    << = otherwise,0 1t0,t )t(f - δ(t-1) δ(t) f ”(t) t –δ’(t-1) 1 -δ(t-1) 0 1 f ‘(t) t f"(t) = δ(t) - δ(t - 1) - δ'(t - 1) Taking the Fourier transform gives -ω 2 F(ω) = 1 - e -jω - jωe -jω F(ω) = 2 j 1e)j1( ω −ω+ ω or F ∫ ω− =ω 1 0 tj dtet)( But ∫ +−= c)1ax( a e dxex 2 ax ax () =−ω− ω− =ω ω− 1 0 2 j )1tj( j e )(F () [ ] 1ej1 1 j 2 −ω+ ω ω− Chapter 18, Solution 3. 2t2, 2 1 )t('f,2t2,t 2 1 )t(f <<−=<<−= ∫ − − ω− ω −ω− ω− ==ω 2 2 2 2 2 tj tj )1tj( )j(2 e dtet 2 1 )(F [] )12j(e)12j(e 2 1 2j2j 2 −ω−−ω− ω −= ωω− ( ) [] 2j2j2j2j 2 eeee2j 2 1 ω−ωωω −++ω− ω −= () ω+ωω− ω 2sin2j2cos4j 2 1 2 −= F( ω ) = )2cos22(sin j 2 ωω−ω ω Chapter 18, Solution 4. 1 –2δ(t–1) 2δ (t+1) –1 g’ 2 0 –2 t –2δ’(t–1) –2δ (t+1) 1 –2δ(t–1) 2δ’ (t+1) –1 g” 4δ(t) 0 –2 t 4sin4cos4 ej2e24ej2e2)(G)j( )1t(2)1t(2)t(4)1t(2)1t(2g jjjj2 +ωω−ω−= ω−−+ω+−=ωω −δ ′ −−δ−δ++δ ′ ++δ−= ′′ ω−ω−ωω )1sin(cos 4 )(G 2 −ωω+ω ω =ω Chapter 18, Solution 5. 1 0 h’(t) –1 –2δ(t) 1 t δ(t+1) –δ(t–1) 1 0 h”(t) –1 –2δ’(t) 1 t ω−ω=ω−−=ωω δ ′ −−δ−+δ= ′′ ω−ω j2sinj2j2ee)(H)j( )t(2)1t()1t()t(h jj2 H(ω) = ω ω − ω sin j2j2 2 Chapter 18, Solution 6. dtetdte)1()(F tj 0 1 1 0 tj ω− − ω− ∫∫ +−=ω )1(cos 1 tsin t tcos 1 tsin 1 tdtcosttdtcos)(F Re 2 1 0 2 0 1 0 1 1 0 −ω ω =         ω ω +ω ω +ω ω −= ω+ω−=ω − − ∫∫ Chapter 18, Solution 7. (a) f 1 is similar to the function f(t) in Fig. 17.6. )1t(f)t(f 1 −= Since ω −ω = 2 F ω j )1(cos )( =ω=ω ω )(Fe)(F j 1 ω −ω ω− j )1(cose2 j Alternatively, ) ) 2t()1t(2)t()t(f ' 1 −δ+−δ−δ= e2e(eee21)(Fj jjj2jj 1 ωωω−ω−ω− +−=+−=ωω )2cos2(e j −ω= ω− F 1 (ω) = ω −ω ω− j )1e j (cos2 (b) f 2 is similar to f(t) in Fig. 17.14. f 2 (t) = 2f(t) F 2 (ω) = 2 )cos1(4 ω ω− Chapter 18, Solution 8. (a) 2 1 tj 2 2 1 tj 1 0 tj 2 1 tj 1 0 tj )1tj(e 2 e j 4 e j 2 dte)t24(dte2)(F −ω− ω− − ω− + ω− = −+=ω ω−ω−ω− ω−ω− ∫∫ ω−ω−ω− ω+ ω − ω − ω + ω + ω =ω 2j 2 2jj 2 e)2j1( 2 e j 4 j 2 e j 22 )(F (b) g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ] ω ω − ω ω =ω sin22sin4 )(G Chapter 18, Solution 9. (a) y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ] ω ω +ω ω =ω sin 4 2sin 2 )(Y (b) )j1( e22 )1tj( e2 dte)t2()( 2 j 2 1 0 1 0 2 tj tj ω+ ω − ω =−ω− ω− − =−=ω ω−ω− ω− ∫ Z Chapter 18, Solution 10. (a) x(t) = e 2t u(t) X(ω) = 1/(2 + j ω ) (b)     < > = − − 0t,e 0t,e e t t )t( ∫∫∫ −− ω−−ωω +==ω 1 1 0 1 1 0 tjttjttj dteedteedte)t(y)(Y 1 0 t)j1( 0 1 t)j1( )j1( e j1 e ω+− + ω− = ω+− − ω−       ω+ ω−ω + ω− ω+ω − ω+ = − j1 sinjcos j1 sinjcos e 1 2 1 2 Y(ω) = [ ] )sin(cose1 1 2 1 2 ωω−ω− ω+ − Chapter 18, Solution 11. f(t) = sin π t [u(t) - u(t - 2)] ( ) ∫∫ ω−π−πω− −=π=ω 2 0 2 0 tjtjtjtj dteee j2 1 dtetsin)(F       + ∫ π+ω−π+ω−+ 2 0 t)(jt)(j dt)ee( j2 1 =       π+ω− + π−ω− π+ω− π−ω− 2 0 t)(j 2 0 t)(j )(j e e )(j 1 j2 1 =         ω+π − + ω−π − ω−ω− 2j2j e1e1 2 1 = () ω− π+π ω−π 2j 22 e22 )(2 1 = F(ω) = ( ) 1e 2j 22 − π−ω π ω− Chapter 18, Solution 12. (a) F = dtedtee)( 0 2 0 t)j1(tjt ∫∫ ∞ ω−ω− =ω = ω− = ω− 2 0 t)j1( e j1 1 ω− − ω− j1 1e 2j2 (b) ∫∫ − ω−ω− −+=ω 0 1 1 0 tjtj dte)1(dte)(H ()( ) )cos22( j 1 1e j 1 e1 j 1 jj ω+− ω =− ω +− ω ω−ω −= = ω ω− = j 2/sin4 2 2 2/ 2/sin j       ω ω ω Chapter 18, Solution 13. (a) We know that )]a()a([]at[cos +ωδ+−ωδπ=F . Using the time shifting property, )a(e)a(e)]a()a([e)]a3/t(a[cos 3/j3/ja3/j +ωδπ+−ωδπ=+ωδ+−ωδπ=π− ππ−ωπ− F (b) sin tsinsintcoscostsin)1t( π−=ππ+ππ=+π g(t) = -u(t+1) sin (t+1) Let x(t) = u(t)sin t, then 22 1 1 1)j( 1 )(X ω− = +ω =ω Using the time shifting property, 1 e e 1 1 )(G 2 j j 2 −ω = ω− −=ω ω ω (c ) Let y(t) = 1 + Asin at, then Y )]a()a([Aj)(2)( −ωδ−+ωδπ+ωπδ=ω h(t) = y(t) cos bt Using the modulation property, )]b(Y)b(Y[ 2 1 )(H −ω++ω=ω [][] )ba()ba()ba()ba( 2 Aj )b()b()(H −−ωδ−−+ωδ++−ωδ−++ωδ π +−ωδ++ωδπ=ω (d) )14j( e j e1 )1tj( e j e dte)t1()( 2 4j4j 2 4 0 2 tj 4 0 tj tj +ω ω − ω − ω =−ω− ω− − ω− =−=ω ω−ω−ω−ω− ω− ∫ I Chapter 18, Solution 14. (a) )t3cos()0(t3sin)1(t3cossint3sincost3cos)t3cos( −=−−=π−π=π+ (f )t(ut3cose)t t − −= F(ω) = ( ) () 9j1 j1 2 +ω+ ω+− (b) [] )1t(u)1t(utcos)t('g −−−ππ= g(t) t 1 -1 -1 1 - π -1 1 g’(t) t π )1t()1t()t(g)t("g 2 −πδ++πδ−π−= ω−ω π+π−ωπ−=ωω− jj22 ee)(G)(G ( ) ωπ−=−π−=ωω−π ω−ω sinj2)ee()(G jj22 G(ω) = 22 sinj2 π−ω ωπ Alternatively, we compare this with Prob. 17.7 f(t) = g(t - 1) F(ω) = G(ω)e -jω () ( ) ωω−ω − π−ω π =ω=ω jj 22 j eee)(FG 22 sin2j π−ω ωπ− = G(ω) = 22 sinj2 ω−π ωπ (c) tcos)0(tsin)1(tcossintsincostcos)1t(cos π−=π+−π=ππ+ππ=−π Let ex = )t(he)1t(u)1t(cos)t( 2)1t(2 −=−−π −− and )t(u)tcos(e)t(y t2 π= − 22 )j2( j2 )(Y π+ω+ ω+ =ω )1t(x)t(y −= ω− ω=ω j e)(X)(Y () () 2 2 j j2 ej2 )(X π+ω+ ω+ =ω ω )(He)(X 2 ω−=ω )(Xe)(H 2 ω−=ω − = () () 2 2 2j j2 ej2 π+ω+ ω+− −ω (d) Let x )t(y)t(u)t4sin(e)t( t2 −=−−= − )t(x)t(p −= where )t(ut4sine)t(y t2 = () 2 2 4j2 j2 )(Y +ω+ ω+ =ω () 16j2 j2 )(Y)(X 2 +ω− ω− =ω−=ω =ω−=ω )(X)(p () 162j 2j 2 +−ω −ω (e) 2j2j e j 1 )(23e j 8 )(Q ω−ω−         ω +ωπδ−+ ω =ω Q(ω) = 2j2j e)(23e j 6 ω−ω ωπδ−+ ω Chapter 18, Solution 15. (a) F =−=ω ω−ω 3j3j ee)( ω 3sinj2 (b) Let g ω− =ω−δ= j e2)(G),1t(2)t( =ω)(F F       ∫ ∞− t dt)t(g )()0(F j )(G ωδπ+ ω ω = )()1(2 j e2 j ωδ−πδ+ ω = ω− = ω ω− j e j 2 (c) F [] 1 2 1 )t2( ⋅=δ =ω−⋅=ω j 2 1 1 3 1 )(F 2 j 3 1 ω −