734604588_7_Ch1-作业解答-2010

734604588_7_Ch1-作业解答-2010

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Unformatted text preview: C h1- ¹8 ë ž * ª y (t ) = sin π t + sin t¹ ëª D ¹ ëª D N≅ B½ªÐ™´F D Ÿ 빪 sin π t y (t )Ÿ 빪 y (t )žD8빪 t =0 2 sin t t 2πD≅ N B½ªÐ™´F Tt ∀t y (t + T ) = y (t ) sin π T + sin T = 0 ….(1) sin(π t + π T ) + sin(t + T ) = sin π t + sin t cos(2π t + π T )sin π T + cos(2t + T ) sin T = 0 y (t + T ) = y (t ) ♠ ≠* (1) sin π T = − sin T sin T = 0 [cos(2π t + π T ) − cos(2t + T )]sin π T = 0 T ∈Z T = kπ ( k ∈ Z ) t ¹ª D sin π T = 0 cos(2π t + π T ) = cos(2t + T ) ¹ª D cos π ≠ cos1 y (tª )ž ¹ 8ë D 2t + T = 1 1-4 1-1 8 Dë ª¹ ž f ( 3t ) f (t ) f ( − 3t − 2) f ( −F tD½ ) ™C ´N ªÐ (1). ) 1-18 ¹ ëž f (t )→ f (t − 2)→ f (3t − 2)→ f (−3t − 2) (2). ) (3). ) f (t )→ f (3t )→ f [3(t − )] →f (−3t − 2) f (t )→f (−t) → f [−(t + 2)] →f (−3t − 2) 2 3 1-5 f (t ) t f (t0 − at )¯ 0 ½ =ª*a 2ˆº t0 t ¹ ë a ¸ª* — t 1) t 3) f ( −at ) f ( at ) t t0 t0 a 2) 4) f ( at ) t f ( −at ) t0 t0 a t 4) f ( −at ) t0 t a 44 œý¼ ) 1-4) 1-5 •ž 1-4 c pâD áµl D 1-5 ªÐ™´F ·u D ½ = 0¯ªc 2ˆºD ta pâD áµl * 1-9·& F ´ ™ Ð ª (1). f (t ) = (2 − e )u (t ) −t 2 1.5 1 0.5 0 0 1 2 3 4 5 6 (2). f (t ) = (3e + 6e −t −2 t )u (t ) 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 (3). f (t ) = (5e − 5e −t −3t )u (t ) g '(t ) = −5e − t + 15e −3t = 0 1 t* = ln 3 ≈ 0.549 2 g (t ) = 5e −t − 5e −3t g (t*) = 5 5 − ≈ 1.925 3 ( 3)3 2 X: 0.55 Y: 1.924 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 (4). f (t ) = e cos(10π t )[u (t − 1) − u (t − 2)] −t 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0 0.5 1 1.5 2 2.5 3 1-14 á * â p µ 5 7 صr ∫ ∫ ∞ −∞ ∞ (e −t + t )δ (t + 2)dt e − jωt [δ (t ) − δ (t − t0 )]dt −∞ ∫ ∫ * ª * º 02 ↓ ªa ˆa = 5 7 ∞ −∞ ∞ −∞ f (t )δ (t )dt = f (0) ∞ −∞ f (t )δ (t − t0 )dt = ∫ f (t0 )δ (t − t0 )dt = f (t0 ) ∫ ∫ ∞ −∞ ∞ (e −t + t )δ (t + 2)dt = e − ( −2) − 2 = e 2 − 2 e − jωt [δ (t ) − δ (t − t0 )]dt = e − jω 0 − e − jωt0 = 1 − e − jωt0 −∞ 1-18ªë “ ¹ * ( 1-18Ê ªº ¸ F ´ ™ Ð * * â p µ á f (t ) = f e (t ) + f o (t ) … (1) f e (t )“ (ë ª ¹ f e (t ) = f e (−t ) … f o (t ) = − f o ( −t ) … 1 ~ ë* 38 ¹ª ’ 2 3 f o (t ) ° º Ê 1 f e (t ) = [ f (t ) + f (−t )] 2 1 f o (t ) = [ f (t ) − f (−t )] 2 a *ª 8 ¹ f (t ) = e − (t − 2) [u (t − 2) − u (t − 3)] f ( −t ) = et + 2 [u (−t − 2) − u (−t − 3)] a-2 a-1 1 f e (t ) = [ f (t ) + f (−t )] 2 1 − ( t − 2) e ,2 ≤ t ≤ 3 2 ={ 1 t +2 e , −3 ≤ t ≤ −2 2 1 f o (t ) = [ f (t ) − f (−t )] 2 1 − ( t − 2) e ,2 ≤ t ≤ 3 2 ={ 1 − e t + 2 , −3 ≤ t ≤ −2 2 a-3 a-4 1 1-8(a) b * ïˆ ª ¹ 1 1 f (t ) = u (t + ) − u (t − ) 2 2 f e (t ) = f (t ) f o (t ) = 0 2 f (t ) = f (−t ) fe(t) 1 0.8 0.6 0.4 0.2 0 -0.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 e2 1-18 b 1-20 á * â p µ 1) ) ) Linearityp™´F D ¾ mÐ ” ª x1 (t ) t x2 (t )•¹ øë D ª y1 (t ) t y2 (t ) t T [ x1 (t )] = y1 (t ) T [c1 x1 (t ) + c2 x2 (t )] = c1 y1 (t ) + c2 y2 (t ) “ — —@ F ´ › T [ x2 (t )] = y2 (t ) @›´FE T c7 @ ≅@jƒnRsâ F ´ › @ ø∠ F ´ š t 2) ª h¹ t t0 ¨ Causalityh D¹ ë • ª 3) ) D¹ ª •ë Dªª º ½ ˆ c D0 =¯ 2 c ¨ D¹ ø ª 1 •ë * D½ª º =c ª 02 c ¯ˆ 2¨ c1 t c2 t ”› F ´ Time-Invariabilityè F D¾ Ð m g ´ª™ t0 ¨ r (t ) = f [e(t )] D ¹t0ë ø• ª r (t − t0 ) = f [e(t − t0 )] t = t0 t < ª t0 D ¹ •ë y (t ) = T [ x(t )]ë η ª ¹ •D = h(t )u (t ) h (t ) = { ≠ h(t )u (t ) ¨ ¨ y (t1 ) = T [ x (t2 )], t1 < t2 3 = h ( n )u ( n ) h( n) = { ≠ h ( n )u ( n ) 7 r (t ) = ∫ e(τ )dτ −∞ t r1 (t ) = ∫ e1 (τ )dτ , r2 (t ) = ∫ e2 (τ )dτ −∞ −∞ t t c1r1 (t ) + c2 r2 (t ) = c1 ∫ e1 (τ )dτ + c2 ∫ e2 (τ )dτ = ∫ [c1e1 (τ ) + c2e2 (τ )]dτ −∞ −∞ −∞ t t t ∫ t −∞ e(τ − t0 )dτ = t τ −t0 =α ∫ t − t0 −∞ e(α )dα = r (t − t0 ) r (t ) = ∫ e(τ )dτ * â p µ á −∞ r (t ) = ∫ e(τ )dτ −∞ 5t 8 r1 (t ) = ∫ e1 (τ )dτ , r2 (t ) = ∫ e2 (τ )dτ −∞ −∞ 5t 5t c1r1 (t ) + c2 r2 (t ) = c1 ∫ e1 (τ )dτ + c2 ∫ e2 (τ )dτ = ∫ [c1e1 (τ ) + c2e2 (τ )]dτ −∞ −∞ −∞ 5t 5t 5t ∫ ÷ ♠ ≠ 5t −∞ e(τ − t0 )dτ = t =1 τ −t0 =α ∫ 5 t −t0 −∞ 5 t e(α )dα = r (t − 0 ) ≠ r (t − t0 ) 5 r (1) = ∫ e(τ )dτ* ªa0º= ˆ a ª ¯½∗ 2 −∞ 2-22 (1) ∫ e jωt dω = 2πδ (t ) −∞ ∞ −− = lim ∫ e jωt d ω = lim x →∞ − x x e jxt − e − jxt 2 sin xt = lim x →∞ x →∞ jt t Q lim sin xt = δ (t ) (−− 2-73 P74 x →∞ π t 2 sin xt ∴ lim = 2πδ (t ) x →∞ t −− ∫ ∞ −∞ e jωt dω = 2πδ (t ) 2 lim[ τ →0 τ ] = δ (t ) π (t + τ 2 ) 2 δ (t ) ∞ ∞ τ τ ∫−∞ lim[π (t 2 + τ 2 ) ]ϕ (t )dt = lim ∫−∞ π (t 2 + τ 2 ) • ϕ (t )dt τ →0 τ →0 ϕ0 δ (t ) ∫ ε ∞ −∞ −ε ε ∞ τ τ τ τ • ϕ (t )dt = ∫ • ϕ (t )dt + ∫ • ϕ (t )dt + ∫ • ϕ (t )dt 2 −∞ π (t 2 + τ 2 ) − ε π (t 2 + τ 2 ) ε π (t 2 + τ 2 ) π (t + τ ) 2 ( −∞, −ε ) ( ε , ∞ ) lim ∫ τ →0 −ε 2 t = 0, ∞ τ τ • ϕ (t )dt = lim ∫ • ϕ (t )dt = 0 2 −∞ π (t + τ ) τ → 0 ε π (t 2 + τ 2 ) lim ∫ τ →0 τ • ϕ (t )dt −∞ π (t + τ 2 ) ∞ 2 = lim ∫ τ ]ϕ (t )dt = ϕ (0) π (t + τ 2 ) τ − lim[ 2 ] = δ (t ) τ → 0 π (t + τ 2 ) −−−−−−−−− Dirac τ −−−− t≠0 lim[ 2 ]=0 τ → 0 π (t + τ 2 ) ∞ ∞ τ τ ∫−∞ lim[ π (t 2 + τ 2 ) ]dt = lim ∫−∞ π (t 2 + τ 2 ) dt τ →0 τ →0 ∞ 1 1 t = lim ∫ d( τ → 0 −∞ t2 π (( ) + 1) τ τ 1 t = lim(arctan ) |∞ = 1 −∞ π τ →0 τ τ lim[ 2 ] = δ (t ) τ → 0 π (t + τ 2 ) τ • ϕ (t )dt π (t + τ 2 ) ε τ ≈ ϕ (0) lim ∫ dt τ → 0 − ε π (t 2 + τ 2 ) ε ϕ (0) 1 t = lim ∫ d π τ →0 −ε 1 + ( t )2 τ τ ϕ (0) t = lim(arctan ) |ε ε = ϕ (0) − τ →0 π τ ε τ →0 − ε 2 ∫ ∞ −∞ τ → 0 lim[ 2 (3) ∑ e − jω n = 2π n =−∞ ∞ k =−∞ ∑ δ (ω − 2π k ) ∞ ∫− ∞ −∞ f (t )δ (t − t0 )dt = f (t0 ) n =−∞ ∑ ∞ e − jω n = n =−∞ ∞ ∑∫ −∞ ∞ ∞ −∞ e − jωtδ (t − n)dt ∞ = ∫ e − jωt [ ∑ δ (t − n)]dt n =−∞ −−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−− −−−−−−−1(−− n)] = ω0 QF [ ∑ δ t − n =−∞ ∞ ∞ n =−∞ −−−−−−− = F [ ∑ δ (t − n)] n =−∞ ∞ ∑e ∞ − jω n k =−∞ ∑ δ (ω − kω0 ) ∞ k =−∞ n =−∞ ∑ δ (t − n ) ∞ ω0 = 2π −−− n =−∞ ∑ ∞ e − jω n = ω0 k =−∞ ∑ δ (ω − kω0 ) = 2π ∞ k jkω1ω ∞ ∑ δ (ω − 2π k ) 2π =1 T −−−−−−−−−−−−−−− − 2π k =−∞ ∑ δ (ω − 2π k ) = ∑ c e k =−∞ ∞ ω1 = − jkω ck = ∴ 2π ∞ 1 2π ∫ π −π (2π m =−∞ ∑ δ (ω − 2π m))e k =−∞ ∞ dω = 1 k =−∞ ∑ δ (ω − 2π k ) = ∑ ∞ e jkω = k =−∞ ∑ ∞ e − jkω = n =−∞ ∑e ∞ − jω n −−−− n =−∞ ∑ e− jωn = lim ∞ N →∞ n =− N ∑ N e − jωn = lim N →∞ n =− N ∑e jω 2 N jω n 1 − e jω ( 2 N +1) = lim e − jω N g N →∞ 1 − e jω = lim e (e 1 − jω ( N + ) 2 jω − 2 −e 1 jω ( N + ) 2 jω 2 ) ω ≠ 2 kπ ω → 2k π e ge − e ) ( 1 sin( N + )ω 2 = lim N →∞ 1 sin( ω ) 2 ∞ 1 lim sin( N + )ω = 0 P81 2 − 100 ∑ e− jωn = 0 N →∞ 2 n =−∞ 1 1 sin( N + )ω sin( N + )(ω − 2kπ ) 2 = lim 2 lim N →∞ N →∞ 1 1 sin( ω ) sin( (ω − 2kπ )) 2 2 1 1 sin( N + )(ω − 2kπ ) / ( (ω − 2kπ )) 2 2 = lim N →∞ 1 1 sin( (ω − 2kπ )) / ( (ω − 2kπ )) 2 2 1 1 = lim sin( N + )(ω − 2kπ ) / ( (ω − 2kπ )) N →∞ 2 2 1 sin( N + )(ω − 2kπ ) 2 = 2π lim = 2πδ (ω − 2π k ) N →∞ (ω − 2kπ ) = 2π k =−∞ N →∞ jω 2 n =−∞ ∑e ∞ − jω n ∑ δ (ω − 2π k ) s 4 7@ ∞ 2 ≅ @@ F´› O DRF n´ ƒ› p âs 3F´›@ ...
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