734604588_7_Ch1-&auml;&frac12;œ&auml;&cedil;š&egrave;&sect;&pound;&ccedil;&shy;”-2010

# 0 0 t t dt t dt t dt

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Unformatted text preview: 1 (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[...
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## This note was uploaded on 03/28/2011 for the course ELECTRICAL 30230104 taught by Professor Yongren during the Fall '11 term at Tsinghua University.

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