Kỹ thuật điện: electric circuits + solution manual nilsson 7th ed (en)

17–26 CHAPTER 17 The Fourier Transform [b] In the phasor domain: Vo − 125 Vo Vo + + =0 j200 j800 120 12Vo − 1500 + 3Vo + j20Vo = Vo = Io = 1500 = 60/− 53.13◦ V 15 + j20 Vo = 75 × 10−3 /− 143.13◦ A j800 io (t) = 75 cos(40,000t − 143.13◦ ) mA P 17.30 [a] Vg s Vg s2 Vo = = 25 + (100/s) + s s + 25s + 100 H(s) = Vo s2 = ; Vg (s + 5)(s + 20) H(ω) = (jω)2 (jω + 5)(jω + 20) vg = 25ig = −450e10t u(−t) − 450e−10t u(t) V Vg = − 450 450 − −jω + 10 jω + 10 Vo (ω) = H(ω)Vg = −450(jω)2 (−jω + 10)(jω + 5)(jω + 20) + −450(jω)2 (jω + 10)(jω + 5)(jω + 20) = K2 K3 K4 K5 K6 K1 + + + + + −jω + 10 jω + jω + 20 jω + jω + 10 jω + 20 Problems K1 = 450(100) = −100 (15)(30) K2 = 450(25) = −50 (15)(15) K3 = 450(400) = 400 (30)(−15) Vo (ω) = K4 = K5 = 17–27 −450(25) = −150 (5)(15) −450(100) = 900 (−5)(10) K6 = −450(400) = −1200 (−15)(−10) −200 −800 900 −100 + + + −jω + 10 jω + jω + 20 jω + 10 vo = −100e10t u(−t) + [900e−10t − 200e−5t − 800e−20t ]u(t) V [b] vo (0− ) = −100 V [c] vo (0+ ) = 900 − 200 − 800 = −100 V [d] At t = 0− the circuit is Therefore, the solution predicts v1 (0− ) will be −350 V Now v1 (0+ ) = v1 (0− ) because the inductor will not let the current in the 25 Ω resistor change instantaneously, and the capacitor will not let the voltage across the 0.01 F capacitor change instantaneously At t = 0+ the circuit is From the circuit at t = 0+ we see that vo must be −100 V, which is consistent with the solution for vo obtained in part (c) CHAPTER 17 The Fourier Transform 17–28 P 17.31 Vo s Vo − Vg 100Vo + =0 + 25 s 100s + 125 × 104 · Vo = Io = s(100s + 125 × 104 )Vg 125(s2 + 12,000s + 25 × 106 ) sVo 100s + 125 × 104 Io s2 H(s) = = Vg 125(s2 + 12,000s + 25 × 106 ) −8 × 10−3 ω H(ω) = (25 × 106 − ω ) + j12,000ω Vg (ω) = 300π[δ(ω + 5000) + δ(ω − 5000)] Io (ω) = H(ω)Vg (ω) = io (t) = −2.4π 2π = −1.2 = 12 −2.4πω [δ(ω + 5000) + δ(ω − 5000)] (25 × 106 − ω ) + j12,000ω ω [δ(ω + 5000) + δ(ω − 5000)] jtω e dω −∞ (25 × 106 − ω ) + j12,000ω ∞ 25 × 106 ej5000t 25 × 106 e−j5000t + −j(12,000)(5000) j(12,000)(5000) e−j5000t ej5000t + −j j ◦ ◦ = 0.5[e−j(5000t+90 ) + ej(5000t+90 ) ] io (t) = cos(5000t + 90◦ ) A Problems 17–29 P 17.32 [a] From the plot of vg note that vg is −10 V for an infinitely long time before t = Therefore · vo (0− ) = −10 V There cannot be an instantaneous change in the voltage across a capacitor, so vo (0+ ) = −10 V [b] io (0− ) = A At t = 0+ the circuit is 30 − (−10) 40 = = 8A 5 [c] The s-domain circuit is io (0+ ) = Vo = Vg + (10/s) Vo = H(s) = Vg s+2 10 2Vg = s s+2 17–30 CHAPTER 17 The Fourier Transform H(ω) = jω + Vg (ω) = jω − 5[2πδ(ω)] + Vo (ω) = H(ω)Vg (ω) = 10 30 30 = − 10πδ(ω) + jω + jω jω + 10 30 − 10πδ(ω) + jω + jω jω + = 20πδ(ω) 60 20 − + jω(jω + 2) jω + (jω + 2)(jω + 5) = K0 K1 K2 K3 20πδ(ω) + + + − jω jω + jω + jω + jω + K0 = 20 = 10; Vo (ω) = K1 = 20 = −10; −2 K2 = 60 = 20; K3 = 60 = −20 −3 10 20 20πδ(ω) 10 10 20 10 + − − = + + − 10πδ(ω) jω jω + jω + jω + jω jω + jω + vo (t) = 5sgn(t) + [10e−2t − 20e−5t ]u(t) − V P 17.33 [a] Vo (Vo − Vg )s Vo + =0 + 10 4s 800 · Vo = s2 Vg s2 + 1250s + 25 × 104 Vo s2 = H(s) = Vg (s + 250)(s + 1000) H(ω) = (jω)2 (jω + 250)(jω + 1000) vg = 45e−500|t| ; Vg (ω) = · Vo (ω) = H(ω)Vg (ω) = = 45,000 (jω + 500)(−jω + 500) 45,000(jω)2 (jω + 250)(jω + 500)(jω + 1000)(−jω + 500) K2 K3 K4 K1 + + + jω + 250 jω + 500 jω + 1000 −jω + 500 Problems K1 = 45,000(−250)2 = 20 (250)(750)(750) K2 = 45,000(−500)2 = −90 (−250)(500)(1000) K3 = 45,000(−1000)2 = 80 (−750)(−500)(1500) K4 = 45,000(500)2 = 10 (750)(1000)(1500) · vo (t) = [20e−250t − 90e−500t + 80e−1000t ]u(t) + 10e500t u(−t) V [b] vo (0− ) = 10 V; Vo (0+ ) = 20 − 90 + 80 = 10 V vo (∞) = V [c] IL = 0.25sVg Vo = 4s (s + 250)(s + 1000) H(s) = 0.25s IL = Vg (s + 250)(s + 1000) H(ω) = 0.25(jω) (jω + 250)(jω + 1000) IL (ω) = 0.25(jω)(45,000) (jω + 250)(jω + 500)(jω + 1000)(−jω + 500) = K4 = K1 K2 K3 K4 + + + jω + 250 jω + 500 jω + 1000 −jω + 500 (0.25)(500)(45,000) = mA (750)(1000)(1500) iL (t) = 5e500t u(−t); · iL (0− ) = mA K1 = (0.25)(−250)(45,000) = −20 mA (250)(750)(750) K2 = (0.25)(−500)(45,000) = 45 mA (−250)(500)(1000) K3 = (0.25)(−1000)(45,000) = −20 mA (−750)(−500)(1500) · iL (0+ ) = K1 + K2 + K3 = −20 + 45 − 20 = mA Checks, i.e., iL (0+ ) = iL (0− ) = mA 17–31 17–32 CHAPTER 17 The Fourier Transform At t = 0− : vC (0− ) = 45 − 10 = 35 V At t = 0+ : vC (0+ ) = 45 − 10 = 35 V [d] We can check the correctness of our solution for t ≥ 0+ by using the Laplace transform Our circuit becomes Vo (Vo − Vg )s Vo × 10−3 −6 + =0 + + 35 × 10 + 106 s 800 4s · (s2 + 1250s + 25 × 104 )Vo = s2 Vg − (35s + 5000) vg (t) = 45e−500t u(t) V; Vg = · (s + 250)(s + 1000)Vo = 45 s + 500 45s2 − (35s + 5000)(s + 500) (s + 500) 10s2 − 22,500s − 250 × 104 · Vo = (s + 250)(s + 500)(s + 1000) = 90 80 20 − + s + 250 s + 500 s + 1000 · vo (t) = [20e−250t − 90e−500t + 80e−1000t ]u(t) V This agrees with our solution for vo (t) for t ≥ 0+ P 17.34 [a] Vg (ω) = 36 36 72jω − = − jω + jω (4 − jω)(4 + jω) Problems Vo (s) = (16/s) Vg (s) 10 + s + (16/s) H(s) = 16 16 Vo (s) = = Vg (s) s + 10s + 16 (s + 2)(s + 8) H(ω) = 16 (jω + 2)(jω + 8) Vo (ω) = H(ω) · Vg (ω) = = 17–33 1152jω (4 − jω)(4 + jω)(2 + jω)(8 + jω) K2 K3 K4 K1 + + + − jω + jω + jω + jω K1 = 1152(4) =8 (8)(6)(12) K2 = 1152(−4) = 72 (8)(−2)(4) K3 = 1152(−2) = −32 (6)(2)(6) K4 = 1152(−8) = −32 (12)(−4)(−6) · Vo (jω) = 72 32 32 + − − − jω + jω + jω + jω · vo (t) = 8e4t u(−t) + [72e−4t − 32e−2t − 32e−8t ]u(t)V [b] vo (0− ) = V [c] vo (0+ ) = 72 − 32 − 32 = V The voltages at 0− and 0+ must be the same since the voltage cannot change instantaneously across a capacitor P 17.35 Vo (s) = 30 40 600(s + 10) 10 + − = s s + 20 s + 30 s(s + 20)(s + 30) Vo (s) = H(s) · 15 s · H(s) = 40(s + 10) (s + 20)(s + 30) · H(ω) = 40(jω + 10) (jω + 20)(jω + 30) 17–34 CHAPTER 17 The Fourier Transform · Vo (ω) = vo (ω) = 40(jω + 10) 1200(jω + 10) 30 · = jω (jω + 20)(jω + 30) jω(jω + 20)(jω + 30) 60 80 20 + − jω jω + 20 jω + 30 vo (t) = 10sgn(t) + [60e−20t − 80e−30t ]u(t) V P 17.36 [a] f (t) = [b] W = [c] W = [d] π π 2π −∞ eω ejtω dω + ∞ (1/π) dt = 2 (1 + t ) π ∞ −2ω 0 ∞ ∞ e−2ω dω = 1e π −2 e−ω ejtω dω = ∞ 1/π + t2 dt J = 2 (1 + t ) 2π = J 2π 0.9 , − e−2ω1 = 0.9, e−2ω dω = 2π ω1 = (1/2) ln 10 ∼ = 1.15 rad/s ω1 e2ω1 = 10 P 17.37 Io = Ig R RCsIg = R + (1/sC) RCs + H(s) = s Io = Ig s + (1/RC) RC = (100 × 103 )(1.25 × 10−6 ) = 125 × 10−3 ; H(s) = s ; s+8 Ig (ω) = 30 × 10−6 jω + H(ω) = Io (ω) = H(ω)Ig (ω) = jω jω + 30 × 10−6 jω (jω + 2)(jω + 8) 1 = =8 RC 0.125 Problems 17–35 ω(30 × 10−6 ) √ |Io (ω)| = √ ( ω + 4)( ω + 64) |Io (ω)|2 = K1 K2 900 × 10−12 ω = + 2 (ω + 4)(ω + 64) ω + ω + 64 K1 = (900 × 10−12 )(−4) = −60 × 10−12 (60) K2 = (900 × 10−12 )(−64) = 960 × 10−12 (−60) |Io (ω)|2 = W1Ω = = = π 960 × 10−12 60 × 10−12 − ω + 64 ω2 + ∞ |Io (ω)|2 dω = 120 × 10−12 ω tan−1 π 960 × 10−12 π ∞ − ∞ 60 × 10−12 dω − ω + 64 π ω 30 × 10−12 tan−1 π %= 120 π 30 π × 10−12 = (60 − 15) × 10−12 = 45 pJ · − · π π Between and rad/s W1Ω = ∞ 120 30 tan−1 − tan−1 × 10−12 = 7.14 pJ π π 7.14 (100) = 15.86% 45 P 17.38 [a] Vg (ω) = 60 (jω + 1)(−jω + 1) 0.4 (jω + 0.5) H(s) = Vo 0.4 ; = Vg s + 0.5 Vo (ω) = 24 (jω + 1)(jω + 0.5)(−jω + 1) Vo (ω) = 32 −24 + + jω + jω + 0.5 −jω + H(ω) = vo (t) = [−24e−t + 32e−t/2 ]u(t) + 8et u(−t) V ∞ dω +4 ω2 17–36 CHAPTER 17 The Fourier Transform [b] |Vg (ω)| = [c] |Vo (ω)| = [d] Wi = [e] Wo = 60 + 1) (ω 24 √ (ω + 1) ω + 0.25 ∞ 0 −∞ = 32 + −2t 900e e−2t dt = 1800 −2 64e2t dt + ∞ −2t [576e ∞ ∞ = 900 J (−24e−t + 32e−t/2 )2 dt − 1536e−3t/2 + 1024e−t ] dt = 32 + 288 − 1024 + 1024 = 320 J Problems [f] |Vg (ω)| = 60 , +1 ω2 |Vg2 (ω)| = 3600 (ω + 1)2 Wg = 3600 π = 3600 π ω + tan−1 ω 2 ω +1 = 1800 + tan−1 = 863.53 J π = Wo = + 1)2 (ω + 0.25) 768 1024 1024 − − ω + 0.25 (ω + 1)2 (ω + 1) 1024 · · tan−1 2ω π −768 P 17.39 Io = ω + tan−1 ω ω +1 2048 384 1024 tan−1 − + tan−1 − tan−1 π π π 319.2 × 100 = 99.75% 320 sIg 0.5sIg = 0.5s + 25 s + 50 H(s) = s Io = Ig s + 50 H(ω) = jω jω + 50 I(ω) = 2 = 319.2 J %= 576 (ω −1024 tan−1 ω = 863.53 × 100 = 95.95% 900 · % = [g] |Vo (ω)|2 = dω (ω + 1)2 12 jω + 10 Io (ω) = H(ω)I(ω) = 12(jω) (jω + 10)(jω + 50) 17–37 17–38 CHAPTER 17 The Fourier Transform 12ω |Io (ω)| = (ω + 100)(ω + 2500) |Io (ω)|2 = = 144ω (ω + 100)(ω + 2500) ω2 Wo (total) = = 150 −6 + + 100 ω + 2500 π 150dω − + 2500 π ∞ ω2 ω tan−1 π 50 ∞ − ∞ ω2 6dω + 100 ω 0.6 tan−1 π 10 ∞ = 1.5 − 0.3 = 1.2 J Wo (0–100 rad/s) = 0.6 tan−1 (2) − tan−1 (10) π π = 1.06 − 0.28 = 0.78 J Therefore, the percent between and 100 rad/s is 0.78 (100) = 64.69% 1.2 P 17.40 [a] |Vi (ω)|2 = × 104 ; ω2 |Vi (100)|2 = × 104 = 4; 1002 |Vi (200)|2 = × 104 =1 2002 Problems [b] Vo = RCVi Vi R = R + (1/sC) RCs + H(s) = s Vo ; = Vi s + (1/RC) H(ω) = jω jω + 100 |Vo (ω)| = × 104 , ω + 104 |Vo (100)|2 = [c] W1Ω = π = 100 ≤ ω ≤ 200 rad/s; × 104 = 2; 104 + 104 200 100 |Vo (200)|2 = × 104 × 104 dω = − ω2 π ω |Vo (ω)|2 = 0, × 104 = 0.8 × 104 200 100 × 104 200 ∼ − = = 63.66 J π 100 200 π = π = 106 10−3 1000 = = = 100 RC (0.5)(20) 10 |ω| 200 200 √ ·√ = |ω| ω + 104 ω + 104 |Vo (ω)|2 = [d] W1Ω 17–39 200 100 × 104 ω × 104 · tan−1 dω = ω + 10 π 100 400 [tan−1 − tan−1 1] ∼ = 40.97 J π 200 100 elsewhere 17–40 CHAPTER 17 The Fourier Transform P 17.41 [a] Vi (ω) = H(s) = A ; a + jω |Vi (ω)| = √ s ; s+α H(ω) = WIN = ∞ = A2 π A2 ; 2a when α = a we have ω dω A2 = π (ω + a2 )2 a dω − a + ω2 a2 dω (a2 + ω )2 a π −1 ∞ A2 π ω2 A2 dω = (a2 + ω )2 4a WOUT (a) = 0.5 − = 0.1817 or WOUT (total) π Therefore ω α2 + ω ω A2 (a2 + ω )(α2 + ω ) a A2 4aπ WOUT (total) = |H(ω)| = √ (a2 + ω )(α2 + ω ) A2 e−2at dt = WOUT (a) = jω ; α + jω ωA Therefore |Vo (ω)| = Therefore |Vo (ω)|2 = A + ω2 a2 18.17% [b] When α = a we have WOUT (α) = = π ω A2 dω (a2 + ω )(α2 + ω ) α A2 π where K1 = α a2 K1 K2 + dω +ω α + ω2 a a2 − α2 and K2 = −α2 a2 − α2 Therefore WOUT (α) = α A2 απ a tan−1 − 2 π(a − α ) a WOUT (total) = Therefore π A2 π A2 − α a = π(a2 − α2 ) 2 2(a + α) α απ WOUT (α) · a tan−1 − = WOUT (total) π(a − α) a √ For α = a 3, this ratio is 0.2723, √ or 27.23% of the output energy lies in the frequency band between and a √ [c] For α = a/ 3, the ratio is 0.1057, √ or 10.57% of the output energy lies in the frequency band between and a/