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# Chapter 14 solutions_to_exercises(engineering circuit analysis 7th)

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### Chapter 14 solutions_to_exercises(engineering circuit analysis 7th)

1. 1. Engineering Circuit Analysis, 7th Edition 1. Chapter Fourteen Solutions 10 March 2006 (a) s = 0; (b) s = ± j9 s-1; (c) s = -8 s-1; (d) s = -1000 ± j1000 s-1; (e) v(t) = 8 + 2 cos t mV cannot be attributed a single complex frequency. In a circuit analysis problem, superposition will need to be invoked, where the original function v(t) is expressed as v(t) = v1(t) + v2(t), with v1(t) = 8 mV and v2(t) = 2 cos t mV. The complex frequency of v1(t) is s = 0, and the complex frequency of v2(t) is s = ± 2 s-1. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
2. 2. Engineering Circuit Analysis, 7th Edition 2. Chapter Fourteen Solutions 10 March 2006 (a) s = 0 (b) s = ± j77 s–1 (c) s = –5 s–1 (d) s = 0.5 s–1, –5 ± j8 s–1 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
3. 3. Engineering Circuit Analysis, 7th Edition 3. Chapter Fourteen Solutions 10 March 2006 (a) 8e–t (b) 19 (c) 9 + j7 = 11.4∠37.87o (d) e− jωt → 1∠0o (e) cos 4t → 1∠0o (f) sin 4t → 1∠0o (g) 88∠9o PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4. 4. Engineering Circuit Analysis, 7th Edition 4. Chapter Fourteen Solutions 10 March 2006 (a) (6 – j)* = 6 + j (b) (9)* = 9 (c) (-j30)* = +j30 (d) (5 e-j6)* = 5 e+j6 (e) (24 ∠ -45o )* = 24 ∠ 45o * ⎛ 4 − j18 ⎞ ⎛ 4 + j18 ⎞ (f) ⎜ ⎜ 3.33 + j ⎟ = ⎜ 3.33 − j ⎟ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = 18.44 ∠77.47o = 5.303 ∠ 94.19o 3.477 ∠ - 16.72o * * ⎛ 5 ∠0.1o ⎞ ⎛ ⎞ 5 ∠0.1o o * (g) ⎜ = 0.6202 ∠ − 60.36o ⎜ 4 − j 7 ⎟ = ⎜ 8.062∠ − 60.26o ⎟ = 0.6202 ∠60.36 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ( ) (h) (4 – 22 ∠ 92.5o)* = (4 + 0.9596 – j21.98)* = (4.9596 – j21.98)* = 4.9596 + j21.98 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
5. 5. Engineering Circuit Analysis, 7th Edition 5. Chapter Fourteen Solutions 10 March 2006 Q = 9∠43o μ C, s = j 20π . Thus, q = 9 cos(20π t + 43o ) μ C. (a) At t = 1, q(1) = q (1) = 9 cos(20π + 43o ) μ C = 6.582 μ C. (b) Maximum = 9 μC (c) NO. The indication would be a negative real part in the complex frequency. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
6. 6. Engineering Circuit Analysis, 7th Edition 6. Chapter Fourteen Solutions 10 March 2006 (a) The missing term is Vx*e( −2− j 60)t = (8 + j100)e( −2− j 60)t . We can tell it is missing since vx(t) is not purely real as written; the complex conjugate term above was omitted. (b) s = –2 ± j60 s–1 (c) This means simply that the sine term amplitude is larger than the cosine term amplitude. (d) This indicates that the source is oscillating more strongly than it is decaying. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
7. 7. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 7. Re { i (t )} = i (t ) . No units provided. (a) 10 March 2006 ix (t ) = (4 − j 7) e( −3+ j15)t = (8.062∠ − 60.26°) e−3t e j15t = 8.062e−3t e j (15t −60.26° ) ∴ ix (t ) = Re ix (t ) = 8.062e−3t cos(15t − 60.26°) (b) iy (t ) = (4 + j 7)e−3t (cos15t − j sin15t ) = 8.062e−3t e− j15t + j 60.26° ∴ i y (t ) = 8.062e−3t cos(15t − 60.26°) (c) iA (t ) = (5 − j8)e( −1.5t + j12)t = 9.434e− j 57.99°e−1.5t e j12t = 9.434e−1.5t e j (125−57.99°) ∴ Re iA (0.4) = 9.434e−0.6 cos(4.8rad − 57.99°) = −4.134 (d) iB (t ) = (5 + j8)e( −1.5+ j12) t = 9.434e j 57.99°e −1.5t e− j12t = 9.434e−1.5t e− j (12t −57.99°) ∴ Re iB (0.4) = −4.134 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
8. 8. Engineering Circuit Analysis, 7th Edition 8. Chapter Fourteen Solutions 10 March 2006 (a) ω = 279 Mrad/s, and ω = 2 πf. Thus, f = ω/2π = 44.4 MHz (b) If the current i(t) = 2.33 cos (279×106 t) fA flows through a precision 1-TΩ resistor, the voltage across the resistor will be 1012 i(t) = 2.33 cos (279×106 t) mV. We may write this as 0.5(2.33) cos (279×106 t) + j (0.5)2.33 sin (279×106 t) + 0.5(2.33) cos (279×106 t) - j (0.5)2.33 sin (279×106 t) mV = 1.165 e j279×106 t + 1.165 e -j279×106 t mV PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9. 9. Engineering Circuit Analysis, 7th Edition 9. Chapter Fourteen Solutions 10 March 2006 (a) vs(0.1) = (20 – j30) e(-2 + j50)(0.1) = (36.06 ∠ -56.31o) e(-0.2 + j5) = 36.06e-0.2 ∠ [-56.31o + j5(180)/ π] = 29.52 ∠230.2o V (or 29.52 ∠-129.8o V). (b) Re{ vs } = 36.06 e-2t cos (50t – 56.31o) V. (c) Re{ vs(0.1) } = 29.52 cos (230.2o) = -18.89 V. (d) The complex frequency of this waveform is s = -2 + j50 s-1 (e) s* = (-2 + j50)* = -2 – j50 s-1 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10. 10. Engineering Circuit Analysis, 7th Edition 10. ( Chapter Fourteen Solutions 10 March 2006 ) Let vS forced = 10∠3o est . Let i forced = I m est . di di , so vS forced (t ) = Ri forced + L forced , a superposition of our actual dt dt voltages and currents with corresponding imaginary components. (a) vS (t ) = Ri + L Substituting, 10∠3o est = RIest + Lsest I [1] o o 10∠3 10∠3 or I = = = 0.1∠2.99o R + sL 100 + ( −2 + j10 ) 2 × 10−3 Thus, i(t) = Re{Iest} = 0.1e–2t cos (10t + 2.99o) A. (b) By Ohm’s law, v1(t) = 100i(t) = 10 e–2t cos (10t + 2.99o) V. We obtain v2(t) by recognising from Eq. [1] that V2 est = Lsest I , or V2 = (2×10–3)(–2 + j10) 0.1∠2.99o = 2.04∠104.3o mV ( ) Thus, v2(t) = 2.04e–2t cos (10t + 104.3o) mV PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11. 11. Engineering Circuit Analysis, 7th Edition 11. Chapter Fourteen Solutions 10 March 2006 (a) Let the complex frequency be σ + jω. V = Vm ∠θ . I = I m ∠θ RESISTOR v = Ri Vm eσ t e j (ωt +θ ) = RI m eσ t e j (ωt +θ ) Thus, Vm ∠θ = RI m ∠θ or which defines an impedance R. V = RI di . Let i = I m est = I m eσ t e j (ωt +θ ) . dt v(t ) = (σ + jω ) LI m eσ t e j (ωt +θ ) = Vm eσ t e j (ωt +θ ) INDUCTOR v(t ) = L Thus, Vm ∠θ = (σ + jω ) LI m ∠θ V = Z LI or which defines an impedance ZL = sL = (σ + jω ) L. dv . Let v = Vm est = Vm eσ t e j (ωt +φ ) . dt i (t ) = (σ + jω )CVm eσ t e j (ωt +θ ) = I m eσ t e j (ωt +θ ) CAPACITOR i (t ) = C Thus, I m ∠θ = ⎡(σ + jω ) C ⎤ (Vm ∠θ ) ⎣ ⎦ which defines an impedance ZC = or V = ZC I 1 1 = (σ + jω )C sC (b) ZR = 100 Ω. ZL = (–2 + j10)(0.002) = 20.4 ∠101.3o Ω . (c) Yes. Z R → R; Z L → jω L; ZC → 1 jωC PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12. 12. Engineering Circuit Analysis, 7th Edition 12. Chapter Fourteen Solutions 10 March 2006 (a) s = 0 + j120π = + j120π (b) We first construct an s-domain voltage V(s) = 179 ∠ 0o with s given above. The equation for the circuit is di di v(t) = 100 i(t) + L = 100 i(t) + 500×10-6 dt dt and we assume a response of the form Iest. Substituting, we write (179 ∠ 0o) est = 100 Iest + sL Iest Supressing the exponential factor, we may write 179∠0o 179∠0o 179∠0o = = = 1.79 ∠ -0.108o A I = -6 -6 o 100 + j120π (500 × 10 ) 100∠0.108 100 + s500 × 10 Converting back to the time domain, we find that i(t) = 1.79 cos (120πt – 0.108o) A. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13. 13. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 13. (a) vs = 10e −2t cos(10t + 30°) V ∴ s = −2 + j10, Vs = 10∠30° V 10 5 −1 − j 5 −5 − j 25 (−25 − j125) / 26 = = , Zc 5 = −2 + j10 −1 + j 5 26 26 (−5 − j 25 + 130) / 26 −25 − j125 −1 − j 5 ∴ Zc 5 = = = − j1 ∴ Zin = 5 + 0.5(−2 + j10) − j1 = 4 + j 4 Ω 125 − j 25 5 − j1 10∠30° (−5 − j 25) / 26 10∠30° −5 − j 25 5∠30° −5 − j 25 1∠30° −1 − j5 = = = ∴ Ix = × 4 + j 4 5 + (−5 − j 25) / 26 4 + j 4 130 − 5 − j 25 2 + j 2 125 − j 25 2 + j 2 5 − j1 1∠30° ∴ Ix = (− j1) = 0.3536∠ − 105° A 2 2∠45° Zc = (b) ix (t ) = 0.3536e −2t cos(10t − 105°) A PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14. 14. Engineering Circuit Analysis, 7th Edition 14. Chapter Fourteen Solutions 10 March 2006 (a) s = 0 + j100π = + j100π (b) We first construct an s-domain voltage V(s) = 339 ∠ 0o with s given above. The equation for the circuit is dv dv v(t) = 2000 i(t) + vC(t) = 2000 C C + vC(t) = 0.2 C + vC(t) dt dt and we assume a response of the form VCest. Substituting, we write (339 ∠ 0o) est = 0.2s VCest + VCest Supressing the exponential factor, we may write VC = 339∠0o 339∠0o 339∠0o = = 5.395 ∠ -89.09o A = o 1 + 0.2s 1 + j100π (0.2) 62.84∠89.09 Converting back to the time domain, we find that vC(t) = 5.395 cos (100πt – 89.09o) V. and so the current is i(t) = C dvC = –0.1695 sin(100πt – 89.09o) A dt = 169.5 cos (100πt + 0.91o) mA. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15. 15. Engineering Circuit Analysis, 7th Edition 15. 10 March 2006 iS 1 = 20e −3t cos 4t A, iS 2 = 30e−3t sin 4t A (a) Chapter Fourteen Solutions IS 1 = 20∠0°, IS 2 = − j 30, s = −3 + j 4 10 −3 − j 4 = 0.4(−3 − j 4) = −1.2 − j1.6, Z L = −6 + j8 −3 + j 4 − 3 − j 4 −6 + j 8 5(7.2 + j 6.4) (−6 + j8)(3.8 − j1.6) ∴ Vx = 20 × − j 30 −2.2 + j 6.4 −7.2 + j 6.4 −2.2 + j 6.4 −600 + j800 − j 30(−22.8 + 12.8 + j 30.4 + j 9.6) −600 + j800 − j 30(−10 + j 40) = = −2.2 + j 6.4 −2.2 + j 6.4 −600 + 1200 + j1000 600 + j1000 = = = 185.15− ∠ − 47.58° V −2.2 + j 6.4 −2.2 + j 6.4 ∴ Zc = (b) vx (t ) = 185.15− e−3t cos(4t − 47.58°) V PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16. 16. Engineering Circuit Analysis, 7th Edition 16. Chapter Fourteen Solutions 10 March 2006 (a) If v(t) = 240 2 e-2t cos 120πt V, then V = 240 2 ∠0o V where s = -2 + j120π. 240 2 ∠0o = 113.1 ∠0o kA. Thus, Since R = 3 mΩ, the current is simply I = −3 3 × 10 i(t) = 113.1e-2t cos 120πt kA (b) Working in the time domain, we may directly compute i(t) = v(t) / 3×10-3 = (240 2 e-2t cos 120πt ) / 3×10-3 = 113.1e-2t cos 120πt kA (c) A 1000-mF capacitor added to this circuit corresponds to an impedance 1 1 1 = = Ω in parallel with the 3-mΩ -3 (-2 + j120π )(1000 × 10 ) - 2 + j120π sC resistor. However, since the capacitor has been added in parallel (it would have been more interesting if the connection were in series), the same voltage still appears across its terminals, and so i(t) = 113.1e-2t cos 120πt kA as before. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17. 17. Engineering Circuit Analysis, 7th Edition 17. L {K u (t )} = Chapter Fourteen Solutions ∞ ∞ ∞ 0 0 0 − st −st −st ∫ - Ke u (t )dt = K ∫ - e u (t )dt = K ∫ e dt = 10 March 2006 − K − st e s ∞ 0 ⎛ − K − st ⎞ ⎛K ⎞ = lim ⎜ e ⎟ + lim ⎜ e − st ⎟ t →∞ t →0 ⎝ s ⎠ ⎝s ⎠ If the integral is going to converge, then lim (e − st ) = 0 (i.e. s must be finite). This leads t →∞ to the first term dropping out (l’Hospital’s rule assures us of this), and so L {K u (t )} = K s PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
18. 18. Engineering Circuit Analysis, 7th Edition 18. (a) L {3 u (t )} = Chapter Fourteen Solutions ∞ ∞ 0 − 3 −st e s ∞ 0 10 March 2006 0 − st − st − st ∫ - 3e u (t )dt = 3∫ - e u (t )dt = 3∫ e dt = ∞ 0 ⎛ − 3 − st ⎞ ⎛3 ⎞ e ⎟ + lim ⎜ e −st ⎟ = lim ⎜ t →∞ ⎝ s ⎠ t →0 ⎝ s ⎠ If the integral is going to converge, then lim (e − st ) = 0 (i.e. s must be finite). This leads t →∞ to the first term dropping out (l’Hospital’s rule assures us of this), and so L {3 u (t )} = (b) L {3 u (t − 3)} 3 s − 3 −st = ∫ - 3e u (t − 3)dt = 3∫ e dt = e 0 3 s ∞ ∞ − st ∞ −st 3 ⎛ − 3 −st ⎞ ⎛3 ⎞ e ⎟ + ⎜ e − 3s ⎟ = lim ⎜ t →∞ ⎠ ⎝s ⎠ ⎝ s If the integral is going to converge, then lim (e − st ) = 0 (i.e. s must be finite). This leads t →∞ to the first term dropping out (l’Hospital’s rule assures us of this), and so L {3 u (t − 3)} = 3 − 3s e s (c) L {3 u (t − 3) − 3} = = − st ∫0 [3u(t − 3) − 3]e dt ∞ - − 3 −st e s ∞ 3 − 3 −st e s ∞ ∞ 3 0 = 3∫ e −st dt - 3∫ - e −st dt ∞ 0− Based on our answers to parts (a) and (b), we may write ( ) 3 − 3s 3 3 − 3s = e − e -1 s s s L {3 u (t − 3) − 3} = (d) L {3 u (3 − t )} − 3 −st = 3∫ - e u (3 − t )dt = 3∫ - e dt = e 0 0 s ∞ = 3 −st − 3 − 3s e −1 s ( ) = ( 3 − st 3 1 − e − 3s s 0- ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
19. 19. Engineering Circuit Analysis, 7th Edition 19. (a) L {2 + 3 u (t )} = Chapter Fourteen Solutions − st ∫ - e [2 + 3u (t )]dt = ∞ 0 ∫ ∞ 0 10 March 2006 ∞ 5e − st dt = − 5 − st e s 0 ⎛ − 5 − st ⎞ ⎛5 ⎞ e ⎟ + lim ⎜ e − st ⎟ = lim ⎜ t →∞ ⎝ s ⎠ t →0 ⎝ s ⎠ If the integral is going to converge, then lim (e − st ) = 0 (i.e. s must be finite). This leads t →∞ to the first term dropping out (l’Hospital’s rule assures us of this), and so L {2 + 3 u (t )} = (b) L {3 e -8t } ∫ = ∞ 0- - 8t − st 3 e e dt = ∫ ∞ 0- 3e − (8 + s ) t 5 s − 3 − (8 + s ) t dt = e s+8 ∞ 0− 3 3 ⎛ − 3 − ( s + 8) t ⎞ ⎛ 3 − ( s + 8)t ⎞ e e = = lim ⎜ ⎟ + lim ⎜ ⎟ = 0+ t →∞ s+8 s+8 ⎝s+8 ⎠ t →0 ⎝ s + 8 ⎠ (c) L { u (−t )} = (d) L {K } = ∫ ∞ 0- ∞ − st ∫ - e u (−t )dt = 0 ∞ 0 − st ∫ - e u(−t )dt = 0 Ke dt = K ∫ - e dt = K ∫ − st − st 0 ∞ 0 ∫ 0- 0- (0) e − st dt = − K − st e dt = e s 0 ∞ − st 0 ⎞ ⎛K ⎛ − K − st ⎞ = lim ⎜ e ⎟ + lim ⎜ e − st ⎟ t →∞ t →0 ⎠ ⎝s ⎠ ⎝ s If the integral is going to converge, then lim (e − st ) = 0 (i.e. s must be finite). This leads t →∞ to the first term dropping out (l’Hospital’s rule assures us of this), and so L {K } = K s PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
20. 20. Engineering Circuit Analysis, 7th Edition 20. Chapter Fourteen Solutions 10 March 2006 (a) The frequency-domain representation of the voltage across the resistor is (1)I(s) 4 4 where I(s) = L 4e-t u (t ) = A . Thus, the voltage is V. s +1 s +1 { } (b) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
21. 21. Engineering Circuit Analysis, 7th Edition 21. Chapter Fourteen Solutions 10 March 2006 (a) L {5 u (t ) − 5 u (t − 2)} = ∞ ∞ 0 = 2 5∫ e − st dt − 5∫ e− st dt ∞ ∫ [5 u(t ) − 5 u(t − 2)] e − st 0- −5 − st e s = ∞ + 0 5 −st e s ∞ 2 ⎛ − 5 − st ⎞ ⎛5 ⎞ −5 e ⎟ + lim ⎜ e −st ⎟ + lim ⎛ e −st ⎞ − ⎜ ⎟ t →∞ ⎝ s ⎠ t →0 ⎝ s ⎠ ⎝ s ⎠ = lim ⎜ t →∞ dt ⎛ 5 −2 s ⎞ ⎜ e ⎟ ⎝s ⎠ If the integral is going to converge, then lim (e − st ) = 0 (i.e. s must be finite). This leads t →∞ to the first and third terms dropping out (l’Hospital’s rule assures us of this), and so L {5 u (t ) − 5 u (t − 2)} = 5 1 − e −2 s s ( ) (b) The frequency domain current is simply one ohm times the frequency domain voltage, or 5 1 − e −2s s ( ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
22. 22. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 22. (a) ∞ f(t) = t + 1 ∴ F( s ) = ∫ − (t + 1) e − ( σ+ jω)t dt ∴ σ > 0 0 ∞ (b) f (t ) = (t + 1) u (t ) ∴ F( s ) = ∫ − (t + 1) e − ( σ+ jω) t dt ∴ σ > 0 (c) f (t ) = e50t u (t ) ∴ F( s ) = ∫ − e50t e − ( σ+ jω)t dt ∴σ > 50 (d) f (t ) = e50t u (t − 5) ∴ F( s ) = ∫ − e50t u (t − 5) e − ( σ+ jω)t dt ∴ σ > 50 (e) f (t ) = e −50t u (t − 5) ∴ F( s ) = ∫ − e −50t u (t − 5) e − ( σ+ jω)t dt ∴ σ < 50 0 ∞ 0 ∞ 0 ∞ 0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
23. 23. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 23. (a) f (t ) = 8e −2t [u (t + 3) − u (t − 3)] ∞ 3 0 0 F( s ) = ∫ f (t )e − st dt = ∫ 8e( −2+s )t dt = 8 [1 − e −6−3s ] 2+s (b) f (t ) = 8e 2t [u (t + 3) − u (t − 3)] ∞ 3 0 0 F(s) = ∫ − f (t )e − st dt = ∫ 8e(2−s )t dt = 8 8 ⎡1 − e6 e −3s ⎤ [e6−3s − 1] = ⎦ s−2⎣ 2−s (c) f (t ) = 8e ∞ −2 t [u (t + 3) − u (t − 3)] 3 F(s) = ∫ − f (t )e − st dt = ∫ − 8e( −2−s )t dt = 0 0 8 ⎡1 − e −6−3s ⎤ ⎦ s+2 ⎣ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
24. 24. Engineering Circuit Analysis, 7th Edition 24. Chapter Fourteen Solutions 10 March 2006 ⎧ ⎛ 1 ⎞⎫ 1 (a) L ⎨L-1⎜ ⎟⎬ = s ⎩ ⎝ s ⎠⎭ (b) L 1 + u (t ) + [u (t )]2 { } (c) L {t u (t ) − 3} = 1 3 − s2 s = 1 1 1 + + = s s s (d) L { - δ (t ) + δ (t − 1) − δ (t − 2)} = 1 3 s 1 − 1 + e − s − e − 2s s PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
25. 25. Engineering Circuit Analysis, 7th Edition 25. Chapter Fourteen Solutions 10 March 2006 (a) f(t) = e-3t u(t) (b) f(t) = δ(t) (c) f(t) = t u(t) (d) f(t) = 275 δ(t) (e) f(t) = u(t) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
26. 26. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 26. L { f1 (t ) + f 2 (t )} = ∫ [ f (t ) + ∞ 0 - 1 f 2 (t )]e − st dt = = L { f1 (t )} + L { f 2 (t )} ∫ ∞ 0 - f1 (t )e − st dt + ∫ ∞ 0- f 2 (t )e − st dt PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
27. 27. Engineering Circuit Analysis, 7th Edition 27. (a) Chapter Fourteen Solutions ∞ f (t ) = 2u (t − 2) ∴ F( s ) = 2∫ e − st dt + 2 ∴ F(1 + j 2) = −2 st e s ∞ = 2 10 March 2006 2 −2 s e ; s = 1+ j2 s 2 e−2 e − j 4 = 0.04655+ + j 0.11174 1+ j2 (b) f (t ) = 2δ (t − 2) ∴ F( s ) = 2e −2 s , F (1 + j 2) = 2e −2 e − j 4 = −0.17692 + j 0.2048 (c) f (t ) = e u (t − 2) ∴ F( s) = ∫ e −t −∞ 2 ∴ F(1 + j 2) = − ( s +1) t 1 dt = e − ( s +1)t −s + 1 ∞ = 2 1 −2 s − 2 e s +1 4 1 e−2 e −2 e− j = (0.4724 + j 6.458)10−3 2 + j2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
28. 28. Engineering Circuit Analysis, 7th Edition 28. (a) (b) (c) (d) ∫ ∞ −∞ ∞ 8 sin 5t δ (t − 1) dt ∫ (t − 5) δ (t − 2)dt 2 −∞ ∫ ∞ ∫ ∞ −∞ −∞ Chapter Fourteen Solutions = 8 sin 5 × 1 = - 7.671 = (2 − 5) 2 5e − 3000tδ (t − 3.333 × 10− 4 )dt Kδ (t − 2)dt 10 March 2006 = 9 = 5e − 3000( 3.333×10 −4 ) = 1.840 = K PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
29. 29. Engineering Circuit Analysis, 7th Edition 29. (a) Chapter Fourteen Solutions 10 March 2006 ∞ f (t ) = [u (5 − t )] [u (t − 2)] u (t ), ∴ F( s ) ∫ − [u (5 − t )] [u (t − 2)] u (t ) e − st dt 0 5 1 ∴ F( s) = ∫ e − st dt = − e− st 2 s 5 = 2 1 −2 s −5 s (e − e ) s (b) ∞ 4 f (t ) = 4u (t − 2) ∴ F( s ) = 4∫ e − st dt = e −2 s 2 s (c) f (t ) = 4e−3t u (t − 2) ∴ F( s ) = 4 ∫ e− ( s +3)t dt = ∞ 2 ∴ F( s) = −4 − ( s +3) t e s+3 ∞ 2 4 −2 s −6 e s+3 2+ ∞ (d) f (t ) = 4δ (t − 2) ∴ F( s) = 4∫ − δ (t − 2) e − st dt = 4 ∫ e−2 s δ (t − 2) dt = 4e−2 s (e) f (t ) = 5δ (t ) sin (10t + 0.2π) ∴ F( s ) = 5∫ − δ(t ) [sin 0.2π] X 1dt = 5sin 36° 0 2 0+ 0 ∴ F( s ) = 2.939 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
30. 30. Engineering Circuit Analysis, 7th Edition 30. (a) (b) (c) (d) ∫ ∞ −∞ ∞ cos 500t δ (t ) dt ∫ (t ) δ (t − 2)dt 5 −∞ ∫ ∞ −∞ ∫ ∞ −∞ = = (2) 5 2.5e − 0.001tδ (t − 1000)dt − K 2δ (t − c)dt Chapter Fourteen Solutions 10 March 2006 cos 500 × 0 = 1 = 32 = 2.5e − 0.001(1000 ) = 0.9197 = - K2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
31. 31. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 31. (a) f(t) = 2 u(t – 1) u(3 – t) u(t3) 3 3 2 F(s) = ∫ e − st dt = - e − st = 1 s 1 ∞ 2 −s (e - e−3s ) s −2 2 (0 − e−4 s ) = e−4 s s s (b) f (t ) = 2u (t − 4) ∴ F( s ) = 2 ∫ e − st dt = (c) f (t ) = 3e−2t u (t − 4) ∴ F( s) = 3∫ e − ( s + 2)t dt = (d) f (t ) = 3δ (t − 5) ∴ F( s ) = 3∫ − δ(t − 5) e − st dt = 3e −5 s (e) f (t ) = 4δ (t − 1) [cos πt − sin πt ] 4 ∞ 4 3 −4 s −8 e s+2 ∞ 0 ∞ ∴ F( s) = 4∫ − δ (t − 1) [cos πt − sin πt ] e− st dt ∴ F( s) = −4e− s 0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
32. 32. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions (a) F(s) = 3 + 1/s; f(t) = 3δ(t) + u(t) (b) F(s) = 3 + 1/s2; 32. 10 March 2006 f(t) = 3δ(t) + tu(t) (c) F(s) = (d) F(s) = 1 ( s + 3)( s + 4 ) = 1 1 ; − ( s + 3) ( s + 4 ) f(t) = ⎡ e−3t − e −4t ⎤ u (t ) ⎣ ⎦ 1 1/ 2 1 1/ 2 ; = − + ( s + 3)( s + 4 )( s + 5) ( s + 3) ( s + 4 ) ( s + 5 ) 1 ⎡1 ⎤ f(t) = ⎢ e−3t − e−4t + e−5t ⎥ u (t ) 2 ⎣2 ⎦ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
33. 33. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions (a) G(s) = 90 – 4.5/s; g(t) = 90δ(t) – 4.5u(t) (b) G(s) = 11 + 2/s; 33. 10 March 2006 g(t) = 11δ(t) + 2u(t) (c) G(s) = (d) G(s) = 1 ( s + 1) 2 ; g(t) = te− t u (t ) 1 1/ 2 1 1/ 2 1 ⎡1 ⎤ = − + ; g(t) = ⎢ e− t − e−2t + e−3t ⎥ u (t ) 2 ( s + 1)( s + 2 )( s + 3) ( s + 1) ( s + 2 ) ( s + 3) ⎣2 ⎦ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
34. 34. Engineering Circuit Analysis, 7th Edition 34. Chapter Fourteen Solutions 10 March 2006 (a) f(t) = 5 u(t) – 16 δ(t) + e-4.4t u(t) (b) f(t) = δ(t) – u(t) + t u(t) 5 88 a b + + + s+7 s s+6 s +1 17 17 where a = = - 3.4 and b = = 3.4 . s + 1 s = -6 s + 6 s = -1 Thus, f(t) = 5 e-7t u(t) + 88 u(t) –3.4 e-6t u(t) + 3.4 e-t u(t) (c) F(s) = Check with MATLAB: EDU» T1 = '5/(s+7)'; EDU» T2 = '88/s'; EDU» T3 = '17/(s^2 + 7*s + 6)'; EDU» T = symadd(T1,T2); EDU» P = symadd(T,T3); EDU» p = ilaplace(P) p= 5*exp(-7*t)+88-17/5*exp(-6*t)+17/5*exp(-t) EDU» pretty(p) 5 exp(-7 t) + 88 - 17/5 exp(-6 t) + 17/5 exp(-t) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
35. 35. Engineering Circuit Analysis, 7th Edition 35. Chapter Fourteen Solutions 10 March 2006 5 , then v(t) = 5 u(t) V. The voltage at t = 1 ms is then simply 5 V, and the s current through the 2-kΩ resistor at that instant in time is 2.5 mA. If V(s) = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
36. 36. Engineering Circuit Analysis, 7th Edition 36. Chapter Fourteen Solutions 10 March 2006 5 pA, so i(t) = 5 e-10t u(t) pA. The voltage across the 100-MΩ resistor is s + 10 therefore 500 e-10t u(t) μV. I( s) = (a) The voltage as specified has zero value for t < 0, and a peak value of 500 μV. (b) i(0.1 s) = 1.839 pA, so the power absorbed by the resistor at that instant = i2R = 338.2 aW. (A pretty small number). (c) 500 e-10t1% = 5 Taking the natural log of both sides, we find t1% = 460.5 ms PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
37. 37. Engineering Circuit Analysis, 7th Edition 37. Chapter Fourteen Solutions 10 March 2006 s +1 2 1 2 + = 1+ + ⇔ δ(t ) + u (t ) + 2e− t u (t ) s s +1 s s +1 (a) F(s) = (b) F(s) = (e − s + 1) 2 = e−2s + 2e− s + 1 ⇔ δ (t − 2) + 2δ (t − 1) + δ(t ) (c) F(s) = 2e− (s +1) = 2e −1 e−2s ⇔ 2e−1 δ (t − 1) (d) F(s) = 2e-3s cosh 2s = e-3s (e2s + e-2s) = e-s + e-5s ⇔ δ(t – 1) + δ(t – 5) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
38. 38. Engineering Circuit Analysis, 7th Edition 38. Chapter Fourteen Solutions 10 March 2006 N(s) = 5s. (a) D(s) = s2 – 9 so where a = N(s) 5s 5s a b = 2 = = + (s + 3)(s - 3) (s + 3) (s − 3) D(s) s -9 5s - 15 = = 2.5 and b = (s − 3) s = -3 - 6 5s 15 = = 2.5 . Thus, (s + 3) s = 3 6 f(t) = [2.5 e-3t + 2.5 e3t] u(t) (b) D(s) = (s + 3)(s2 + 19s + 90) = (s + 3)(s + 10)(s + 9) so N(s) 5s a b = = + + (s + 3) (s + 10) D(s) (s + 3)(s + 10)(s + 9) a = c = 5s (s + 10)(s + 9) 5s (s + 3)(s + 10) c (s + 9) = - 15 5s - 50 = - 0.3571, b = = = - 7.143 (s + 3)(s + 9) s = -10 (-7)(-1) (7)(6) = - 45 = 7.5. (-6)(1) s = -3 s = -9 ∴ f(t) = [-0.3571 e-3t - 7.143 e-10t + 7.5 e-9t] u(t) (c) D(s) = (4s + 12)(8s2 + 6s + 1) = 32(s + 3)(s + 0.5)(s + 0.25) so a b N(s) s ⎛ 5 ⎞ = + + = ⎜ ⎟ (s + 3) (s + 0.5) D(s) ⎝ 32 ⎠ (s + 3)(s + 0.5)(s + 0.25) s ⎛ 5 ⎞ a = ⎜ ⎟ ⎝ 32 ⎠ (s + 0.5)(s + 0.25) s ⎛ 5 ⎞ c =⎜ ⎟ ⎝ 32 ⎠ (s + 3)(s + 0.5) c (s + 0.25) s ⎛ 5 ⎞ = − 0.06818, b = ⎜ ⎟ = 0.125 ⎝ 32 ⎠ (s + 3)(s + 0.25) s = - 0.5 s = -3 = - 0.05682 s = - -0.25 ∴ f(t) = [-0.06818 e-3t + 0.125 e-0.5t – 0.05682e-0.25t] u(t) (d) Part (a): EDU» N = [5 0]; EDU» D = [1 0 -9]; EDU» [r p y] = residue(N,D) Part (b): Part (c): EDU» N = [5 0]; EDU» D = [1 22 147 270]; EDU» [r p y] = residue(N,D) r= -7.1429 7.5000 -0.3571 EDU» N = [5 0]; EDU» D = [32 120 76 12]; EDU» [r p y] = residue(N,D) r= -0.0682 0.1250 -0.0568 p= 3 -3 p= -10.0000 -9.0000 -3.0000 p= -3.0000 -0.5000 -0.2500 y= y= y= r= 2.5000 2.5000 [] [] [] PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
39. 39. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 39. (a) F( s) = 5 ↔ 5e −t u (t ) s +1 (b) F( s) = 5 2 − ↔ (5e− t − 2e −4t ) u (t ) s +1 s + 4 (c) F( s) = 18 6 6 = − ↔ 6 (e −t − e −4t ) u (t ) ( s + 1) ( s + 4) s + 1 s + 4 (d) F( s ) = 18s −6 24 = + ↔ 6 (4e−4t − e −t ) u (t ) ( s + 1) ( s + 4) s + 1 s + 4 (e) F( s) = 18s 2 6 96 = 18 + − ↔ 18δ (t ) + 6 (e − t − 16e−4t ) u (t ) s +1 s + 4 ( s + 1) ( s + 4) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
40. 40. Engineering Circuit Analysis, 7th Edition 40. Chapter Fourteen Solutions 10 March 2006 N(s) = 2s2. (a) D(s) = s2 – 1 so where a = 2s 2 (s − 1) N(s) 2s 2 2s 2 a b +2 = 2 = = + (s + 1)(s - 1) (s + 1) (s − 1) D(s) s -1 = s = -1 2 = - 1 and b = -2 2s 2 2 = = 1 . Thus, (s + 1) s = 1 2 f(t) = [2δ(t) – e–t + et] u(t) (b) D(s) = (s + 3)(s2 + 19s + 90) = (s + 3)(s + 10)(s + 9) so N(s) 2s 2 a b = = + + (s + 3) (s + 10) D(s) (s + 3)(s + 10)(s + 9) 2s 2 a = (s + 10)(s + 9) c = 2s 2 (s + 3)(s + 10) c (s + 9) 18 2s 2 200 = = 0.4286, b = = = 28.57 (s + 3)(s + 9) s = -10 (-7)(-1) (7)(6) s = -3 = s = -9 162 = - 27. (-6)(1) ∴ f(t) = [0.4286 e-3t + 28.57 e-10t - 27 e-9t] u(t) (c) D(s) = (8s + 12)(16s2 + 12s + 2) = 128(s + 1.5)(s + 0.5)(s + 0.25) so N(s) s2 a b ⎛ 2 ⎞ = ⎜ = + + ⎟ (s + 1.5) (s + 0.5) D(s) ⎝ 128 ⎠ (s + 1.5)(s + 0.5)(s + 0.25) c (s + 0.25) s2 s2 ⎛ 2 ⎞ ⎛ 2 ⎞ a=⎜ = 0.02813, b = ⎜ = - 0.01563 ⎟ ⎟ ⎝ 128 ⎠ (s + 0.5)(s + 0.25) s = -1.5 ⎝ 128 ⎠ (s + 1.5)(s + 0.25) s = - 0.5 s2 ⎛ 2 ⎞ c =⎜ = 0.003125 ⎟ ⎝ 128 ⎠ (s + 1.5)(s + 0.5) s = - 0.25 ∴ f(t) = 0.02813 e-1.5t – 0.01563 e-0.5t + 0.003125e-0.25t] u(t) (d) Part (a): EDU» N = [2 0 0]; EDU» D = [1 0 -1]; EDU» [r p y] = residue(N,D) Part (b): Part (c): EDU» N = [2 0 0]; EDU» D = [1 22 147 270]; EDU» [r p y] = residue(N,D) r= 28.5714 -27.0000 0.4286 EDU» N = [2 0 0]; EDU» D = [128 288 160 24]; EDU» [r p y] = residue(N,D) r= 0.0281 -0.0156 0.0031 p= -1.0000 1.0000 p= -10.0000 -9.0000 -3.0000 p= -1.5000 -0.5000 -0.2500 y= y= y= r= -1.0000 1.0000 2 [] [] PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
41. 41. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 41. (a) F( s ) = 2 3 ∴ f(t) = 2 u(t) – 3 e-t u(t) so − s s +1 (b) F( s ) = 2 s + 10 4 ↔ 2δ(t ) + 4e−3t u (t ) = 2+ s+3 s+3 (c) F( s ) = 3e−0.8 s ↔ 3δ (t − 0.8) (d) F( s) = 12 3 3 ↔ 3(e −2t − e−6t ) u (t ) = − ( s + 2) ( s + 6) s + 2 s + 6 (e) F( s ) = 12 3 A 0.75 = + + 2 2 ( s + 2) ( s + 6) ( s + 2) s + 2 s + 6 12 3 A 0.75 = + + ∴ A = −0.75 4× 6 4 2 6 3 0.75 0.75 ∴ F( s ) = − + ↔ (3te −2t − 0.75e −2t + 0.75e −6t ) u (t ) 2 ( s + 2) s + 2 s + 6 Let s = 0 ∴ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
42. 42. Engineering Circuit Analysis, 7th Edition 42. F(s) Chapter Fourteen Solutions 10 March 2006 1 π + 3 2 s s + 4s + 5s + 2 1 π = 2 − + s (s + 2)(s + 1) 2 1 a b c = 2 − + + + 2 s (s + 2) (s + 1) (s + 1) = 2 − where a = b = and c = π (s + 1) 2 π (s + 2) s = π s = −2 = π = −1 ⎤ d ⎡ π 2 ⎢( s + 1) ⎥ 2 ds ⎢ (s + 2) ( s + 1) ⎥ ⎣ ⎦s = = −1 d ⎡ π ⎤ ⎢ ⎥ ds ⎣ (s + 2) ⎦ s = −1 ⎡ π ⎤ = ⎢− = −π 2⎥ ⎣ (s + 2) ⎦ s = −1 Thus, we may write f(t) = 2 δ(t) – u(t) + πe–2t u(t) + πte–t u(t) – πe–t u(t) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
43. 43. Engineering Circuit Analysis, 7th Edition 43. (a) F(s) = (s + 1)(s + 2) s(s + 3) (s + 1)(s + 2) (s + 3) s = 0 a = = = Chapter Fourteen Solutions 1+ 2 3 and a s + b = 10 March 2006 b (s + 3) (s + 1)(s + 2) s s = -3 = (-2)(-1) 2 = -3 3 so 2 2 −3t f(t) = δ (t ) + u (t ) − e u (t ) 3 3 (b) F(s) = (s + 2) (s 2 + 4) a = b = c (s + 2) = s (s 2 + 4) 2 d ds = = s =0 a b c c* + + + 2 s s (s + j 2) (s − j 2) 2 = 0.5 4 ⎡ (s 2 + 4) − 2s(s + 2) ⎤ ⎡ (s + 2) ⎤ 4 = ⎢ = 2 = 0.25 ⎥ ⎢ 2 ⎥ 2 2 (s + 4) ⎣ (s + 4) ⎦ s = 0 ⎣ ⎦s = 0 4 (s + 2) s (s − j 2) = 0.1768∠ − 135o 2 (c* = 0.1768∠135o) s = − j2 so o o f(t) = 0.5 t u(t) + 0.25 u(t) + 0.1768 e–j135 e-j2t u(t) + 0.1768 ej135 ej2t u(t) The last two terms may be combined so that f(t) = 0.5 t u(t) + 0.25 u(t) + 0.3536 cos (2t + 135o) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
44. 44. Engineering Circuit Analysis, 7th Edition 44. (a) Chapter Fourteen Solutions 10 March 2006 G(s) is not a rational function, so first we perform polynomial long division (some intermediate steps are not shown): 12s − 36 −36 2 3 2 2 s + 3s + 2 ) 12s ( ( s + 3s + 2 ) −36s − 24s − 36s 2 − 24s , so G(s) = 12s − 36 + Hence, g(t) = 12 and 84s +72 84s + 72 12 96 + = 12s − 36 − (s + 1)(s + 2) s +1 s + 2 d δ (t ) − 36δ (t ) − 12e− t u (t ) + 96e−2t u (t ) dt (b) G(s) is not a rational function, so first we perform polynomial long division (some intermediate steps are not shown): 12 3 2 3 ( s + 4s + 5s + 2 ) 12s 12s3 + 48s 2 + 60s + 24 , − 48s 2 − 60s − 24 so G(s) = 12 − 48s 2 + 60s + 24 A B C = 12 + + + 2 2 (s + 2s + 1)(s + 2) ( s + 1) s + 1 s + 2 Where A = –12, B = 48 and C = –96. Hence, g(t) = 12δ (t ) − 12te−t u (t ) + 48e − t u (t ) − 96e−2t u (t ) (c) G(s) is not a rational function, so first we perform polynomial long division on the second term (some intermediate steps are not shown): 12 3 2 3 s + 6s + 11s + 6 12s ( ) 12s3 + 72s 2 + 132s + 72 , − 72s − 132s − 72 2 so G(s) = 3s − 12 + 72s 2 + 132s + 72 A B C = 3s − 12 + + + (s + 1)(s + 2)(s + 3) s +1 s + 2 s + 3 Where A = 6, B = –96 and C = 162. Hence, g(t) = 3 d δ (t ) − 12δ (t ) + 6e−t u (t ) − 96e −2t u (t ) + 162e−3t u (t ) dt PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
45. 45. Engineering Circuit Analysis, 7th Edition 45. (a) H(s) = (b) H(s) = Chapter Fourteen Solutions 10 March 2006 s +1 1 , hence h(t) = δ(t) – e–2t u(t) = 1− s+2 s+2 s+3 2 1 , hence h(t ) = ⎡ 2e − t − e −2t ⎤ u (t ) = − ⎣ ⎦ ( s + 1)( s + 2 ) s + 1 s + 2 (c) We need to perform long division on the second term prior to applying the method of residues (some intermediate steps are not shown): s−5 3 2 4 s + 5s + 7s + 3 s ( ) 18s 2 + 32s + 15 Thus, H(s) = 3s − s + 5 − 18s 2 + 32s + 15 A B C + 1 = 2s + 6 + + + 2 2 (s + 1) (s + 3) ( s + 1) s + 1 s + 3 where A = –1/2, B = 9/4, and C = –81/4. Thus, h(t) = 2 d 1 9 81 δ (t ) + 6δ (t ) − te −t u (t ) + e−t u (t ) − e−3t u (t ) dt 2 4 4 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
46. 46. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 46. (a) 5[sI(s) – i(0-)] – 7[s2I(s) – si(0-) – i'(0-)] + 9I(s) = 4 s (b) m[s2P(s) – sp(0-) – p'(0-)] + μf [sP(s) – p(0-)] + kP(s) = 0 (c) [s ΔNp(s) – Δnp(0-)] = − ΔN p (s ) τ + GL s PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
47. 47. Engineering Circuit Analysis, 7th Edition 47. Chapter Fourteen Solutions 10 March 2006 15u (t ) − 4δ(t ) = 8 f (t ) + 6 f ′(t ), f (0) = −3 15 15 − 4s 15 − 4 s ∴ − 4 = 8F( s ) + 6sF( s) + 18 = ∴ F( s) (6 s + 8) = 18 + s s s −22s + 15 15 / 8 ∴ F( s ) = = ∴ f (t ) = (1.875 − 5.542e −4t / 3 ) u (t ) 6s ( s + 4 / 30) s + 4 / 3 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
48. 48. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 48. (a) (b) (c) -5 u(t – 2) + 10 iL(t) + 5 diL = 0 dt − 5 − 2s e + 10 I L (s) + 5 [sI L (s) - iL (0- )] = 0 s 5 − 2s e + 5 iL (0- ) e −2s + 5 × 10-3 s s IL(s) = = 5s + 10 s (s + 2 ) 5 × 10-3 b ⎤ ⎡a IL(s) = e ⎢ + ⎥ + s+2 ⎣s s + 2 ⎦ 1 1 1 = , and b = where a = s s + 2 s=0 2 − 2s s = -2 1 = - , so that we may write 2 1 − 2s ⎡ 1 1 ⎤ 5 × 10-3 IL(s) = e ⎢ − ⎥ + s+2 2 ⎣s s + 2⎦ Thus, iL(t) = = [ ] 1 u (t − 2) − e − 2(t − 2 ) u (t − 2) + 5 × 10-3 e - 2t u (t ) 2 [ ] 1 1 − e − 2 (t − 2 ) u (t − 2) + 5 × 10-3 e- 2t u (t ) 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
49. 49. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 49. (a) vc (0− ) = 50 V, vc (0+ ) = 50 V (b) ′ 0.1 vc + 0.2 vc + 0.1(vc − 20) = 0 (c) ′ ∴ 0.1 vc + 0.3 vc = 2, 0.1sVc − 5 + 0.3Vc = 2 s 2 5s + 2 = s s 5s + 2 20 / 3 130 / 3 ⎛ 20 130 −3t ⎞ e ⎟ u (t ) V ∴ Vc ( s ) = = + ∴ vc (t ) = ⎜ + s (0.1s + 0.3) s s+3 3 ⎝ 3 ⎠ ∴ Vc (0.1s + 0.3) = 5 + PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
50. 50. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 50. (a) (b) (c) 5 u(t) – 5 u(t – 2) + 10 iL(t) + 5 diL = 0 dt 5 5 − e − 2s + 10 I L (s) + 5 [sI L (s) - iL (0- )] = 0 s s 5 −2 s 5 + 5 iL (0- ) e − e −2s + 5 ×10−3s − 1 s = IL(s) = s 5s + 10 s (s + 2) b ⎤ c d ⎡a IL(s) = e − 2s ⎢ + ⎥ + s + s + 2 where ⎣s s + 2 ⎦ 1 1 5 × 10−3 s − 1 1 1 a = = ,b = =- , c = s s = -2 2 s+2 s + 2 s=0 2 −3 d = 5 × 10 s − 1 s = s = −2 = − s =0 1 , and 2 −3 −10 × 10 − 1 = 0.505 , −2 so that we may write 1 1 ⎤ 0.505 ⎡1 ⎛ 1 ⎞1 IL(s) = e −2s ⎢ − ⎟ ⎥ + s+2 − ⎜ 2⎠s 2 ⎣s s + 2⎦ ⎝ Thus, iL(t) = 1 ⎡ 1 −2 ( t − 2 ) u (t − 2) ⎤ + 0.505e-2t u (t ) − u (t ) ⎣u (t − 2) − e ⎦ 2 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
51. 51. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 51. 12 = 20sF2 − 20 (2) + 3F2 s 12 12 + 40 s 2s + 0.6 ∴ + 40 = (20s + 3) F2 = ∴ F2 ( s ) = s s s ( s + 0.15) 4 2 ∴ F2 ( s ) = − ↔ (4 − 2e −0.15t ) u (t ) s s + 0.15 12 u (t ) = 20 f 2′ (t ) + 3 f 2 (0− ) = 2 ∴ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
52. 52. Engineering Circuit Analysis, 7th Edition 52. Chapter Fourteen Solutions 10 March 2006 (a) f(t) = 2 u(t) - 4δ(t) (b) f(t) = cos (c) F(s) = where a = ( 99 t ) 1 - 5 s + 5s + 6 2 1 s−2 Thus, (d) f(t) = δ '(t) = = 1 and s=3 a b + - 5 s−3 s−2 b = 1 s−3 = -1 s=2 f(t) = e-3t u(t) – e-2t u(t) – 5δ(t) (a “doublet”) (e) f(t) = δ'"(t) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
53. 53. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 53. x′ + y = 2u (t ), y′ − 2 x + 3 y = 8u (t ), x(0− ) = 5, y (0− ) = 8 2 8 1⎛ 2 ⎞ 2 5 Y sX − 5 + Y = , sY − 8 − 2X + 3Y = ∴ X = ⎜ + 5 − Y ⎟ = 2 + − s s s s s⎝s ⎠ s 4 10 2Y 8 2 ⎞ 4 18 ⎛ − + = 8+ ∴ Y⎜s + 3+ ⎟ = 2 + +8 2 s s s s s⎠ s s ⎝ ⎛ s 2 + 3s + 2 ⎞ 4 + 18s + 8s 2 0 8s 2 + 18s + 4 2 6 + = = + Y⎜ , Y( s ) + ⎟ 2 s s s ( s + 1) ( s + 2) s s + 1 s + 2 ⎝ ⎠ ∴ sY + 3Y − ∴ y (t ) = (2 + 6e − t ) u (t ); x(t ) = 1 1 [ y′ + 3 y − 8u (t )] = y′ + 1.5 y − 4u (t ) 2 2 1 [−6e −t u (t )] + 1.5 [2 + 6e − t ] u (t ) − 4u (t ) 2 ∴ x(t ) = 6e − t u (t ) − u (t ) = (6e − t − 1) u (t ) ∴ x(t ) = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
54. 54. Engineering Circuit Analysis, 7th Edition 54. Chapter Fourteen Solutions 10 March 2006 8 (a) F(s) = 8s + 8 + , with f(0-) = 0. Thus, we may write: s f(t) = 8 δ(t) + 8 u(t) + 8δ ' (t) (b) F(s) = s2 -s + 2. (s + 2) f(t) = δ ' (t) - 2δ(t) + 4e-2t u(t) - δ ' (t) + 2δ(t) = 4e-2t u(t) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
55. 55. Engineering Circuit Analysis, 7th Edition 55. Chapter Fourteen Solutions 10 March 2006 40 − 100 = −0.6 A 100 (a) ic (0− ) = 0, vc (0) = 100 V, ∴ ic (0+ ) = (b) 40 = 100 ic + 50 ∫ − ic dt + 100 (c) 60 50 = 100 Ic ( s ) + Ic ( s) s s −6 −0.6 6 10 s + 5 ∴ = Ic = ↔ ic (t ) = −0.6e −0.5t u (t ) , Ic ( s) = s s 10 s + 5 s + 0.5 ∞ 0 − PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
56. 56. Engineering Circuit Analysis, 7th Edition 56. (a) 4 cos 100t ↔ Chapter Fourteen Solutions 4s s + 1002 2 (b) 2 sin 103t – 3 cos 100t ↔ 2 × 103 3s - 2 2 6 s + 1002 s + 10 (c) 14 cos 8t - 2 sin 8o ↔ (d) δ(t) + [sin 6t ]u(t) ↔ 1 + (e) cos 5t sin 3t 10 March 2006 14s 2 sin 8o s 2 + 64 s 6 s + 36 2 = ½ sin 8t + ½ sin (-2t) = ½ (sin 8t – sin 2t) ↔ 4 1 - 2 s + 64 s + 4 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
57. 57. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 t 57. is = 100e −5t u (t ) A; is = v′ + 4v + 3∫ − vdt (a) is = (b) 100 3 = sV( s ) + 4V( s ) + V( s ) s+5 s 3⎞ 100s s 2 + 4 s + 3 100 ⎛ V( s ) ⎜ s + 4 + ⎟ = V( s ) = , V( s ) = 4⎠ ( s + 1) ( s + 3) ( s + 5) s s+5 ⎝ 0 v 1 t 1 1 + Cv′ + ∫ − vdt ; R = Ω, C = 1F, L = H R L 0 4 3 ∴ V( s ) = −12.5 75 62.5 , v(t ) = (75e −3t − 12.5e − t − 62.5e−5t ) u (t ) V + − s +1 s + 3 s + 5 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
58. 58. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 58. (a) V(s) = 7 e −2 s + V s s (b) V(s) = e −2s V s +1 (c) V(s) = 48e-s V PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
59. 59. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 59. ∞ 4 u (t ) + ic + 10 ∫ − ic dt + 4 [ic − 0.5δ (t )] = 0 0 4 10 4 2s − 4 ⎛ 10 ⎞ ∴ + Ic + Ic + 4Ic = 2, Ic ⎜ 5 + ⎟ = 2 − + s s s ⎠ s s ⎝ 2s − 4 1.6 ∴ Ic = = 0.4 − 5s + 10 s+2 ∴ ic (t ) + 0.4δ (t ) − 1.6e −2t u (t ) A PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
60. 60. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 60. t v′ + 6v + 9 ∫ − v( z ) dz = 24 (t − 2) u (t − 2), v′(0) = 0 0 9 1 s 2 + 6s + 9 ( s + 3) 2 = V( s ) V( s) = 24e −2 s 2 = V( s ) s s s s ⎡1/ 9 1/ 9 1 s 1/ 3 ⎤ = 24e −2 s ⎢ − − 2 2 s ( s + 3) s + 3 ( s + 3) 2 ⎥ ⎣ s ⎦ ∴sV( s ) − 0 + 6 V( s ) + ∴ V( s ) = 24e −2 s ⎡8 / 3 8 8 ⎤ 8 ∴ V( s) = e −2 s ⎢ − − ↔ [u (t − 2) − e −3( t − 2) u (t − 2)] 2⎥ s + 3 ( s + 3) ⎦ 3 ⎣ s ⎡8 8 ⎤ −8(t − 2) e −3(t − 2) u (t − 2) ∴ v(t ) = ⎢ − e−3( t − 2) − 8(t − 2) e−3(t − 2) ⎥ u (t − 2) ⎣3 3 ⎦ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
61. 61. Engineering Circuit Analysis, 7th Edition 61. Chapter Fourteen Solutions 10 March 2006 (a) All coefficients of the denominator are positive and non-zero, so we may apply the Routh test: 1 13 44.308 35 47 35 0 [(13)(47) – 35]/13 [35(44.308) – 0]/44.308 No sign changes, so STABLE. (b) All coefficients of the denominator are positive and non-zero, so we may apply the Routh test: 1 13 –1.69 1 35 0 [13 – 35]/13 No need to proceed further: we see a sign change, so UNSTABLE. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
62. 62. Engineering Circuit Analysis, 7th Edition 62. Chapter Fourteen Solutions 10 March 2006 (a) All coefficients of the denominator are positive and non-zero, so we may apply the Routh test: 1 3 8 8 0 [(3)(8) – 0]/3 No sign changes, so STABLE. 2 3 3 ⎛3⎞ Verification: roots of D(s) = − ± ⎜ ⎟ − 8 = − ± 2 2 ⎝2⎠ parts, so the function is indeed stable. 1 ⎛ 23 ⎞ 2 j ⎜ ⎟ , which have negative real ⎝ 4 ⎠ (b) All coefficients of the denominator are positive and non-zero, so we may apply the Routh test: 1 2 1 1 0 [(2)(1) – 0]/2 No sign changes, so STABLE. Verification: roots of D(s) = –1, –1, which have negative real parts, so the function is indeed stable. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
63. 63. Engineering Circuit Analysis, 7th Edition 63. Chapter Fourteen Solutions 10 March 2006 (a) All coefficients of the denominator are positive and non-zero, so we may apply the Routh test: 1 3 2 1.5 3 3 1 1 0 [(3)(3) – 3]/3 [6 – 3]/2 No sign changes, so STABLE. (b) All coefficients of the denominator are positive and non-zero, so we may apply the Routh test: 1 3 No sign changes, so STABLE. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
64. 64. Engineering Circuit Analysis, 7th Edition 64. Chapter Fourteen Solutions 10 March 2006 (a) v(t ) = 7u (t ) + 8e −3t u (t ) Therefore V (s) = 7 8 15s + 21 . + = s s + 3 s(s + 3) 21 15s + 21 s = 15 V = lim lim sV (s) = lim s →∞ s →∞ s →∞ 3 s+3 1+ s 15 + (b) v(0) = 7 + 8 = 15 V (verified) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
65. 65. Engineering Circuit Analysis, 7th Edition 65. Chapter Fourteen Solutions 10 March 2006 (a) v(t ) = 7u (t ) + 8e −3t u (t ) Therefore V (s) = 7 8 15s + 21 . + = s s + 3 s(s + 3) 15s + 21 = 7V s →0 s+3 lim sV (s) = lim s→0 (b) v(∞) = 7 + 0 = 7 V (verified) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
66. 66. Engineering Circuit Analysis, 7th Edition 66. (a) Chapter Fourteen Solutions 10 March 2006 lim 5s ( s 2 + 1) 5( s 2 + 1) + F( s) = 3 ∴ f (0 ) = =5 s →∞ ( s + 1) s3 + 1 5s ( s 2 + 1) , but 1 pole in RHP ∴ indeterminate f (∞ ) = s →0 s3 + 1 lim lim 5s ( s 2 + 1) 5( s 2 + 1) ∴ f (0+ ) = =0 s→∞ s 4 + 16 s 3 + 16 f (∞) is indeterminate since poles on jω axis (b) F( s) = (c) F( s) = lim s ( s + 1) (1 + e −4 s ) ( s + 1) (1 + e−4 s ) ∴ f (0+ ) = =1 s →∞ s2 + 2 s2 + 2 f (∞) is indeterminate since poles on jω axis PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
67. 67. Engineering Circuit Analysis, 7th Edition 67. Chapter Fourteen Solutions 10 March 2006 ⎛ 2s 2 + 6 ⎞ ⎟ = 2 (a) f(0+) = lim[s F(s)] = lim⎜ 2 s →∞ s → ∞ ⎜ s + 5s + 2 ⎟ ⎝ ⎠ ⎛ 2s 2 + 6 ⎞ 6 f(∞) = lim[s F(s)] = lim⎜ 2 ⎜ s + 5s + 2 ⎟ = 2 = 3 ⎟ s →0 s →0 ⎝ ⎠ ⎛ 2se − s ⎞ ⎟ = 0 (b) f(0+) = lim[s F(s)] = lim⎜ s →∞ s → ∞⎜ s + 3 ⎟ ⎝ ⎠ ⎛ 2se − s ⎞ ⎟ = 0 f(∞) = lim[s F(s)] = lim⎜ s →0 s →0⎜ s + 3 ⎟ ⎝ ⎠ ⎡ s(s 2 + 1)⎤ (c) f(0+) = lim[s F(s)] = lim ⎢ 2 ⎥ = ∞ s →∞ s →∞ ⎣ s +5 ⎦ f(∞) : This function has poles on the jω axis, so we may not apply the final value theorem to determine f(∞). PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
68. 68. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 68. (a) lim 5s ( s 2 + 1) 5( s 2 + 1) ∴ f (0+ ) = =5 s →∞ ( s + 1)3 ( s + 1)3 F( s ) = ⎡ 5(s 2 + 1) ⎤ f (∞) = lim ⎢s = 0 (pole OK) s→0 ⎣ (s + 1)3 ⎥ ⎦ (b) F( s ) = 5( s 2 + 1) 5( s 2 + 1) ∴ f (0+ ) = lim =0 s→∞ ( s + 1)3 s ( s + 1)3 5( s 2 + 1) = 5 (pole OK) s→0 ( s + 1)3 f (∞) = lim (c) (1 − e − 3 s ) 1 − e −3s ∴ f (0 + ) = l im =0 s →∞ s2 s ⎡ 1 − e −3s ⎤ f ( ∞ ) = l im ⎢ = (using L'Hospital's rule) l im ( 3e − 3 s ) = 3 s→ 0 ⎣ s→ 0 s ⎥ ⎦ F(s ) = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
69. 69. Engineering Circuit Analysis, 7th Edition Chapter Fourteen Solutions 10 March 2006 69. 1 f (t ) = (eat − e− bt ) u (t ) t (a) Now, 1 f (t ) ↔ t ∫ ∞ s F( s) ds ∴ e− at u (t ) ↔ 1 ∴ (e− at − e− bt u (t ) ↔ t (b) ∫ ∞ s 1 1 , − e− bt u (t ) ↔ − s+a s+b s+a 1 ⎞ ⎛ 1 − ⎜ ⎟ ds = ln s+b ⎝ s+a s+b⎠ ∞ s s+a = ln s+b ∞ = ln s s+b s+a lim 1 − at + ... − 1 + bt 1 − at −bt (e − e ) u (t ) = =b−a t →0+ t t →0+ t lim s + b lim ln ( s + b) − ln ( s + a) sln = s→∞ s + a s→∞ 1/ s lim 1/( s + b) − 1/( s + a) lim ⎡ 2 ( a − b) ⎤ Use l′ Hospital. ∴ sF( s) = = −s =b−a 2 s→∞ s→∞ ⎢ −1/ s ( s + b) ( s + a ) ⎥ ⎣ ⎦ lim PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
70. 70. Engineering Circuit Analysis, 7th Edition 70. (a) F( s) = (b) F( s) = 10 March 2006 F( s) = (c) Chapter Fourteen Solutions lim s (8s − 2) 8s − 2 ∴ f (0+ ) = =8 s →∞ s 2 + 6 s + 10 s + 6s + 10 lim s (8s − 2) ⎛ ⎞ −6 ± 36 − 40 , LHP, ∴ OK ⎟ f (∞ ) = = 0 ⎜ poles: s = 2 ⎜ ⎟ s →0 s + 6 s + 10 2 ⎝ ⎠ 2 lim 2 s 3 − s 2 − 3s − 5 2 s 3 − s 2 − 3s − 5 ∴ f (0+ ) = =∞ s →∞ s 3 + 6 s 2 + 10s s 2 + 6 s + 10 lim 2 s 3 − s 2 − 3s − 5 f (∞ ) = = −0.5 (poles OK) s →0 s 2 + 6 s + 10 lim s (8s − 2) 8s − 2 ∴ f (0+ ) = =8 s →∞ s 2 − 6 s + 10 s − 6 s + 10 f (∞ ) = (d) 2 s (8s − 2) 6 ± 36 − 40 , s= RHP ∴ indeterminate s →0 s − 6 s + 10 2 lim 2 8s 2 − 2 F(s) = ∴ f (0+ ) = lim sF(s) = 0 2 2 s→∞ (s + 2) (s + 1) (s + 6s + 10) s (8s 2 − 2) f ( ∞ ) = lim = 0 (pole OK) s→0 (s + 2) 2 (s + 1) (s 2 + 6s + 10) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.