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.
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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
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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
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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
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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.
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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.
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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
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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
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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
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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
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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
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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.
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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
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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.
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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
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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.
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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
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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-
)
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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)
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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
(
)
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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
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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 ⎣
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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
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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)
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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. 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
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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
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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
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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
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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
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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
⎦
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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
⎦
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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)
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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
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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. 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)
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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. 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)
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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
[]
[]
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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 ∴
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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)
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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)
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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
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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
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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
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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
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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
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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 +
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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
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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 ∴
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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)
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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 ) =
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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)
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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
−
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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
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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
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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
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and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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
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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
⎦
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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.
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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.
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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. 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. 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. 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. 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. 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. 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. 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.