Chapter 14 solutions_to_exercises(engineering circuit analysis 7th)

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.


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



3.


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



4.


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



5.


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.



6.


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.




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



8.


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



9.


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



10.

(


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



11.


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



12.


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.




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



14.


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.



15.

10 March 2006

iS 1 = 20e −3t cos 4t A, iS 2 = 30e−3t sin 4t A

(a)


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



16.


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.



17.

L {K u (t )} =


∞

∞

∞

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



18.

(a) L {3 u (t )} =


∞

∞

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
⎠

t →∞

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

t →∞

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-

)



19.

(a) L {2 + 3 u (t )} =


− 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
⎠

t →∞


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

t →∞

L {K } =

K
s



20.


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)



21.


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
⎠

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

(

)




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




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 ⎣



24.


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



25.


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)




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



27.
(a)


∞

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



28.

(a)
(b)
(c)
(d)

∫

∞

−∞

∞

8 sin 5t δ (t − 1) dt

∫ (t − 5) δ (t − 2)dt
2

−∞

∫

∞

∫

∞

−∞

−∞


= 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



29.
(a)


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



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


10 March 2006

cos 500 × 0 = 1
= 32
= 2.5e − 0.001(1000 ) = 0.9197

= - K2




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




(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
⎦




(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
⎦



34.


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)



35.


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) =



36.


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



37.


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)



38.


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];
r=
-7.1429
7.5000
-0.3571

EDU» N = [5 0];
EDU» D = [32 120 76 12];
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

[]

[]

[]




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)



40.


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];

Part (b):

Part (c):

EDU» N = [2 0 0];
EDU» D = [1 22 147 270];
r=
28.5714
-27.0000
0.4286

EDU» N = [2 0 0];
EDU» D = [128 288 160 24];
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

[]

[]




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 ∴



42.

F(s)


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)



43.

(a) F(s) =

(s + 1)(s + 2)
s(s + 3)

(s + 1)(s + 2)
(s + 3) s = 0

a =

=

=


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)



44.

(a)


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



45.

(a) H(s) =

(b) H(s) =


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




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



47.


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




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




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 +




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




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 ∴



52.


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)




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 ) =



54.


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)



55.


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

−



56.

(a) 4 cos 100t ↔


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




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




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




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




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
⎦



61.


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.



62.


10 March 2006

Routh test:
1
3
8

8
0
[(3)(8) – 0]/3

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 ⎠

Routh test:
1
2
1

1
0
[(2)(1) – 0]/2

Verification: roots of D(s) = –1, –1, which have negative real parts, so the function is
indeed stable.



63.


10 March 2006

Routh test:
1
3
2
1.5

3
3
1

1
0
[(3)(3) – 3]/3
[6 – 3]/2

Routh test:
1
3



64.


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)



65.


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)



66.
(a)


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



67.


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(∞).




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 ) =




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



70.
(a)

F( s) =

(b)

F( s) =

10 March 2006

F( s) =

(c)


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)


Chapter 14 solutions_to_exercises(engineering circuit analysis 7th)

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Chapter 14 solutions_to_exercises(engineering circuit analysis 7th)

Similar to Chapter 14 solutions_to_exercises(engineering circuit analysis 7th) (20)

Recently uploaded

Recently uploaded (20)

Chapter 14 solutions_to_exercises(engineering circuit analysis 7th)