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Chapter 16: Problem 14, 24, 26
Chapter 14: Problem 35
Chapter 16, Problem 14. (compares exact and approx. values of impedance)
Let ω 0 = 30 krad/s, Q 0 = 10, and R = 600 Ω for a certain parallel resonant circuit. ( a ) Find the
bandwidth. ( b ) Calculate N at ω = 28 krad/s. ( c ) Use approximate methods to determine
Z in ( j 28,000). ( d ) Find the true value of Z in ( j 28,000). ( e ) State the percentage error incurred by
using the approximate relationships to calculate | Z in | and ang Z in at 28 krad/s.
ω o = 30 krad/s, Q o = 10, R = 600 Ω,
(a) B 3 krad/s
Q
o
o
(b)
N 1.
B / 2 1.
ω − ω o −
(c) Z in ( j 28 000) = 600 / (1 – j 1.333) = 360 ∠ 53.
o Ω
(d)
(e)
(As expected: The error should be less than 5% for Q 0 ≥ 5 and 0.9ω 0 ≤ ω ≤ 1.1ω 0 .)
1
1
1 1 Q 10
Z ( 28, 000) 28, 000C , C
600 28, 000L R 30, 000 600
R 600 1 30, 000 10 1 28 10 30 10
L , Z
Q 30, 000 10 L 600 600 30 600 28 600
Z 351.906 54.
−
−
×
×
= = = ∴ = + × −
×
o in o
in o o
in
j j j
j
j
ω
ω
approx-true 360 351. magnitude: 100% 100% 2.300% true 351.
angle: 100% 1.7752%
Chapter 16, Problem 24. (high quality circuits produce high voltages/currents)
Inspect the circuit of Fig. 16.51, noting the amplitude of the source voltage. Now decide whether
you would be willing to put your bare hands across the capacitor if the circuit were actually built
in the lab.
Find the Thévenin equivalent seen by the inductor-capacitor combination:
1 1 1 1
1
6
max
V
SC : 1.5 V 10 0.105 V V 50 V
I 0.4 A
OC :V 0 V 1.5 V R 3.
1/ 4 0.25 10 1000, Q 1066.
B / Q 0.9375, B 0.4688 rad/s 1066.7 2
V Q V 1066.7 1.5 1600 V
−
×
∴ = × × = = =
= = × =
SC
OC th
o o
o o
C o th
ω
ω
Therefore, keep your hands off!
(To generate a plot of | V C | vs. frequency, note that V C ( j ω) =
C
j j L
C
j
Chapter 14, Problem 35. (focus on bringing results from s-domain back to t-domain)
Obtain partial-fraction expansions for each of the following rational functions, and then
determine the corresponding time functions:
( a ) F ( s ) = [( s +1) ( s +2) ]/[ s ( s +3) ]; ( b ) F ( s ) = ( s +2) /[ s
2 ( s 2+4 )].
(a) F ( s ) = ( 3 ) ( 3 )
ss s s
s s a b
0
s =
s
s s a and 3
s s =
s s b
so
f ( t ) = δ(t) + ( 1 ) ()
u t e ut e ut
− t − t = − + δ(t)
(b) F ( s ) = ( 2 )
2 2 2 j
c
j
a b c
s s s s s s
s
0
2
s =
s
s a
2 0
2 2
2
0
2
s = s =
s
s ss
s
s
d s
d b
c = (s+2) / s
2 (s-j2) | (^) s = -j2 = (2 - j2) / -4(-j4) = -0.125 - j 0.125 = -0.1768 ∠ 45 °
and c
= -0.1768 ∠-45°
so
f ( t ) = 0.5 t u ( t ) + 0.25 u ( t ) − 0.1768 e
j45o e
-j45o e
j 2 t u ( t )
The last two terms may be combined so that
f(t) = 0.5 t u(t) + 0.25 u(t) − 0.25 cos(2t)u(t) − 0.25 sin(2t)u(t)
or f ( t ) = 0.5 t u ( t ) + 0.25 u ( t ) − 0.3536 cos (2 t - 45
o )
