Problem set #9, EE 223, 4/01/2003 – 4/08/2003
Chapter 16, Problem 1.
A parallel RLC circuit has
R=1 kΩ, C =47 µF, and L=11 mH.
(a) Compute Q0.
(b) Determine the resonant frequency
(in Hz).
(c) Sketch the voltage response as a
function of frequency if the circuit is
excited by a steady-state 1-mA
sinusoidal current source.
We have a parallel RLC with R = 1 kΩ, C = 47 µF and L = 11 mH.
(a) Qo = R(C/L)½ =
65.37
(b) fo = ωo/ 2π = (LC)-½ / 2π =
221.3 Hz
(c) The circuit is excited by a steady-state 1-mA sinusoidal source:
The admittance Y(s) facing the source is Y(s) = 1/R + 1/sL + sC
= C(s2 + s/RC + 1/LC)/ s so Z(s) = (s/C) / (s2 + s/RC + 1/LC) and
Z(jω) = (1/C) (jω) / (1/LC – ω2 + jω/RC).
Since V = 10-3 Z, we note that |V| > 0 as ω → 0 and also as ω → ∞.
Bandwidth B = ω0 / Q0 = 21.27 s-1 (≅3.4 Hz) ⇒ f1 = 219.6 Hz and f2 = 223.0 Hz
Chapter 16, Problem 2.
A parallel RLC circuit is measured to have a Q0 of 200. Determine the remaining
component value if (a)R=1Ω and C=1 µF; (b)L=12 fH and C=2.4nF; (c)R
=121.7 kΩ and L=100 pH.
(a) R = 1 Ω and C = 1 µF.
Qo = R(C/L)½ = 200 so L = C(R/ Qo)2 =
(b) L = 12 fH and C = 2.4 nF
R = Qo (L/ C)½
=
(c) R = 121.7 kΩ and L = 100 pH
C = (Qo / R)2 L =
447.2 mΩ
270 aF
25 pH
Chapter 16, Problem 5.
A parellel resonant circuit has parameter values of α =80 Np/s and ωd=1200rad/s. If the impedance at s
=-2α + jωd has a magnitude of 400Ω , calculate Q0, R , L , and C .
α = 80Np/s, ω d = 1200 rad/s, Z(−2α + jω d ) = 400 Ω
ω o = 12002 + 802 = 1202.66 rad/s ∴ Qo =
Now, Y( s ) = C
( s + α − jω d )( s + α + jω d )
(−α )(−α + j 2ω d )
∴ Y(−2α + jω d ) = C
s
−2α + jω d
∴ Y(−160 + j1200) = C
∴C =
ωo
= 7.517
2α
−80(−80 + j 2400)
1
−1 + j 30
∴ Y(−160 + j1200) =
= 80C
−160 + j1200
400
−2 + j15
1
229
1
1
= 15.775− µ F; L = 2 = 43.88 mH; R =
= 396.7 Ω
32, 000 901
ωo C
2α C
Chapter 16, Problem 7.
Let R=1 MΩ, L=1 H,C=1 µF, and I =10
µA in the circuit of
Fig. 16.1. (Parallel Resonant Circuit)
(a)Find ω0 and Q0. (b) Plot |V | as a function of ω, 995 <ω<1005 rad/s.
Parallel: R = 106 , L = 1, C = 10−6 , Is = 10∠0° µ A
1
= 1000 rad/s; Qo = ω o RC = 103+ 6−6 = 1000
LC
(a)
ωo =
(b)
V = I / Y = 10-5 / (10-6 + j (10-6ω - 1/ω)) = 10-5 / (10-3(10-3 + j (10-3ω - 1000/ω))
|V| = 10-2 / √(10-6 + (10-3ω - 1000/ω)2)
B = ω0 / Q0 = 1 s-1 ⇒ ω1 = 999.5 s-1 and ω2 = 1000.5 s-1
ω
V
995
996
997
998
999
999.5
1000
1000.5
1001
1002
1003
1004
1005
0.993
1.238
1.642
2.423
4.47
7.070
10.0
7.072
4.47
2.428
1.646
1.243
0.997