ENSC 320
Problem of the Week 1 March, 2004
De Carlo and Lin, Problem 17-60
This amplifier circuit contains a real-world coil, one with
resistance. The combination of 1 mH ideal inductor and 0.1
µF capacitor puts the resonant frequency near ω = 105 rad/s.
(a) Find the Q of the coil (not of the resonant circuit) at ω =
105 rad/s.
(b) Represent the coil by a parallel RL circuit that is valid
for frequencies near ω = 105 rad/s.
(c) Obtain the transfer function H(s) = Vout(s)/Vin(s).
(d) From the transfer function, obtain the resonant
frequency, the bandwidth and the Q of the circuit.
Solution
(a) The Q of the coil is the ratio of wanted part of the
impedance to the unwanted part.
5
ω := 10 sec
−1
ω⋅ L
QL :=
RL
L := 1 ⋅ mH
RL := 2 ⋅ Ω
QL = 50
(b) To convert coil to a parallel model, use page 4.4.25 of the
class notes or Table 17.2 of the text.
Gp :=
RL
RL + ( ω ⋅ L )
2
Gp = 1.999 × 10
−4
1
Gp
2
Rp :=
S
Rp = 5.002 × 10 Ω
3
Note DC & L solution for Rp is incorrect.
Bp :=
−ω⋅ L
RL + ( ω ⋅ L )
2
Bp = −9.996 × 10
Lp :=
2
−3
S
1
−ω⋅ Bp
Lp = 1 × 10
−3
H
(c) To obtain the H(s), first represent the impedance of the
parallel combination following the amp as
Zpar =
1
Ypar
=
1
1
1
−6
+
+ s ⋅ 0.1 ⋅ 10
2500 s⋅ 10 − 3
7
=
10 ⋅ s
2
s + 4000 ⋅ s + 10
10
Then
4
1
−4 ⋅ 10 ⋅ s
H ( s) =
⋅ ( −10) ⋅ Zpar =
2
10
500 + 2000
s + 4000 ⋅ s + 10
From this, we get
5 rad
ωo := 10 ⋅
sec
rad
β := 4000 ⋅
sec
Q :=
ωo
β
Q = 25
** circuit file for profile: bandwidth
Date/Time run: 03/01/04 20:59:21
Temperature: 27.0
(A) pow-mar-1-SCHEMATIC1-bandwidth.dat (active)
20
0
-20
SEL>>
-40
DB(V(Vout)/V(Vin))
-0d
-100d
-200d
-300d
1.0KHz
3.0KHz
P(V(Vout)/V(Vin))
10KHz
30KHz
100KHz
300KHz
1.0MHz
Frequency
A1:(15.849K,-168.765)
Date: March 01, 2004
A2:(1.0000K,-107.801)
DIFF(A):(14.849K,-60.964)
Page 1
Time: 21:50:09