ECE 453: Homework Assignment 5 Solution
413 Solution
At 10 MHz, the reactance of the capacitors is
−
j
5Ω
.
The branch consisting of
C
1
in parallel with
R
= 50 Ω
has
Q
= 50
/
5 = 10
,andtrans
formsinto
−
j
5
in series with
50
/
10
2
=0
.
, as shown in Figure a.
jX
L
−
j
5Ω
−
j
5Ω
0
.
5Ω
The two capacitors can be combined in series to yield a branch consisting of
−
j
10 Ω
in series with
0
.
.
This branch transforms into
−
j
10 Ω
in parallel with
200 Ω
as shown in Figure b.
L
−
j
10
200Ω
a. The inductor needs to have impedance
+
j
10 Ω
to resonate with the parallel capacitance.
b. At resonance, the impedance
Z
= 200 Ω
.
c.
Q
p
= 200
/
10 = 20
.
420 Solution
If a parallel LC circuit is used, then
Q
=
Q
p
= 100

100
/
(
ω
o
L
)
.
If a series LC circuit is used, then
Q
=
Q
s
=
ω
o
L/
(100 + 100)
.
In part a, the largest Q results when a parallel LC is used with the smallest possible
inductance. Answer for part a: Parallel LC with
L
= 10
nH,
C
= 253
.
3
pF. In part b, the largest Q reults
when a series LC is used with the largest possible inductance. Answer for part b: Series LC with
L
= 100
nH,
C
.
253
pF.
421 Solution
a, For the series RLC case with
R
S
=
R
L
=
R
, the loaded Q of the system will be
Q
o
=
X
o
/
(2
R
)
.F
o
rth
e
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Staff
 RLC, LC circuit, center frequency

Click to edit the document details