ECE 453: Homework Assignment 5 Solution
413 Solution
At 10 MHz, the reactance of the capacitors is
−
j
5
Ω
.
The branch consisting of
C
1
in parallel with
R
= 50
Ω
has
Q
= 50
/
5 = 10
, and transforms into
−
j
5
in series with
50
/
10
2
= 0
.
5
Ω
, as shown in Figure a.
jX
L
−
j
5
Ω
−
j
5
Ω
0
.
5
Ω
The two capacitors can be combined in series to yield a branch consisting of
−
j
10
Ω
in series with
0
.
5
Ω
.
This branch transforms into
−
j
10
Ω
in parallel with
200
Ω
as shown in Figure b.
jX
L
−
j
10
200
Ω
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
= 0
.
253
pF.
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 Spring '08
 Staff
 RLC, LC circuit, center frequency

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