Page 80 - Circuit Analysis II with MATLAB Applications

Chapter 2 Resonance

Next, we want to find the bandwidth Z – Z 1 in terms of the quality factor Q 0P . At the half-power
2
points, the magnitude of the admittance is 22 Y p and, if we use the half-power points as refer-

e
ence, then to obtain the admittance value of

|Y max = 2G
we must set

------ –
------
Q 0P § © Z 2 Z 0 · = 1
Z
0 Z ¹ 2
for Z Z = 2 .
We must also set

Q 0P § © Z 1 0 Z 0 · 1 = – 1
------
------ –
Z ¹
Z
for Z Z = 1 .
Recalling that 1 r j1 = 2 and solving the above expressions for Z 1 and Z 2 , we get

1
1
Z = 1 + § © ------------- · ¹ 2 + ------------- (2.31)
2
2Q
2Q
0P
0P
and
1
1
Z = 1 + § © ------------- · ¹ 2 – ------------- (2.32)
1
2Q
2Q
0P
0P
Subtraction of (2.32) from (2.31) yields
Z
0
BW = Z – Z = --------- (2.33)
1
2
Q
0P
or
f
0
BW = f – f = --------- (2.34)
1
2
Q
0P
As mentioned earlier, Z and Z are not equidistant from Z In fact, the resonant frequency Z is
1 2 0 0
*
the geometric mean of Z 1 and Z 2 , that is,
Z = Z Z (2.35)
0 1 2
* The geometric mean of n positive numbers a 1 , a 2 ,..., a n is the nth root of the product.a ˜ 1 a ˜ 2 } a n
˜

2-14 Circuit Analysis II with MATLAB Applications

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