Chapter 6
BROADBAND
IMPEDANCE-MATCHING NETWORKS
6.1 NETWORK MODEL FOR IMPEDANCE MATCHING
As a two-port model an impedance-matching network (IMN) between a real source
(ZS = RS ) and a real load (ZL = RL) is represented in Figure 6.1, where
Z1 = R1 (ω) + jX1 (ω)
(6.1)
Z2 = R2 (ω) + jX2 (ω)
(6.2)
at the input, and
at the output. From the input side of the network, the load RL is contained as part of
Z1 , and the real part of Z1 may be taken as the impedance to which power actually
is delivered from the source. Hence, from the input side, the impedance-matching
network may be represented as shown in Figure 6.2, where all impedances have
been divided by R1 (ω) to normalize with respect to the real part of Z1 . For a given
frequency ω0 , an impedance match is achieved when the source impedance RS is
equal to the input impedance of the matching network R1 . This requires that
X1 (ω0 ) = 0
(6.3)
R1 (ω0 ) = RS
(6.4)
and
or
z1 =
Z1 (ω0 )
=R=1
R1 (ω0 )
173
(6.5)
174
MICROWAVE TRANSMISSION LINE CIRCUITS
ZS = RS
VS
IMN
ZL = RL
Z 2 = R2 (ω ) + jX 2 (ω )
Z1 = R1 (ω ) + jX 1 (ω )
Figure 6.1 Two-port model of an impedance-matching network showing the impedance at the input
(Z1 ) and at the output (Z2 ) of the network.
RS
=R
R1 (ω )
j
X 1 (ω )
R1 (ω )
RL (ω )
=1
R1 (ω )
VS
Z1
X (ω )
= z1 = 1 + j 1
R1 (ω )
R1 (ω )
Figure 6.2 Model of the impedance-matching network from port 1, the input side. The matching
network is chosen such that R1 (ω) = RL (ω).
The frequency-selective network represented by jX1 (ω)/R1 (ω) in Figure 6.2
can be composed of lumped or distributed elements, or any combination of the
two, which preferably are lossless and hence purely reactive. There can be few or
many such elements, which then can be grouped in sections. Since the network is
terminated in unity source and load resistances, the sections within the network
may be designed by using the Q-tapering techniques presented in Chapter 5.
Thus, frequency-selective sections within the impedance-matching network may be
related to lowpass prototype elements in such a way that a specified response versus
frequency is achieved along with the desired impedance match at f = f0 .
The method of Q-tapering is very useful for designing impedance-matching
networks, and since R = 1 at f = f0 in Figure 6.2, the Q of the k th section is, as
in Chapter 5, related to the total Q by
Qk =
gk
QT
2
(6.6)
BROADBAND IMPEDANCE-MATCHING NETWORKS
175
where gk is a lowpass prototype element chosen to yield a desired frequency
response. The exact design procedure will be made clear through example solutions.
However, for later use, we need to determine the frequency selectivity (Qk ) of an
impedance-transforming section of transmission line.
6.2 THE Q OF λ/4 AND λ/2 TRANSFORMER SECTIONS
The impedance-matching network in Figure 6.1 can be as simple as a length of
transmission line. Recalling from Chapter 2 the expression for the input impedance
of a loaded transmission line, we can write the input impedance of the transformation section between the load RL and the source RS in Figure 6.3 as
Z1 = Z0t
RL + jZ0t tan θt
Z0t + jRL tan θt
At resonant frequencies we require an impedance match such that:
2
Z0t /RL for θt = π2 (2p − 1)
Z1 = RS =
RL
for θt = pπ
(6.7)
(6.8)
where p is a positive integer. Note that for p = 1, the upper equation represents a
quarter-wave line and the lower equation represents a half-wave line.
Solving for the real and imaginary parts of (6.7) yields,
Z1 =
2
2
Z0t
RL(1 + tan2 θt ) + jZ0t (Z0t
− R2L ) tan θt
= R1 + jX1
2 + R2 tan2 θ
Z0t
t
L
The frequency selectivity or Q of the circuit in Figure 6.3 at ω = ω0 is,
ω ∂X1
Q=
R ∂ω ω=ω0
where R = RS + R1 . Carrying out the operation indicated yields,
θt Z0t
RL π
Q=
−
, θt = p
4 RL
Z0t 2
(6.9)
(6.10)
(6.11)
Note that (6.11) is valid for both quarter-wavelength and half-wavelength transformers of integer length p. For the shortest quarter-wavelength transformer (p =
176
MICROWAVE TRANSMISSION LINE CIRCUITS
1, θt = π/2):
π
Q=
8
r
r
Z0t
π RS
π RS
R
Z
R
L
0t
L
RL − Z0t = 8 Z0t − RS = 8 RL − RS (6.12)
2
where Z0t
/RL = RS , or Z0t /RL = RS /Z0t , has been used to express the Q of the
λ/4 transformer in the three equivalent ways.
Applying the Q-tapering concept to the λ/4 transformer/shorted-stub impedance-matching network considered earlier,
√ as shown in Figure 6.4, a Butterworth
response requires that Q1 = Q2 = QT / 2. Thus,
Q1 =
π RS
1
= √ QT
8 Z01
2
(6.13)
and
π
Q2 =
8
or, equating Q1 and Q2 ,
Z02
RS 1
RS − Z02 = √2 QT
Z02
RS RS
RS − Z02 − Z01 = 0
(6.14)
(6.15)
Also, an impedance match at θ = θ0 = π/2 requires,
2
RS = Z02
/RL
(6.16)
From (6.15) and (6.16),
Z01 =
RS p
RS RL
RL − RS
(6.17)
and
Z02 =
p
RS RL
(6.18)
These solutions for Z01 and Z02 yield a Butterworth response for any RS
and RL , and as shown in an earlier example, the shorted stub serves to broaden the
impedance match over what it would be with the transformer alone.
BROADBAND IMPEDANCE-MATCHING NETWORKS
RS
VS
177
θt
Z0t
RL
Ζ1
Figure 6.3 Impedance-transforming section of transmission line.
RS
Ζ1
θ
RL>RS
θ
01
Z02
Z
VS
Figure 6.4 Impedance-matching network using a shorted stub and λ/4 transformer.
6.3 MULTIPLE QUARTER-WAVELENGTH TRANSFORMERS IN
CASCADE
To broaden the frequency band of an impedance match and also achieve a specified
response versus frequency, multiple quarter-wavelength transformers may be used.
At the center-band frequency (f = f0 ) the impedance at each junction of the
cascaded transformer is as shown in Figure 6.5, where Z1 = RS for the impedance
match.
Using the impedances in Figure 6.5, the Q of each cascaded section is
178
MICROWAVE TRANSMISSION LINE CIRCUITS
RS
VS
Z1 =
Z 012
Z2
λ0/4
λ0/4
λ0/4
Z01
Z02
Z0n
Z2 =
Z 022
Z3
Z3
Zn =
RL
Z 02n
RL
Figure 6.5 Impedance-matching network using cascaded λ/4 transformers.
Q1
=
Q2
.
.
.
=
=
=
=
Qn
=
π
8
RS
π Z01
=
Z01 − ZR01
8 Z2 −
2 S
π Z01 /RS
02
− Z 2Z/R
8 Z02
S
01
.
.
.
RL π Z0n
8 RL − Z0n Z2 Z01 (6.19)
The design proceeds by setting Z1 = RS for a match at f = f0 , and tapering
Q1 , Q2 , ..., Qn as
gk
Qk = QT , k = 1, 2, ..., n
(6.20)
2
for a desired frequency response using any number of sections. Note that each
section will have a different input impedance and so there will be multiple internal
reflections within the network. Rigorous mathematical analysis shows that these
internal reflections actually cancel at the input, yielding the desired impedance
match.
6.3.1 Two Cascaded Sections
Two cascaded quarter-wavelength line sections between RS and RL are shown in
Figure 6.6, and Q1 and Q2 are
π
Q1 =
8
RS
Z01 π Z01
Z2 Z01 − RS = 8 Z2 − Z01 (6.21)
BROADBAND IMPEDANCE-MATCHING NETWORKS
RS
VS
λ0/4
λ0/4
Z01
Z02
Z2 =
2
179
RL
Z 022
RL
Z2  Z 
Z1 = 01 =  01  RL = RS
Z 2  Z 02 
Figure 6.6 Two cascaded λ/4 transformers as an impedance-matching network.
and
Z02
RL −
(6.22)
RL
Z02 √
√
If a Butterworth response is required then (g1 = 2, g2 = 2), and Q1 = Q2 =
√1 QT . Equating (6.21) and (6.22) yields
2
Q2 =
π
8
RS
Z01 Z02
RL −
=
−
Z01
RS RL
Z02 (6.23)
From Figure 6.6 the input impedance is
Z2
Z1 = 01 =
Z2
Z01
Z02
2
RL = RS
(6.24)
Solving (6.23) and (6.24) for Z01 and Z02 :
3
1
Z01 = RS4 RL4
and
1
(6.25)
3
Z02 = RS4 RL4
(6.26)
Thus, if RS = 300 Ω and RL = 50 Ω, then Z01 = 191.68√
Ω and Z02 = 78.25 Ω.
If RS = 200
Ω
and
R
=
50
Ω,
then
Z
=
100
2 = 141.42 Ω and
L
01
√
Z02 = 100/ 2 = 70.71 Ω.
180
MICROWAVE TRANSMISSION LINE CIRCUITS
6.3.2 Three Cascaded Sections
With n = 3, or three cascaded λ/4 transformers, the input impedance for a match
at f = f0 is,
2 2
Z01
Z03
Z1 = RS =
RL
(6.27)
Z02
RL
as indicated in Figure 6.7. Again, if a Butterworth response is required (Q1 =
QT /2, Q2 = QT , Q3 = QT /2), the section Qs are tapered accordingly, as:
π RS
Z01 1
Q1 = −
= QT
(6.28)
8 Z01
RS 2
2
π Z02 RL
Z03
= QT
Q2 = −
(6.29)
2
8 Z03
Z02 RL π Z03
RL 1
Q3 = −
= QT
(6.30)
8 RL
Z03 2
Using (6.27) through (6.30) yields:
r
2 r
2
RS Z01
RL RS
Z01
RS
−
+2
−
=0
RL RS
RS Z01
RS
Z01
p
Z02 = RS RL
(6.31)
(6.32)
and
RS RL
(6.33)
Z01
For given values of RS and RL, Z01 , Z02 , and Z03 are determined from (6.31),
(6.32), and (6.33). Clearly, the solutions to (6.32) and (6.33) for Z02 and Z03 follow
quite easily once (6.31) is solved for Z01 . Finding Z01 from (6.31) can be done
numerically or by trial and error. For example, if RS = 75 Ω and RL = 300 Ω:
Z01 = 89.25 Ω, Z02 = 150 Ω, and Z03 = 252 Ω. Also, if RS = 50 Ω and
RL = 200 Ω: Z01 = 59.5 Ω, Z02 = 100 Ω, and Z03 = 168 Ω.
Z03 =
6.4 MORE COMPACT IMPEDANCE-MATCHING NETWORKS
As an impedance-matching element, a single λ/4 transformer generally is broader
in bandwidth than the stub tuners and the two-element (LC) matching networks
181
BROADBAND IMPEDANCE-MATCHING NETWORKS
RS
VS
2
λ0/4
λ0/4
λ0/4
Z01
Z02
Z03
RL
2
Z
Z
Z1 =  01   03  RL = RS
 Z 02   RL 
Figure 6.7 Three cascaded λ/4 transformers as an impedance-matching network.
considered earlier. Multiple λ/4 transformers extend the impedance match over
even greater bandwidths. However, cascaded λ/4 transformers, or even a single
λ/4 transformer, may become excessively long unless the operating frequency is
quite high. We might ask: Can we simulate the properties of the λ/4 transformer by
using more compact structures? To find the answer, we will compare the ABCD
matrix of the λ/4 transformer with the ABCD matrices of candidate equivalent
circuits consisting of lumped elements and transmission line segments.
6.4.1 Lumped-Element Equivalent of the Quarter-Wavelength Transformer
√
With θt = π2 ff0 and Z0t = RS RL, the ABCD matrix of the λ/4 transformer in
Figure 6.8(a) is
A B
cos θt
jZ0t sin θt
=
(6.34)
C D
jY0t sin θt
cos θt
and the ABCD matrix of the LC network in Figure 6.8(b) is
A B
1 − ω2 LC
jωL
=
C D
jωC(2 − ω2 LC) 1 − ω2 LC
(6.35)
√
These two matrices are equal term by term at f = f0 if f0 = 1/(2π LC), thus
yielding the design equation
ω0 L = Z0t =
1
ω0 C
(6.36)
This same design equation relates the other three networks in (c), (d), and (e) of
Figure 6.8 to the λ/4 transformer by making all of the ABCD matrices equal at
f = f0 .
182
MICROWAVE TRANSMISSION LINE CIRCUITS
The LC networks in Figure 6.8 may be used as direct replacements for the
λ/4 transformer if the operating frequency is not so high that the component size
becomes comparable to a wavelength, as mentioned in Chapter 3. Alternatively, the
LC elements may be replaced by appropriate transmission line segments that are
less than λ/4 in length.
6.4.2 The Eighth-Wavelength Transformer
To explore another method that replaces the λ/4 transformer with shorter segments
of transmission line, consider the relatively simple networks in Figure 6.9. Each
segment in Figure 6.9(a) has the same characteristic impedance and phase length,
symbolized by Z0 and θ, respectively. These parameters have the same symbols in
Figure 6.9(b), but the actual values of Z0 may differ from those in Figure 6.9(a).
The ABCD matrix of the network in Figure 6.9(a) is
A B
(1 − tan2 θ) cos θ
jZ0 sin θ
=
(6.37)
C D
jY0 (3 − tan2 θ) sin θ (1 − tan2 θ) cos θ
This matrix is equal to the matrix of the λ/4 transformer at f = f0 , if
θ=
and
Z0 =
π f
4 f0
(6.38)
√
2Z0t
(6.39)
Thus, the three line-segment lengths in Figure 6.9(a) are λ/8 at f = f0 , and this
network will be referred to as an λ/8 transformer.
From the ABCD√matrix of the λ/4 transformer in (6.34), the transmission
coefficient with Z0t = RS RL is
√
2 RS RL
2
q t=
= q
ARL + B + CRS RL + DRS
RL
RS
RS +
RL cos θt + j2 sin θt
(6.40)
and t = −j = 16 −90◦ at f = f0 , indicating perfect transmission and a 90◦
phase delay, as expected. For comparison, using the ABCD parameters
of
√
√ the λ/8
transformer in (6.37), the transmission coefficient with Z0 = 2Z0t = 2RS RL
is
√
2 RS RL
t =
ARL + B + CRS RL + DRS
183
BROADBAND IMPEDANCE-MATCHING NETWORKS
RS
θt
VS
Z0t
RL
(a)
L
C
C
L
(b)
L
(c)
C
C
L
L
C
(d)
C
L
(e)
Figure 6.8 Comparison of the λ/4 transformer in (a) with LC equivalents in (b), (c), (d), and (e).
MICROWAVE TRANSMISSION LINE CIRCUITS
RS
λ0/8
VS
(a)
RL
Z 0 = 2 Z 0t
/8
0
λ
0
/8
0
λ
0
0
λ
Z
0
λ
/8
0
Z
0
/8
Z
Z0
Z
184
RS
VS
Z0
λ0/8
Figure 6.9 λ/8 transformers that are equivalent to the λ/4 transformer.
RL
Z0 =
(b)
Z 0t
2
185
BROADBAND IMPEDANCE-MATCHING NETWORKS
=
2
q
RL
RS
+
q
RS
RL
(1 − tan2 θ) cos θ + j √12 (5 − tan2 θ) sin θ
(6.41)
and at f = f0 , where θ = 45◦ , t = −j = 16 −90◦ .
The ABCD matrix of the network in Figure 6.9(b) is
A
C
B
D
=
(1 − tan2 θ) cos θ
jY0 sin θ
j(3 − tan2 θ)Z0 sin θ
(1 − tan2 θ) cos θ
(6.42)
This matrix is equal to the matrix of the λ/4 transformer at f = f0 , if
θ=
and
π f
4 f0
√
Z0 = Z0t / 2
(6.43)
(6.44)
Thus again, the three line-segment lengths in Figure 6.9(b) are λ/8 at f =
f0 , and this network, which has the same transmission coefficient as given in
(6.41), will also be referred to as a λ/8 transformer. The λ/8 transformer in
Figure 6.9(a) may be constructed conveniently using stripline or microstrip, whereas
the transformer in Figure 6.9(b) is well suited for coplanar waveguide (CPW) or
twin lead construction.
Example 6.1: For RS = 50 Ω and RL = 100 Ω, use two cascaded
λ/4 transformers to form an input impedance match that has a Butterworth response function, and then use lumped-element networks of
the type shown in Figure 6.8(b) to replace the λ/4 transformers at
f = 500 MHz.
Solution: From (6.25) and (6.26), the λ/4 transformer impedances are
Z01 = 500.751000.25 = 59.46 Ω and Z02 = 500.251000.75 = 84.09 Ω.
In the replacement network there will be five lumped elements. If these
are numbered with subscripts 1 through 5, the central element is the
sum of two adjacent elements as C3 = C1 + C5 . The calculations
proceed using (6.36) to obtain:
186
MICROWAVE TRANSMISSION LINE CIRCUITS
C1 =
1
= 5.35 pF
59.46ω0
L2 =
59.46
= 18.93 nH
ω0
C3 = C1 + C5 = 9.14 pF
L4 =
84.09
= 26.77 nH
ω0
C5 =
1
= 3.79 pF
84.09ω0
and
The resulting networks are shown in Figure 6.10(a) and (b).
The input reflection coefficient (S11 in dB) is plotted versus frequency in
Figure 6.11 for the transmission line network in Figure 6.10(a) and for the lumpedelement network in Figure 6.10(b). As illustrated, both networks are perfectly
matched (S11 ∼
= −∞ dB) at 500 MHz (the design frequency). The bandwidth for
S11 = −15 dB (an acceptable impedance match) is 507 MHz for the transmission
line circuit while the bandwidth for the lumped-element circuit is considerably less
at 246 MHz. The reduced bandwidth would not be a problem unless an extremely
wide bandwidth is required. Notice particularly in this example and the plot of S11
versus frequency that impedance matching networks are also frequency filters. In
many applications this is a desirable feature that may be used to prevent out-ofband signals from being transmitted to the load.
Example 6.2: (a) Replace the two cascaded λ/4 transformers in the
previous example with λ/8 transformers of the type shown in Figure 6.9(a). (b) Repeat using λ/8 transformers of the type shown in
Figure 6.9(b).
Solution: In both parts (a) and (b), the result will be five transmission
line segments each of length λ/8, and the central segments will be
the combination of two adjacent segments. In part (a), before combining the central segments, the√characteristic impedance of each of
the first three segments is 59.46 2 = 84.09 Ω and for the last three
BROADBAND IMPEDANCE-MATCHING NETWORKS
ZS = 50 Ω
VS
λ0/4
λ0/4
Z01 = 59.46 Ω
Z02 = 84.09 Ω
ZS = 50 Ω
VS
L2
C1
187
ZL = 100 Ω
(a)
ZL = 100 Ω
(b)
L4
C3
C5
Figure 6.10 Lumped-element replacement of cascaded λ/4 transformers. (a) Cascaded transformers
yielding a Butterworth response. (b) The lumped-element replacement network where: C1 = 5.35 pF,
L2 = 18.93 nH, C3 = C1 + C5 = 9.14 pF, L4 = 26.77 nH, and C5 = 3.79 pF.
Figure 6.11 Input reflection coefficient (S11 in dB) versus frequency for the two equivalent networks
in Figure 6.10(a) and (b).
MICROWAVE TRANSMISSION LINE CIRCUITS
λ0/8
92
Ω
VS
59.46 Ω
λ0/8
λ0/8
.4
6
/8
0
λ
0
42.04 Ω
59
/8
50
1.
10
/8
0
Ω
Ω
0
0
λ
λ
42
.0
4
Ω
0
λ
/8
/8
49
.4
6
Ω
9
/8
84
.0
50 Ω
100 Ω (a)
118.92 Ω
Ω
84.09 Ω
λ
VS
λ0/8
λ
50 Ω
11
8.
188
100 Ω (b)
Figure 6.12 Cascaded λ/8 transformers that are equivalent to the cascaded λ/4 transformers in
Figure 6.10(a).
√
it is 84.09 2 = 118.92 Ω. The completed network is shown in Figure 6.12(a), where the 49.26 Ω characteristic impedance is the parallel
combination of 84.09 Ω and 118.92 Ω. In part (b), before combining the central segments, the characteristic
impedance of each of the
√
first three√segments is 59.46/ 2 = 42.04 Ω and for the last three it
is 84.09/ 2 = 59.46 Ω. The completed network is shown in Figure 6.12(b), where the 101.50 Ω characteristic impedance is the series
combination of 42.04 Ω and 59.46 Ω.
Example 6.3: A network consisting of a lossless transmission line of
characteristic impedance Z0 = 10 Ω and electrical length θ = 90◦
transmits power from a 50 Ω source to a 2 Ω load. (a) Find the transducer gain (Gt ) at the frequency where θ = 90◦ . (b) If VS = 200 V
(RMS), find I1 , V1 , I2 , and V2 . (c) Find P1 (the power extracted from
189
BROADBAND IMPEDANCE-MATCHING NETWORKS
the source) and PL (the power delivered to the 2 Ω load).
Solution: (a) From (4.36), Gt at f = f0 is
Gt
=
=
=
4RS RL
|ARL + B + CRS RL + DRS |2
400
|0 + j10 + j100/10 + 0|2
1
(6.45)
(b) The input voltage and current are related to the output (load) voltage
and current by
V1 = AV2 + BI2
(6.46)
I1 = CV2 + DI2
(6.47)
With I1 = VS /100 = 2 A and V1 = 2 × 50 = 100 V, then,
I2 = 100/(j10) = −j10 A and V2 = −j10 × 2 = −j20 V.
(c) The power input to the transmission line from the source is, P1 =
|V1 I1 | = 100 W, and the power delivered to the load is, P2 = |V2 I2 | =
100 W.
Example 6.4: The transmission line of characteristic impedance Z0 =
10 Ω and electrical length θ = 90◦ in Example 6.3 is replaced by the
LCL network in Figure 6.8 (c). (a) Find the numerical values of ω0 L
and ω0 C as functions of Z0 = 10 Ω that make this replacement valid.
(b) Find Gt for the LCL circuit. (c) The LCL network is now replaced
by smaller sections of transmission lines as shown in Table 3.1 in Chapter 3. If Z01 = 100 Ω for the inductive lines and Y02 = 0.1 S = 101 Ω
for the capacitive line, find the associated lengths l1 and l2 in fractional
wavelengths.
Solution: (a) For the θ = 90◦ , Z0 = 10 Ω line:
A
C
B
D
=
=
cos θ
jY0 sin θ
0
jY0
jZ0
0
jZ0 sin θ
cos θ
(6.48)
190
MICROWAVE TRANSMISSION LINE CIRCUITS
and the ABCD matrix of the LCL network is
A B
1 − ω2 LC jωL(2 − ω2 LC)
=
C D
jωC
1 − ω2 LC
0
jω0 L
=
(6.49)
jω0 C
0
These two matrices are equal at f = f0 if ω02 LC = 1, to yield
ω0 L = Z0 = 1/ω0 C.
(b) As in Example 6.3
Gt
=
=
=
4RS RL
|ARL + B + CRS RL + DRS |2
400
|0 + j10 + j100/10 + 0|2
1
(6.50)
Note that since Gt = |S21 |2 , then
√
◦
20
2 RS RL
S21 =
=
= −j1 = 1 × e−j90
ARL + B + CRS RL + DRS
j20
(6.51)
(c) The inductances are replaced by series sections of transmission line
using
2πl1
jω0 L = jZ01 tan
, or l1 = 0.0159λ
(6.52)
λ
and the shunt capacitance is replaced by an open-ended section of line
using
2πl2
jω0 C = jY02 tan
, or l2 = 0.125λ
(6.53)
λ
Example 6.5: If in Example 6.4, ZL = RL + jXL = 2 + j10 Ω, match
the 50 Ω source to this ZL by using the same LCL network.
Solution: When the j10 Ω inductive reactance of the load was absent,
the match to RL = 2 Ω was performed using series jω0 L = j10 Ω,
shunt jω10 C = −j10 Ω and series jω0 L = j10 Ω. Therefore, all we
need do here is to omit the last series jω0 L in the network, since the
exact value of series inductive reactance is already supplied by the load
ZL .
BROADBAND IMPEDANCE-MATCHING NETWORKS
191
6.4.3 Impedance Matching a Real Source to a Complex Load
The procedure introduced in Example 6.5 for impedance matching a real source
(RS ) to a complex load that has series elements (ZL = RL + jXL ) or that has
parallel elements (YL = GL + jBL ) may be made more generally useful by
the following steps. First, since only the real part of the load impedance absorbs
power, √
we can ignore the imaginary part and use a λ/4 transformer of impedance
Z0t = RS RL to match RS to RL. For series load elements (ZL = RL + jXL ),
if jXL = jω0 LL , use the network in Figure 6.8(c) and let the reactance of the third
element of the network be
jX3 = jω0 L − jω0 LL
(6.54)
If jXL = −j ω01CL , use the network in Figure 6.8(e) and let the reactance of the
third element of the network be
1
1
−j
ω0 C
jX3 = −j
+j
=
1−
(6.55)
ω0 C
ω0 C L
ω0 C
ω0 C L
For load elements connected in parallel (YL = GL + jBL ), if jBL = 1/jXL =
jω0 CL, use the network in Figure 6.8(b) and let the susceptance of the third element
of the network be
ω0 C L
jB3 = jω0 C − jω0 CL = jω0 C 1 −
(6.56)
ω0 C
If jBL = 1/jXL = −j ω01LL , use the network in Figure 6.8(d) and let the
susceptance of the third element of the network be
jB3 =
−j
j
−j
+
=
ω0 L ω0 LL
ω0 L
1−
ω0 L
ω0 LL
(6.57)
Note that the third element in each matching network (jX3 or jB3 ) decreases in
value as the imaginary part of the load increases from zero in (6.54), (6.55), (6.56)
and (6.57), so that the third element is diminished to zero when ω0 LL = ω0 L
or ω0 CL = ω0 C. Also note that if the imaginary part of the load is greater than
the original value of the third element in (6.54), (6.55), (6.56) and (6.57), these
equations correctly yield the value and sign of the new impedance or admittance
that becomes jX3 or jB3 of the third element.
192
MICROWAVE TRANSMISSION LINE CIRCUITS
Example 6.6: As in Example 6.5, matching a 50 Ω source to a 2 Ω load
requires √
a quarter-wavelength transformer of characteristic impedance
Z0t = 50 × 2 = 10 Ω, and each lumped element in Figure 6.8
will have the impedance jω0 L = jZ0t = j10 Ω, or j ω−1
=
0C
−j10 Ω (jω0 C = j0.1 S).
(a) If the load impedance is ZL = 2 + j5 Ω, choose the appropriate
network in Figure 6.8 and determine jX3 or jB3 for an impedance
match to the 50 Ω source.
(b) Repeat part (a) if ZL = 2 + j15 Ω.
(c) Repeat part (a) if the load is 2 Ω in parallel with −j5 Ω.
(d) Repeat part (a) if the load is 2 Ω in parallel with −j15 Ω.
Solution: (a) Since the load has a series inductance we choose Figure 6.8(c), and use (6.54) to find the new value of the third element as
jX3 = j10 − j5 = j5 Ω.
(b) We again use Figure 6.8(c) and (6.54) to find the new value of the
third element as jX3 = j10 − j15 = −j5 Ω. Thus, the original j10 Ω
inductive reactance of the third element must be replaced by −j5 Ω
of capacitive reactance. This capacitive reactance of −j5 Ω resonates
with the extra j5 Ω in the load to maintain the impedance match, and
the overall bandwidth will be decreased by a small amount.
(c) Here we want to combine parallel capacitive elements, so we choose
Figure 6.8(b) and use (6.56) to find the new value of the third element as
jB3 = j0.1−j0.2 = −j0.1 S. Thus, the third element is changed from
j0.1 S of capacitive susceptance to −j0.1 S of inductive susceptance.
Viewed another way, the new value of jX3 = 1/(−j0.1) = j10 Ω is
combined in parallel with −j5 Ω of the load to yield −j10 Ω which
was the original value of jX3 .
(d) Here again, the new value of jX3 must be combined in parallel with
−j15 Ω of capacitive reactance in the load to yield the original value of
jX3 = −j10 Ω. We again choose Figure 6.8(b) and use (6.56) to find
the new value of jB3 = j0.1 − j0.0667 = j0.0333 S. This is the same
as adding the new value of jX3 = 1/(j0.0333) = −j30 Ω in parallel
with −j15 Ω in the load to yield the original value of jX3 = −j10 Ω.
BROADBAND IMPEDANCE-MATCHING NETWORKS
M Z0
θ=
VS
VS
π f
2 f0
Z0
M Z0
M Z0
193
L
C
C
Z0
Figure 6.13 Show that the lumped-element circuit is equivalent to the λ/4 transformer by equating
ABCD matrices. Determine the transducer gain of each circuit.
PROBLEMS
1. (a) In Figure 6.13, by equating ABCD matrices, show that the
√ lumpedelement circuit
is
equivalent
to
the
λ/4
transformer
if
ω
L
=
M Z0 and
0
√
ω0 C = 1/( M Z0 ), where the multiplier M is some positive
number
and Z0
√
is a real load impedance. (b) If Z0 = 50 Ω, M = 36 or M Z0 = 300 Ω,
determine the transducer gain (in dB) of each circuit at f = f0 and f = 2f0 .
2. The equivalent networks shown in Figure 6.14 will provide a broadband
impedance match of a 300 Ω TV set to a 75 Ω antenna at a center frequency
of 166 MHz. For the lumped-element equivalent circuit shown, determine the
values of L and C in µH and pF, respectively.
3. The network shown in Figure 6.14 is to provide a broadband impedance
match of a 300 Ω TV set to a 75 Ω antenna at a center frequency of 166 MHz.
If all transmission lines are twin-lead in air, (a) determine the total length
l = λ0 /2 in cm. (b) If the physical length is to be halved by replacing each
λ/4 transformer by a λ/8 transformer, sketch the replacement network and
determine all characteristic impedances.
4. The broadband impedance-matching network between RS = 200 Ω and
RL = 50 Ω in Figure 6.15 consists of the transmission line segments of
characteristic impedance Z01 , Z02 , and Z03 , all λ0 /4 in length at f0 . (a) If
the network is to have a Butterworth response versus frequency, determine
194
MICROWAVE TRANSMISSION LINE CIRCUITS
75 Ω
VS
λ0/4
λ0/4
Z01 = 106.066 Ω
Z02 = 212.132 Ω
300 Ω
ANTENNA
TV SET
75 Ω
VS
L2
C1
L4
C3
C5
300 Ω
Figure 6.14 Transmission line and lumped-element matching networks for Problem 2 and Problem 3.
Z01 , Z02 , and Z03 . (b) Three cascaded λ0 /4 transformers could have been
used instead of the shorted stub and two transformers. Which choice would
yield the lowest QT , and therefore the broadest bandwidth?
λ0/4
Z01
Z02
RL = 50 Ω
0
λ
/4
Z
VS
λ0/4
03
RS = 200 Ω
Figure 6.15 Broadband impedance-matching network for Problem 4.
BROADBAND IMPEDANCE-MATCHING NETWORKS
195
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