1

Chapter. 5 Impedance Matching and Tuning

Maximum power is delivered when the load is matched to the

line (assuming the generator is matched), and power loss in

the feed line is minimized.

Impedance matching sensitive receiver components (antenna,

low-noise amplifier, etc.) improves the signal-to-noise ratio of

the system.

Impedance matching in a power distribution network (such as

an antenna array feed network) will reduce amplitude and

phase errors.

FIGURE 5.1 A lossless network matching an arbitrary load impedance to a transmission line. - Factors that may be important in the selection of a particular

matching network:

1) Complexity: simple

2) Bandwidth: restricted frequency band

3) Implementation: easy

4) Adjustability: tunable

2

5.1 MATCHING WITH LUMPED ELEMENTS (L NETWORKS) - L-section

- Simplest type including two reactive elements

- zL = ZL / Z0

- X, B: lumped or distributed element depending on the operating

frequency.

(a) (b) FIGURE 5.2 L section matching networks. (a) Network for zL inside the 1 + jx circle. (b)

Network for zL outside the 1 + jx circle

Analytic Solutions

- ZL = RL + jXL (assumption: Z0 < RL, inside of 1 + jx circle)

- Impedance seen looking into the matching network:

)1.5()/(1

1

0

in

ZjXRjB

jXZLL

=++

+=

3

- Rearranging and separating into real and imaginary parts:

)2.5()( 00 aZRZXXRB LLL −=−

)2.5()1( 0 bXRBZBXX LLL −=−

- Solving Eq. (5.2a) for X and substituting into Eq. (5.2b)

)3.5(1

)2.5(

)(0)(where

)3.5(/

00

00

0022

220

220

bBRZ

RZX

B

aBR

ZRZBXX

ZRRZXR

aXR

RZXRZRXB

LL

L

L

LL

LLLL

LL

LLLLL

−+=

←−+

=

>>−+

+−+±

=

Two solutions: dual valued components (B, X)

- ZL = RL + jXL (assumption: Z0 > RL, outside of 1 + jx circle)

- Admittance seen looking into the matching network:

)4.5(1)(

1

0

in

Z

XXjRjBY

LL

=

+++=

4

- Rearranging and separating into real and imaginary parts:

)5.5()( 00 aRZXXBZ LL −=+

)5.5()( 0 bRBZXX LL =+

- Solving for X and B:

)6.5()( 0 aXRZRX LLL −−±=

)6.5(/)(

0

0 bZ

RRZB LL−

±=

Two solutions: dual valued components (B, X)

Smith Chart Solutions

[Ex. 5.1] Design an L-section matching network to match a series

RC load with an impedance ZL = 200 - j100 Ω to a 100 Ω line, at a

frequency of 500 MHz.

Sol. 1) zL = 2 - j1

yL = 0.4 + j0.2

y = yL + jb = 0.4 + j0.5 ← jb = j0.3

Interconnection with admittance circle of 1 - jx

z = 1 - j1.2

x = 1.2

5

(a)

Using Eq.(5.3a, b), b = 0.29 and x = 1.22

000

0

,pF92.02 Z

bCCZbbYB

fZbC

ωω

π====←==

0 0038.8 nH ,

2xZ xZL X xZ ωL Lπf ω

= = ← = = =

6

Sol. 2) yL = 0.4 + j0.2

y = yL + jb = 0.4 - j0.5 ← jb = -j0.7

Interconnection with admittance circle of 1 + jx

z = 1 + j1.2

x = -1.2

Using Eq.(5.3a, b), b = -0.69 and x = -1.22

00 0

1 1 12.61 pF ,2

C jX jxZ j CfxZ C xZπ ω ω

= = ← = = − = −

0 00

0

146.1 nH ,2

Z ZjbL jB jbY j Lfb Z L bπ ω ω

= = ← = = = − = −

(b)

(c)

FIGURE5.3 Continued. (b) The two possible L section matching circuits. (c) Reflection.

7

coefficient magnitudes versus frequency for the matching circuits of (b).

5.2 SINGLE-STUB TUNING

- Single open-circuited or short circuited length of transmission line

(a “stub”), connected either in parallel or in series with the

transmission feed line at a certain distance from the load

The shunt tuning stub is especially easy to fabricate in microstrip

or stripline form.

- Shunt-stub case

YL → Y0 + jB using trasmission line having length d

→ Y0 using trasmission line having length l

8

- Series-stub case

ZL → Z0 + jX using trasmission line having length d

→ Z0 using trasmission line having length l

(a)

(b)

FIGURE 5.4 Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.

- λ/4 transmission line with a shunt stub

- Open-circuited stubs are easier to fabricate since a via hole through

the substrate to the ground plane is not needed.

9

Shunt Stubs

[Ex. 5.2] Single-Stub Shunt Tuning

- ZL = 15 + j10 Ω, f = 2 GHz

- Design two single-stub shunt tuning networks to match this load to

a 50 Ω line.

- Plot the reflection coefficient magnitude from 1 GHz to 3 GHz for

each solution.

Solution

- zL = 0.3 + j0.2

- SWR circle

- Convert to the load admittance, yL.

- SWR circle intersects the 1 + jb circle at two points, denoted as y1

and y2 in Figure 5.5a. → d1 or d2

- d1 = (0.328 – 0.284)λ = 0.044 λ

d2 = (0.5 – 0.284) + 0.171λ = 0.387 λ

- The matching stub is kept as close as possible to the load, to

improve the bandwidth of the match and to reduce losses caused by

a possibly large standing wave ratio on the line between the stub

and the load.

- y1 = 1 – j1.33, y2 = 1 + j1.33

10

- l1 = 0.147 λ for jb = j1.33

l2 = 0.353 λ for jb = -j1.33

(a)

(b)

0.328

0.147

0.171

0.284

0.353

11

(c)

FIGURE 5.5 Solution to Example 5.2. (a) Smith chart for the shunt-stub tuners. (b) The two shunt-stub tuning solutions. (c) Reflection coefficient magnitudes versus frequency for the tuning circuits of (b).

- Derivation of formulas for d and l

ZL = 1/YL = RL + jXL

Impedance Z down a length, d, of line from the load:

dt

tjXRjZtjZjXRZZ

LL

LL

βtanwhere

)7.5()(

)(

0

00

=++++

=

ZjBGY 1

=+=

12

02

02

2 1)8.5()(

)1(whereZ

atZXR

tRGLL

L =+++

=

)8.5(])([

))((2

02

0

002

btZXRZ

tZXtXZtRBLL

LLL

+++−−

=

d (which implies t) is chosen so that G = Y0 = 1 / Z0

0)(2)( 2200

200 =−−+−− LLLLL XRZRtZXtZRZ

)9.5(for,/])[(

00

022

0 ZRZR

ZXRZRXt L

L

LLLL ≠−

+−±=

If RL = Z0, then t = -XL / 2Z0.

If RL ≠ Z0,

<+

≥=

==

−

−

.0for),tan(21

0for,tan21

2tantan

1

1

tt

ttd

ddt

ππ

πλ

λπβ

(5.10)

To find the required stub lengths, BS = -B.

1) For an open-circuited stub,

jBS = jY0tanβlo = jY0tan(2πlo/λ) =- jB

)11.5(tan2

1tan21

0

1

0

1 aYB

YBl so

−=

= −−

ππλ

13

2) For a short-circuited stub,

jBS = -jY0cotβls = - jY0cot(2πls/λ) = -jB

)11.5(tan21tan

21 0101 b

BY

BYl

s

s

=

−= −−

ππλ

If the lengths (lo, ls) are negative, λ/2 can be added to give a

positive result.

Shunt Stubs

[Ex. 5.2] Single-Stub Series Tuning

- ZL = 100 + j80 Ω, f = 2 GHz

- Match a load to a 50 Ω using a single series open-circuited stub.

- Plot the reflection coefficient magnitude from 1 GHz to 3 GHz for

each solution.

Solution

- zL = 2 + j1.6

- SWR circle

- The SWR circle intersects the 1 + jx circle at two points, denoted

as z1 and z2.

- The shortest distance, d1, from the load to the stub:

d1 = (0.328 – 0.208) λ = 0.120 λ

The second distance:

d2 = (0.5 – 0.208) + 0.172 λ = 0.463 λ

14

- z1 = 1 – j1.33, z2 = 1 + j1.33

- l1 = 0.397 λ for jx = j1.33 (open stub)

l2 = 0.103 λ for jx = -j1.33 (open stub)

(a)

(b)

0.328

0.208

0.172

0.353 → 0.353 – 0.25 = 0.103

0.284

0.147 ← 0.25 + 0.147 = 0.397

15

(c)

FIGURE 5.6 Solution to Example 5.3. (a) Smith chart for the series-stub tuners. (b) The two series-stub tuning solutions. (c) Reflection coefficient magnitudes versus frequency for the tuning circuits of (b).

- Derivation of formulas for d and l for series-stub tuner

YL = 1 / ZL = GL + jBL

Impedance Y down a length, d, of line from the load:

00

0

00

/1,tanwhere

)12.5()(

)(

ZYdtjBGjtYjtYjBGYY

LL

LL

==++++

=

β

YjXRZ 1

=+=

02

02

2 1)13.5()(

)1(whereY

atYBG

tGRLL

L =+++

=

16

)13.5(])([

))((2

02

0

002

btYBGY

tYBtBYtGXLL

LLL

+++−−

=

Now d (which implies t) is chosen so that R = Z0 = 1 / Y0.

0

2200

200

/1)13.5(.Eq0)(2)(

YaBGYGtYBtYGY LLLLL

=←=−−+−−

00

022

0 for,/])[(

YGYG

YBGYGBt L

L

LLLL ≠−

+−±=

If GL = Y0, then t = -BL / 2Y0.

If GL ≠ Y0,

<+

≥=

==

−

−

.0for),tan(21

0for,tan21

2tantan

1

1

tt

ttd

ddt

ππ

πλ

λπβ

(5.15)

To find the required stub lengths, XS = -X.

1) For a short-circuited stub,

jXS = jZ0tanβls = jZ0tan(2πls/λ) = -jX

)16.5(tan2

1tan21

0

1

0

1 aZX

ZXl ss

−=

= −−

ππλ

17

2) For a open-circuited stub,

jXs = -jZ0cotβlo = -jZ0cot(2πlo/λ) = - jX

)16.5(tan21tan

21 0101 b

XZ

XZl

s

o

=

−= −

ππλ

If the lengths (lo, ls) are negative, λ/2 can be added to give a

positive result.

5.3 DOUBLE-STUB TUNING - Disadvantage of single-stub tuning: requiring a variable length of

line between the load and the stub.

Difficult if an adjustable tuner was desired

(a) (b)

FIGURE 5.7 Double-stub tuning. (a) Original circuit with the load an arbitrary distance from the first stub. (b) Equivalent circuit with load at the first stub.

18

Smith Chart Solution

- The susceptance of the first stub, b1 (or b1’, for the second solution),

moves the load admittance (yL = g0 + jb) to y1 (or y1’).

- The amount of rotation is d wavelengths toward the load

→ Transforming y1 (or y1’) to y2 (or y2’): 1 + jb circle

- The second stub then adds a susceptance b2 (or b2’), which brings

us to the center of the chart, and completes the match.

FIGURE 5.8 Smith chart diagram for the operation of a double-stub tuner.

19

[Ex. 5.4] Double-Stub Tuning

- Design a double-stub shunt tuner to match a load impedance ZL =

60 - j80 Ω to a 50 Ω line at 2 GHz.

- The stubs are to be open-circuited stubs, and are spaced λ/8 apart.

- Plot the reflection coefficient magnitude versus frequency from 1 ~

3 GHz.

yL = 0.3 + j0.4

By moving every point on the g = 1 circle λ/8 toward the load, we

then find the susceptance of the first stub, which can be one of

two possible values: b1 = 1.314 or b1’ = -0.114

→ y2 = 1 – j3.38 or y2‘ = 1 + j1.38

Susceptance of the second stub: b2 = 3.38 or b2’ = -1.38

Lengths of the short-circuited stubs are then found as,

l1 = 0.146 λ, l2 = 0.482 λ,

or l1’ = 0.204 λ, l2’ = 0.350 λ

20

(a)

(b)

90o rotation Of unit circle

21

(c)

FIGURE 5.9 Solution to Example 5.4. (a) Smith chart for the double-stub tuners. (b) The two double-stub tuning solutions. (c) Reflection coefficient magnitudes versus frequency for the tuning circuits of (b).

Analytic Solution

- Load admittance with the first stub:

)17.5()( 111 BBjGjBYY LLL ++=+=

- Admittance just to the right of the second stub:

00

10

0102

/1,tan where

)18.5()(

)(

ZYdtjBjBGjtYtYBBjGYY

LL

LL

==++++++

=

β

- Re(Y2) = Y0:

22

)19.5(0)(12

210

2

2

02 =

−−+

+−

ttBtBY

ttYGG L

LL

)20.5()1(

)(4112

1222

0

210

2

2

2

0

+−−

−±+

=tY

tBtBYtttYG L

L

- Since GL is real,

1)1(

)(40 2220

210

2

≤+

−−≤

tYtBtBYt L

)21.5(sin

10 20

2

2

0 dY

ttYGL β=

+≤≤⇒

- After d has been fixed, the first stub susceptance can be determined

from (5.19) as:

)22.5()1( 22

02

01 t

tGYGtYBB LL

L−+±

+−=

- jB2 = -jIm(Y2):

)23.5()1( 0

22200

2 tGYGtGtGYY

BL

LLL +−+±=

- Open-circuited stub length: )24.5(tan21

0

1 aYBlo

= −

πλ

23

- Short-circuited stub length: )24.5(tan2

1 01 bBYls

−

= −

πλ

Where B = B1, B2

5.4 THE QUARTER-WAVE TRANSFROMER

- A single-section λ/4 transformer: narrow band impedance match

A multi-section λ/4 transformer: broad band impedance match

- One drawback of the quarter-wave transformer: match a real load

impedance.

FIGURE 5.10 A single-section quarter-wave matching transformer. l=λ0/4 at the design

frequency f0.

)25.5(01 LZZZ =

where l = λ0/4 : electrical length at operating frequency, f0.

24

- Input impedance seen looking into the matching section:

2/ and tan where

)26.5(1

11in

πθββ ===++

=

llttjZZtjZZZZ

L

L

- Reflection coefficient:

L

LL

LL

ZZZZZZjtZZZZZZjtZZZ

ZZZZ

02

1

02

101

02

101

0in

0in )27.5()()()()(

=←

+++−+−

=+−

=Γ

)28.5(2 00

0

LL

L

ZZtjZZZZ

++−

=Γ

- Reflection coefficient magnitude:

2/10

220

0

]4)[( LL

L

ZZtZZZZ

++−

=Γ

2/1200

2200 ])/(4[)]/()[(

1ZZZZtZZZZ LLLL −+−+

=

2/120

20

200 ])/(4[])/(4[1

1ZZtZZZZZZ LLLL −+−+

=

)29.5(sec])/(4[1

12/122

00 θZZZZ LL −+=

)sectan11( 222 θθ =+=+ t

25

- If the frequency is near the design frequency, f0, then l ≈ λ0/4 and θ

≈ π/2. Then sec2θ >> 1.

)30.5(2/nearfor,cos2 0

0 πθθL

L

ZZZZ −

≅Γ

FIGURE 5.11 Approximate behavior of the reflection coefficient magnitude for a single-

section quarter-wave transformer operation near its design frequency.

- If we set a maximum value, Γm, of the reflection coefficient

magnitude that can be tolerated, then we can define the bandwidth

of the matching transformer as

)31.5(2

2

−=∆ mθπθ

2

0

02 sec

211

−+=

Γ mL

L

m ZZZZ

θ

26

)32.5(2

1cos

0

0

2 ZZZZ

L

L

m

mm −Γ−

Γ=θ

- If we assume TEM line,

00 242

ff

fv

vfl p

p

ππβθ ===

πθ 02 ff m

m =

- Fractional bandwidth:

πθmmm

ff

fff

ff 4222)(2

00

0

0

−=−=−

=∆

)33.5(2

1cos42

0

0

2

1

−Γ−

Γ−= −

ZZZZ

L

L

m

m

π

FIGURE 5.12 Reflection coefficient magnitude versus frequency for a single-section

quarter-wave matching transformer with various load mismatches.

27

- When non-TEM lines (such as waveguides) are used, the

propagation constant is no longer a linear function of frequency,

and the wave impedance will be frequency dependent.

- The effect of reactances associated with discontinuities of

transmission line must be considered.

[Ex. 5.5] Quarter-Wave Transformer Bandwidth

Design a single-section quarter-wave matching transformer to match

a 10 Ω load to a 50 Ω line, at f0 = 3 GHz. Determine the percent

bandwidth for which the SWR ≤ 1.5.

Ω=== 36.22)10)(50(01 LZZZ

The length of the matching section is λ/4 at 3 GHz.

2.015.115.1

1SWR1SWR

=+−

=+−

=Γm

Fractional bandwidth:

−Γ−

Γ−=

∆ −

0

0

2

1

0

2

1cos42

ZZZZ

ff

L

L

m

m

π

−−−= −

5010)10)(50(2

)2.0(12.0cos42

2

1

π

%.29or29.0=

28

5.5 THE THEORY OF SMALL REFLECTIONS

- Theory of small reflections: total reflection coefficient caused by

the partial reflections from several small discontinuities

Single-Section Transformer

- Derivation an approximate expression for the overall reflection

coefficient Γ

FIGURE 5.13 Partial reflections and transmissions on a single-section matching

transformer.

)34.5(12

121 ZZ

ZZ+−

=Γ

29

)35.5(12 Γ−=Γ

)36.5(2

23 ZZ

ZZ

L

L

+−

=Γ

)37.5(2121

2121 ZZ

ZT+

=Γ+=

)38.5(2121

1212 ZZ

ZT+

=Γ+=

- Expression of the total reflection as an infinite sum of partial

reflections and transmissions:

...42

232112

2321121 +ΓΓ+Γ+Γ=Γ −− θθ jj eTTeTT

)39.5(0

232

2321121 ∑

∞

=

−− ΓΓΓ+Γ=n

jnnnj eeTT θθ

- Using the geometric series and small reflection condition,

1for,1

10

<−

=∑∞

=n

n xx

x

)41.5(1

1)1)(1(

11,1,

)40.5(1

231

231

231

2311

23211

121212112

232

232112

1

θ

θ

θ

θθ

θ

θ

j

j

j

jj

j

j

ee

eee

TTeeTT

−

−

−

−−

−

−

ΓΓ+Γ+Γ

≅

ΓΓ+ΓΓ+Γ−+ΓΓΓ−Γ

=

Γ−=Γ+=Γ+=Γ−=Γ←ΓΓ−Γ

+Γ=Γ

30

)42.5(231

θje−Γ+Γ≅Γ

- Intuitive ideas:

1) The total reflection is dominated by the reflection from the

initial discontinuity between Z1 and Z2 (Γ1), and the first

reflection from the discontinuity between Z2 and ZL (Γ3e-2jθ).

2) The e-2jθ term accounts for the phase delay when the incident

wave travels up and down the line.

Multi-Section Transformer

- N equal-length (commensurate) sections of transmission lines.

FIGURE 5.14 Partial reflection coefficients for a multisection matching transformer.

- Derivation an approximate expression for the total reflection coefficient Γ.

31

)43.5(,01

010 a

ZZZZ

+−

=Γ

)43.5(,1

1 bZZZZ

nn

nnn +

−=Γ

+

+

)43.5(. cZZZZ

NL

NLN +

−=Γ

- Assumptions:

1) All Zn increase or decrease monotonically across the transformer.

2) ZL is real.

0

0

if0if0

ZZZZ

Ln

Ln

<<Γ>>Γ⇒

- Overall reflection coefficient:

)42.5()44.5()( 242

210 ←Γ++Γ+Γ+Γ=Γ −−− θθθθ jN

Njj eee

- If the transformer be made symmetrical, so that Γ0 = ΓN, Γ1 = ΓN-1,

Γ2 = ΓN-2, etc. Then (5.44) can be written as;

)45.5(][][)( )2()2(10 ++Γ++Γ=Γ −−−−− θθθθθθ NjNjjNjNjN eeeee

Where if N is odd, the last term is Γ(N-1)/2(ejθ+e-jθ), while if N is

even the last term is ΓN/2.

32

)46.5(evenfor ]21

)2cos(...)2cos(cos[2)(

2/

10

aN

nNNNe

N

njN

Γ++

−Γ++−Γ+Γ=Γ −

θθθθ θ

)46.5( odd for ]cos)2cos(...)2cos(cos[2)(

2/)1(

10

bNnNNNe

N

njN

θθθθθ θ

−

−

Γ++−Γ++−Γ+Γ=Γ