1 EKT 441 MICROWAVE COMMUNICATIONS CHAPTER 1: TRANSMISSION LINE THEORY (PART II)

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EKT 441

MICROWAVE COMMUNICATIONS

CHAPTER 1:

TRANSMISSION LINE THEORY (PART II)

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•The Smith Chart – Intro

•Using the Smith Chart

•Reflection Coefficient Mag & Angle

•VSWR

•Impedance Matching

•Quarter-Wave Transformer

•Single/double stub Tuner

•Lumped element tuner

•Multi-section transformer

TRANSMISSION LINE THEORY PART II

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THE SMITH CHART - INTRO

Figure 8: The Smith Chart

The Smith Chart parameters:

• The reflection coefficient: Γ = |Γ|ejθ

(|Γ| ≤ 1), (-180°≤ θ ≤180°)

• The normalized impedance:z = Z / Z0

• The normalized admittance:y = 1 / z

• SWR and RL scale

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• Graphical tool for use with transmission line circuits and microwave circuit elements.

• Only lossless transmission line will be considered.

• Two graphs in one ;

Plots normalized impedance at any point.

Plots reflection coefficient at any point.

THE SMITH CHART - INTRO

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The transmission

line calculator,

commonly

referred as the

Smith Chart

THE SMITH CHART - INTRO

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USING THE SMITH CHART

The Smith Chart is a plot of normalized

impedance. For example, if a Z0 = 50 Ω

transmission line is terminated in a load

ZL = 50 + j100 Ω as below:

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To locate this point on Smith Chart, normalize the

load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2

Ω

USING THE SMITH CHART

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The normalized load

impedance is located

at the intersection of

the r =1 circle and the

x =+2 circle.

USING THE SMITH CHART

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The reflection coefficient has a magnitude

and an angle : j

L e

L

Where the magnitude can be measured using a

scale for magnitude of reflection coefficient

provided below the Smith Chart, and the angle

is indicated on the angle of reflection

coefficient scale shown outside the

circle on chart.

1L

USING THE SMITH CHART

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USING THE SMITH CHART

Scale for magnitude of reflection coefficient

Scale for angle of reflection coefficient

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For this example,

0457.0 j

jL

e

e

USING THE SMITH CHART

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After locating the normalized impedance

point, draw the constant circle. For

example, the line is 0.3λ length:

jL e

USING THE SMITH CHART

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• Move along the constant circle is

akin to moving along the transmission line.

Moving away from the load (towards

generator) corresponds to moving in the

clockwise direction on the Smith Chart.

Moving towards the load corresponds to

moving in the anti-clockwise direction on

the Smith Chart.

jL e

USING THE SMITH CHART

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• To find ZIN, move towards the generator by:

Drawing a line from the center of chart to

outside Wavelengths Toward Generator

(WTG) scale, to get starting point a at

0.188λ

Adding 0.3λ moves along the constant

circle to 0.488λ on the WTG scale.

Read the corresponding normalized input

impedance point c, ZNIN = 0.175 - j0.08Ω

jL e

USING THE SMITH CHART

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Denormalizing, to find an input impedance,

475.80

jZ

ZZZ

IN

NININ

USING THE SMITH CHART

VSWR is at point b,

9.5VSWR

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For Z0 = 50Ω ,

a ZL = 0 (short cct)

b ZL = ∞ (open cct)

c ZL = 100 + j100

Ω

d ZL = 100 - j100 Ω

e ZL = 50 Ω

USING THE SMITH CHART

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EXAMPLE 1.5

ZL= 50 - j25 and Z0=50 Ohm. Find Zin, VSWR and ΓL using the Smith Chart.

LL

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SOLUTION TO EXAMPLE 1.5

(i) Locate the normalized load, and label it as point a, where it corresponds to

(ii) Draw constant circle.

(iii)It can be seen that

5.01 jZNL

jL e

076245.0 jL e and 66.1VSWR

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(iv) Move from point a (at 0.356λ) on the WTG

scale, clockwise toward generator a distance

λ/8 or 0.125λ to point b, which is at 0.481λ.

We could find that at this point, it corresponds

to

Denormalizing it,

07.062.0 jZNIN

5.331 jZ IN

SOLUTION TO EXAMPLE 1.5 (Cont’d)

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EXAMPLE 1.6

The input impedance for a 100 Ω

lossless transmission line of length

1.162 λ is measured as 12 + j42Ω.

Determine the load impedance.

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SOLUTION TO EXAMPLE 1.6

(i) Normalize the input impedance:

(ii) Locate the normalized input impedance and

label it as point a

42.012.0100

4212

0

jj

Z

Zz in

in

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(iii) Take note the value of wavelength for point a at

WTL scale.

At point a, WTL = 0.436λ

(iv) Move a distance 1.162λ towards the load to point b

WTL = 0.436λ + 1.162λ

= 1.598λ

But, to plot point b, 1.598λ – 1.500λ = 0.098λ Note: One complete rotation of WTL/WTG =

0.5λ

SOLUTION TO EXAMPLE 1.6 (Cont’d)

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(v) Read the point b:

7.015.0 jZNL

Denormalized it:

70150

j

ZZZ NLL

SOLUTION TO EXAMPLE 1.6 (Cont’d)

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EXAMPLE 1.7

On a 50 lossless transmission line,

the VSWR is measured as 3.4. A

voltage maximum is located 0.079λ

away from the load (towards

generator). Determine the load.

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(i) Use the given VSWR to draw a constant

circle.

(ii) Then move from maximum voltage at

WTG = 0.250λ (towards the load) to point

a at WTG = 0.250λ - 0.079λ = 0.171λ.

(iii)At this point we have ZNL = 1 + j1.3 Ω,

or ZL = 50 + j65 Ω.

SOLUTION TO EXAMPLE 1.7

jL e

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THE IMPEDANCE MATCHING

The important of impedance matching or tuning:

Maximum power is delivered when the load is matched to the line. The power loss in the feed line is minimized. (increase power handling

capability by optimizing VSWR) Improved the signal-to-noise ratio (SNR). (e.g with controlled

mismatch, an amplifier can operate with minimum noise generation) Reduced the amplitude and phase errors.

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THE IMPEDANCE MATCHING

Factors in the matching network selection: Complexity Bandwidth Implementation Adjustability

Matching Network

LoadZL

Z0

Figure 9: A lossless network matching an arbitrary load impedance to a transmission line.

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IMPEDANCE MATCHING

• The transmission line is said to be matched

when Z0 = ZL which no reflection occurs.

• The purpose of matching network is to

transform the load impedance ZL such that

the input impedance Zin looking into the

network is equal to Z0 of the transmission

line.

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Adding an impedance matching networks ensures that all power make it or delivered to the load.

IMPEDANCE MATCHING (Cont’d)

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Techniques of impedance matching :

Quarter-wave transformer

Single / double stub tuner

Lumped element tuner

Multi-section transformer

IMPEDANCE MATCHING (Cont’d)

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• It much more convenient to add shunt elements rather than series elements Easier to work in terms of admittances.

• Admittance:Z

Y1

IMPEDANCE MATCHING (Cont’d)

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Adding shunt elements using admittances:

With Smith chart, it is easy to find normalized

admittance – move to a point on the opposite

side of the constant circle. jL e

IMPEDANCE MATCHING (Cont’d)

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QUARTER WAVE TRANSFORMER

The quarter wave transformer matching

network only can be constructed if the load

impedance is all real (no reactive component)

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To find the impedance looking into the quarter

wave long section of lossless ZS impedance

line terminated in a resistive load RL:

ljRZ

ljZRZZ

LS

SLSin

tan

tan

, 24

2 lBut, for quarter wavelength,

l tan

QUARTER WAVE TRANSF. (Cont’d)

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So,

0

2

ZR

ZZ

L

Sin

Rearrange to get impedance matched line,

LS RZZ 0

QUARTER WAVE TRANSF. (Cont’d)

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EXAMPLE 1.8.5

Calculate the position and

characteristic impedance of a

quarter wave transformer that will

match a load impedance, RL =

15Ω; to a 50 Ω line.

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To get the transformer’s impedance, use

SOLUTION TO EXAMPLE 1.8.5

39.27)15)(50(0 LS RZZ

To find the position of quarter-wave transformer

from the load:

d = 0.25 λ

ZT = 27.39 Ω

ZT = 27.39 Ω 15 Ω

Qwave transformer

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EXAMPLE 1.8.6

A transistor has an input impedance of ZL = 25

Ω, at an operating frequency of 500 MHz. Find:

a)The length, l

b)Width, w

c)characteristic impedance of the quarter-wave

parallel plate line transformer for which

matching is achieved.

Assume thickness of the dielectric is d = 1 mm,

and the relative dielectric constant is εr = 4.

Assume that surface resistance, R and shunt

conductance, G, can be neglected.

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EXAMPLE 1.8.6

We can directly apply that

355.35)25)(50(0 LS RZZ

To transistor Zin = 25Ωw Zline

ZL

Zin

Z0=50Ω

ℓ=λ/4

The characteristic of the line is

w

d

C

LZ p

Line

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EXAMPLE 1.8.6 (cont)

Paramete

r2-wire line Coaxial Line

Parallel Plate Line

Unit

R Ω/m

L H/m

G S/m

C F/m

conda

1

a

Da

2cosh

aDa

diel

2cosh

aDa 2cosh

bacond

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2

1

a

bln

2

ab

diel

ln

2

abln

2

condw

2

w

d

d

wdiel

d

w

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EXAMPLE 1.8.6 (Cont)

Thus, the width of the line is

From previous table, we find the values for

capacitance and inductance as;

mmZ

dw

rline

p 329.50

0

mnHw

dL p /8.235

mpFd

wC

p

/6.188

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EXAMPLE 1.8.6 (Cont)

The line length l follows from the condition

The input impedance of the combined transmission line and

the load is:

mmLCf

l 967.744

1

4

)(1

)(1

tan

tan

d

dZ

djZZ

djZZZZ line

Lline

lineLlinein

Where d = l = λ/4, and the reflection coefficient is given

by

dv

fj

ZZ

ZZed

plineL

lineLdj 22exp)( 2

0

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EXAMPLE 1.8.6 (Cont)

Quarter Wave Impedance

Matching

• Designed to achieve matching at a

single frequency/narrow bandwidth

• Easy to build and use

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STUB MATCHING

Stub Matching

• Single stub or Double Stub

• Parallel Stub or Series Stub

• Open stub or Shorted Stub

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SINGLE STUB TUNING

YLY0

d

lY=1/Z

Y0

Open or

shorted stub

(a)

YLZ0

d

l Z=1/Y

Z0

Open or

shorted stub

(b)

Z0

Figure 11: Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.

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SHUNT STUB MATCHING NETWORK

The matching network has to transform the real part

of load impedance, RL to Z0 and reactive part, XL to

zero Use two adjustable parameters – e.g. shunt-stub.

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Thus, the main idea of shunt stub matching

network is to:

(i) Find length d and l in order to get yd and yl .

(ii) Ensure total admittance ytot = yd + yl = 1 for

complete matching network.

SHUNT STUB MATCHING NET. (Cont’d)

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• Locate the normalized load impedance ZNL.

• Draw constant SWR circle and locate YNL.

• Move clockwise (WTG) along circle to intersect with 1 ± jB value of yd.

• The length moved from YNL towards yd is the through line length, d.

• Locate yl at the point jB .

• Depends on the shorted/open stub, move along the periphery of the chart towards yl (WTG).

• The distance traveled is the length of stub, l .

SHUNT STUB USING SMITH CHART

jL e

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SHORTED SHUNT STUB MATCHING

Generic layout of the shorted shunt stub

matching network:

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EXAMPLE 1.9

Construct the shorted shunt stub

matching network for a 50Ω line

terminated in a load ZL = 20 –

j55Ω

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1. Locate the normalized load impedance,

ZNL = ZL/Z0 = 0.4 – j1.1Ω

2. Draw constant circle.

3. Locate YNL. (0.112λ at WTG)

4. Moving to the first intersection with the

1 ± jB circle, which is at 1 + j2.0 yd

5. Get the value of through line length, d from 0.112λ to 0.187λ on the WTG

scale, so d = 0.075λ

SOLUTION TO EXAMPLE 1.9

jL e

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6. Locate the location of short on the Smith Chart

(note: when short circuit, ZL = 0, hence YL =

∞)

on the right side of the chart with

WTG=0.25λ

7. Move clockwise (WTG) until point jB, which

is at 0 - j2.0, located at WTG= 0.324λ yl

8. Determine the stub length, l

0.324λ – 0.25λ = 0.074 λ

SOLUTION TO EXAMPLE 1.9 (Cont’d)

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Thus, the values are:

d = 0.075 λ

l = 0.074 λ

yd = 1 + j2.0 Ω

yl = -j2.0 Ω

Where YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1

SOLUTION TO EXAMPLE 1.9 (Cont’d)

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OPEN END SHUNT STUB MATCHING

Generic layout of the open ended shunt

stub matching network:

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EXAMPLE 1.10

Construct an open ended shunt stub

matching network for a 50Ω line

terminated in a load ZL = 150 + j100

Ω

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SOLUTION TO EXAMPLE 1.10

1. Locate the normalized load impedance,

ZNL = ZL/Z0 = 3.0 + j2.0Ω

2. Draw constant circle.

3. Locate YNL. (0.474λ at WTG)

4. Moving to the first intersection with the

1 ± jB circle, which is at 1 + j1.6 yd

5. Get the value of through line length, d from 0.474λ to 0.178λ on the WTG

scale, so d = 0.204λ

jL e

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6. Locate the location of open end on the Smith

Chart (note: when open circuit, ZL = ∞, hence

YL = 0) on the left side with WTG = 0.00λ

7. Move clockwise (WTG) until point jB, which

is at 0 – j1.6, located at WTG= 0.339λ yl

8. Determine the stub length, l

0.339λ – 0.00λ = 0.339 λ

SOLUTION TO EXAMPLE 1.10 (Cont’d)

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Thus, the values are:

d = 0.204 λ

l = 0.339 λ

yd = 1 + j1.6 Ω

yl = -j1.6 Ω

Where YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1

SOLUTION TO EXAMPLE 1.10 (Cont’d)

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In both previous example, we chose the first

intersection with the1 ± jB circle in designing

our matching network. We could also have

continued on to the second intersection.

Thus, try both intersection to determine which

solution produces max/min length of through

line, d or length of stub, l.

IMPORTANT!!

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Determine the through line length and stub

length for both example above by using

second intersection.

For shorted shunt stub (example 1.9):

d = 0.2 λ and l = 0.426 λ

For open ended shunt stub (example 1.10):

d = 0.348 λ and l = 0.161 λ

EXERCISE (TRY THIS!)