lecture outline - eskisehir.edu.tr 509/icerik...microwave engineering january 29, 2003 dr. wolfgang...

Microwave Engineering January 29, 2003

Dr. Wolfgang J.R. Hoefer 1

Microwave Engineering

University of VictoriaDr. Wolfgang J.R. HoeferLayout by Dr. Poman P.M. So

Lecture 5

Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 1

Lecture Outline

Impedance Transformation Techniques

Impedance-Admittance Conversion

Matching with Lumped Elements

Stub Admittances and Shunt Matching

Stub Impedances and Series Matching

Double Stub Matching




Reasons for Impedance Transformation

Maximum power is delivered when the load and generator are matched to the line.Proper input impedance transformation of sensitive receiver components (antenna, LNA, etc.) improves the S/N ratio of the system.Impedance matching in a power distribution network (such as antenna array feed network) will reduce amplitude and phase errors.


Transforming Network Selection Criteria

Complexity — A simpler impedance transformation network is usually cheaper, more reliable, and less lossy than a more complex design.

Bandwidth — larger BW → increase in complexity.

Implementation — Short-circuited stubs in coax and waveguide. Open-circuited stubs in stripline and microstrip.

Adjustability — tuning screws in waveguides.




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

To avoid unnecessary power loss, matching network is ideally lossless.The impedance looking in to the matching network is Zo.Reflections are eliminated on the transmission line to the left of the matching network.There will be multiple reflections between the matching network and the load.

MatchingNetwork

LoadZL

Zo

Zo

A Lossless Matching Network


Matching with Lumped Elements

L section matching network.(a) Network for zL inside the 1+jx circle (i.e. RL>Zo).(b) Network for zL outside the 1+jx circle (i.e. RL<Zo).ZL must have non-zero real part.

ZL

jX

jBZo

(a)

L

o

L

oL

LL

LoLLoLL

BRZ

RZX

BX

XRRZXRZRX

B

−+=

+−+±

=

1

22

22

ZL

jX

jBZo

(b)

( )

( ) LLoL

o

LLo

XRZRX

ZRRZ

B

−−±=

−±=

D.M. Pozar, Microwave Engineering, 2nd Edition, section 5.1, p.p. 252, John Wiley & Sons, 1998

Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 11/52

Matching with lumped elements (L networks)

Simplest type matching is L-section with 2 reactive elementsTwo possible configurations:

(a): network for zL within 1+jx circle(b): network for zL outside 1+jx circle

Reactive elements: capacitor or inductorParasitics limit usable frequency rangeSolutions analytical or using Smith Chart

ELEC344, Kevin Chen, HKUST 1

Design an L section matching network to match a series RCload with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, ata frequency of 500 MHz.

Solution:

Step 1: Convert the load impedance to admittance bydrawing the SWR circle through the load, and a straight linefrom the load through the center of the Smith Chart.

Step 2: Move the load impedance to the impedance circle of1+ jx (done in admittance Smith Chart) -- add j 0.3 insusceptance ELEC344, Kevin Chen, HKUST 2

Step 1

Step 2

Step 3: Convert back toimpedance.

Step 4: Move to the centerof the Smith Chart byadding an series inductor

Step 3

Step 4


pFfZ

bC 92.0

2 0

==π

Therefore we have b = 0.3, x = 1.2 (check this result with theanalytic solution). Then for a frequency at f = 500 MHz,

we have

nHf

xZL 8.38

20 ==

π

Is there another solution?


There are two solutions for the matching networks. In thiscase, there is no substantial difference in bandwidthbetween the two solutions.


Lect. 13: Impedance Matching (2)

Smith Chart Solution 2 (not using combined ZY SmithChart)Example 5.1 on Page 254 of Pozar


Solution:



Step 1

Step 2



Step 3

Step 4


pFfZ

bC 92.0

2 0

==π


we have

nHf

xZL 8.38

20 ==

π







Smith Chart Solution 1

ZL=200–j100Ω, Zo=100Ω, fo=500MHz.

D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998

ZL

jX

jBZo

Plot zL =2–j1

Draw SWR and y =1 circles

Convert zL to yL

Add shunt susceptance to yL

Add series reactance

Convert y to z



ZL=200–j100Ω, Zo=100Ω, fo=500MHz.


ZL

jX

jBZo




Convert y to z

Convert zL to yL

Plot zL =2–j1





Solution:



Step 1

Step 2



Step 3

Step 4


pFfZ

bC 92.0

2 0

==π


we have

nHf

xZL 8.38

20 ==

π








ZL=200–j100Ω, Zo=100Ω, fo=500MHz.


ZL

jX

jBZo

Plot zL =2–j1


Convert zL to yL



Convert y to z



ZL=200–j100Ω, Zo=100Ω, fo=500MHz.


ZL

jX

jBZo




Convert y to z

Convert zL to yL

Plot zL =2–j1




Smith Chart Solutions 1&2

nH 8.382

pF 92.02

===

===

fπxZ

ωxZL

fZπb

ωbYC

oo

o

o

→= jXZ

pF 61.22

11

1

=−=−=

==−=

ofxZπωXC

ωCjjBX

jY

→= jBY

nH 1.462

1

1

=−=−=

==−=

fbπZ

ωBL

ωLjjXB

jZ

o

2.1 ,3.0 == xb 2.1 ,7.0 −=−= xb

ZL

L

CZo ZL

C

LZo


Single Stub MatchingProblems of Matching with Lumped Elements:

Lumped element impedance matching is not always possible or easily realizable.

Solutions:A section of open-circuited or short-circuited transmission line (a “stub”) connected in parallel or in series with the feed line at a distance from the load can be used.The tuning parameters are the distance from the load (d) and the length of the stub (l).





Solution:



Step 1

Step 2



Step 3

Step 4


pFfZ

bC 92.0

2 0

==π


we have

nHf

xZL 8.38

20 ==

π







Smith Chart Solutions 1&2

nH 8.382

pF 92.02

===

===

fπxZ

ωxZL

fZπb

ωbYC

oo

o

o

→= jXZ

pF 61.22

11

1

=−=−=

==−=

ofxZπωXC

ωCjjBX

jY

→= jBY

nH 1.462

1

1

=−=−=

==−=

fbπZ

ωBL

ωLjjXB

jZ

o

2.1 ,3.0 == xb 2.1 ,7.0 −=−= xb

ZL

L

CZo ZL

C

LZo


Single Stub MatchingProblems of Matching with Lumped Elements:

Lumped element impedance matching is not always possible or easily realizable.

Solutions:A section of open-circuited or short-circuited transmission line (a “stub”) connected in parallel or in series with the feed line at a distance from the load can be used.The tuning parameters are the distance from the load (d) and the length of the stub (l).


Single-stub matching

Single open or short circuited length TL (stub) connected either in shunt or series at certain distance from loadNo lumped components are required, convenient for MICShunt stub often preferred (especially in stripline or µstrip)Parameters: distance series line and value susceptance (or reactance) provided by shunt or series stubShunt: d choosen that after line admittance into line (Y) is Y0+jB, then stub susceptance = -jB; for series d such that Z into line has form Z0+jX

shunt stub series stub

Single-stub tuningA single open-circuited or short-circuited length of transmission line (i.e. a stub) can connect in either parallel or series with the main feed line to achieve impedance matching.

The two adjustable parameters are the distance, d, from the load to the stub position, and the value of susceptance or reactance provided by the shunt or series stub.

( = Z0+jX @ the stub )

-jB

( = Y0+jB @ the stub )

-jX

Single-stub tuningProper length of both open or shored transmission line can provide any desired value of reactance or susceptance.

For a given susceptance or reactance, the difference in lengthsof an open- or short-circuited stub is λ/4.

For microstrip or stripline, open-circuited stubs are easier to fabricate.

For lines like coax or waveguide, however, short-circuited stubsare usually preferred. (open-circuited stubs tend to radiate)




Shunt Stub Matching

YL

d

YoYo

Yo

Open orshortedstub

ZY 1=

l

D.M. Pozar, Microwave Engineering, 2nd Edition, Figure 5.4(a), p.p. 258, John Wiley & Sons, 1998

Matching Operations:Select d, so that y= 1+jb.Select l, so that the stub susceptance is –jb.


Example 5.2 — Solution 1

ZL=15+j10Ω, Zo=50Ω, fo=2GHz.


Convert zL to yL

Transform yL to y1, d=0.044λAdd shunt susceptance to y1

YL

d

YoYo

Yo

Open orshortedstub

ZY 1=

l

Plot zL =0.3+j0.2

b=–1.33

l=0.147λ b=1.33

The stub length is: l=0.147λ

y=0


Single-stub Shunt Matching

For shunt stub in microstrip or stripline open stub is preferred (no VIA hole needed), for waveguide, coax and also to apply DC-bias short stub often more convenient

? For ZL=15+j10 Ω, design two single-stub shunt tuning networks to match load to 50Ω line (3rd Ed. book has different example)First plot normalized load impedanceConvert to admittance by imagingPlot 1+jb circle on Y-chart (1+jx on Z)Turn on SWR circle leads to two intersections (y1 & y2)Distance d from load to stub given by WTG scaleSusceptance stub given by normalized admittance y1 & y2Length stub for given b, determined on Smith chart (start from short!).




Shunt Stub Matching

YL

d

YoYo

Yo

Open orshortedstub

ZY 1=

l

D.M. Pozar, Microwave Engineering, 2nd Edition, Figure 5.4(a), p.p. 258, John Wiley & Sons, 1998

Matching Operations:Select d, so that y= 1+jb.Select l, so that the stub susceptance is –jb.





Convert zL to yL


YL

d

YoYo

Yo

Open orshortedstub

ZY 1=

l

Plot zL =0.3+j0.2

b=–1.33

l=0.147λ b=1.33


y=0







Convert zL to yL


YL

d

YoYo

Yo

Open orshortedstub

ZY 1=

l

Plot zL =0.3+j0.2

l=0.353λ b=–1.33

b=1.33


y=0


Series Stub Matching

ZL

d

ZoZo

Zo

Open orshortedstub

YZ 1=l

D.M. Pozar, Microwave Engineering, 2nd Edition, Figure 5.4(b), p.p. 258, John Wiley & Sons, 1998

Matching Operations:Select d, so that z= 1+jx.Select l, so that the stub susceptance is –jx.


Single-stub shunt matching on Smith Chart

2.03.01015

jzjZ

L

L+=

Ω+=

1+jb circle

λλ

387.0171.0284.05.0044.0284.0328.0

2

1=+−=

=−=

dd

33.1133.11

2

1jyjy

+=

−=

λλ

353.0147.0

2

1=

=

ll


Two solutions single-stub shunt matching

Solution leading to shortest length transmission lines has clearly better bandwidthShorter TL reduce frequency variation match and also in practice will reduce lossesAnalytical solution in book (p. 231-232)

Single-stub tuningShunt stubs – Smith chart

For a load impedance ZL=60-j80, design two single-stub shunt tuningnetworks to match this load to a 50Ω line at 2 GHz.

1.2 1.6Lz j= −

0.3 0.4Ly j= +

1

2

1

2

1

2

0.176 0.065 0.1100.325 0.065 0.260

1.00 1.471.00 1.47

0.0950.405

dd

y jy j

ll

λλ

λλ

= − == − =

= += −

==

d1d2

y1=1+j1.47

y2=1-j1.47zL

yL

-jb=1+jb

s.c.

Single-stub tuningAt 2 GHz, ZL=60-j80 can be modeled as a series combination of R=60Ωand C=0.995 pF. The frequency response is then

1

2

1

2

0.110 0.260

0.095 0.405

dd

ll

λλ

λλ

==

==







Convert zL to yL


YL

d

YoYo

Yo

Open orshortedstub

ZY 1=

l

Plot zL =0.3+j0.2

l=0.353λ b=–1.33

b=1.33


y=0


Series Stub Matching

ZL

d

ZoZo

Zo

Open orshortedstub

YZ 1=l

D.M. Pozar, Microwave Engineering, 2nd Edition, Figure 5.4(b), p.p. 258, John Wiley & Sons, 1998

Matching Operations:Select d, so that z= 1+jx.Select l, so that the stub susceptance is –jx.


Single-stub Series Tuning

? Match ZL=100+j80 Ω to 50Ω line using single series open-circuit stub

First plot normalized load impedance

Plot 1+jx circle on Z-chart

Turn on SWR circle gives 2 intersections (z1 & z2)

Distance d1 & d2 from load to stub given by WTG scale

Reactance stub given by normalized impedance z1 & z2

Length stub for given b, determined on Smith chart starting from open point on Z-chart.






Draw SWR and z =1 circles

Transform zL to z1, d=0.120λAdd series reactance to z1

Plot zL =2+j1.6

ZL

d

ZoZo

Zo

Open orshortedstub

YZ 1=l

b=–1.33

l=0.397λ b=1.33


z=∞




Draw SWR and z =1 circles

Transform zL to z1, d=0.463λAdd series reactance to z1

Plot zL =2+j1.6

ZL

d

ZoZo

Zo

Open orshortedstub

YZ 1=l

b=–1.33 l=0.103λ

b=1.33


z=∞


Single-stub series matching on Smith Chart

1+jx circle6.12

80100jz

jZ

L

L

+=

Ω+=

λλ

463.0172.0208.05.0120.0208.0328.0

2

1

=+−=

=−=

dd

33.1133.11

2

1

jzjz

+=

−=

λλ

103.0397.0

2

1

=

=

ll


Two solutions single-stub series matching

Lengths for both solutions approx. similar, so no big difference in bandwidthSeries matching not so convenient, requires seperate connection to conductor and groundAnalytical solution: see Pozarp.234-235

Single-stub tuningSeries stubs Series stubs –– Smith chart Smith chart

For a load impedance ZL=100+j80, design two single-stub series tuningnetworks to match this load to a 50Ω line at 2 GHz.

2 1.6Lz j= +

1

2

1

2

1

2

(0.328 0.208) 0.120

(0.5 0.208) 0.172 0.463

1.00 1.331.00 1.33

0.3970.103

d

d

z jz j

ll

λ

λ

λλ

= −== − +=

= −= +

==

d1

d2

z1=1-j1.33

z2=1+j1.33

zL

o.c.

=1+jx-jx

Single-stub tuningAt 2 GHz, ZL can be modeled as a series combination of R=100Ω and L=6.37 nH. The frequency response is then

1

2

1

2

0.120 0.463

0.397 0.103

dd

ll

λλ

λλ

==

==

Single-stub tuningStunt stubs Stunt stubs –– analytical solutionanalytical solution

To derive formulas for d and l, let the load impedance be written as .

The impedance Z down a length, d, of line from the load is

where . The admittance at this point is thus

For matching reason, the d (which implies t) is chosen so that .

LLLL XRYZ +== /1

dt βtan=

00 /1 ZYG ==

Single-stub tuningThis results in a quadratic equation for t:

Solving for t gives

Thus, the two principal solutions for d are

02/ ZXt L−= 0for LR Z=

Single-stub tuningFor the required stub lengths, first use t in (5.8b) to find the stub susceptance, and Bs = -B.

Then, for an open-circuit stub,

While for a short-circuited stub,

If the resultant length is negative, λ/2 can be added to give a positive result.

Single-stub tuningSeries stubs Series stubs –– analytical solutionanalytical solution

The input impedance Zin = Rin +j Xin down a length d from the load can be first evaluated.

The matching condition is Rin= Z0 at the stub (let ),

and the two principal solutions for d are then

The required stub lengths arefor an open-circuited stub.

for a short-circuited stub. where X was obtained by substituting t into (5.13b)

0for LG Y=02/ YBt L−=

dt βtan=




Double Stub MatchingDisadvantages of Matching with Single Stub:

A variable length of line between the load and the stub is needed.This would be a problem if an adjustable tuner was desired.

Solution:Double stub matching — two tuning stubs in fixed positions.Adjustable stubs are usually connected in parallel to the main feed line.Double stub tuner cannot match all load impedances.


Matching Operations

Double Shunt Stub Matching

Select l1, so that y1 lies on the rotated 1+jb circle; the amount of rotation is dwavelengths towards the load. In practice, d=λ/8 or 3λ/8.Transform y1 toward the generator through a length d; the new admittance, y2 =1+jb2, lies on the 1+jb circle.Select l2, so that the stub susceptance is –b2.

d

Y’LYoYo

Yo

Open orshortedstub

l2

Yo

Yo

Open orshortedstub

LL Z

Y 1=

l1

jB2 jB1

Y1Yo





ZL=60–j80Ω, Zo=50Ω, d=λ/8,fo=2GHz.Plot yL =0.3+j0.4

d

YLYoYo

Yo

Shortedstub l2

Yo

Shortedstub l1

jB2 jB1Y1Yo

Draw conductance circles

y1 → y2; via the SWR circle.

yL → y1; b1=1.314

y2 → 1; b2=3.38Forbidden Region


Example 5.4 — Solution 2d

YLYoYo

Yo

Shortedstub l2

Yo

Shortedstub l1

jB2 jB1Y1Yo

ZL=60–j80Ω, Zo=50Ω, d=λ/8,fo=2GHz.

Draw conductance circles

yL → y1; b1=–0.114y1 → y2; via the SWR circle.

Plot yL =0.3+j0.4

y2 → 1; b2=–1.38Forbidden Region

lecture outline - eskisehir.edu.tr 509/icerik...microwave engineering january 29, 2003 dr. wolfgang...

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