Berkeley

The Smith Chart

Prof. Ali M. Niknejad

U.C. BerkeleyCopyright c© 2017 by Ali M. Niknejad

September 14, 2017

1 / 29

The Smith Chart

The Smith Chart is simply a graphical calculator forcomputing impedance as a function of reflection coefficientz = f (ρ)

More importantly, many problems can be easily visualized withthe Smith Chart

This visualization leads to a insight about the behavior oftransmission lines

All the knowledge is coherently and compactly represented bythe Smith Chart

Why else study the Smith Chart? It’s beautiful!

Aside: There are deep mathematical connections in the SmithChart. It’s the tip of the iceberg! Study complex analysis tolearn more.

2 / 29

An Impedance Smith Chart

Without further ado, here it is!

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1.0

1.0

1.0

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1.2

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1.4

1.4

1.4

1.61.6

1.6

1.81.8

1.8

2.02.0

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3.0

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4.0

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1.01.0

90-9

085

-85

80-8

0

75-7

5

70-7

0

65-6

5

60-6

0

55-55

50-50

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-45

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-25

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-10

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AN

GLE O

F TRAN

SMISSIO

N CO

EFFICIENT IN

DEG

REES

AN

GLE O

F REFLECTION

COEFFICIEN

T IN D

EGREES

—>

WA

VEL

ENG

THS

TOW

ARD

GEN

ERA

TOR

—>

<— W

AVE

LEN

GTH

S TO

WA

RD L

OA

D <

—

IND

UCT

IVE

REAC

TAN

CE C

OMPO

NENT (+jX/Zo), O

R CAPACITIVE SUSCEPTANCE (+jB/Yo)

CAPACITIVE REACTANCE COMPONENT (-j

X/Zo), O

R INDUCTI

VE SU

SCEP

TAN

CE (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

3 / 29

Generalized Reflection Coefficient

We have found that in sinusoidal steady-state, the voltage onthe line is a T-line

v(z) = v+(z) + v−(z) = V+(e−γz + ρLeγz)

Recall that we can define the reflection coefficient anywhereby taking the ratio of the reflected wave to the forward wave

ρ(z) =v−(z)

v+(z)=ρLe

γz

e−γz= ρLe

2γz

Therefore the impedance on the line ...

Z (z) =v+e−γz(1 + ρLe

2γz)v+

Z0e−γz(1− ρLe2γz)

4 / 29

Normalized Impedance

...can be expressed in terms of ρ(z)

Z (z) = Z01 + ρ(z)

1− ρ(z)

It is extremely fruitful to work with normalized impedancevalues z = Z/Z0

z(z) =Z (z)

Z0=

1 + ρ(z)

1− ρ(z)

Let the normalized impedance be written as z = r + jx (notesmall case)

The reflection coefficient is “normalized” by default since forpassive loads |ρ| ≤ 1. Let ρ = u + jv

5 / 29

Dissection of the Transformation

Now simply equate the real (<) and imaginary (=)components in the above equaiton

r + jx =(1 + u) + jv

(1− u)− jv=

(1 + u + jv)(1− u + jv)

(1− u)2 + v2

To obtain the relationship between the (r , x) plane and the(u, v) plane

r =1− u2 − v2

(1− u)2 + v2

x =v(1− u) + v(1 + u)

(1− u)2 + v2

The above equations can be simplified and put into a niceform

6 / 29

Completing Your Squares...

If you remember your high school algebra, you can derive thefollowing equivalent equations(

u − r

1 + r

)2

+ v2 =1

(1 + r)2

(u − 1)2 +

(v − 1

x

)2

=1

x2

These are circles in the (u, v) plane! Circles are good!

We see that vertical and horizontal lines in the (r , x) plane(complex impedance plane) are transformed to circles in the(u, v) plane (complex reflection coefficient)

7 / 29

Resistance Transformations

u

r=0

r=0.5

r=1

r=2

v

r=.5 r=1 r=2r=0

r

r = 0 maps to u2 + v2 = 1 (unit circle)

r = 1 maps to (u − 1/2)2 + v2 = (1/2)2 (matched real part)

r = .5 maps to (u− 1/3)2 + v2 = (2/3)2 (load R less than Z0)

r = 2 maps to (u − 2/3)2 + v2 = (1/3)2 (load R greater thanZ0)

8 / 29

Reactance Transformations

x

r

u

x=

.5

x=

1 x=2

x=-.5

x=

-1

x=-2

v

x = 1

x = 2

x = -1

x = -2

x = 0

x = ±1 maps to (u − 1)2 + (v ∓ 1)2 = 1

x = ±2 maps to (u − 1)2 + (v ∓ 1/2)2 = (1/2)2

x = ±1/2 maps to (u − 1)2 + (v ∓ 2)2 = 22

Inductive reactance maps to upper half of unit circle

Capacitive reactance maps to lower half of unit circle

9 / 29

Complete Smith Chart

u

r=

0

r=

0.5

r=

1

x=

.5

x=

1 x=2

x=-

.5

x=

-1 x=-2

v

shortopen

match

rr=

22

r > 1

inductive

capacitive

10 / 29

Reading the Smith Chart

11 / 29

Reading the Smith Chart

load

First map zL on the Smith Chart as ρL

To read off the impedance on the T-line at any point on alossless line, simply move on a circle of constant radius since

ρ(z) = ρLe2jβz

12 / 29

Motion Towards Generator

loadSWR

C I R CL Emo

vement to

wards generator

Moving towards generator means ρ(−`) = ρLe−2jβ`, or

clockwise motion

We’re back to where we started when 2β` = 2π, or ` = λ/2

Thus the impedance is periodic (as we know)

Aside: For a lossy line, this corresponds to a spiral motion andso the Smith Chart is more useful for low-loss lines

13 / 29

SWR Circle

load

movement to

wards generatorSince SWR is a function of|ρ|, a circle at origin in (u, v)plane is called an SWR circle

Recall the voltage maxoccurs when the reflectedwave is in phase with theforward wave, soρ(zmin) = |ρL|

This corresponds to the intersection of the SWR circle withthe positive real axis (read off SWR by just reading the valueof r)Likewise, the intersection with the negative real axis is thelocation of the voltage min

14 / 29

Example of Smith Chart Visualization

load

Prove that if ZL has an inductance reactance, then theposition of the first voltage maximum occurs before thevoltage minimum as we move towards the generatorProof: On the Smith Chart start at any point in the upper halfof the unit circle. Moving towards the generator correspondsto clockwise motion on a circle. Therefore we will always crossthe positive real axis first and then the negative real axis.

15 / 29

Admittance Chart

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1.81.8

1.8

2.02.0

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1.0

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-20

30-30

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-50

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-60

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-70

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-80

90

-90

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-100

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-110

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-170

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±

90-9

085

-85

80-8

0

75-7

5

70-7

0

65-6

5

60-6

0

55-5

5

50-5

0

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-45

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-40

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-35

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-30

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-15

10

-10

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0.389999

0.389999

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0.4

0.41

0.41

0.419999

0.419999

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0.449

999

0.449

999

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0.47

9999

0.47

9999

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0.49

0.0

0.0

AN

GLE O

F TRA

NSM

ISSION

CO

EFFIC

IENT IN

DEG

REES

AN

GLE O

F REFL

ECT

ION

CO

EFFICIEN

T IN D

EGR

EES

Ð>

WA

VEL

ENG

THS

TOW

AR

D G

ENER

ATO

R Ð

><Ð

WA

VEL

ENG

THS

TOW

AR

D L

OA

D <

Ð

IND

UC

TIV

E RE

AC

TAN

CE C

OM

PON

ENT (+

jX/Zo), O

R CAPACITIVE SUSCEPTANCE (+jB/Yo)

CAPACITIVE REACTANCE COMPONENT (-jX

/Zo), O

R IND

UCTI

VE

SUSC

EPTA

NC

E (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

Since y = 1/z = 1−ρ1+ρ , you can

imagine that an AdmittanceSmith Chart looks very similar

In fact everything is switchedaround a bit and you canconstruct a combinedadmittance/impedance SmithChart. You can also use animpedance chart for admittanceif you simply map x → b andr → g

Be careful ... the caps are now on the top of the chart andthe inductors on the bottom

The short and open likewise swap positions

16 / 29

Admittance on Smith Chart

Sometimes you may need to work with both impedances andadmittances.

This is “easy” on the Smith Chart due to the impedanceinversion property of a λ/4 line (it actually can get prettyconfusing)

Z ′ =Z 20

Z

If we normalize Z ′ we get y

Z ′

Z0=

Z0

Z=

1

z= y

17 / 29

Admittance Conversion

z

y

Thus if we simplyrotate π degrees onthe Smith Chartand read off theimpedance, we’reactually reading offthe admittance!

Rotating π degreesis easy. Simplydraw a line throughorigin and zL andread off the secondpoint of intersectionon the SWR circle

18 / 29

Example Calculation

19 / 29

HW Problem

V0

ZS

ZL

Z0

z = 0z = -d

z

1 Consider the transmission line circuit shown below. A voltagesource generating 10V amplitude of sinewave at 10 GHz isdriving a transmission line terminated with load ZL = 80 - j40ohm. The transmission line has a characteristic impedance ofZ0 (= 100Ω), effective dielectric constant of 4, and lengthd =22.5 mm.

1 Find the reflection coefficient at the load (z = 0) and at thesource (z = −d). [ Note this is 1.5λ ]

2 Find the input impedance at the source (z = -d) and at z =18.75 mm. [Note this is 1.25λ ]

3 Plot the magnitude of the voltage along the line. Find voltagemaximum, voltage minimum, and standing wave ratio.

20 / 29

Homework Problem with Aid from Mr. Smith

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1.01.0

90-9

085

-85

80-8

0

75-7

5

70-7

0

65-6

5

60-6

0

55-55

50-50

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-45

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-40

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0.0

AN

GLE O

F TRAN

SMISSIO

N C

OEFFIC

IENT IN

DEG

REES

AN

GLE O

F REFLECTIO

N C

OEFFIC

IENT IN

DEG

REES

—>

WA

VEL

ENG

THS

TOW

ARD

GEN

ERA

TOR

—>

<— W

AVE

LEN

GTH

S TO

WA

RD L

OA

D <

—

IND

UCT

IVE

REA

CTAN

CE C

OMPO

NENT (+jX/Zo), O

R CAPACITIVE SUSCEPTANCE (+jB/Yo)

CAPACITIVE REACTANCE COMPONENT (-j

X/Zo), O

R INDUCTI

VE SU

SCEP

TAN

CE (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

|ρ| = .25

ρ = 257

|ρ|ej ρ = −.056− .24j

21 / 29

Impedance Matching with Smith Chart

22 / 29

Impedance Matching Example

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1.6

1.81.8

1.8

2.02.0

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90-9

085

-85

80-8

0

75-7

5

70-7

0

65-6

5

60-6

0

55-55

50-50

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-45

40

-40

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-35

30

-30

25

-25

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-20

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-15

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-10

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0.130.14

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0.49

0.0

0.0

AN

GLE O

F TRAN

SMISSIO

N C

OEFFIC

IENT IN

DEG

REES

AN

GLE O

F REFLECTIO

N C

OEFFIC

IENT IN

DEG

REES

—>

WA

VEL

ENG

THS

TOW

ARD

GEN

ERA

TOR

—>

<— W

AVE

LEN

GTH

S TO

WA

RD L

OA

D <

—

IND

UCT

IVE

REA

CTAN

CE C

OMPO

NENT (+jX/Zo), O

R CAPACITIVE SUSCEPTANCE (+jB/Yo)

CAPACITIVE REACTANCE COMPONENT (-j

X/Zo), O

R INDUCTI

VE SU

SCEP

TAN

CE (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

SWR CIRCLE

r=1 CIRCLE

LOAD

Single stub impedancematching is easy to dowith the Smith Chart

Simply find theintersection of the SWRcircle with the r = 1circle

The match is at thecenter of the circle. Graba reactance in series orshunt to move you there!

23 / 29

Series Stub Match

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1.6

1.81.8

1.8

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1.01.0

90-9

085

-85

80-8

0

75-7

5

70-7

0

65-6

5

60-6

0

55-55

50-50

45

-45

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-40

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-35

30

-30

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-25

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-20

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-15

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-10

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0.20.2

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0.220.22

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0.240.25

0.25

0.26

0.26

0.27

0.27

0.28

0.28

0.29

0.29

0.3

0.3

0.31

0.31

0.32

0.320.33

0.33

0.34

0.34

0.35

0.35

0.36

0.36

0.37

0.37

0.38

0.38

0.39

0.39

0.4

0.4

0.41

0.41

0.42

0.42

0.43

0.43

0.44

0.44

0.45

0.45

0.46

0.46

0.47

0.47

0.48

0.48

0.49

0.49

0.0

0.0

AN

GLE O

F TRAN

SMISSIO

N C

OEFFIC

IENT IN

DEG

REES

AN

GLE O

F REFLECTIO

N C

OEFFIC

IENT IN

DEG

REES

—>

WA

VEL

ENG

THS

TOW

ARD

GEN

ERA

TOR

—>

<— W

AVE

LEN

GTH

S TO

WA

RD L

OA

D <

—

IND

UCT

IVE

REA

CTAN

CE C

OMPO

NENT (+jX/Zo), O

R CAPACITIVE SUSCEPTANCE (+jB/Yo)

CAPACITIVE REACTANCE COMPONENT (-j

X/Zo), O

R INDUCTI

VE SU

SCEP

TAN

CE (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

SWR CIRCLE

r=1 CIRCLE

LOADOpen or Short

Z0

Z0

d

ZL

24 / 29

Shunt Stub Match

0.1

0.1

0.1

0.2

0.2

0.2

0.3

0.3

0.3

0.4

0.4

0.4

0.50.5

0.5

0.60.6

0.6

0.7

0.7

0.7

0.8

0.8

0.8

0.9

0.9

0.9

1.0

1.0

1.0

1.2

1.2

1.2

1.4

1.4

1.4

1.61.6

1.6

1.81.8

1.8

2.02.0

2.0

3.0

3.0

3.0

4.0

4.0

4.0

5.0

5.0

5.0

10

10

10

20

20

20

50

50

50

0.2

0.2

0.2

0.2

0.4

0.4

0.4

0.4

0.6

0.6

0.6

0.6

0.8

0.8

0.8

0.8

1.0

1.0

1.01.0

90-9

085

-85

80-8

0

75-7

5

70-7

0

65-6

5

60-6

0

55-55

50-50

45

-45

40

-40

35

-35

30

-30

25

-25

20

-20

15

-15

10

-10

0.04

0.04

0.05

0.05

0.06

0.06

0.07

0.07

0.08

0.08

0.09

0.09

0.1

0.1

0.11

0.11

0.12

0.120.13

0.130.14

0.140.15

0.150.16

0.160.17

0.17

0.18

0.18

0.190.19

0.20.2

0.210.21

0.220.22

0.23

0.230.24

0.240.25

0.25

0.26

0.26

0.27

0.27

0.28

0.28

0.29

0.29

0.3

0.3

0.31

0.31

0.32

0.320.33

0.33

0.34

0.34

0.35

0.35

0.36

0.36

0.37

0.37

0.38

0.38

0.39

0.39

0.4

0.4

0.41

0.41

0.42

0.42

0.43

0.43

0.44

0.44

0.45

0.45

0.46

0.46

0.47

0.47

0.48

0.48

0.49

0.49

0.0

0.0

AN

GLE O

F TRAN

SMISSIO

N C

OEFFIC

IENT IN

DEG

REES

AN

GLE O

F REFLECTIO

N C

OEFFIC

IENT IN

DEG

REES

—>

WA

VEL

ENG

THS

TOW

ARD

GEN

ERA

TOR

—>

<— W

AVE

LEN

GTH

S TO

WA

RD L

OA

D <

—

IND

UCT

IVE

REA

CTAN

CE C

OMPO

NENT (+jX/Zo), O

R CAPACITIVE SUSCEPTANCE (+jB/Yo)

CAPACITIVE REACTANCE COMPONENT (-j

X/Zo), O

R INDUCTI

VE SU

SCEP

TAN

CE (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

SWR CIRCLE

r=1 CIRCLE

LOAD

Open or Short

Z0

Z0

Z0 d

ZL

Let’s now solve the samematching problem with ashunt stub.

To find the shunt stubvalue, simply convert thevalue of z = 1 + jx toy = 1 + jb and place areactance of −jb in shunt

25 / 29

Lumped Components On Smith Chart

LL

CC

g-circle r-circle

Adding an inductor inseries moves us onthe constant r circleclock-wise (CW).Adding a capacitor inseries moves counterclock-wise (CCW).

On the Y-plane, to toCW, add a shunt C.To move CCW, add ashunt L.

26 / 29

Matching with Lumped Components (Inside)

0.10.1

0.1

0.20.2

0.2

0.30.3

0.3

0.40.4

0.4

0.50.5

0.5

0.60.6

0.6

0.7

0.7

0.7

0.8

0.8

0.8

0.9

0.9

0.9

1.0

1.0

1.0

1.2

1.2

1.2

1.4

1.4

1.4

1.61.6

1.6

1.81.8

1.8

2.02.0

2.0

3.03.0

3.0

4.04.0

4.0

5.05.0

5.0

1010

10

2020

20

5050

50

0.20.2

0.20.2

0.40.4

0.40.4

0.60.6

0.60.6

0.80.8

0.80.8

1.01.0

1.01.0

90-9

085

-85

80-8

0

75-7

5

70-7

0

65-6

5

60-6

0

55-55

50-50

45

-45

40-40

35-35

30-30

25-25

20-20

15-15

10-10

0.04

0.04

0.05

0.05

0.06

0.06

0.07

0.070.08

0.080.09

0.090.1

0.10.11

0.11

0.12

0.12

0.13

0.13

0.14

0.14

0.15

0.15

0.16

0.16

0.17

0.17

0.18

0.18

0.190.19

0.20.2

0.210.21

0.220.22

0.23

0.230.24

0.240.25

0.25

0.26

0.26

0.27

0.27

0.28

0.28

0.29

0.29

0.3

0.3

0.31

0.31

0.320.32

0.330.33

0.340.34

0.350.35

0.360.36

0.370.370.38

0.38

0.390.39

0.40.4

0.410.41

0.420.42

0.430.43

0.44

0.44

0.45

0.45

0.46

0.46

0.47

0.47

0.48

0.48

0.49

0.49

0.0

0.0

AN

GLE O

F TRAN

SMISSIO

N C

OEFFIC

IENT IN

DEG

REES

AN

GLE O

F REFLECTIO

N C

OEFFIC

IENT IN

DEG

REES

—>

WA

VEL

ENG

THS

TOW

ARD

GEN

ERA

TOR

—>

<— W

AVE

LEN

GTH

S TO

WA

RD L

OA

D <

—

IND

UCT

IVE

REA

CTAN

CE C

OMPO

NENT (+jX/Zo), O

R CAPACITIVE SUSCEPTANCE (+jB/Yo)CAPACITIVE REACTANCE COMPONENT (-j

X/Zo), O

R INDUCTI

VE SU

SCEP

TAN

CE (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

r=1 CIRCLE

LOAD (z)

g=1 CIRCLE

LOAD (y)

Suppose the load is inside the 1 + jx circle.27 / 29

Escaping from “Insdie”

L

L

C

g-circle

r-circle

C

Notice that there are two paths to get to the center. Thedifference is one is AC coupled versus DC coupled, so oftenthe application will determine the choice.

28 / 29

Matching with Lumped Components (Outside)

0.10.1

0.1

0.20.2

0.2

0.30.3

0.3

0.40.4

0.4

0.50.5

0.5

0.60.6

0.6

0.7

0.7

0.7

0.8

0.8

0.8

0.9

0.9

0.9

1.0

1.0

1.0

1.2

1.2

1.2

1.4

1.4

1.4

1.61.6

1.6

1.81.8

1.8

2.02.0

2.0

3.03.0

3.0

4.04.0

4.0

5.05.0

5.0

1010

10

2020

20

5050

50

0.20.2

0.20.2

0.40.4

0.40.4

0.60.6

0.60.6

0.80.8

0.80.8

1.01.0

1.01.0

90-9

085

-85

80-8

0

75-7

5

70-7

0

65-6

5

60-6

0

55-55

50-50

45

-45

40-40

35-35

30-30

25-25

20-20

15-15

10-10

0.04

0.04

0.05

0.05

0.06

0.06

0.07

0.070.08

0.080.09

0.090.1

0.10.11

0.11

0.12

0.12

0.13

0.13

0.14

0.14

0.15

0.15

0.16

0.16

0.17

0.17

0.18

0.18

0.190.19

0.20.2

0.210.21

0.220.22

0.23

0.230.24

0.240.25

0.25

0.26

0.26

0.27

0.27

0.28

0.28

0.29

0.29

0.3

0.3

0.31

0.31

0.320.32

0.330.33

0.340.34

0.350.35

0.360.36

0.370.370.38

0.38

0.390.39

0.40.4

0.410.41

0.420.42

0.430.43

0.44

0.44

0.45

0.45

0.46

0.46

0.47

0.47

0.48

0.48

0.49

0.49

0.0

0.0

AN

GLE O

F TRAN

SMISSIO

N C

OEFFIC

IENT IN

DEG

REES

AN

GLE O

F REFLECTIO

N C

OEFFIC

IENT IN

DEG

REES

—>

WA

VEL

ENG

THS

TOW

ARD

GEN

ERA

TOR

—>

<— W

AVE

LEN

GTH

S TO

WA

RD L

OA

D <

—

IND

UCT

IVE

REA

CTAN

CE C

OMPO

NENT (+jX/Zo), O

R CAPACITIVE SUSCEPTANCE (+jB/Yo)CAPACITIVE REACTANCE COMPONENT (-j

X/Zo), O

R INDUCTI

VE SU

SCEP

TAN

CE (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

r=1 CIRCLE

LOAD (z)

g=1 CIRCLE

Suppose the load is outside the 1 + jx circle.29 / 29