Bài giảng: Kỹ thuật siêu cao tần (CT389)Chương II: Đồ thị Smith(Smith Chart)

3/21/2012

Bài giảng: Kỹ

thuật siêu cao tần (CT389)

Chương II : Đồ

thị Smith

(Smith Chart)

Giảng viên: GVC.TS. Lương Vinh Quốc Danh
Bộ môn Điện tử Viễn thông, Khoa Công Nghệ
E-mail:

Luong Vinh Quoc Danh

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Introduction:
 The Smith chart is one of the most useful graphical tools for high frequency
circuit applications.
 From a mathematical point of view, the Smith chart is a representation of all
possible complex impedances with respect to coordinates defined by the complex
reflection coefficient.
 The domain of definition of the reflection coefficient for a lossless line is a circle
of unitary radius (vòng tròn đơn vị) in the complex plane (mặt phẳng phức). This is
also the domain of the Smith chart.

The goal of the Smith chart is to identify all possible

impedances on the domain of existence of the
reflection coefficient.

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Introduction: (Cont’d)
We start from the general definition of line impedance:

1  ( x )
1  ( x )

Z ( x)  Z 0

(2.1)

or normalized line impedance:

Reflection coefficient:

z( x) 

1  ( x )
1  ( x )

( x ) 

Z ( x)  Z 0
Z ( x)  Z 0

(2.3)

z( x)  1
z( x)  1

(2.4)

( x ) 

(2.2)

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Introduction: (Cont’d)

Let’s represent the reflection coefficient in terms of its coordinates

(x) = r(x) + j i(x)
or  = r + ji
Where

(2.6)

r = Re() and i = Im()

Normalized line impedance: z = r + jx

(2.8)

r = R/Z0: normalized resistance
x = X/Z0: normalized reactance

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Introduction: (Cont’d)
Now we can rewrite (2.2) as follows:

r  jx 

The real part r gives:

r

1  r  ji
1  r  ji

(2.10)

1  r2  i2
(1  r ) 2  i2
2

r 

 1 
2
 r 
  i  

1 r 

1 r 

(2.11)

2

(2.13)

Equation of a circle

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Introduction: (Cont’d)
The result for the real part indicates that on the complex plane with coordinates
(Re(Γ), Im(Γ)) all the possible impedances with a given normalized resistance r
are found on a circle with

As the normalized resistance r varies from 0 to ∞ , we obtain a family of circles
completely contained inside the domain of the reflection coefficient | Γ | ≤ 1.

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Introduction: (Cont’d)
The imaginary part x gives:

x

2i
(1  r ) 2  i2

r  12   i


(2.12)

2

1
1
   
x
 x

2

(2.15)

Equation of a circle
The result for the imaginary part indicates that on the complex plane with
coordinates (Re(Γ), Im(Γ)) all the possible impedances with a given normalized
reactance x are found on a circle with

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Introduction: (Cont’d)
As the normalized reactance x varies from -∞ to ∞ , we obtain a family of arcs
contained inside the domain of the reflection coefficient | Γ | ≤ 1 .

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Introduction (Cont’d)
Basic Smith Chart techniques for loss-less transmission lines

 Given Z(d)  Find Γ(d)
Given Γ(d)  Find Z(d)
 Find dmax and dmin (maximum and minimum locations for the voltage
standing wave pattern)
 Find the Voltage Standing Wave Ratio (VSWR)

 Given Z(d)  Find Y(d)
Given Y(d)  Find Z(d)

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Smith Chart
Phillip Hagar Smith (1905–1987): graduated from Tufts
College in 1928, invented the Smith Chart in 1939 while
he was working for the Bell Telephone Laboratories.

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Smith Chart (Cont’d)
Toward
Generator

Constant
Reflection
Coefficient Circle

Away
From
Generator

Scale in
Wavelengths

Full Circle is One Half
Wavelength Since
Everything Repeats
K. A. Connor
Luong
VinhDepartment
Quoc Danh
RPI ECSE

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Smith Chart (Cont’d)
(2.20)

where

d=l–x

 = 0 (lossless)

(2.21)

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Smith Chart (Cont’d)

SWR: Standing Wave Ratio
RFL. COEFF. P : Reflection Coefficient
Power ||2
RFL. COEFF. E or I : Reflection
Coefficient Voltage ||
RTN. LOSS [dB] : Return Loss S11

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Smith Chart : Admittances

j

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Smith Chart : Admittances (Cont’d)
Since related impedance and admittance are on opposite sides of the
same Smith chart, the imaginary parts always have different sign.
Therefore, a positive (inductive) reactance corresponds to a negative
(inductive) susceptance, while a negative (capacitive) reactance
corresponds to a positive (capacitive) susceptance.
Analytically, the normalized impedance and admittance are related as

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Smith Chart : Nodes and Anti-nodes

(anti-node)
r max = S

Load impedance ZL can
be deduced graphically
if we know:
+ SWR
+ Distance from node
(anti-node) to load.

(node)
r min = 1/S

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Basic Applications


Given Z(d) ⇒ Find Γ(d)

1. Normalize the impedance z
2. Find the circle of constant normalized resistance r
3. Find the arc of constant normalized reactance x

4. The intersection of the two curves indicates the reflection
coefficient in the complex plane. The chart provides directly the
magnitude and the phase angle of Γ(d)

Example: Find Γ(d), given Z(d)=25+ j100 [Ω] with Z0 = 50Ω

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Basic Applications (Cont’d)

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Basic Applications (Cont’d)
• Given Γ(d) ⇒ Find Z(d)

1. Determine the complex point representing the given
reflection coefficient Γ(d) on the chart.
2. Read the values of the normalized resistance r coefficient
point. and of the normalized reactance x that correspond to the
reflection coefficient point.
3. The normalized impedance is

and the actual impedance is

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Basic Applications (Cont’d)


Given ΓR and ZR ⇐⇒ Find Γ(d) and Z(d)

1. Identify the load reflection coefficient ΓR and the normalized load
impedance ZR on the Smith chart.
2. Draw the circle of constant reflection coefficient amplitude |Γ(d)|
= |ΓR|.
3. Starting from the point representing the load, travel on
the circle in the clockwise direction, by an angle

4. The new location on the chart corresponds to location d on the
transmission line. Here, the values of Γ(d) and Z(d) can be read
from the chart as before.
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Basic Applications (Cont’d)
Example: Given ZR = 25+ j 100 (Ω) with Z0=50 (Ω)
Find Z(d) and Γ(d) for d = 0.18λ



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Basic Applications (Cont’d)


Given ΓR and ZR ⇒ Find dmax and dmin

1. Identify on the Smith chart the load reflection coefficient ΓR or
the normalized load impedance ZR.
2. Draw the circle of constant reflection coefficient amplitude |Γ(d)|
= |ΓR|. The circle intersects the real axis of the reflection
coefficient at two points which identify dmax (when Γ(d) = Real
positive) and dmin (when Γ(d) = Real negative)
3. A commercial Smith chart provides an outer graduation where
the distances normalized to the wavelength can be read directly.
The angles, between the vector ΓR and the real axis, also provide
a way to compute dmax and dmin.

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Basic Applications (Cont’d)
Example: Find dmax and dmin for ZR =25+j100 (Ω); and ZR =
25−j100 (Ω) with Z0 = 50 (Ω).

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Basic Applications (Cont’d)

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Basic Applications (Cont’d)


Given ΓR and ZR ⇒ Find the Voltage Standing Wave Ratio
(VSWR)

1. Identify the load reflection coefficient ΓR and the normalized
load impedance ZR on the Smith chart.
2. Draw the circle of constant reflection coefficient amplitude |Γ(d)|
= |ΓR|.
3. Find the intersection of this circle with the real positive axis for
the reflection coefficient (corresponding to the transmission line
location dmax).
4. A circle of constant normalized resistance will also intersect this
point. Read or interpolate the value of the normalized resistance
to determine the VSWR.
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Basic Applications (Cont’d)
Example: Find the VSWR for ZR1= 25 + j100 (Ω) ; ZR2= 25 − j100
(Ω) with Z0 = 50 Ω

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Basic Applications (Cont’d)


Given Z(d) ⇐⇒ Find Y(d)

1. Identify the load reflection coefficient ΓR and the normalized
load impedance ZR on the Smith chart.
2. Draw the circle of constant reflection coefficient amplitude
|Γ(d)| =|ΓR|.
3. The normalized admittance is located at a point on the circle of
constant |Γ| which is diametrically opposite to the normalized
impedance.

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Basic Applications (Cont’d)
Example: Given ZR = 25+ j 100 (Ω) with Z0 = 50 Ω. Find YR.

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Basic Applications (Cont’d)


Find circuit impedance

Given R = 50 ; C1 = 10 pF; C2 = 12 pF; L = 22.5 nH;
 = 109 rad/s. Find Z.

Z

Choose Z0 = 50 
(point A)
(point B)
(point C)
(point D)

(point E)

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Impedance matching
Power transferred to a voltage anti-node on the lossless TX line:
(2.52)
(2.55)
where

: Power generated by signal source
: Reflected power

When || = 0, there is no reflected waves, therefore, no reflected power. Power generated
by signal source is totally received by load.

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Impedance matching (Cont’d)


For lossy transmission line:
P L

: incident power at load

P  : incident power at source
S

P L

: reflected power at load

PS : reflected power at source

PL  PS . e  2  l

(2.57)

PL  |  |2 PL  |  |2 . PS . e  2  l

(2.58)

PS  PL . e  2  l  |  |2 . P  . e  4  l (2.59)
S

Power generated by source:
PS  PS  PS  PS ( 1  |  |2 . e  4  l )

(2.60)

Power consumption at load:
2
PL  PL  PL  PS . e  2  l ( 1   )

(2.61)
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Impedance matching (Cont’d)
Power efficiency:


PL
PS



e  2  l ( 1  |  |2 )
1  |  |2 . e  4  l

When   0    e  2  l

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(2.62)

(2.63)

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Impedance matching using lumped elements

Matching
circuit

Z0

ZL

-circuit

-circuit
X1
Z0

X2

X1
ZL

Z0

X2

ZL

X1, X2 : inductance, or capacitance
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Impedance matching using lumped elements (Cont’d)

Chip capacitors

Chip resistors
(size = 2 mm x 1 mm)
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Impedance matching: -circuit
Example: ZL = 10 - j40 (); Z0 = 50 ;  = 109 rad/s

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Impedance matching: -circuit (cont’d)

Zt’ = ZL + jx1

Zt = ZL + jx1

ZL = 0.2 – j0.8
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Impedance matching: -circuit (cont’d)

Zt’ = ZL + jx1

Zt = ZL + jx1

yt = 1 + j2

yt‘= 1 - j2

y = yt + jb2
jx1 = zt – zL = 0.2 –j0.4 – (0.2 – j0.8)

jx1 = zt‘ – zL = 0.2 + j0.4 – (0.2 – j0.8)
= j1.2

= j0.4  L1 =0.4R0 / = 20 nH

 L1 = 1.2R0/ = 60 nH

jb2 = 1- yt = - j2  L2 = R0/2 = 25 nH

 jb2 = 1- yt‘ = j2  C2 = 2/R0 =
40 pF

R0 = 50 
ZL = 10 – j40 []

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Impedance matching: -circuit


Example: ZL = 10 - j40 (); Z0 = 50 ;  = 109 rad/s

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Impedance matching: -circuit (cont’d)

yt = yL + jb2

zL = 0.2 – j0.8

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Impedance matching: -circuit (cont’d)
yt’ = 0.3 – j0.46

yt = 0.3 + j0.46

zt = 1 – j1.55

zt‘= 1 + j1.55

z = zt + jx1

jb2 = yt‘ – yL = 0.3 - j0.46 – (0.3 +
j1.18) = -j1.64

jb2 = yt – yL = 0.3 + j0.46 – (0.3 + j1.18)

 L2 = R0/1.64 = 30.5 nH

= - j0.72  L2 = R0 /0.72  = 70 nH

 jx1 = 1- zt‘ = - j1.55  C1 =
1/1.5R0 = 13 pF

jx1 = 1- zt = j1.55  L1 = 1.55R0/ =
77.5 nH

R0 = 50 

R0 = 50 
ZL = 10 – j40 []

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ZL = 10 – j40 []

40

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Impedance matching: forbidden regions
Forbidden region
of -circuit

Forbidden region
of  -circuit

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Impedance matching: Single stub method


Example: Z0 = 50 , ZL = 50/[2+j(2+3)] (). Short-circuited stub:
length = l, Rs0 = 100 (). Find l, d.

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Impedance matching: Single stub (cont’d)

yL = Z0/ZL = 2+ j3.732

yL  moving toward Gen.  yd = 1- j2.6

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Impedance matching: Single stub (cont’d)
 yd = 1 – j2.6
Yd = yd .G0 = (1- j2.6)/50 = 0.02 – j0.052 [S]
Therefore Bd = - 0.052
Bs = - Bd = 0.052
Normalized susceptance bs = Bs/Gs0 = Bs . Rs0 = 0.052 x 100 = 5.2
Short-circuited stub:
l = (0.5 – 0.031)  = 0.469
y = j5.2

d = (0.302 – 0.215)  = 0.087 
y=∞
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Impedance matching: Single stub (cont’d)
An RF Filter (1.1 GHz)

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Impedance matching: Double stub method


Example: Z0 = 50 ; ZL = (100 + j100) []. Stub #1: l1, shortcircuited line, d = 0.4. Stub #2: l2; d12 = 3/8. Find l1, l2.

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Impedance matching: Double stub (cont’d)

zL = 2+j2

yd1= 0.55 – j1.08

yt1 = yd1 + ys1
yt1’ = yd1’ + ys1

yL rotated CW 0.4 yd1

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Impedance matching: Double stub (cont’d)

ys1 = yt1 – yd1 = j0.97
ys1’ = yt1’ – yd1’ = -j0.8
ys1  l1 = 0.373
ys1’  l2 = 0.143

yt1 = 0.55 – j0.11
yt1’ = 0.55 – j1.88
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Impedance matching: Double stub (cont’d)

ys2 = yt2 – yd2 = j0.61

yd2’ = 1+ j2.6

ys2  l2 = 0.337
yt1’ rotated CW 3/8  yd2’

ys2’ = yt2’ – yd2’ = -j2.6
ys2’  l2’ = 0.058

yt1 rotated CW 3/8  yd2

yd2 = 1- j0.61

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Impedance matching: Double stub (cont’d)






The advantage of this technique is the position of stubs
(d12 and x) are fixed. The matching are done by
changing the length of stubs.
d12 = /8, /4, or 3/8
The disadvantage of this technique is not all impedances
can be matched (forbidden regions depend on values of

d12 and x).

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