Lecture 6 Lossy Transmission Lines and the Smith Chart

8/2/2019 Lecture 6 Lossy Transmission Lines and the Smith Chart

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EECS 117

Lecture 6: Lossy Transmission Lines and the SmithChart

Prof. Niknejad

University of California, Berkeley

University of California, Berkeley EECS 117 Lecture 6 p. 1/


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Dispersionless Line

To find the conditions for the transmission line to bedispersionless in terms of the R, L, C, G, expand

=

(jL + R)(jC + G)

=

(j)2LC(1 +

R

jL+

G

jC+

RG

(j)2LC)

=

(j)2LC

Suppose that R/L = G/Cand simplify the

term

= 1 +2R

jL+

R2

(j)2L2



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Dispersionless Line (II)

For R/L = G/C the propagation constant simplifies

=

1 +

R

jL2

= jLC1 + RjLBreaking into real and imaginary components

= RCLjLC= + j

The attenauation constant is independent of

frequency. For low loss lines, RZ0 The propagation constant is a linear function of

frequencyUniversity of California, Berkeley EECS 117 Lecture 6 p. 3/


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Lossy Transmission Line Attenuation

The power delivered into the line at a point z is nownon-constant and decaying exponentially

Pav(z) = 12 (v(z)i(z)) = |v

+

|2

2|Z0|2 e2z (Z0)

For instance, if = .01m1, then a transmission line oflength = 10m will attenuate the signal by 10 log(e2) or

2 dB. At = 100m will attenuate the signal by 10 log(e2)or 20 dB.



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Lossy Transmission Line Impedance

Using the same methods to calculate the impedance forthe low-loss line, we arrive at the following linevoltage/current

v(z) = v+ez(1 + Le2z) = v+ez(1 + L(z))

i(z) =v+

Z0 ez

(1 L(z))Where L(z) is the complex reflection coefficient atposition z and the load reflection coefficient is unalteredfrom before

The input impedance is therefore

Zin(z) = Z0 ez + Lezez Lez



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Lossy T-Line Impedance (cont)

Substituting the value of L we arrive at a similarequation (now a hyperbolic tangent)

Zin() = Z0ZL + Z0 tanh()Z0 + ZL tanh()

For a short line, if

1, we may safely assume that

Zin() = Z0 tanh() Z0

Recall that Z0= Z/YZYAs expected, input impedance is therefore the seriesimpedance of the line (where R = R and L = L)

Zin() = Z = R + jLUniversity of California, Berkeley EECS 117 Lecture 6 p. 6/


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Review of Resonance (I)

Wed like to find the impedance of a series resonator

near resonance Z() = jL + 1jC + R

Recall the definition of the circuit Q

Q = 0time average energy stored

energy lost per cycle

For a series resonator, Q = 0L/R. For a smallfrequency shift from resonance 0

Z(0 + ) = j0L + jL +1

j0C

1

1 + 0

+ R



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Review of Resonance (II)

Which can be simplified using the fact that 0L =1

0C

Z(0 + ) = j2L + R

Using the definition of Q

Z(0 + ) = R

1 + j2Q

0

For a parallel line, the same formula applies to the

admittance

Y(0 + ) = G

1 + j2Q

0

Where Q = 0C/GUniversity of California, Berkeley EECS 117 Lecture 6 p. 8/


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/2 T-Line Resonators (Series)

A shorted transmission line of length has inputimpedance of Zin = Z0 tanh()

For a low-loss line, Z0 is almost real

Expanding the tanh term into real and imaginary parts

tanh(+j) =

sinh(2)

cos(2) + cosh(2) +

j sin(2)

cos(2) + cosh(2)

Since 0f0 = c and = 0/2 (near the resonantfrequency), we have = 2/ = 2f/c = + 2f/c = + /0

If the lines are low loss, then 1



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/2 Series Resonance

Simplifying the above relation we come to

Zin = Z0 + j 0

The above form for the input impedance of the seriesresonant T-line has the same form as that of the series

LRC circuit

We can define equivalent elements

Req = Z0 = Z0/2

Leq =

Z0

20 Ceq =

2

Z00



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/2 Series Resonance Q

The equivalent Q factor is given by

Q =1

0ReqCeq=

0=

0

2

For a low-loss line, this Q factor can be made very large.A good T-line might have a Q of 1000 or 10,000 or more

Its difficult to build a lumped circuit resonator with sucha high Q factor



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/4 T-Line Resonators (Parallel)

For a short-circuited /4 line

Zin = Z0 tanh( + j) = Z0tanh + j tan

1 + j tan tanh

Multiply numerator and denominator by j cot

Zin = Z01j tanh cot tanhj cot

For = /4 at = 0 and = 0 +

=0

v+

v=

2+

20


/


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/4 T-Line Resonators (Parallel)

So cot = tan20

20

and tanh

Zin = Z0

1 + j/20

+ j/20 Z0

+ j/20

This has the same form for a parallel resonant RLCcircuit

Zin =1

1/R + 2jC

The equivalent circuit elements are

Req =Z0

Ceq =

40Z0Leq =

1

20Ceq


/


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/4 T-Line Resonators Q Factor

The quality factor is thus

Q = 0RC=

4

=

2



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The Smith Chart

The Smith Chart is simply a graphical calculator forcomputing impedance as a function of reflectioncoefficient z = f()

More importantly, many problems can be easilyvisualized with the Smith Chart

This visualization leads to a insight about the behavior

of transmission lines

All the knowledge is coherently and compactlyrepresented by the Smith Chart

Why else study the Smith Chart? Its beautiful!

There are deep mathematical connections in the SmithChart. Its the tip of the iceberg! Study complex analysis

to learn more.



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Generalized Reflection Coefficient

In sinusoidal steady-state, the voltage on the line is aT-line

v(z) = v+

(z) + v

(z) = V+

(ez

+ Lez

)

Recall that we can define the reflection coefficientanywhere by taking the ratio of the reflected wave to the

forward wave

(z) =v(z)

v+(z)=

Lez

ez= Le

2z

Therefore the impedance on the line ...

Z(z) =v+ez(1 + Le

2z)v+

Z0ez(1 Le2z)


N li d I d


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Normalized Impedance

...can be expressed in terms of (z)

Z(z) = Z01 + (z)

1 (z)It is extremely fruitful to work with normalizedimpedance values z = Z/Z0

z(z) =Z(z)

Z0=

1 + (z)

1 (z)Let the normalized impedance be written as z = r + jx(note small case)

The reflection coefficient is normalized by default

since for passive loads || 1. Let = u + jvUniversity of California, Berkeley EECS 117 Lecture 6 p. 18/

Di ti f th T f ti


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Dissection of the Transformation

Now simply equate the and components in theabove 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 anice formUniversity of California, Berkeley EECS 117 Lecture 6 p. 19/

C l ti Y S


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Completing Your Squares...

If you remember your high school algebra, you canderive the following equivalent equations

u r

1 + r

2+ v2 = 1

(1 + r)2

(u 1)2 + v 1x2

= 1x2

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 tocircles in the (u,v) plane (complex reflection coefficient)


R i t T f ti


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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 realpart)r = .5 maps to (u 1/3)2 + v2 = (2/3)2 (load R less thanZ0)

r = 2 maps to (u 2/3)2 + v2 = (1/3)2 (load R greaterthan Z0)


R t T f ti


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Reactance Transformations

x

r

u

x=.5

x

=

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

x = 1/2 maps to (u 1)2 + (v 2)2 = 22Inductive reactance maps to upper half of unit circle

Capacitive reactance maps to lower half of unit circle


Complete Smith Chart


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Complete Smith Chart

u

r=0

r

=0

.5

r

=

1

x=

.5

x

=

1 x=2

x=-.5

x

=

-1

x=-2

v

short

open

match

r

=2

r > 1

inductive

capacitive


Load on Smith Chart


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Load on Smith Chart

load

First map zL on the Smith Chart as L

To read off the impedance on the T-line at any point ona lossless line, simply move on a circle of constant

radius since (z) = Le2j


Motion Towards Generator


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Motion Towards Generator

load

mov

emen

t towards generator

Moving towardsgenerator means

(

) = Le

2j, or

clockwise motionFor a lossy line, thiscorresponds to a

spiral motionWere back to wherewe started when

2 = 2, or = /2Thus the impedanceis periodic (as we

know)


SWR Circle


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SWR Circle

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

SW

RC I R

C L E

voltage min voltage max

L = |L|ej

=

|L|ej(2)

Recall the voltage max

occurs when the reflectedwave is in phase with theforward wave, so(zmin) =

|L|This corresponds to the

intersection of the SWRcircle with the positive real

axis

Likewise, the intersectionwith the negative real axis

is the location of the voltgemin


Example of Smith Chart Visualization


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Example of Smith Chart Visualization

Prove that if ZL has an inductance reactance, then theposition of the first voltage maximum occurs before thevoltage minimum as we move towards the generator

A visual proof is easy using Smith Chart

On the Smith Chart start at any point in the upper halfof the unit circle. Moving towards the generator

corresponds to clockwise motion on a circle. Thereforewe will always cross the positive real axis first and thenthe negative real axis.


Impedance Matching Example


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Impedance Matching Example

Single stub impedance matching is easy to do with theSmith Chart

Simply find the intersection of the SWR circle with the

r = 1 circle

The match is at the center of the circle. Grab areactance in series or shunt to move you there!


Series Stub Match


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Series Stub Match


Admittance Chart


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Admittance Chart

Since y = 1/z = 11+ , you can imagine that an

Admittance Smith Chart looks very similar

In fact everything is switched around a bit and you canbuy or construct a combined admittance/impedancesmith chart. You can also use an impedance chart foradmittance if you simply map x

b and r

g

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

The short and open likewise swap positions


Admittance on Smith Chart


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Admittance on Smith Chart

Sometimes you may need to work with bothimpedances and admittances.

This is easy on the Smith Chart due to the impedance

inversion property of a /4 line

Z =Z20

ZIf we normalize Z we get y

Z

Z0= Z0Z = 1z = y


Admittance Conversion


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Admittance Conversion

Thus if we simply rotate degrees on the Smith Chartand read off the impedance, were actually reading offthe admittance!

Rotating degrees is easy. Simply draw a line throughorigin and zL and read off the second point ofintersection on the SWR circle


Shunt Stub Match


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Shunt Stub Match

Lets now solve the same matching problem with ashunt stub.

To find the shunt stub value, simply convert the value of

z = 1 + jx to y = 1 + jb and place a reactance of jb inshunt


Lecture 6 Lossy Transmission Lines and the Smith Chart

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