Lecture 4b -- Transmission Line Behavior · Transmission Line Behavior Slide 33 Example: Impedance...

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Transmission Line Behavior Slide 1

EE 4347

Applied Electromagnetics

Topic 4b

Transmission Line Behavior These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited

Course InstructorDr. Raymond C. RumpfOffice: A‐337Phone: (915) 747‐6958E‐Mail: [email protected]

Lecture Outline


• Scattering at an Impedance Discontinuity

• Power on a Transmission Line

• Voltage Standing Wave Ratio (VSWR)

• Input Impedance, Zin

• Parameter Relations

• Special Cases of Terminated Transmission Lines

– Shorted line (ZL = 0)

– Open‐circuit line (ZL = )–Matched line (ZL = Z0)

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Scattering at an Impedance Discontinuity


Problem Setup

Transmission Line 1

z

1 1, Z

Transmission Line 2

2 2, Z

We will get a reflection

?

0z

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Incorporate Reflected Wave

Transmission Line 1

z

1 1, Z

Transmission Line 2

2 2, Z

0z

1 1

1 1

1 1 1

1 11

1 1

z z

z z

V z V e V e

V VI z e e

Z Z

2

2

2 2

22

2

z

z

V z V e

VI z e

Z


Enforce Boundary Conditions (1 of 2)

Transmission Line 1

z

1 1, Z

Transmission Line 2

2 2, Z

0z

1 1 2

1 1 2

1 2

1 1 2

1 2

1 1 2

1 1 2

z z z

z z z

V z V z

V e V e V e

I z I z

V V Ve e e

Z Z Z

Boundary conditions require the voltage and current on either side of the interface to be equal.

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Enforce Boundary Conditions (2 of 2)

Transmission Line 1

z

1 1, Z

Transmission Line 2

2 2, Z

0z

1 2

1 1 2

1 2

1 1 2

1 1 2

0 0

0 0

V V

V V V

I I

V V V

Z Z Z

The interface occurs at z = 0.


Reflection Coefficient,

1 1 2 Eq. 1V V V

Enforcing the boundary conditions at z = 0 gave us

1 1 2

1 1 2

Eq. 2V V V

Z Z Z

Substitute Eq. (1) into Eq. (2) to eliminate . 2V

1 1 1 1

1 1 2

V V V V

Z Z Z

Solve this new expression for . 1 1V V

1 1 1 11 1 2 2

1 1 1 11 2 1 2

1 11 2 1 2

2 1 1 2 1 1

1 2 1

1 2 1

1 1 1 1

1 1 1 1

1 1 1 1

V V V VZ Z Z Z

V V V VZ Z Z Z

V VZ Z Z Z

Z Z V Z Z V

V Z Z

V Z Z

1 2 1

1 2 1

V Z Z

V Z Z

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Revised Equations for V(z) and I(z)

The total voltage and current in any section of line was written as

0 0z zV z V e V e 0 0

0 0

z zV VI z e e

Z Z

Using the concept of the reflection coefficient , these equations can now be written as

0 0 0z z z zV z V e V e V e e

0 0 0

0 0 0

z z z zV V VI z e e e e

Z Z Z

0 2 1

0 2 1

V Z Z

V Z Z

Reflection coefficient at the load


Power on a Transmission Line

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Power Flowing Along Length of Line

The RMS power flowing at a distance z from the load is

*avg

1Re

2P z V z I z

* is complex conjugate

For lossless lines (not lossless loads), we have

00

0

j z j z j z j zL L

VV z V e e I z e e

Z

Substituting these equations into our expression for Pavg(z) gives

*

*0

avg 00

1Re

2j z j z j z j z

L L

VP z V e e e e

Z

20

avg0

12 L

VP

Z

Notice that the z dependence vanished. This is because power flows uniformly without decay in lossless lines.

This equation is valid for any line, even those with loss.


Voltage Standing Wave Ratio (VSWR)

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Voltage Standing Wave Ratio (VSWR)

The VSWR is essentially the same concept as the standing wave ratio (SWR) discussed along with waves. The only difference is that it describes voltage and current instead of electromagnetic fields.

max maxVSWR

min min

V z I z

V z I z


Derivation of VSWR (1 of 2)

We start with our expression for waves travelling in opposite directions on a transmission line. We will assume a lossless line.

00

0

j z j z j z j zL L

VV z V e e I z e e

Z

The magnitude of the voltage signal V(z) is

20 0 1j z j z j z

L LV z V e e V e

By inspection of this equation, we determine the maximum and minimum values of this function.

max 0

min 0

max 1

min 1

L

L

V V z V

V V z V

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Derivation of VSWR (2 of 2)

The VSWR is therefore

0

0

1maxVSWR

min 1

L

L

VV z

V z V

1 VSWR

1L

L

The VSWR is an easily measured quantity and we can calculate the magnitude of the reflection coefficient || from the VSWR.

VSWR 1

VSWR 1L


Animation of VSWR (1 of 6)

Case 1: 50 transmission line terminated with a short‐circuit load.

L 1

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Case 2: 50 transmission line terminated with an open‐circuit load.

L 1



Case 3: 50 transmission line terminated with a 16.5 load.

0 L L 0.5Z Z

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Case 4: 50 transmission line terminated with a 150 load.

0 L L 0.5Z Z



Case 5: 50 transmission line terminated with an RL load.

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Case 6: 50 transmission line terminated with an RC load.


Input Impedance, Zin

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Problem Setup

gZ

LZ

Generator Transmission Line Load

z0z z

0, ZinZ

gZ

inZ

Generator Input Impedance

zz

inZ

The input impedance Zin is the impedance observed by the generator.

The input impedance Zin is NOT necessarily the line’s characteristic impedance Z0 or the load impedance ZL.

gV

gV


Animation of Impedance Transformation

in L2Z m Z

20

inL4 2

ZZ m

Z

Input impedance inverts Input impedance repeats

in 10 10 Z j in 125 125 Z j

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Derivation of Input Impedance, Zin (1 of 2)

The reflection coefficient at any point z from the load is

0

0

z

z

V ez

V e

This means that from the perspective of the generator, the reflection going into the transmission line will change depending on the length of the transmission line. This can only happen of the input impedance to the transmission line is changing.

Backward Wave

Forward Wave

20

0

zVe

V


Derivation of Input Impedance, Zin (2 of 2)

We define the impedance of the line at position z to be

V z

Z zI z

We previously wrote V(z) and I(z) as

0

0

0

z zL

z zL

V z V e e

VI z e e

Z

0 0

0 0

LL

L

V Z Z

V Z Z

Substituting in our expressions for V(z) and I(z) gives

0

00

0

z z z zL L

z zz z L

L

V e e e eZ z Z

V e ee e

Z

It makes sense that the impedance is not a function of voltage in a linear system.

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Sanity Check: Input Impedance at Load

The input impedance at the load can be determined by setting z = 0 in our previous equation.

0 0

in 0 0 0

0

L 0

L 00

L 0

L 0

L 0 L 00

L 0 L 0

L0

0

L

0

1

1

1

1

2

2

L

L

L

L

e eZ Z

e e

Z

Z Z

Z ZZ

Z Z

Z Z

Z Z Z ZZZ Z Z Z

ZZ

Z

Z


Input Impedance at z The input impedance at location isz

in 0 0L L

LL

e e e eZ Z Z

e ee e

A Note About Sign: Backing away from the load, z becomes negative. However, we defined so stays positive in this equation and for equations that follow.

z

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Impedance Transformation Formula (1 of 2)

Recall that

0L

L

e eZ Z

e e

0

0

LL

L

Z Z

Z Z

We can eliminate L from the input impedance equation by substituting in our expression for L.

0

00in 0 0

0 0

0

L

LL

L L

L

Z Ze e

Z e e Z e eZ ZZ Z Z

Z Z Z e e Z e ee eZ Z


Impedance Transformation Formula (2 of 2)

Now recall the definitions of hyperbolic sine and cosine functions.

sinh2

z ze ez

cosh

2

z ze ez

This lets us write the input impedance expression as

00

in 0 00

0

sinh

2cosh 2sinh cosh

sinh2sinh 2cosh

cosh

LL

LL

Z ZZ Z

Z Z ZZ Z

Z Z

0in 0

0

tanh

tanhL

L

Z ZZ Z

Z Z

Recognizing that tanh(z) = sinh(z)/cosh(z), our expression reduces to

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Input Transformation for Lossless Line

The lossless line has

0

j

Putting these values into our impedance transformation formula gives

0in 0

0

tanh

tanhL

L

Z Z jZ Z

Z Z j

Recognizing that tanh(jz) = jtan(z), our expression for lossless lines becomes

0in 0

0

tan

tanL

L

Z jZZ Z

Z jZ


Input Impedance Repeats for Lossless Lines

For lossless lines, the function in the impedance transformation equation tells us that the function is periodic and repeats.

The function repeats every integer multiple of .

tan

Recognizing that = 2/, the above expression leads to

, , 3, 2, 1,0,1,2,3, ,m m

2m

This means the input impedance repeats for every half‐wavelength long the transmission line is.

We will revisit this when we cover Smith charts, which will give you a way to visualize the impedance transformation phenomenon.

Note: is the wavelength in the transmission line, not the free space wavelength 0.

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Example: Impedance Transformation (1 of 3)

A transmission line with 50 characteristic impedance is connected to a 10 nF capacitor as the load. If the phase constant of the transmission line is = 60 m-1, what is the input impedance Zin of a 1 inch section of line operating at 4.0 GHz? What equivalent circuit would the source see?

Transmission Line Load

z0z z

50 inZ 10 nF

1 inch



Loss was not specified so we assume a lossless transmission line. Our impedance transformation equation is therefore

0in 0

0

tan

tanL

L

Z jZZ Z

Z jZ

The variables in this equation are

0

1

9 1 9

50

2.54 cm 1 m60 m 1 inch 1.524

1 inch 100 cm

1 1 10.004

2 2 4.0 10 s 10 10 FL

Z

Z jj C j fC j

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Substituting in the values of our variables gives

3in

0.004 50 tan 60 0.025450 1.07 10

50 0.004 tan 60 0.0254

j jZ j

j j

The input impedance is purely imaginary and positive. Thus, the input impedance looks like an inductor to the generator.

in eq

38in in

eq 9 1

1.07 10 4.24 10 H 42.4 nH

2 2 4.0 10 s

Z j L

Z Z jL

j j f j


Parameter Relations

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Vmax, Vmin, Imax & Imin in Terms of VSWR

Vmax and Vmin

max 0 0

min 0 0

2VSWR1

VSWR 12

1VSWR 1

L

L

V V V

V V V

Imax and Imin

0 0

max0 0

0 0

min0 0

2VSWR1

VSWR 1

21

VSWR 1

L

L

V VI

Z Z

V VI

Z Z


Z0 in Terms of VSWR

The characteristic impedance Z0 can be calculated from Vmax and Imax

or Vmin and Imin.

max min0

max min

V VZ

I I

The input impedance Zin repeats as you back away from the load. We can calculate the maximum and minimum impedance as

maxin 0

min

0minin

max

max VSWR

minVSWR

VZ Z

I

ZVZ

I

in in inmin maxZ Z Z

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Example (1 of 3)

A 50 impedance transmission line is connected to an antenna with a 72 input impedance. A source provides an input signal of 24 V peak‐to‐peak.

What is the reflection coefficient at the antenna?

L 0L

L 0

72 50 0.1803

72 50

Z Z

Z Z

In this case, the antenna is the load.

What fraction of the input power is delivered to the antenna?2 2

L 0.1803 0.0325

1 1 0.0325 0.9675 96.7%

R

T R

Despite the mismatch, almost all power is still delivered to the antenna. This still does not mean the antenna will radiate!

What is the VSWR on the line feeding the antenna?

L

L

dB 10 10

1 1 0.1803VSWR 1.44

1 1 0.1803

VSWR 20log VSWR 20log 1.44 3.17 dB


Example (2 of 3)

What is the minimum and maximum voltage on the line?

p-p0

24 V12 V

2 2

VV

First, we need to convert voltage peak‐to‐peak Vp-p to voltage magnitude V0.

Now we are in a position to calculate Vmin and Vmax.

min 0 L

max 0 L

V 1 12 V 1 0.1803 9.84 V

V 1 12 V 1 0.1803 14.16 V

V

V

When we are utilizing high voltages, we want to be sure Vmax will not cause arcing or any other breakdown problems.

What is the minimum and maximum current on the line?

minmin

0

maxmax

0

V 9.84 VI 0.1967 A

50

V 14.16 VI 0.2833 A

50

Z

Z

At high power, we want to be sure Imax will not cause heating problems.

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Example (3 of 3)

What is the total range of input impedances a source could see?

minin

max

maxin

min

9.84 Vmin 34.72

0.2833 A

14.16 Vmax 72

0.1967 A

VZ

I

VZ

I

in in in

in

min max

34.72 72

Z Z Z

Z


Special Cases of Terminated

Transmission Lines

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Shorted Line, ZL = 0

Reflection from Load

L 1

Input Impedance

0in

0

tanh lossy

tan lossless

ZZ

jZ

Vmin and Vmax

min

max 0

0

2

V

V V

Imin and Imax

min

0

max0

0

2

I

VI

Z

VSWR

Voltage Standing Wave Ratio

min[Zin] and max[Zin]

in

in

min 0

max

Z

Z

There exists V(z) = 0.

short circuit

open circuit

Note 1: Zin for the lossless line is purely imaginary. This means it is purely reactive and no dissipation occurs in the line. The input impedance alternates between being capacitive and inductive as you back away from the load.

Note 2: The shorted line behaves much the same way as the open‐circuit line. We also observe that

2in,short in,open 0Z Z Z


Open‐Circuit Line, ZL = Reflection from Load

L 1

Input Impedance

0in

0

coth lossy

cot lossless

ZZ

jZ

Vmin and Vmax

min

max 0

0

2

V

V V

Imin and Imax

min

0

max0

0

2

I

VI

Z

VSWR



in

in

min 0

max

Z

Z

There exists V(z) = 0.

short circuit

open circuit

Note 1: Zin for the lossless line is purely imaginary. This means it is purely reactive and no dissipation occurs in the line. The input impedance alternates between being capacitive and inductive as you back away from the load.

Note 2: The open‐circuit line behaves much the same way as the shorted line. We also observe that

2in,short in,open 0Z Z Z

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Matched Line, ZL = Z0

Reflection from Load

L 0

Input Impedance

in 0Z Z

Vmin and Vmax

min max 0V V V

Imin and Imax

min max 0 0I I V Z

VSWR 1



in in 0min maxZ Z Z

because Vmax = Vmin

Note: F the matched line, there are no reflections and all of the power is delivered to the load.

Lecture 4b -- Transmission Line Behavior · Transmission Line Behavior Slide 33 Example: Impedance...

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Transcript of Lecture 4b -- Transmission Line Behavior · Transmission Line Behavior Slide 33 Example: Impedance...