Fields and Waves I - Rensselaer Polytechnic Institutehibp.ecse.rpi.edu/~connor/education/Fields/FWI-Lec2.pdf · 2006. 9. 4. · 4 September 2006 Fields and Waves I 10 Computer-Based

Fields and Waves I

Lecture 2Sine Waves on Transmission Lines

K. A. ConnorElectrical, Computer, and Systems Engineering Department

Rensselaer Polytechnic Institute, Troy, NY

4 September 2006 Fields and Waves I 2

These Slides Were Prepared by Prof. Kenneth A. Connor Using Original Materials Written Mostly by the Following:

Kenneth A. Connor – ECSE Department, Rensselaer Polytechnic Institute, Troy, NYJ. Darryl Michael – GE Global Research Center, Niskayuna, NY Thomas P. Crowley – National Institute of Standards and Technology, Boulder, COSheppard J. Salon – ECSE Department, Rensselaer Polytechnic Institute, Troy, NYLale Ergene – ITU Informatics Institute, Istanbul, TurkeyJeffrey Braunstein – Chung-Ang University, Seoul, Korea

Materials from other sources are referenced where they are used.Those listed as Ulaby are figures from Ulaby’s textbook.


Overview

ReviewVoltages and Currents on Transmission LinesStanding WavesInput ImpedanceLossy Transmission LinesLow Loss Transmission Lines

Henry Farny Song of the Talking Wire


Transmission Line Representation

As 0z ⇒Δ limit

)z(V

I

)zz(V Δ+

tIL)z(V)zz(V∂∂

−=−+ Δ

zl Δ⋅zV

zV

tIl

∂∂

=ΔΔ

=∂∂⋅−



Similarly,tVc

zI

∂∂⋅−=

∂∂

2

2

2

2

tVlc)

zI(

tl)

tIl(

zzV

∂∂

=∂∂

∂∂

⋅−=∂∂⋅−

∂∂

=∂∂

Obtain the following PDE:2

2

2

2

tVlc

zV

∂∂

=∂∂

Solutions are: )uzt(f ±

These are functions that move with

velocity u

tIl

zV

∂∂⋅−=

∂∂looks like



Example:

Functions that move with velocity u

)zu

tcos( ωω ±

At t=0, )zu

cos( ω−

Wave moving to the right

At ωt =1 )zu

1cos( ω− ωt =0

ωt =1

zuω


Workspace – look at the general form of the solution


Some Numerical Experiments

PSpice can be used to do simple numerical experiments that demonstrate how transmission lines work

VVT1

0

R2

50

0

V1

FREQ = 1MegHzVAMPL = 10VOFF = 0

R1

50

Simple T-Line


PSpice

VVT1

0

R2

50

0

V1

FREQ = 1MegHzVAMPL = 10VOFF = 0

R1

50

Time

2.0us 2.5us 3.0us 3.5us 4.0us 4.5us 5.0usV(R1:2) V(T1:B+)

-5.0V

0V

5.0V

INPUT OUTPUT

t


Computer-Based Tools

When you use a program like PSpice, applets, or any handy tools available online … remain skeptical.Do not assume that the answers are correct. Apply crude plausibility checks.Know the assumptions and limitations of the tools you are using.Test all tools on problems you can solve other ways or with tools you have already tested.Use even sometimes incorrect tools as long as errors are recognized.

Pig from http://www.cincinnatiskeptics.org


PSpice Example

Let us return to the configuration shown above and simulate it using PSpiceList some conclusions from this exercise.• ?• ?• ?


Sine Waves

The form of the wave solution

First check to see that these solutions have the properties we expect by plotting them using a tool like Matlab

( )A tu

z A t zcos cosω ω ω βm m⎛⎝⎜

⎞⎠⎟=


Sine Waves

The positive wave

u

cos cosω ω π πtu

z ft fu

z−⎛⎝⎜

⎞⎠⎟= −⎛

⎝⎜⎞⎠⎟

2 2

Apply frequency and wavelength analogy argument to show this is reasonable


Solutions to the Wave Equation

Thus, our sine wave is a solution to the voltage or current equation

if or

u = the speed of wave propagation = the speed of light

∂∂

∂∂

2

2

2

2

Vz

lc Vt

=

β ω ω= =u

lc ulc

=1


Sine Waves

Consider one other property. What is the distance required to change the phase of this expression by ? We just did this qualitatively.

This distance is called the wavelength or

cos cosω ω π πtu

z ft fu

z−⎛⎝⎜

⎞⎠⎟= −⎛

⎝⎜⎞⎠⎟

2 2

2π

β ω π πzu

z fu

z= = =2 2

βλ ω λ π λ π= = =u

fu

2 2 λ πβ

= =2 u

f


Sine Waves -- Summary

Solutions look like

β ω ω ω με πλ

= = = =u

lc 2

( )A t zcos ω βm

ω π π= =2 2f

Tu

lc= =

1 1με

λ πβ

= =2 u

fε ε ε= r o μ μ μ= r o

Figure from http://www.emc.maricopa.edu/


Phasor Notation

For ease of analysis (changes second order partial differential equation into a second order ordinary differential equation), we use phasor notation.

The term in the brackets is the phasor.

( ) { }( )f z t A t z Ae ej z j t( , ) cos Re= =ω β β ωm m

f z Ae j z( ) = m β


Phasor Notation

To convert to space-time form from the phasor form, multiply by and take the real part.

If A is complex

e j tω

( )f z t Ae e A t zj z j t( , ) Re( ) cos= =m mβ ω ω β

A Ae j A= θ

( )f z t Ae e e A t zj j z j tA

A( , ) Re( ) cos= = +θ β ω ω β θm m


Transmission Lines

Incident Wave

Reflected Wave

Mismatched load

Standing wave due to interference


Standing Waves

Besser AssociatesNo Standing Wave

http://www.bessernet.com


Standing Waves

Besser AssociatesStanding Wave


Standing Waves

Besser Associates

This may be wrong

We will see shortly

Standing Wave


Standing Waves

Besser Associates

This may be wrong

We will see shortly

Standing Wave


Standing Waves

From Agilent


Transmission Lines - Standing Wave Derivation

Phasor Form of the Wave Equation:

VclzV

tVcl

zV

22

2

2

2

2

2

⋅⋅⋅−=∂∂

⇒

∂∂⋅⋅=

∂∂

ω

where:

zjeVV ⋅⋅±⋅= βm

General Solution: zjzj eVeVV ⋅⋅+−⋅⋅−+ += ββ


Workspace


Workspace



zjzj eVeVV ⋅⋅+−⋅⋅−+ += ββ

Forward Wave

)ztcos( ⋅−⋅ βω

Backward Wave

)ztcos( ⋅+⋅ βω

Vmax occurs when Forward and Backward Waves are in Phase

Vmin occurs when Forward and Backward Waves are out of Phase

CONSTRUCTIVE INTERFERENCE

DESTRUCTIVE INTERFERENCE

TIME DOMAIN


Transmission Lines Formulas

Fields and Waves I Quiz Formula SheetIn the class notes

Note:

All are used in various handouts, texts, etc. There is no standard notation.

V V Vm++ += =

v z V e V ej z j z( ) = ++−

−+β β

i zV e V e

ZVZ

eVZ

ej z j z

o o

j z

o

j z( ) =−

= −+−

−+

+ − − +β β

β β

V V Vm−− −= =


RG58/U Cable

Assume 2 VP-P 1.5MHz sine wave is launched on such a line. Find and

Answers?

β ω ω ω με πλ

= = = =u

lc 2 λ


Reflection Coefficient Derivation

Define the Reflection Coefficient:

+− ⋅= mLm VV Γ

Maximum Amplitude when in Phase:−+ += mmmax VVV

)1(VV Lmmax Γ+⋅=∴ +

Similarly: )1(VV Lmmin Γ−⋅= +

Standing Wave Ratio (SWR) =L

L

min

max

11

VV

ΓΓ

−+

=



Distance between Max and Min is λ/2

Forward Phase is = zj ⋅⋅− β

Backward Phase is = zj ⋅⋅+ β

Difference in Phase is = zj2 ⋅⋅⋅− β

Varies by 2π (distance between maxima)

Show this

πΔβ ⋅=⋅⋅∴ 2z2 22z λ

λππ

βπΔ =

⋅==

V ±Assume are real (will be complex if the load is complex)



Let z=0 at the LOAD

)1(

00

L

jjload

V

VV

eVeVV

Γ+⋅=

+=

⋅+⋅=⇒

+

−+

⋅⋅+−⋅⋅−+ ββ

Need a relationship between current and voltage:

tIl

zV

∂∂⋅−=

∂∂ Ilj

zV

⋅⋅⋅−=∂∂

⇒ ω



zV

lj1I

∂∂⋅

⋅⋅−=

ω

)(1 zjzj eVjeVjlj

⋅⋅+−⋅⋅−+ ⋅⋅⋅+⋅⋅⋅−⋅⋅⋅

−= ββ ββω

)( zjzj eVeVl

⋅⋅+−⋅⋅−+ ⋅−⋅⋅⋅

= ββ

ωβ

)( zjzj eVeVl

clI ⋅⋅+−⋅⋅−+ ⋅−⋅⋅⋅⋅⋅

=∴ ββ

ωω

zjzj e

cl

Ve

cl

V ⋅⋅+−

⋅⋅−+

⋅−⋅= ββ zjLzj e

cl

Ve

cl

V ⋅⋅++

⋅⋅−+

⋅Γ⋅

−⋅= ββ



At LOAD: LZIV=

Use derived terms of V and I at z=0 (position of the LOAD)

( ) LLL ZVV

cl

V

cl

V=⋅Γ+⋅

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛Γ⋅

− ++

−++

1

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

⋅=L

LL 1

1clZ

ΓΓ

OR0

0

ZZZZ

L

LL +

−=Γ

Note thatclZ0 =


Short Circuit Load

For ZL=0, we have ΓLL o

L o

o

o

Z ZZ Z

ZZ

=−+

=−+

= −00

1

( )v z V e V e V e ej zL

j z j z j z( ) = + = −+ − + + + − +β β β βΓ

e z j zj z+ = +β β βcos sine z j zj z− = −β β βcos sin

( )v z V j z( ) sin= − + 2 β


Short Circuit Load

Convert to space-time form

This is a standing wave

( ) ( )( )v z t v z e V j z ej t j t( , ) Re ( ) Re sin= = −+ω ωβ2

( )( ) ( )( )Re sin Re sin cos sin− = − −j z e z j z zj t2 2β β β βω

v z t V z t( , ) sin sin= +2 β ω


Short Circuit Load

What are the voltage maxima and minima?

Where are they?

The standing wave pattern is the envelope of this function.

v z t V z t( , ) sin sin= +2 β ω


Lumped Transmission Line

0

0

L1

1uH

L2

1uH

L3

1uH

L4

1uH

L5

1uH

L6

1uH

L7

1uH

L8

1uH

L9

1uH

L10

1uH

L11

1uH

L12

1uH

L13

1uH

L14

1uH

L15

1uH

L16

1uH

L17

1uH

L18

1uH

L19

1uH

L20

1uH

C1

390pF

C2

390pF

C3

390pF

C4

390pF

C5

390pF

C6

390pF

C7

390pF

C8

390pF

C9

390pF

C10

390pF

C11

390pF

C12

390pF

C13

390pF

C14

390pF

C15

390pF

C16

390pF

C17

390pF

C18

390pF

C19

390pF

C20

390pF

R1

50V1

R2

93

Function Generator

Load

1 2 3 4 5

15

15

5

20

10



Input Not Shown

Both Outputs Shown



In Out

1

15

0 5

Input is BNC Output is both BNC and Banana Plugs (for some loads)


Lumped Transmission Line Experiment

Treat the lumped version just like the reel of cable. (Connectors are opposite so you will need connector cables.) Monitor the output of the function generator on one channel Monitor the voltages on each node (one at a time) on the other channel. You can use just the signal (red) lead, since the ground (black) lead is connected through the other cables. Use the voltage cursors to obtain VP-P for each node. Record your values and plot with Excel, Matlab, etc.


Use PSpice to set up the standard transmission line, matched and notLook at the output for a variety of frequenciesSet up the lumped line in PSpice (more work) and repeatUse the lumped line model to show the standing wave patternWill there be any obvious differences between the physical and numerical experiments?

Lumped Transmission Line Numerical Experiment(Not required)


Workspace


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Workspace


Workspace

Documents

Fields and Waves I - Rensselaer Polytechnic Institutehibp.ecse.rpi.edu/~connor/education/Fields/FWI-Lec2.pdf · 2006. 9. 4. · 4 September 2006 Fields and Waves I 10 Computer-Based