transmission linesranga/courses/en4620_2008/... · transmission lines ranga rodrigo january 13,...

Transmission Lines

Ranga Rodrigo

January 13, 2009

Antennas and Propagation: Transmission Lines 1/46

1 Basic Transmission Line Properties

2 Standing Waves

Antennas and Propagation: Transmission Lines Outline 2/46

Outline


2 Standing Waves

Antennas and Propagation: Transmission Lines Basic Transmission Line Properties 3/46

Introduction

A transmission line is a circuit element thattransfers energy in the form of electromagneticwaves from one place to the other.

Transmission Lines

Balanced

E.g., flat twin

Unbalanced

E.g., coaxial cable


The construction of the line will vary dependingon its end use:

Copper wires for low-frequency audio applications.Copper dielectric mixtures for VHF, UHF, andmicrowave.Solid dielectric such as plastic or glass for opticaluse.

By carefully exploiting the properties of a givenline configuration, useful circuit componentssuch as filters and impedance matchingnetworks can be built.



Copper wires for low-frequency audio applications.

Copper dielectric mixtures for VHF, UHF, andmicrowave.Solid dielectric such as plastic or glass for opticaluse.




Copper wires for low-frequency audio applications.Copper dielectric mixtures for VHF, UHF, andmicrowave.

Solid dielectric such as plastic or glass for opticaluse.




Copper wires for low-frequency audio applications.Copper dielectric mixtures for VHF, UHF, andmicrowave.Solid dielectric such as plastic or glass for opticaluse.



Distributed Nature of Trans. Lines10 MHz and Below

Wavelengths are long (> 30 m).

Standard components, capacitors, inductors,etc., appear very short.Gives rise to the notion of lumped circuits.

Higher Frequencies

Wavelength is comparable with that ofcomponents.Distributed circuit techniques must be used.Division occurs when the component dimensionis not greater than 1/20 of the wavelength.



Wavelengths are long (> 30 m).Standard components, capacitors, inductors,etc., appear very short.

Gives rise to the notion of lumped circuits.

Higher Frequencies




Wavelengths are long (> 30 m).Standard components, capacitors, inductors,etc., appear very short.Gives rise to the notion of lumped circuits.

Higher Frequencies





Higher FrequenciesWavelength is comparable with that ofcomponents.

Distributed circuit techniques must be used.Division occurs when the component dimensionis not greater than 1/20 of the wavelength.




Higher FrequenciesWavelength is comparable with that ofcomponents.Distributed circuit techniques must be used.

Division occurs when the component dimensionis not greater than 1/20 of the wavelength.




Higher FrequenciesWavelength is comparable with that ofcomponents.Distributed circuit techniques must be used.Division occurs when the component dimensionis not greater than 1/20 of the wavelength.


Z0

CG

R L

CG

R L

CG

R L

Z0

R: series loss resistance (copper losses).G: shunt loss conductance (losses in dielectric).L: series inductance representing energystorage within the line.C : shunt capacitance representing energystorage within the line.


I1

V1

I2

V2

I3

V3

l l

If an infinitely long pair of wires were consideredand the voltage and current were somehowmeasured at uniform spaced points along theline, then

V1

I1= V2

I2= ·· · = Vk

Ik= constant = Z0Ω. (1)

This is termed the characteristic impedance ofthe line and is denoted by Z0.


∆Y

Z0

∆Z

Zin = Z0

V

Figure: Characteristic line impedance.

The input impedance Zin for the L-section shownabove is

Zin = Z0 = (Z0 +∆Z )/∆Y

Z0 +∆Z +1/∆Y= Z0 +∆Z

1+Z0∆Y∆Y ,∆Z −→ 0.

(2)


∆Y

Z0

∆Z

Zin = Z0

V

Figure: Characteristic line impedance.

The input impedance Zin for the L-section shownabove is

Zin = Z0 = (Z0 +∆Z )/∆Y

Z0 +∆Z +1/∆Y= Z0 +∆Z

1+Z0∆Y∆Y ,∆Z −→ 0.

(2)Antennas and Propagation: Transmission Lines Basic Transmission Line Properties 9/46

Solving Equation 2 for Z0 gives

Z0 =(∆Z

∆Y

)1/2

, (3)

where ∆Z = R + jωL and ∆Y =G jωC hence,

Z0 =(

R + jωL

G + jωC

)1/2

Ω . (4)

This expression is important as it relates the lumpedcircuit model for the transmission line to one of theprimary line constants, characteristic impedance.

At very low frequencies:

Z0 =(

R

G

)1/2

Ω. (5)

For high frequencies (ωL À R and ωC ÀG):

Z0 =(

L

C

)1/2

Ω (6)


Let’s consider Equation 4,

Z0 =(

R + jωL

G + jωC

)1/2

Ω.

At very low frequencies:

Z0 =(

R

G

)1/2

Ω. (7)

For high frequencies (ωL À R and ωC ÀG):

Z0 =(

L

C

)1/2

Ω (8)

Since transmission line circuit designs are done athigh frequencies, Equation 8 is often used. If lossescannot be neglected, then the line is said to be lossy.


ExampleA lossless transmission line has a characteristicimpedance of 50 Ω and s self-inductance of 0.08µH/m. Calculate the capacitance of a 4-meter lengthof line.

Solution

The line is lossless. Therefore, R = 0, G =∞.

Z0 =(

L

C

)1/2

⇒C = L

Z 2o

= 0.08×10−6

50×50= 32 pF/m.

The total capacitance for a 4-meter length is4×32 pF/m= 128 pF/m. This capacitance will limit thehigh frequency response of the cable.


Propagation ConstantConsider the lumped equivalent circuit again.

Z0

CG

R L

CG

R L

CG

R L

Z0

The voltage drop ∆V across one lumped section is

∆V =−I (R + jωL)∆x. (9)

here ∆x is the incremental length of the linerepresented by the lumped section.


Dividing Equation 9 by ∆x and since ∆x → 0,

dV

d x=−(R + jωL)I . (10)

Similarly we can obtain

d I

d x=−(G + jωC )V. (11)

By differentiating and substitution

d 2V

d x2= (R + jωL)(G + jωC )V. (12)


Equation 12 can be written as

d 2V

d x2= γ2V , (13)

where γ=√(R + jωL)(G + jωC ) is termed the

propagation constant. This is usually expressed as

γ=α+ jβ, (14)

where α represents the attenuation per unit lengthand β the phase shift per unit length.


Equation 13 has a solution of the form

V (x) =

Ae−γx +Beγx . (15)

This suggests that the line will contain two waves,one traveling in the positive x-direction (e−γx) and theother traveling in the negative x-direction (eγx).To evaluate A, let’s consider an infinite-length lineexcited by sine wave of amplitude Vin. If the linecontains resistive elements, then at x =∞, anyvoltage would have decayed to zero.

0 = Ae−γ∞+Beγ∞ ⇒ B = 0. (16)

Therefore, we have

V (x) = Ae−γx . (17)


Equation 13 has a solution of the form

V (x) = Ae−γx +Beγx . (15)

This suggests that the line will contain two waves,one traveling in the positive x-direction (e−γx) and theother traveling in the negative x-direction (eγx).To evaluate A, let’s consider an infinite-length lineexcited by sine wave of amplitude Vin. If the linecontains resistive elements, then at x =∞, anyvoltage would have decayed to zero.

0 = Ae−γ∞+Beγ∞ ⇒ B = 0. (16)

Therefore, we have

V (x) = Ae−γx . (17)


When we consider that at the driving end of the line(x = 0), the voltage is Vin, we get A =Vin. Therefore,

V (x) =Vine−γx . (18)

Since γ=α+ jβ,

V (x) =Vine−αxe− jβx . (19)


e−αx TermLet’s consider the attenuation term, e−αx. For a linecontaining uniform loss per unit length,

V2 = kV1,

V3 = kV2,...= ...

Vn+1 = kVn.

(20)

where k = e−αx is a constant less than one (k = 1 forlossless). Therefore,

Vn+1 =V1kn. (21)


Taking the natural logarithm, we have

ln

(Vn+1

V1

)=−nαx . (22)

The term nαx is defined as the total line attenuationand is measured in units called nepers (Np). It canbe shown that 1 Np is equal to 8.686 dB.


e− jβx TermNow let’s consider the term e− jβx. Taking the currentI as a reference, we find that the voltage drop acrossa single incremental inductance element L isjωLI∆x, while the voltage across the element Z0 isI Z0.

I Z0

ωLI∆x

∆β

∆β= tan−1

(ωLI∆x

I Z0

). (23)


For small angles, tanθ ≈ θ, so

∆β= ωL∆x

Z0. (24)

For a lossless line,

∆β= ωL∆x

(L/C )1/2=ω(LC )1/2∆x. (25)

The quantity ∆β/∆x is called the phase-shift-changeper unit length or the wave number β.

β=ω(LC )1/2. (26)

We have

vp

= f λ= ω

2πλ, β= 2π

λ, vp = ω

β, Z0 = vpL = 1

vpC.

(27)



∆β= ωL∆x

Z0. (24)


∆β= ωL∆x

(L/C )1/2=ω(LC )1/2∆x. (25)


β=ω(LC )1/2. (26)

We have

vp = f λ= ω

2πλ, β

= 2π

λ, vp = ω

β, Z0 = vpL = 1

vpC.

(27)



∆β= ωL∆x

Z0. (24)


∆β= ωL∆x

(L/C )1/2=ω(LC )1/2∆x. (25)


β=ω(LC )1/2. (26)

We have

vp = f λ= ω

2πλ, β= 2π

λ, vp

= ω

β, Z0 = vpL = 1

vpC.

(27)



∆β= ωL∆x

Z0. (24)


∆β= ωL∆x

(L/C )1/2=ω(LC )1/2∆x. (25)


β=ω(LC )1/2. (26)

We have

vp = f λ= ω

2πλ, β= 2π

λ, vp = ω

β, Z0

= vpL = 1

vpC.

(27)



∆β= ωL∆x

Z0. (24)


∆β= ωL∆x

(L/C )1/2=ω(LC )1/2∆x. (25)


β=ω(LC )1/2. (26)

We have

vp = f λ= ω

2πλ, β= 2π

λ, vp = ω

β, Z0 = vpL = 1

vpC.

(27)Antennas and Propagation: Transmission Lines Basic Transmission Line Properties 21/46

Sending End ImpedanceWhen a load is placed at the end of a section oftransmission line the line will produce atransformation effect. The load impedance willappear different when viewed from the sendingend of the line.

Using this effect, the impedance transformationcaused by inserting a section of transmissionline between a given impedance and ameasurement point can be carefully controlledto allow impedance matching.If measurements are to be made on an unknownload impedance, the impedance transformationcaused by inserting a connecting section oftransmission line needs to be taken into account.


Sending End ImpedanceWhen a load is placed at the end of a section oftransmission line the line will produce atransformation effect. The load impedance willappear different when viewed from the sendingend of the line.Using this effect, the impedance transformationcaused by inserting a section of transmissionline between a given impedance and ameasurement point can be carefully controlledto allow impedance matching.

If measurements are to be made on an unknownload impedance, the impedance transformationcaused by inserting a connecting section oftransmission line needs to be taken into account.


Sending End ImpedanceWhen a load is placed at the end of a section oftransmission line the line will produce atransformation effect. The load impedance willappear different when viewed from the sendingend of the line.Using this effect, the impedance transformationcaused by inserting a section of transmissionline between a given impedance and ameasurement point can be carefully controlledto allow impedance matching.If measurements are to be made on an unknownload impedance, the impedance transformationcaused by inserting a connecting section oftransmission line needs to be taken into account.Antennas and Propagation: Transmission Lines Basic Transmission Line Properties 22/46

Let’s consider Equation 15,

V (x) = Ae−γx +Be+γx .

ZTZ0

ZS

l

x = 0

IincIref

The total current flowing in the line IT is the sum ofcurrent traveling in the forward direction from x = 0 tox = l , I inc, and the amount of current that is reflectedfrom the load termination ZT , Iref.


This reflected current is the portion of the incident(forward) current that is not absorbed by the load.

IT = I inc− Iref, (28)

IT = A

Z0e−γx − B

Z0e+γx . (29)

At position x = l the impedance is ZT .

ZT = VT

IT=

Ae−γl +Be+γl

Ae−γl −Be+γlZ0. (30)

Therefore,B

A= e−2γl ZT −Z0

ZT +Z0. (31)

The term e−2γl is the loss encountered by the signaltraversing from source to termination and back.


This reflected current is the portion of the incident(forward) current that is not absorbed by the load.

IT = I inc− Iref, (28)

IT = A

Z0e−γx − B

Z0e+γx . (29)

At position x = l the impedance is ZT .

ZT = VT

IT= Ae−γl +Be+γl

Ae−γl −Be+γlZ0. (30)

Therefore,B

A= e−2γl ZT −Z0

ZT +Z0. (31)

The term e−2γl is the loss encountered by the signaltraversing from source to termination and back.


Considering the situation at x = 0, we get,

ZS = VS

IS= A+B

A−BZ0. (32)

ZS

Z0= 1+B/A

1−B/A. (33)

Here ZS is the sending end impedance, i.e., theimpedance seen looking into the line toward theload. Substituting,

ZS

Z0=

1+e−2γl ZT −Z0ZT +Z0

1−e−2γl ZT −Z0ZT +Z0

=

ZT (1+e−2γl )+Z0(1−e−2γl )

ZT (1−e−2γl )+Z0(1+e−2γl ). (34)


Considering the situation at x = 0, we get,

ZS = VS

IS= A+B

A−BZ0. (32)

ZS

Z0= 1+B/A

1−B/A. (33)

Here ZS is the sending end impedance, i.e., theimpedance seen looking into the line toward theload. Substituting,

ZS

Z0=

1+e−2γl ZT −Z0ZT +Z0

1−e−2γl ZT −Z0ZT +Z0

= ZT (1+e−2γl )+Z0(1−e−2γl )

ZT (1−e−2γl )+Z0(1+e−2γl ). (34)


sinhγl = 1

2

[eγl −e−γl] ,

1−e−2γl = e−γl [eγl −e−γl] ,

= 2sinh(γl )e−γl .

1+e−2γl = 2cosh(γl )e−γl .

(35)

ZS

Z0=

ZT cosh(γl )+Z0 sinh(γl )

ZT sinh(γl )+Z0 cosh(γl ). (36)

Equation 36 is the desired result. It enables us toevaluate the sending end impedance ZS, in terms ofthe termination impedance ZT and characteristicimpedance Z0.


sinhγl = 1

2

[eγl −e−γl] ,

1−e−2γl = e−γl [eγl −e−γl] ,

= 2sinh(γl )e−γl .

1+e−2γl = 2cosh(γl )e−γl .

(35)

ZS

Z0= ZT cosh(γl )+Z0 sinh(γl )

ZT sinh(γl )+Z0 cosh(γl ). (36)

Equation 36 is the desired result. It enables us toevaluate the sending end impedance ZS, in terms ofthe termination impedance ZT and characteristicimpedance Z0.


Sometimes, it is convenient to use the followingequation as the governing equation (be dividing eachterm by Z0 cosh(γl )):

ZS


ZT sinh(γl )+Z0 cosh(γl )

ZS

Z0=

ZTZ0

+ tanh(γl )

1+ ZTZ0

tanh(γl ). (37)


Sometimes, it is convenient to use the followingequation as the governing equation (be dividing eachterm by Z0 cosh(γl )): ZS


ZT sinh(γl )+Z0 cosh(γl )

ZS

Z0=

ZTZ0

+ tanh(γl )

1+ ZTZ0

tanh(γl ). (37)


From Equation 36, we see that if the load terminationZT = Z0, then the sending end impedance ZS willequal to Z0. In this case, no reflection will occur fromthe load, and the line is said to be matched.

ExampleA transmission line of length 100 m is operated at10.0 MHz, and has an attenuation per unit length of0.002 nepers/m. The phase velocity of the line is2.7×108 m/s, and the line has a characteristicimpedance of 50 Ω. What is the value of the terminalload impedance, if the input impedance looking intothe line is measured to be 30− j 10Ω?


Solution

Following relations are useful:

tanh( jβl ) = j tan(βl ).

tanh(γl ) = tanh(αl + jβl ),

= tanh(αl )+ j tan(βl )

1− tanh(αl ) tan(βl )

tanh(α± jβl ) = sinh(2αl )± j sin(2βl )

cosh(2αl )+cos(2βl ).

(38)


We need to find ZT , given ZS and line parameters.From Equation 36,

ZS


ZT sinh(γl )+Z0 cosh(γl ),

we obtainZS

Z0=

ZTZ0

+ tanh(γl )

1+ ZTZ0

tanh(γl ). (39)

Rearranging,

ZT

ZS=

ZSZ0

− tanh(γl )

1− ZSZ0

tanh(γl ). (40)


Equation 40 is written for traveling from load togenerator. However, as we need to find ZT , we needto travel toward the load. Therefore, we need to use−l in place of +l . We note that tanh(−γl ) =− tanh(γl ).

ZT = quantity to be found,

ZS = 30− j 10Ω,

Z0 = 50+ j 0Ω,

l = 100 m,

α= 0.002 nepers/m,

vp = 2.7×108 m/s.


ZS

Z0= 0.6−0.2 j ,

αl = 0.002×100 = 0.2 nepers,

∴ 2αl = 0.4 nepers,

βl = ωl

vp= 2π×10×106 ×100

2.7×108= 23.27 radians,

∴ 2βl = 46.54 radians,


= tanh(0.2+ j 23.27).

tanh(γl ) = tanh(0.2+ j 23.27),

= sinh0.4+ j sin46.54

cosh0.4+cos46.54.

sinh0.4 = e0.4 −e−0.4

2= 0.411.

cosh0.4 = 1.081.

sin46.54 = 0.5513.

cos46.54 =−0.8343.

tanh(γl ) = 0.411+ j 0.5513

1.081−0.8343.

tanh(γl ) = 0.411+ j 0.5513

1.081−0.8343,

= 1.666+ j 2.2347.


ZS

Z0= 0.6−0.2 j ,

αl = 0.002×100 = 0.2 nepers,

∴ 2αl = 0.4 nepers,

βl = ωl

vp= 2π×10×106 ×100

2.7×108= 23.27 radians,

∴ 2βl = 46.54 radians,


= tanh(0.2+ j 23.27).

tanh(γl ) = tanh(0.2+ j 23.27),

= sinh0.4+ j sin46.54

cosh0.4+cos46.54.

sinh0.4 = e0.4 −e−0.4

2= 0.411.

cosh0.4 = 1.081.

sin46.54 = 0.5513.

cos46.54 =−0.8343.

tanh(γl ) = 0.411+ j 0.5513

1.081−0.8343.

tanh(γl ) = 0.411+ j 0.5513

1.081−0.8343,

= 1.666+ j 2.2347.Antennas and Propagation: Transmission Lines Basic Transmission Line Properties 32/46

Substituting this into the governing equation gives

ZS

Z0=

ZTZ0

+ tanh(γl )

1+ ZTZ0

tanh(γl ).

ZT

Z0= 0.6− j 0.2− (1.666+ j 2.2347)

1− (0.6− j 0.2)(1.666+ j 2.2347),

= 4.55− j 0.7063

3.0.

ZS

Z0=

ZTZ0

+ tanh(γl )

1+ ZTZ0

tanh(γl ).

and finally, the desired result

ZT = 76− j 12Ω.


Short Circuit Termination

If a short circuit is used as the load termination, thenthe sending-end impedance becomes,

ZS

Z0=

ZTZ0

+ tanh(γl )

1+ ZTZ0

tanh(γl ).

ZS|SC = Z0 tanh(γl ), (41)

and, neglecting line losses for short lengths we get

ZS|SC = Z0 tanh( jβl ) = j Z0 tan(βl ). (42)


Open Circuit Termination

When the termination is open circuit, then thesending-end impedance becomes,

ZS

Z0=

ZTZ0

+ tanh(γl )

1+ ZTZ0

tanh(γl ).

ZS|OC = Z0

tanh(γl ), (43)

and, neglecting line losses for short lengths we get

ZS|OC = − j Z0

tanh( jβl )=− j Z0 cot(βl ). (44)


RemarksEquations 42 and 44 indicate that, by correctlychoosing l , we can make the line (stub) to havethe sending end impedance equivalent to alumped capacitor or an inductor.

When a transmission line is terminated in aperfect open or a short circuit, the incidentenergy is reflected back into the line.Product of Equations 42 and 44 gives

ZS|OC×ZS|SC = Z 20 (45)

This is a useful method of measuring thecharacteristic impedance of a transmission line.For this the line length selected should be avalue close to an odd number of λ/8.


RemarksEquations 42 and 44 indicate that, by correctlychoosing l , we can make the line (stub) to havethe sending end impedance equivalent to alumped capacitor or an inductor.When a transmission line is terminated in aperfect open or a short circuit, the incidentenergy is reflected back into the line.

Product of Equations 42 and 44 gives

ZS|OC×ZS|SC = Z 20 (45)

This is a useful method of measuring thecharacteristic impedance of a transmission line.For this the line length selected should be avalue close to an odd number of λ/8.


RemarksEquations 42 and 44 indicate that, by correctlychoosing l , we can make the line (stub) to havethe sending end impedance equivalent to alumped capacitor or an inductor.When a transmission line is terminated in aperfect open or a short circuit, the incidentenergy is reflected back into the line.Product of Equations 42 and 44 gives

ZS|OC×ZS|SC = Z 20 (45)

This is a useful method of measuring thecharacteristic impedance of a transmission line.For this the line length selected should be avalue close to an odd number of λ/8.Antennas and Propagation: Transmission Lines Basic Transmission Line Properties 36/46

βl = 2πlλ

− tan(βl )

π

2

π 3π

2

2π 5π

2

−6

−4

−2

0

2

4

6

Normalized input resistance versus βl for ashort-circuited, lossless, transmission line.Antennas and Propagation: Transmission Lines Basic Transmission Line Properties 37/46

βl = 2πlλ

−cot(βl )

π

2

π 3π

2

2π 5π

2

−6

−4

−2

0

2

4

6

Normalized input resistance versus βl for aopen-circuited, lossless, transmission line.Antennas and Propagation: Transmission Lines Basic Transmission Line Properties 38/46

ExampleThe design of a microwave amplifier requires aninductor to series resonate a capacitive reactance of15,000 Ω. Calculate the residual resistance if theinductor consists of a short-circuited line withZ0 = 75Ω, α= 0.002 dB/cm, and λ= 10 cm.

Solution

To series resonate the − j 15000Ω, the inputimpedance of the shorted line must be + j 15000Ω. Ifthe line were lossless, Zin = j 15000 = j 75tan(βl ), andtherefore

βl = tan−1 200 ≈ π

2− 1

200rad.



Solution


βl =

tan−1 200 ≈ π

2− 1

200rad.



Solution


βl = tan−1 200 ≈ π

2− 1

200rad.


Since β= 2π/λ,

l =

2.5

(π

2− 1

100π

)cm.

Thus a shorted lossless line about 0.3% less than aquarter wavelength (2.5 cm) would provide thedesired reactance without adding any resistance.However, in this line is not lossless and αl = 0.005 dBor 5.76×10−4 Np (1 Np = 8.686 dB). Sincetanh(αl ) ≈αl , substituting Z0 = 75Ω and tan(βl ) = 200in to Equation 37 yields

Zin = 755.76×10−4 + j 200

1+ j 0.115= 1700+ j 14800Ω.


Since β= 2π/λ,

l = 2.5

(π

2− 1

100π

)cm.


Zin =

755.76×10−4 + j 200

1+ j 0.115= 1700+ j 14800Ω.


Since β= 2π/λ,

l = 2.5

(π

2− 1

100π

)cm.


Zin = 755.76×10−4 + j 200

1+ j 0.115= 1700+ j 14800Ω.


Note the effect of line loss. The slight loss in thereactive portion can be adjusted by slightlyincreasing the line length. The resistive componentis significant. If we adjust the length, the residualresistance will actually be 1748 Ω.


Outline


2 Standing Waves

Antennas and Propagation: Transmission Lines Standing Waves 42/46

Standing WavesFor any transmission line, a sinusoidal signal ofappropriate frequency introduced at the sendingend by a generator will propagate along thelength of the transmission line.

If the line has infinite length, then the signalnever reaches the end of the line.If the signal is viewed at some distance downthe line away from the generator, then it willappear to have the same frequency, but willexhibit smaller peak to peak voltage swing thanat the generator.The signal, therefore, has been attenuated bythe line losses due to the conductor resistanceand dielectric imperfections.


Standing WavesFor any transmission line, a sinusoidal signal ofappropriate frequency introduced at the sendingend by a generator will propagate along thelength of the transmission line.If the line has infinite length, then the signalnever reaches the end of the line.

If the signal is viewed at some distance downthe line away from the generator, then it willappear to have the same frequency, but willexhibit smaller peak to peak voltage swing thanat the generator.The signal, therefore, has been attenuated bythe line losses due to the conductor resistanceand dielectric imperfections.


Standing WavesFor any transmission line, a sinusoidal signal ofappropriate frequency introduced at the sendingend by a generator will propagate along thelength of the transmission line.If the line has infinite length, then the signalnever reaches the end of the line.If the signal is viewed at some distance downthe line away from the generator, then it willappear to have the same frequency, but willexhibit smaller peak to peak voltage swing thanat the generator.

The signal, therefore, has been attenuated bythe line losses due to the conductor resistanceand dielectric imperfections.


Standing WavesFor any transmission line, a sinusoidal signal ofappropriate frequency introduced at the sendingend by a generator will propagate along thelength of the transmission line.If the line has infinite length, then the signalnever reaches the end of the line.If the signal is viewed at some distance downthe line away from the generator, then it willappear to have the same frequency, but willexhibit smaller peak to peak voltage swing thanat the generator.The signal, therefore, has been attenuated bythe line losses due to the conductor resistanceand dielectric imperfections.Antennas and Propagation: Transmission Lines Standing Waves 43/46

ReflectionIf the line is lossless, then the signal viewed atsome remote point will be identical to that at thegenerator but time delayed by an amountdependant on the position.

When a sinusoidal signal reaches the open endof a section of a lossless transmission line, itcan dissipate no energy.This means that all the energy propagatingalong the line in the forward direction (incident)will be reflected completely, on reaching theopen circuit termination.The reflected wave (backward wave) must besuch that the total current at the open circuit iszero.


ReflectionIf the line is lossless, then the signal viewed atsome remote point will be identical to that at thegenerator but time delayed by an amountdependant on the position.When a sinusoidal signal reaches the open endof a section of a lossless transmission line, itcan dissipate no energy.

This means that all the energy propagatingalong the line in the forward direction (incident)will be reflected completely, on reaching theopen circuit termination.The reflected wave (backward wave) must besuch that the total current at the open circuit iszero.


ReflectionIf the line is lossless, then the signal viewed atsome remote point will be identical to that at thegenerator but time delayed by an amountdependant on the position.When a sinusoidal signal reaches the open endof a section of a lossless transmission line, itcan dissipate no energy.This means that all the energy propagatingalong the line in the forward direction (incident)will be reflected completely, on reaching theopen circuit termination.

The reflected wave (backward wave) must besuch that the total current at the open circuit iszero.


ReflectionIf the line is lossless, then the signal viewed atsome remote point will be identical to that at thegenerator but time delayed by an amountdependant on the position.When a sinusoidal signal reaches the open endof a section of a lossless transmission line, itcan dissipate no energy.This means that all the energy propagatingalong the line in the forward direction (incident)will be reflected completely, on reaching theopen circuit termination.The reflected wave (backward wave) must besuch that the total current at the open circuit iszero.Antennas and Propagation: Transmission Lines Standing Waves 44/46

ReflectionAs the reflected signal travels back along theline toward the generator, it reinforces theincident waveform at certain points formingmaxima (nodes).

Similarly, it can cancel the incident waveform atcertain other points producing minima(antinodes).In an open circuit line, node voltage points willoccur at the same position as the antinodecurrent points.These waves do not represent traveling waves.They are standing waves, implying that there isno net power flow from generator to the(open-circuit) load.


ReflectionAs the reflected signal travels back along theline toward the generator, it reinforces theincident waveform at certain points formingmaxima (nodes).Similarly, it can cancel the incident waveform atcertain other points producing minima(antinodes).

In an open circuit line, node voltage points willoccur at the same position as the antinodecurrent points.These waves do not represent traveling waves.They are standing waves, implying that there isno net power flow from generator to the(open-circuit) load.


ReflectionAs the reflected signal travels back along theline toward the generator, it reinforces theincident waveform at certain points formingmaxima (nodes).Similarly, it can cancel the incident waveform atcertain other points producing minima(antinodes).In an open circuit line, node voltage points willoccur at the same position as the antinodecurrent points.

These waves do not represent traveling waves.They are standing waves, implying that there isno net power flow from generator to the(open-circuit) load.


ReflectionAs the reflected signal travels back along theline toward the generator, it reinforces theincident waveform at certain points formingmaxima (nodes).Similarly, it can cancel the incident waveform atcertain other points producing minima(antinodes).In an open circuit line, node voltage points willoccur at the same position as the antinodecurrent points.These waves do not represent traveling waves.They are standing waves, implying that there isno net power flow from generator to the(open-circuit) load.Antennas and Propagation: Transmission Lines Standing Waves 45/46

A point one-quarter wavelength away from ashort circuit will have voltage and currentmagnitudes equivalent to those obtained for anopen circuit.

For a lossless line the peak value of the standingwave envelope is twice that of the incident wave.There are nulls as well due to the completecancelation.For lossy (practical) line, the peak will be lessthan twice that of the incident wave, andcomplete cancelation rarely results.


A point one-quarter wavelength away from ashort circuit will have voltage and currentmagnitudes equivalent to those obtained for anopen circuit.For a lossless line the peak value of the standingwave envelope is twice that of the incident wave.

There are nulls as well due to the completecancelation.For lossy (practical) line, the peak will be lessthan twice that of the incident wave, andcomplete cancelation rarely results.


A point one-quarter wavelength away from ashort circuit will have voltage and currentmagnitudes equivalent to those obtained for anopen circuit.For a lossless line the peak value of the standingwave envelope is twice that of the incident wave.There are nulls as well due to the completecancelation.

For lossy (practical) line, the peak will be lessthan twice that of the incident wave, andcomplete cancelation rarely results.


A point one-quarter wavelength away from ashort circuit will have voltage and currentmagnitudes equivalent to those obtained for anopen circuit.For a lossless line the peak value of the standingwave envelope is twice that of the incident wave.There are nulls as well due to the completecancelation.For lossy (practical) line, the peak will be lessthan twice that of the incident wave, andcomplete cancelation rarely results.


transmission linesranga/courses/en4620_2008/... · transmission lines ranga rodrigo january 13,...

Documents