Chapter 2: Transmission Lines

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Bài giảng:

TRƯỜNG ĐIỆN TỪ (CT361)

(ELECTROMAGNETICS)

Chapter 2:

Transmission Lines

(Đường dây truyền sóng)

Giảng viên: GVC.TS. Lương Vinh Quốc Danh

Bộ môn Điện tử Viễn thông, Khoa Cơng Nghệ

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• Lumped-Element (phần tử tập trung) Model of

Transmission Lines

• Transmission-Line Equations

• Wave Propagation on a Transmission Line

• Input Impedance (trở kháng ngõ vào) of the Lossless Line

• Special Cases of Lossless Line

• Power Flow on a Lossless Transmission Line

• Transients (Quá độ) on Transmission Lines

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 Transmission line parameters, equations

Vg(t) VAA’(t) VBB’(t)

A’ B’

L

VBB’(t) = VAA’(t-td) = VAA’(t-L/c)

= V0cos((t-L/c))
= V0cos(t- 2L/),

Recall: =c/f, and  = 2

If  >> L, VBB’(t)  V0cos(t) = VAA’(t),

If  <= L, VBB’(t) VAA’(t)  The circuit theory has to be replaced.

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Propagation Modes

Electric and magnetic fields

oscillate together but perpendicular 
to each other and the

electromagnetic wave moves in a

direction perpendicular to both of 
the fields.  TEM modes

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TEM transmission lines

Propagation Modes (cont.)

d) Strip line e) Microstrip line

Metal

Dielectric spacing Metal ground

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Lumped-Element Model

• Represent transmission lines as parallel-wire configuration

Vg(t) VAA’(t) VBB’(t)

A’ B’

z z z

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Transmission line equations

• Represent transmission lines as parallel-wire configuration

V(z,t) = R’z i(z,t) + L’z  i(z,t)/ t + V(z+ z,t), (2.12)

V(z,t)

R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

i(z,t) = G’z V(z+ z,t) + C’z V(z+ z,t)/t + i(z+z,t), (2.15)

FOUR transmission line parameters: Resistance R’ (Ω/m); Inductance L’ (H/m);

Capacitance C’ (F/m); Conductance G’ (S/m)

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• Transmission line equations

V(z,t) = R’z i(z,t) + L’z  i(z,t)/ t + V(z+ z,t), (2.12)

V(z,t)

R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

-V(z+ z,t) + V(z,t) = R’z i(z,t) + L’z  i(z,t)/ t

- V(z,t)/z = R’ i(z,t) + L’  i(z,t)/ t, (2.14)

Rewrite V(z,t) and i(z,t) as phasors, for sinusoidal V(z,t) and i(z,t):
V(z,t) = Re( V(z) ejt), i (z,t) = Re( i (z) ejt),

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• Transmission line equations

V(z,t)

R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

Recall:

di(t)/dt = Re(d i e jt )/dt = Re(i je jt ),

- V(z,t)/z = R’ i(z,t) + L’  i(z,t)/ t, (2.14)

- dV(z)/dz = R’ i(z) + jL’ i(z), (2.18a)

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• Transmission line equations

V(z,t)

R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

- i (z+ z,t) + i (z,t) = G’z V(z + z ,t) + C’z  V(z + z,t)/ t

-  i(z,t)/z = G’ V(z,t) + C’  V(z,t)/ t, (2.16)

Rewrite V(z,t) and i(z,t) as phasors, for sinusoidal V(z,t) and i(z,t):
V(z,t) = Re( V(z) ejt), i (z,t) = Re( i (z) ejt),

i(z,t) = G’z V(z+ z,t) + C’z V(z+ z,t)/t + i(z+z,t), (2.15)

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• Transmission line equations

V(z,t)

R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

Recall:

dV(t)/dt = Re(d V e jt )/dt = Re(Vje jt ),
- i(z,t)/z = G’ V(z,t) + C’  V(z,t)/ t,

- d i(z)/dz = G’ V(z) + jC’ V(z), (2.18b)

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Telegrapher’s equation

V(z,t)

R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

- d i(z)/dz = G’ V(z) + jC’ V(z), (2.18b)

- dV(z)/dz = R’ i(z) + jL’ i(z), (2.18a)

• Telegrapher’s equation in phasor domain

Take d /dz on both sides of eq. (2.18a)

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- d i(z)/dz = G’ V(z) + jC’ V(z), (2.18b)

- dV(z)/dz = R’ i(z) + jL’ i(z), (2.18a)

• Telegrapher’s equation in phasor domain

substitute (2.18b) to (2.19)

d²V(z)/dz² = (R’ + jL’) (G’+ jC’)V(z),

- d²V(z)/dz² = R’ di(z)/dz + jL’ di(z)/dz, (2.19)

d²V(z)/dz² - (R’ + jL’) (G’+ jC’)V(z) = 0, (2.20)

d²V(z)/dz² - ²V(z) = 0, (2.21)

² = (R’ + jL’) (G’+ jC’),

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V(z,t)

R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

- d i(z)/dz = G’ V(z) + jC’ V(z), (2.18b)

- dV(z)/dz = R’ i(z) + jL’ i(z), (2.18a)

• Telegrapher’s equation in phasor domain

Take d /dz on both sides of eq. (2.18b)

- d² i(z)/dz² = G’ dV(z)/dz + jC’ dV(z)/dz, (*)

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substitute (2.18a) to (*)

d² i(z)/dz² = (R’ + jL’) (G’+ jC’)i(z), 
d² i(z)/dz² - (R’ + jL’) (G’+ jC’) i(z) = 0,

d² i(z)/dz² - ²i(z) = 0, (2.23)

² = (R’ + jL’) (G’+ jC’),

- d² i(z)/dz² = G’ dV(z)/dz + jC’ dV(z)/dz, (*)

- d i(z)/dz = G’ V(z) + jC’ V(z), (2.18b)

- dV(z)/dz = R’ i(z) + jL’ i(z), (2.18a)

• Telegrapher’s equation in phasor domain

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• Wave equations

d² i(z)/dz² - ²i(z) = 0, (2.23)

d²V(z)/dz² - ²V(z) = 0, (2.21)

 =  + j, Complex propagation constant

 = Re (R’ + jL’) (G’+

jC’) ,

 = Im (R’ + jL’) (G’+

jC’) ,

Wave Equations

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Telegrapher’s equation (cont.)

V(z) = V0 (2.26a)

Solving the second order differential equation (2.21), (2.23):

e

-z

+

V-0

e

z

i(z) = I+0

e

-z

+

I0-

e

z (2.26b)

where:

V0 and V-0 are determined by boundary conditions.

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• Characteristic impedance Z0

I0+ =

(R’ + jL’)



V+0

I0- =

(R’ + jL’)

-

V-0

= (R’ + jL’)


Z0 

Define characteristic impedance Z0

I0+

 = (R’ + jL’) (G’+ jC’)
(R’ + jL’)

(G’+j C’)

recall:

Characteristic impedance Z

0

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• Example 2-1, an air line :

Solution:

R’ = G’ = 0, Z0 = 50 ,  = 20 rad/m, f = 700 MHz

L’ = ? and C’ = ?

Z0 = (G’+j C’)(R’ + jL’) =

L’

C’ = 50

 = (R’ + jL’) (G’+ jC’) = j L’C’
 =  + j,

 =  L’C’ = 20 rad/m

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• Lossless transmission line :

= (R’ + jL’) (G’+ jC’)

 =  + j,

If R’<< j L’ and G’ << jC’,

Lossless Transmission Lines

Then:

(2.35)

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Phase velocity

Wavelength

Lossless Transmission Lines (cont.)

(2.39)
(2.40)

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Reflection Coefficient

i(z) =

V(z) = V+0

+

V0-

e

jz

-e

-jz

e

-jz

+
V0

e

jz

-V0

VL = = V+0

+

V-0

V(z)

z = 0

-+
V0
Z0

-V0
Z0
i(z)

z = 0

iL = =

ZL = VL

iL
=

+

-V0

+
V0

-+
V0
Z

-V0
Z
+
V0

-V0 =

ZL - Z0

ZL + Z0

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• Voltage reflection coefficient :

• Current reflection coefficient :

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• Example 2-2 :

A’
z = 0
A
Z0 = 100 

RL = 50 

CL = 10pF

f = 100 MHz

ZL = RL + j/CL = 50 – j159 (Ohms)

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Standing Waves

• From (2.44), we have:

|V(z)| = |V+0| |

e

-jz

+

||

e

jr

e

jz|

= |V+0| [1+ | |² + 2||cos(2z + r)]
1/2

-V0

+
V0

with  =

(2.52)
 |V(z)| is a function of ~ z.

• A similar expression can be derived for |I(z)|

~

|I(z)| = |V+0|/|Z0| [1+ | |² - 2||cos(2z + r)]

1/2

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Standing Waves (cont.)

The repetition period

of the standing wave

pattern is /2.

When voltage is a

maximum

, current is a

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• Special cases

1. ZL= Z0, = 0

= |V+0|

|V(z)|

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2. ZL= 0, short circuit, = -1

= |V+0| [2 + 2cos(2z + )]

1/2

|V(z)|

Standing Waves (cont.)

3. ZL= ,open circuit, = 1

= |V+0| [2 + 2cos(2z )]

1/2

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• Voltage maximum

= |V+0| [1+ | |² + 2||cos(2z + r)] 1/2

|V(z)|

|V0+| [1+ | |],

|V(z)|max =

when 2z + r = - 2n.

Standing Waves (cont.)

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• Voltage minimum

Standing Waves (cont.)

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Voltage standing-wave ratio VSWR

1 - | |

|V(z)|min

S  |V(z)|max

=

1 + | |
S = 1, when  = 0,

S = , when || = 1,

(2.59)

Example 2-4: 50-Ohm transmission line, ZL = 100 + j50 (Ω). Find , SWR.

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Input Impedance of the Lossless Line

Vg(t) VL

z = 0
B

l

z = - l

Zg Ii

Zin(z) = Vi(z)

Ii(z)
=

+



e

jz

e

-jz

V0( )

-



e

jz

e

-jz

V0( )

Z0 =

+

(1 

e

j2z )

-(1 

e

j2z ) Z0

+

(1 

e

-j2 )l

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A 1.05-GHz generator circuit with series impedance Zg = 10- and voltage source
given by Vg(t) = 10 sin(t +30º) is connected to a load ZL = 100 +j5 () through
a 50-, 67-cm long lossless transmission line. The phase velocity is 0.7c. Find
V(z,t) and i(z,t) on the line.

Solution:

Since, Vp = ƒ,  = Vp/f = 0.7c/(1.05 x 109) = 0.2 m.

 = 2/,  = 10 .

 = (ZL-Z0)/(ZL+Z0),  = 0.45exp(j26.6º)

+

(1 

e

-j2 )l

-(1 

e

-j2 )l Z0

Zin(-l) = = 21.9 + j17.4 

V+0[exp(-jl)+ exp(jl)] =

Zin(-l) + Zg

Zin(-l)

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Short-circuited Line

ZL= 0,  = -1, S = 

Vg(t) VL

z = 0
B

l

ZL = 0

z = - l

Zg Ii

Zin

sc

I(z) =

V(z) = V0

- e

)jz

+

(

e

-jz
(

e

-jz

+
V0

e

jz )

= -2jV+0sin(z)

= 2V+0cos(z)/Z0

Zin = V(-l)

I(-l) = jZ0.tan(l)

(2.67)

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Short-circuited Line (cont.)

Zin = V(-l)

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Zin = V(-l)

i(-l) = jZ0tan(l)

• If tan(l) >= 0, the line appears inductive, jLeq = jZ0tan(l),

• If tan(l) <= 0, the line appears capacitive, 1/(jCeq)= jZ0tan(l),

l = 1/[- tan (1/C-1 eqZ0)],

• The minimum length results in transmission line as a capacitor:

Short-circuited Line (cont.)

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tan (l) = - 1/CeqZ0 = -0.354,

Choose the length of a shorted 50- lossless line such that its input impedance
at 2.25 GHz is equivalent to the reactance of a capacitor with capacitance
Ceq = 4pF. The wave phase velocity on the line is 0.75c.

Solution:

Vp = ƒ,   = 2/ = 2ƒ/Vp = 62.8 (rad/m)

l = 2l /  = tan (-0.354) + n = - 0.34 + n -1

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ZL = 0, = 1, S = 

z = 0
z = - l

Vg(t) VL

A B

l

ZL = 

Zg Ii

Zin

I(z) =

V(z) = V0

+ e

)jz

e

-jz
(

e

-jz

+
V0

e

jz )

= 2V+0cos(z)

= 2jV+0sin(z)/Z0

Zin = V(-l)

I(-l) = -jZ0.cot(l)

Open-circuited Line (cont.)

(2.72)

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Open-circuited Line (cont.)

Zin = V(-l)

I(-l) = -jZ0.cot(l)

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Line of Length n/2

= ZL

tan(l) = tan[(2/)(n/2)] = 0,

+

(1 

e

-j2 )l

-(1 

e

-j2 )l Z0
Zin(-l) =

Any multiple of half-wavelength line doesn’t modify the load impedance.

Vg(t)

z = 0
B

l = n/2

ZL

z = - l

Zin

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Quarter-wave transformer: l = /4 + n/2

l = (2/)(/4 + n/2) = /2 ,

= Z0²/ZL

+

(1 

e

-j2 )l

-(1 

e

-j2 )l Z0

Zin(-l) = (1 

+

e

-j 
)

-(1 

e

-j  ) Z0
=

(1 + ) Z0
(1 - )
=

Quarter-wave Transformer

Example 2-9:

A 50- lossless transmission is matched to a resistive load impedance with ZL = 100  via a
quarter-wave section, thereby eliminating reflections along the feed line. Find the
characteristic impedance of the quarter-wave transformer.

Zin = Z0²/ZL= 50 

Z0 = (Zin.ZL) = (50*100) = 70.71 ½ ½ 

ZL= 100 

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• Instantaneous power

P(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)] i i

= Re[|V0|exp(j )exp(jt)] Re[|V+ +0|/Z0 exp(j )exp(jt)] +

= (|V+0|²/Z0) cos²(t +  ) +

Power Flow

-r

P(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)] r r

= Re[|V0|exp(j )exp(jt)] Re[|V+ -0|/Z0 exp(j )exp(jt)] +

= - ||²(|V0+|²/Z0) cos²(t +  + + r)

(2.81)

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Time-Average Power

Time-domain approach:

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Phasor-domain approach:

Time-Average Power (cont.)

= (1-||²) (|V+0|²/2Z0)

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Transient on transmission line

• Step function U(t)

U(t) = 1, if t >= 0; U(t) = 0, if t < 0 U(t)

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Transient of a step function

At t = 0+, immediately after closing the

switch in the circuit in (a), we have:

Rg = 4 Z0

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Voltage distributions on a lossless transmission line at t = T/2, t = 3T/2 and t =

5T/2

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Transient of a step function (cont.)

As t approaches , the ultimate value of V(z,t) is the same at all locations on 
the transmission line, and is given by:

Where x = Lg

By applying (2.132), equation (2.131) can be written as follows:

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Current distributions on a lossless transmission line at t = T/2, t = 3T/2 and t =

5T/2

Transient of a step function (cont.)

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Bounce Diagram

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Bounce Diagram (cont.)

g = 3/5, L = 1/3, z = l/4

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Chapter 2: Transmission Lines

<b>TRƯỜNG ĐIỆN TỪ (CT361)</b>

<b> (ELECTROMAGNETICS)</b>

<b>Transmission Lines</b>

<b> (Đường dây truyền sóng)</b>

•

<i>Lumped-Element (phần tử tập trung) Model of </i>

Transmission Lines

•

Transmission-Line Equations

•

Wave Propagation on a Transmission Line

•

<i>Input Impedance (trở kháng ngõ vào) of the Lossless Line </i>

•

Special Cases of Lossless Line

•

Power Flow on a Lossless Transmission Line

•

<i>Transients (Quá độ) on Transmission Lines</i>

<b>Propagation Modes</b>

<b>TEM transmission lines</b>

<b>Propagation Modes (cont.)</b>

<b>Lumped-Element Model</b>

<b>Transmission line equations</b>

<b>Telegrapher’s equation</b>

<b>Wave Equations</b>

<b>Telegrapher’s equation (cont.)</b>

<i><sub>e</sub></i>

+

<i>e</i>

<i>e</i>

+

<i>e</i>

<b>Characteristic impedance Z</b>

<b>Lossless Transmission Lines </b>

<b>Phase velocity</b>

<b>Wavelength</b>

<b>Lossless Transmission Lines (cont.)</b>

<b>Reflection Coefficient </b>

+

<i>e</i>

<i>-e</i>

<i>e</i>

<i>e</i>

<sub>+</sub>

+

<b>Standing Waves </b>

• From (2.44), we have:

<i>e</i>

+

<i>e</i>

<i>e</i>

<i>• A similar expression can be derived for |I(z)| </i>

<b>Standing Waves (cont.) </b>

The repetition period

of the standing wave

pattern is /2.

When voltage is a

<i>maximum</i>

, current is a

•

<b>Special cases</b>

<b>Standing Waves (cont.) </b>

•

<b>Voltage maximum</b>

<b>Standing Waves (cont.) </b>

•

<b>Voltage minimum</b>

<b>Standing Waves (cont.) </b>

<b>Voltage standing-wave ratio VSWR </b>

=

<b>Input Impedance of the Lossless Line</b>

+

<i>e</i>

<i>e</i>

-

<i>e</i>

<i>e</i>

+