Indraprastha Institute of

Information Technology Delhi ECE230

Lecture – 3 Date: 11.01.2016

• Examples
• Transmission Lines
• Transmission Line Equations
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Example – 1
• A laser beam traveling through fog was observed to have an intensity of
𝜇𝑊
1 ( 2 ) at a distance of 2𝑚 from the laser gun and an intensity of
𝑚
𝜇𝑊
0.2 ( 2 ) at a distance of 3𝑚. Given that the intensity of an
𝑚
electromagnetic wave is proportional to the square of its electric field
amplitude, find the attenuation constant 𝛼 of fog.
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Example – 2
• If 𝑧 = 3𝑒 𝑗𝜋/6 , find the value of 𝑒 𝑧 .
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Example – 3
• Find the instantaneous time sinusoidal functions corresponding to the
following phasors:
𝑗𝜋
𝑎 𝑉= −5𝑒 3 𝑉
𝑏 𝑉 = 𝑗6𝑒 −𝑗𝜋/4 𝑉
𝑐 𝐼 = 6 + 𝑗8 𝐴
𝑑 𝐼 = −3 + 𝑗2 𝐴
𝑒 𝐼= 𝑗 𝐴
𝑓 𝐼 = 2𝑒 𝑗𝜋/6 𝐴
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Example – 4
• The voltage source of the circuit shown below is given by:
𝑣𝑠 𝑡 = 25𝑐𝑜𝑠 4 × 104 𝑡 − 45° 𝑉
Obtain an expression for 𝑖𝐿 𝑡 , the current flowing through the inductor.

𝑣
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Transmission Line (TL)

• It encompasses all structures and media that serve to transfer energy or
information between two points.
• In this course we talk about TL that guide EM waves.
• Such TLs include telephone wires, coaxial cable carrying audio and video
information to TV sets or digital data to computers, microstrips printed on
microwave circuit boards, and optical fibers carrying light waves.

Fundamentally, TL is a two port network with each port

consisting of two terminals.
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Transmission Line (TL) (contd.)

Source end may be any circuit Load end may be:

generating an output voltage e.g., • Antenna in the case of Radar
• Radar transmitter • Input terminal of Amplifier
• Amplifier • Computer terminal in
• Computer terminal in transmitting receiving mode
mode
• In the case of dc: source is represented by Thevenin equivalent generator
𝑉𝑔 and 𝑅𝑔 .
• In the case of ac: the corresponding terms are 𝑉𝑔 and 𝑍𝑔 .
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Transmission Line (TL) (contd.)

The role of wavelength
• In low frequency circuits,
circuit elements usually
are interconnected using
simple wires.

The pertinent questions:

• Is the pair of wires between terminals 𝐴𝐴′ and 𝐵𝐵′ a transmission line?
• If so, under what set of circumstances should we explicitly treat the pair
of wires as a transmission line?
Answer:
• Yes
• The answer to second question depends on the length of the line 𝑙 and
the frequency 𝑓 of the wave provided by the generator.
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Transmission Line (TL) (contd.)

• Essentially, the determining factor is the ratio of length 𝑙 and wavelength
λ of the wave propagating on the transmission line.
′
• Let: 𝑉𝐴𝐴 = 𝑉𝑔 𝑡 = 𝑉0 𝑐𝑜𝑠𝜔𝑡, and assume that the current flowing
through the wires travel at the speed of light.
′
• Then 𝑉𝐵𝐵 (t) will appear after a delay of 𝑙/𝑐.
′ ′
• Therefore, if the wires are lossless: 𝑉𝐵𝐵 t = 𝑉𝐵𝐵 t − 𝑙/𝑐
′ ′ Where:
𝑉𝐵𝐵 t = 𝑉0 cos(𝜔𝑡 − 𝑙/𝑐) 𝑉𝐵𝐵 t = 𝑉0 cos(𝜔𝑡 − 𝜑0 )
𝝋𝟎 = 𝝎𝒍/𝒄

• At t=0 and f=1kHz, 𝑙 = 5𝑐𝑚

′ ′
2𝜋𝑓𝑙
𝑉𝐴𝐴 = 𝑉0 𝑉𝐵𝐵
= 𝑉0 cos = 0.999999998𝑉0
𝑐
′
• At t=0 and f=1kHz, 𝑙 = 20𝑘𝑚 The value of 𝑉𝐵𝐵
′ 2𝜋𝑓𝑙 is controlled by
𝑉𝐴𝐴 = 𝑉0 ′
𝑉𝐵𝐵 = 𝑉0 cos = 0.91𝑉0 𝝋𝟎 = 𝝎𝒍/𝒄
𝑐
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Transmission Line (TL) (contd.)

• The velocity of propagation is related to frequency as 𝑢𝑝 = 𝑓λ and in
present case 𝑢𝑝 = 𝑐.
𝝎𝒍 𝟐𝝅𝒇𝒍 𝒍
∴ 𝝋𝟎 = = = 𝟐𝝅 radians
𝒄 𝒄 𝝀

When 𝑙/λ is small, transmission-line effects may be ignored, but when

𝑙
≥ 0.01, it may be necessary to account not only for the phase shift
λ
due to the time delay, but also for the presence of reflected signals
that may have been bounced back by the load toward the generator.
• Power loss on the line and dispersive effects may need to be considered
as well.
• A dispersive line is one on which the wave velocity is not constant as a
function of the frequency 𝑓.
• Therefore, the shape of a rectangular pulse, which can be decomposed
into many sinusoidal waves of different frequencies, will be distorted on a
dispersive TL.
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Transmission Line (TL) (contd.)

• Preservation of pulse shape is very important in high speed data

transmission.
• For example, at 10GHz the wavelength is 3cm in air but only about 1cm
in semiconductor.
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Transmission Lines (TLs) (contd.)
• Variations in current and voltage across the circuit dimensions → KCL and
KVL can’t be directly applied → This anomaly can be remedied if the line is
subdivided into elements of small (infinitesimal) length over which the
current and voltage do not vary.

z Dz Dz Dz Dz

Circuit Model: Dz
RDz LDZ RDz LDZ RDz LDZ RDz LDZ
GDz CDz GDz CDz GDz CDz GDz CDz

lim  Infinite number of infinitesimal sections

Dz 0
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Example Transmission Line

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Example Transmission Line (contd.)

• Transverse TEM Transmission Lines: Electric and Magnetic fields are
entirely transverse to the direction of propagation.

• The electric field is in the radial direction between the inner and outer
conductors.
• The magnetic field circles the inner conductor.
• Higher order Transmission Lines: waves propagating along these lines
have at least one significant field component in the direction of
propagation. Hollow conducting waveguides, optical fiber etc belong to
this class of lines.
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Example of Transmission Lines

Coaxial Cable
d = conductivity of dielectric [S/m].

 a m = conductivity of metal [S/m].

r

z b

2 0 r 2 d
C  F/m G  S/m 
b b
ln   ln  
a a
0  b  1  1 1 
L ln    H/m  R  /m
2  a   m  2 a 2 b 
2 (skin depth

 m of metal)
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Example of Transmission Lines (contd.)

 0 r
C  F/m
Twin Line  d 
1
cosh  
a = radius of wires  2a 
0 1  d 
L cosh    H/m
  2a 
r
1 1  d 
Z0  0 cosh 1     
d  r  2a 

 
x 
cosh 1 x  ln x  x 2  1   ln 2 x
1 1 d 
Z0  0 ln     
 r  a 
a d
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Transmission Line Equations

i( z, t ) RDz LDz i( z  Dz, t )
 
v( z , t ) CDz v( z  Dz, t ) Lossy
GDz
Transmission Line
  Circuit Model

Apply KVL:
i ( z , t ) v( z , t )  v( z  Dz , t ) i ( z , t )
v( z , t )  v( z  Dz , t )  RDzi ( z , t )  LDz  Ri ( z , t )  L
t Dz t
Describes the v( z , t ) i ( z , t )
voltage along the   Ri ( z , t )  L For ∆𝑧 → 0
z t
transmission lines

KCL on this line segment gives: i( z, t )  i( z  Dz, t )  GDzv( z  Dz , t )  C Dz v( z  Dz, t )

t
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Transmission Line Equations (contd.)

Simplification results in: i( z, t )  Gv( z, t )  C v( z, t ) For ∆𝑧 → 0
z t

Describes the current along the

transmission lines

These differential equations for

current/voltages were derived by Oliver
Heavyside. These equations are known
as Telegrapher’s Equations.
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Transmission Line Equations (contd.)

• Let us define phasrors: 𝑣 𝑧, 𝑡 = 𝑅𝑒 𝑉(𝑧)𝑒 𝑗𝜔𝑡 and 𝑖 𝑧, 𝑡 = 𝑅𝑒 𝐼(𝑧)𝑒 𝑗𝜔𝑡

• With the substitution

of phasors, the Re

 V(z)e jωt   
= - Re RI(z)e jωt + jωLI(z)e jωt
z
equations of voltage
and current wave
Re

 I(z)e jωt = - Re 
GV(z)e jωt
+ jωCV(z)e jωt

result in: z
• The differential equations for current and voltage along the transmission
line can be expressed in phasor form as:

Re

d V(z)e jωt  = - Re  RI(z)e jωt + jωLI(z)e jωt 
As 𝐼(𝑧) and
dz 𝑉(𝑧) are

d  I(z)e jωt 
function of
= - Re  GV(z)e jωt + jωCV(z)e jωt 
only position
Re
dz
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Transmission Line Equations (contd.)

• The equations can be simplified as:
 d V(z)  jωt 
 Re  + RI(z)+ jωLI(z)  e  = 0
 dz 
 
For further  d  I(z)  jωt 
simplification  Re  + GV(z)+ jωCV(z)  e  = 0
 dz 
 

 d V(z) 
At ωt=0, ejωt=1:  Re  + RI(z)+ jωLI(z)   = 0
 dz 

 d V(z)  
At ωt=π/2, ejωt=j:  Re  + RI(z)+ jωLI(z)  j  = 0
 dz  
 
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Transmission Line Equations (contd.)

• Finally we can write: d V(z) These differential
   R  j L  I(z) equations can be
dz
solved for the
d I(z)
   G  jC  V(z) phasors along the
dz transmission line
Differentiating with respect
to 𝒛 gives d 2 V(z)
2
  2
V ( z)  0
dz Here
d 2 I(z)
Transmission
2
  2
I ( z)  0
Line Wave dz
Equations
  ( R  j L)(G  jC )
Complex Propagation
Constant
    j
Attenuation Constant Phase Constant
(nepers/m) (radians/m)
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Transmission Line Equations (contd.)

d 2 V(z)
2
  2
V ( z)  0 V ( z )  V0 e  z  V0e z
dz
𝑽𝟎 + and 𝑽𝟎 − are complex
d 2 I(z) constants
2
  2
I ( z)  0 I ( z )  I 0 e  z  I 0e z
dz
𝑰𝟎 + and 𝑰𝟎 − are complex constants
𝒆−𝜸𝒛 represents a wave propagating in +𝒛 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 whereas
𝒆+𝜸𝒛 a wave propagating in −𝒛 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏.
• There are four unknowns 
(𝑉0+ , 𝑉0− , 𝐼0+ , 𝐼0− ). These can be I ( z)  V0 e  z  V0e z 
related as: R  j L

V0  z V0  z
I ( z)  e  e
Z0 Z0
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Transmission Line Equations (contd.)

• Comparison of current phasor solutions lead us to:
V0 V0

 Z0   Characteristic Impedance
I0 I0

R  j L
Z0 

• It is equal to the ratio of the voltage amplitude to the current
amplitude for each of the traveling waves individually (with an
additional minus sign in the case of −𝑧 propagating wave.
• It is not equal to the ratio of the total voltage 𝑉(𝑧) to the total
current 𝐼 𝑧 unless one of the two is absent.

Definitely not an impedance in traditional sense ……