Transmission lines

1. Waveform Distortion
• Signal transmitted over lines are normally complex and consists of
many frequency components.
• Such signal ( voice) voltage will not have all frequencies transmitted
with equal attenuation and equal time delay.
• Also, the received waveform will not be identical with the input
waveform at the sending end.
• This variation is known as distortion .
• There are two types of distortions. They are frequency distortion and
delay distortion.

2. Frequency Distortion
• 𝛼 =
𝑅𝐺−𝜔2𝐿𝐶+ 𝑅𝐺−𝜔2𝐿𝐶 2+𝜔2 𝐿𝐺+𝐶𝑅 2
2
• 𝛼 is the function of frequency.
• Distortion associated with 𝛼 is frequency distortion.
• All frequency transmitted on a line will not be attenuated equally.
• Ex. A complex applied voltage such as voice voltage containing many frequencies
will not be transmitted with equal attenuation.
• The received waveform will not be identical with the input waveform at the
sending end this variation is known as frequency distortion.
• Frequency distortion is reduced in the transmission of high quality over wire lines
by the use of equalizers at the line terminals.
• Equalizers are networks whose frequency and phase characteristics are adjusted
to be inverse to those of the lines.

3. • Delay or phase distortion
• 𝛽 =
𝜔2𝐿𝐶−𝑅𝐺+ 𝑅𝐺−𝜔2𝐿𝐶 2+𝜔2 𝐿𝐺+𝐶𝑅 2
2
• 𝛽 is the complicated function of frequency
• 𝑣 = 2𝜋𝑓/ 𝛽
• 𝑣 = 𝜔/ 𝛽
• 𝜔 and 𝛽 do not both involve frequency in the same manner and
that velocity of propagation will be in general some function of
frequency.
• All frequencies applied to a transmission line will not have the
same time of transmission, some frequencies being delayed
more than others.
• This type of distortion can be avoided by the use of co-axial
cables.

4. Distortion less line
• It is a line which has neither frequency distortion nor phase distortion
and it is correctly terminated is called distortionless line.
• If the line is to have neither frequency distortion nor phase
distortion, then attenuation constant and velocity of propagation
constant should not be the function of frequency.
• In other words, if 𝑣 =
𝜔
𝛽
, then 𝛽 must be direct functions of
frequency.
• The condition for distortionless line is that
1. RC=LG
2. 𝛼= RG i.e. 𝛼 should be independent of frequency

5. 3. 𝛽= 𝜔 𝐿𝐶, 𝛽 must be direct function of frequency
4. 𝑣=𝜔/ 𝛽, 𝑣 must be independent of frequency
• To Prove condition no.1
• 𝑅𝐺 − 𝜔2𝐿𝐶 2 + 𝜔2 𝐿𝐺 + 𝐶𝑅 2𝑚𝑢𝑠𝑡 𝑏𝑒 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑅𝐺 + 𝜔2𝐿𝑐 2
• 𝑅2
𝐺2
+ 𝜔4
𝐿2
𝐶2
− 2𝜔2
𝐿𝐶𝑅𝐺 + 𝜔2
Re +𝜔2
𝐶2
𝑅2
+ 2𝜔2
𝐿𝐶𝑅𝐺 =
𝑅2
𝐺2
+ 𝜔4
𝐿2
𝐶2
+ 2𝜔2
𝐿𝐶𝑅𝐺
• 𝜔2
𝐿2
𝐺2
+ 𝜔2
𝐶2
𝑅2
= 2𝜔2
𝐿𝐶𝑅𝐺
• 𝜔2
𝐿2
𝐺2
+ 𝜔2
𝐶2
𝑅2
− 2𝜔2
𝐿𝐶𝑅𝐺 = 0

6. • 𝐿𝐺 − 𝐶𝑅 2 = 0
• 𝐿𝐺 = 𝐶𝑅 this will make 𝛼 and 𝑣 independent of frequency simultaneously.
•
𝑅
𝐺
=
𝐿
𝑐
This requires a very large value of L, since G is small.
• Now if G is intentionally increased, 𝛼 and attenuation are increased ,will result in
poor efficiency of the transmission line.
• To Prove condition no.2
• 𝛼 =
𝑅𝐺−𝜔2𝐿𝐶+ 𝑅𝐺−𝜔2𝐿𝐶 2+𝜔2 𝐿𝐺+𝐶𝑅 2
2
• 𝑅𝐺 − 𝜔2
𝐿𝐶 2
+ 𝜔2
𝐿𝐺 + 𝐶𝑅 2
𝑚𝑢𝑠𝑡 𝑏𝑒 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑅𝐺 + 𝜔2
𝐿𝑐 2
• 𝛼 =
𝑅𝐺−𝜔2𝐿𝐶+ 𝑅𝐺+𝜔2𝐿𝐶 2
2

7. • 𝛼 =
𝑅𝐺−𝜔2𝐿𝐶+𝑅𝐺+ 𝜔2𝐿𝐶
2
• 𝛼= 𝑅𝐺
• To Prove condition 3
• 𝛽 =
𝜔2𝐿𝐶−𝑅𝐺+ 𝑅𝐺−𝜔2𝐿𝐶 2+𝜔2 𝐿𝐺+𝐶𝑅 2
2
• 𝑅𝐺 − 𝜔2𝐿𝐶 2 + 𝜔2 𝐿𝐺 + 𝐶𝑅 2𝑚𝑢𝑠𝑡 𝑏𝑒 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑅𝐺 + 𝜔2𝐿𝑐 2
• 𝛽 =
𝜔2𝐿𝐶−𝑅𝐺+𝑅𝐺+ 𝜔2𝐿𝐶
2

8. • 𝛽= 𝜔 𝐿𝐶
• To Prove condition 4
• 𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝛽= 𝜔 𝐿𝐶
•
𝜔
𝛽
=
1
𝐿𝐶
• 𝑣=
1
𝐿𝐶
which is same for all frequencies and thus eliminating delay
distortion
• Therefore, we conclude that a transmission line will have neither
delay nor frequency distortion only if α is independent of frequency
and β should be a function of frequency.

9. Reflection on a line not terminated in Zo
• 𝐸 =
𝐸𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑅
𝑒𝛾𝑠
+
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
𝑒−𝛾𝑠
=
𝑬𝑹 𝒁𝑹+𝒁𝑶
𝟐𝒁𝑹
𝒆𝜸𝒔
+
𝑬𝑹 𝒁𝑹+𝒁𝑶
𝟐𝒁𝑹
𝒁𝑹−𝒁𝑶
𝒁𝑹+𝒁𝑶
𝒆−𝜸𝒔
• 𝐼 =
𝐼𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑂
𝑒𝛾𝑠
−
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
𝑒−𝛾𝑠
=
𝑰𝑹 𝒁𝑹+𝒁𝑶
𝟐𝒁𝑶
𝒆𝜸𝒔
−
𝑰𝑹 𝒁𝑹+𝒁𝑶
𝟐𝒁𝑶
𝒁𝑹−𝒁𝑶
𝒁𝑹+𝒁𝑶
𝒆−𝜸𝒔
• Consider above voltage and current equations
• If 𝑍𝑅 is terminated in 𝑍𝑂 and hence there will be no reflection.
• If when 𝑍𝑅 is not terminated in 𝑍𝑂, (𝑍𝑅 = ∞) then reflection phenomena
exist on the line i.e. part of wave is reflected back from the receiving end.
• Such reflection is maximum when the line is on open circuit i.e. 𝑍𝑅 = ∞ or
short circuit i.e. 𝑍𝑅 = 0.
• The wave that travels from the sending end to the receiving end can be
identified as component varying with 𝑒𝛾𝑠
.

10. • This wave of voltage or current where amplitude of the wave
decreases and varies exponentially with 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑠 𝑖. 𝑒. 𝑒𝛾𝑠 is known
as the incident wave
• 𝑬𝟏 =
𝑬𝑹 𝒁𝑹+𝒁𝑶
𝟐𝒁𝑹
𝒆𝜸𝒔
= Incident Voltage wave
• 𝑰𝟏 = +
𝑰𝑹 𝒁𝑹+𝒁𝑶
𝟐𝒁𝑶
𝒆𝜸𝒔 = Incident current wave
• The wave that travels from the receiving end to the sending end can
be identified as component varying with 𝑒−𝛾𝑠.
• This wave of voltage or current where amplitude of the wave
decreases and varies exponentially with 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑠 𝑖. 𝑒. 𝑒−𝛾𝑠 is
known as the reflected wave.
• Note: s is minimum (s=0)at the receiving end and maximum (s=l) at
the sending end.

11. • 𝑬𝟐 =
𝑬𝑹 𝒁𝑹−𝒁𝑶
𝟐𝒁𝑹
𝒆−𝜸𝒔 = Reflected Voltage wave
• 𝑰𝟐 = −
𝑰𝑹 𝒁𝑹−𝒁𝑶
𝟐𝒁𝑶
𝒆−𝜸𝒔 = Reflected current wave
• Thus the total instantaneous voltage or current at any point on the line
is the phasor sum of voltage or current of the incident and reflected
waves.
• Considering open circuit line with 𝑍𝑅 = ∞, the incident component of
voltage becomes𝑬𝟏 =
𝑬𝑹
𝟐
𝒆𝜸𝒔 and the reflected component of voltage
becomes 𝑬𝟐 =
𝑬𝑹
𝟐
𝒆−𝜸𝒔
• At s=0 (minimum) at the receiving end, open circuit 𝑬𝟏 =
𝑬𝑹
𝟐
, 𝑬𝟐 =
𝑬𝑹
𝟐
• Thus initial value of the reflected wave is equal to incident voltage at
the load on open circuit.

12. • The solid curve is for incident wave while the dotted curve is for
reflected wave.
• The only difference between the curves for the voltage and current
for open circuit line is the reversed phase of the reflected current
wave.
• Instantaneous current waves Instantaneous voltage waves

15. The two current waves are equal and opposite phases
Thus addition of instantaneous currents is always zero.

17. • The term
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
decides the relative phase angles between incident
and reflected waves.
• Thus magnitudes and phase angles of 𝑍𝑅 𝑎𝑛𝑑 𝑍𝑂 are important to
determine the phase angle between the two waves.
• Thus we can conclude that when no reflection on the line, the line is
uniform and finite and no discontinuity exists to send the reflected
wave back along the line.
• Hence, the wave travel smoothly along the finite line and energy is
absorbed in the load without any reflected wave.

18. • This energy is transmitted through the electric field and magnetic
fields.
• The energy conveyed in the electric field depends on the voltage E
and is given by
• 𝑊
𝑒 =
1
2
𝐶𝐸2 J/𝑚3
• The energy conveyed in the magnetic field depends on the current I
and is given by
• 𝑊
𝑚 =
1
2
𝐿𝐼2J/𝑚3
• For an ideal line which is terminated in Zo, the ratio of E and I is fixed
along the line Zo
• 𝑍𝑜 =
𝐸
𝐼

19. • For such a line, R=G=0 and the value of Zo is given by 𝑍𝑜 =
𝐿
𝐶
• 𝑍𝑜
2 =
𝐿
𝐶
hence C=
𝐿
𝑍𝑜
2
• 𝑊
𝑒 =
1
2
𝐿
𝑍𝑜
2 (𝐼2
𝑍𝑜
2
)=
1
2
𝐿𝐼2
= 𝑊
𝑚
• When line is not terminated in Zo then
• 𝑍𝑅 =
𝐸𝑅
𝐼𝑅
• This ratio is practically required for which the redistribution of energy
between electric and magnetic field takes place.
• This redistribution of energy creates the reflected wave back along
the line.

20. Disadvantages of Reflection
• If the attenuation is not large enough the reflected wave appears as
echo at the sending end.
• Reduction in efficiency
• The part of the received energy is rejected by the load hence ouput
reduces

21. • For an open wire line 𝜷 = 𝟎. 𝟎𝟒 𝒓𝒂𝒅/𝒌𝒎 find the wavelength and
the velocity at a frequency of 1600 Hz. Hence calculate the time
taken by the wave to travelb90 km
• 𝜆 = 2𝜋/ 𝛽= 2𝜋/ 0.04 =157.078 km
• 𝑣 = 𝜔/ 𝛽= 2𝜋f/ 𝛽=2 𝜋 ∗ 1600/0.04= 2.5132*105 km/sec
Distance (x)= 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ∗ 𝑡𝑖𝑚𝑒
• T=
90
2.5132∗105 km/sec
=3.58* 10−4
𝑠𝑒𝑐

22. • The constants of the transmission line are R= 48.75 Ω/km, G=38.75
*10−6Ω/𝑘𝑚 C=0.059* 10−6 F/km, L=1.09 mH/km
• Determine the characteristic impedance and propagation constant at
1600Hz

23. Reflection Coefficient
• It is defined as the ratio of amplitudes of the reflected and incident
voltage waves at the receiving end of the line.
• It is denoted by K
• K =
𝑅𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑡 𝑙𝑜𝑎𝑑
𝐼𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑡 𝑙𝑜𝑎𝑑
at s=0 (Receiving end)
• 𝐾 =
𝑬𝑹 𝒁𝑹−𝒁𝑶
𝟐𝒁𝑹
𝑬𝑹 𝒁𝑹+𝒁𝑶
𝟐𝒁𝑹
• K =
𝑬𝑹 𝒁𝑹−𝒁𝑶
𝑬𝑹 𝒁𝑹+𝒁𝑶
• K =
𝒁𝑹−𝒁𝑶
𝒁𝑹+𝒁𝑶

24. • 𝐸 =
𝐸𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑅
𝑒𝛾𝑠
+
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
𝑒−𝛾𝑠
• 𝐼 =
𝐼𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑂
𝑒𝛾𝑠
−
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
𝑒−𝛾𝑠
• Substituting the value of Reflection Coefficient K in above equations
• 𝐸 =
𝐸𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑅
𝑒𝛾𝑠 + 𝐾𝑒−𝛾𝑠
• 𝐼 =
𝐼𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑂
𝑒𝛾𝑠
− 𝐾𝑒−𝛾𝑠
• The sign of K and hence the polarity of the reflected wave is
dependent on the angles and magnitudes of 𝑍𝑅 and 𝑍𝑂

25. Input and transfer impedance
• The input impedance of a transmission line is given as
• 𝑍𝑠 = 𝑍𝑂[
𝑍𝑅𝐶𝑜𝑠ℎ𝛾𝑙+𝑍𝑂𝑆𝑖𝑛ℎ𝛾𝑙
𝑍𝑂𝐶𝑜𝑠ℎ𝛾𝑙+𝑍𝑅𝑆𝑖𝑛ℎ𝛾𝑙
]
• In terms of exponential functions
• 𝑍𝑠 =
𝐸𝑠
𝐼𝑠
=
𝐸𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑅
𝑒𝛾𝑙+
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
𝑒−𝛾𝑙
𝐼𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑂
𝑒𝛾𝑙−
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
𝑒−𝛾𝑙
𝐸𝑅= 𝐼𝑅𝑍𝑅
• 𝑍𝑠 =
𝐸𝑠
𝐼𝑠
=
𝐸𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑅
𝑒𝛾𝑙+
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
𝑒−𝛾𝑙
𝐸𝑅 𝑍𝑅+𝑍𝑂
2𝑍𝑂𝑍𝑅
𝑒𝛾𝑙−
𝑍𝑅−𝑍𝑂
𝑍𝑅+𝑍𝑂
𝑒−𝛾𝑙

26. Input and transfer impedance
• Input impedance
𝑍𝑠 =
𝐸𝑠
𝐼𝑠
= 𝑍𝑂[
𝑍𝑅𝐶𝑜𝑠ℎ𝛾𝑙+𝑍𝑂𝑆𝑖𝑛ℎ𝛾𝑙
𝑍𝑂𝐶𝑜𝑠ℎ𝛾𝑙+𝑍𝑅𝑆𝑖𝑛ℎ𝛾𝑙
] --------(20)
• 𝑍𝑠 = 𝑍𝑂(
𝑒𝛾𝜄+𝐾𝑒−𝛾𝜄
𝑒𝛾𝜄−𝐾𝑒−𝛾𝜄) ------(21)
• Transfer Impedance
• If the sending voltage is known , it is convenient to have the
transfer impedance so that the received current can be
computed directly.
• The sending end voltage 𝐸𝑠 is
• (13)→ E = 𝐸𝑠 =
𝐸𝑅
2𝑍𝑅
(𝑍𝑅 + 𝑍0)[𝑒𝛾𝜄 + K𝑒−𝛾𝜄]

27. • 𝐸𝑠 =
𝐼𝑅
2
(𝑍𝑅 + 𝑍0)[𝑒𝛾𝜄 + K𝑒−𝛾𝜄]
• therefore the transfer impedance is
• 𝑍𝑇 =
𝐸𝑠
𝐼𝑅
=
(𝑍𝑅 + 𝑍0)
2
[𝑒𝛾𝜄 + K𝑒−𝛾𝜄]
• By substituting the value of K
• 𝑍𝑇 =
(𝑍𝑅 + 𝑍0)
2
[𝑒𝛾𝜄 + (
𝑍𝑅 −𝑍0
𝑍𝑅 +𝑍0
) 𝑒−𝛾𝜄]
• = (
𝑍𝑅 + 𝑍0
2
)𝑒𝛾𝜄 + (
𝑍𝑅 − 𝑍0
2
)𝑒−𝛾𝜄

28. • =
𝑍𝑅
2
𝑒𝛾𝜄 +
𝑍0
2
𝑒𝛾𝜄 +
𝑍𝑅
2
𝑒−𝛾𝜄 -
𝑍0
2
𝑒−𝛾𝜄
• = 𝑍𝑅(
𝑒𝛾𝜄+𝑒−𝛾𝜄
2
) + 𝑍0 (
𝑒𝛾𝜄− 𝑒−𝛾𝜄
2
)
• Therefore 𝑍𝑇 = 𝑍𝑅 cosh𝛾𝜄 + 𝑍0sinh𝛾𝜄

29. Open and Short Circuited Lines
• Open circuited when 𝑍𝑅 = ∞
• Short circuited when 𝑍𝑅 = 0
• The input impedance of a line of length 𝜄 is
• (20)→ 𝑍𝑠 = 𝑍𝑂[
𝑍𝑅𝐶𝑜𝑠ℎ𝛾𝑙+𝑍𝑂𝑆𝑖𝑛ℎ𝛾𝑙
𝑍𝑂𝐶𝑜𝑠ℎ𝛾𝑙+𝑍𝑅𝑆𝑖𝑛ℎ𝛾𝑙
]
• For S.C (𝑍𝑅 = 0)
• (20)→ 𝑍𝑠𝑐 = 𝑍𝑂[
𝑆𝑖𝑛ℎ𝛾𝑙
𝐶𝑜𝑠ℎ𝛾𝑙
] = 𝑍𝑂tanh𝛾𝑙 ----(1)

30. • For O.C (𝑍𝑅 =∞ )
• (20)→ 𝑍𝑠 = 𝑍𝑂[
𝑍𝑅𝐶𝑜𝑠ℎ𝛾𝑙+𝑍𝑂𝑆𝑖𝑛ℎ𝛾𝑙
𝑍𝑂𝐶𝑜𝑠ℎ𝛾𝑙+𝑍𝑅𝑆𝑖𝑛ℎ𝛾𝑙
]
• Divide by 𝑍𝑅
• 𝑍𝑜𝑐 = 𝑍𝑂[
𝐶𝑜𝑠ℎ𝛾𝑙+(𝑍𝑂/𝑍𝑅)𝑆𝑖𝑛ℎ𝛾𝑙
(𝑍𝑂/𝑍𝑅)𝐶𝑜𝑠ℎ𝛾𝑙+𝑆𝑖𝑛ℎ𝛾𝑙
]
• 𝑍𝑜𝑐 = 𝑍𝑂[
𝐶𝑜𝑠ℎ𝛾𝑙
𝑆𝑖𝑛ℎ𝛾𝑙
] = 𝑍𝑂Coth𝛾𝑙 -----(2)
• Multiply (1) and (2)
• 𝑍𝑂
2
= 𝑍𝑠𝑐 𝑍𝑜𝑐
• 𝑍𝑂 = 𝑍𝑠𝑐 𝑍𝑜𝑐

31. • Divide (1) and (2)
•
𝑍𝑠𝑐
𝑍𝑜𝑐
=
𝑍𝑂tanh𝛾𝑙
𝑍𝑂Coth𝛾𝑙
= (𝑡𝑎𝑛ℎ𝛾𝑙)2
• tanh𝛾𝑙 =
𝑍𝑠𝑐
𝑍𝑜𝑐
• 𝛾𝑙 = 𝑡𝑎𝑛−1 𝑍𝑠𝑐
𝑍𝑜𝑐
• If tanh𝛾𝑙 = tan (𝛼 + 𝑗𝛽)𝜄 = U+jV
• ∴ 𝑡𝑎𝑛ℎ2𝛼𝜄 =
2𝑈
1+𝑈2+𝑉2
• 𝑡𝑎𝑛ℎ2𝛽𝜄 =
2𝑉
1−𝑈2−𝑉2