**Analysis of Long Transmission Line ( Rigorous Method of Solution of Long
Transmission Line )**

A long transmission line is considered as infinite length. It is not possible to calculate all the parameters for infinite long length therefore here we consider only small length dx in order to evaluate whole line. Let us consider that length of line is dx at x distance from receiving end of the transmission line.

Let

V_{S} =
Sending end voltage to neutral

V_{R} =
Receiving end voltage to neutral

I_{S} =
Sending end current

I_{R} =
Receiving end current

r = Resistance per
unit length of line

x = Reactance per
unit length of line

g = Conductance
per unit length of line

b = Susceptance
per unit length of line

z = Impedance per
unit length of line = √ ( r^{2} + x^{2} )

y = Admittance per
unit length of line = √ ( g^{2} + b^{2} )

V = Voltage per
phase at the end from the small element dx to receiving end side

V + dV = Voltage
per phase at the end from the small element dx to the sending end side

I = Current
leaving from the small element dx

I + dI= Current
entering to the small element dx

Series impedance
of small length dx is = zdx

Shunt admittance
of small length dx is = ydx

Voltage drop in
the transmission line

V = Iz

Rise in voltage
over the small length dx

dV = Izdx

dV / dx = Iz………. (
1 )

similarly, current
entering transmission line

I = Vy

Change in current
over the small length dx

dI / dx = Vy

dI / dx = Vy…………(
2 )

Differentiating
equation ( 1 ) with respect to x

d^{2}V /
dx^{2} = z ( dI / dx )

from equation ( 2
)

d^{2}V /
dx^{2} = Vzy

This is second
order differential equation and its solution

V = A1 e^{√(yz)
x} + A2 e ^{-√(yz) x} ….. (
3 )

Where

A1 , A2 =
Constants

Differentiate
equation ( 3 ) with respect to x

dV / dx = √(yz) A1
e^{√(yz) x} –√(yz)^{ }A2 e ^{-√(yz) x}

dV / dx = √(yz) {
A1 e^{√(yz) x} – ^{ }A2 e
^{-√(yz) x} }…… ( 4 )

from equation ( 1
)

dV / dx = Iz

I = ( 1 / z ) dV /
dx….. ( 5 )

From equation ( 4
) and ( 5 )

I = ( 1 / z ) √(yz)
{ A1 e^{√(yz) x} – ^{ }A2
e ^{-√(yz) x} }

I = √ ( y /z ) {
A1 e^{√(yz) x} – ^{ }A2 e
^{-√(yz) x} }……. ( 5 )

Now, we find out
A, B, C and D parameters of transmission line

At
the receiving end side x = 0, V = V_{R} and I = I_{R}

Substitute value
of x in the equation ( 3 ) and ( 5 )

V_{R} =
A1 + A2 ……….. ( 6 )

I_{R} = √
( y /z ) ( A1 – ^{ }A2 ) ….. ( 7
)

**Characteristic
constant Z _{C}**

Z_{C} = √ ( z / y )……. ( 8 )

**Propagation
constant **

ℽ = √ ( yz )……. ( 9 )

Relation between
propagation constant and characteristic constant

Z_{C} ℽ = √
( z / y ) √ ( yz ) = 1

**Find out constant
parameters A1 and A2**

Solving equation
( 6 ) and ( 7 ) and find out constant A1 and A2

A1 + A2 = V_{R
}………( 10 )

A1 – ^{ }A2 = I_{R} / √ ( y /z )……… ( 11
)

Addition of
equation ( 10 ) and ( 11 )

2A1 = V_{R }+
I_{R} / √ ( y /z )

A1 = { V_{R }+
I_{R} / √ ( y /z ) } / 2

As Z_{C}
= √ ( z / y )

**A1 = { V _{R }+ I_{R} Z_{C}
} / 2 **

Substitute value
of A1 into equation ( 10 )

A1 + A2 = V_{R}_{}

{ V_{R }+
I_{R} / Z_{C} } / 2 + A2
= V_{R}_{}

A2 = V_{R }–
{ V_{R }+ I_{R} Z_{C} } / 2

A2 = V_{R }/
2 – I_{R} Z_{C} / 2

**A2 = { V _{R }– I_{R} Z_{C}
} / 2**

Substitute value
of A1 and A2 in the equation ( 3 )

**Voltage Equation **

V = A1 e^{√(yz)
x} + A2 e ^{-√(yz) x}

V = [{ V_{R }+
I_{R} Z_{C} } / 2 ] ( e ^{ℽ x} ) + [ { V_{R }–
I_{R} Z_{C} } / 2 ]( e ^{–ℽ x} )

= V_{R }/ 2 ( e ^{ℽ x} ) +
( I_{R} Z_{C} ) / 2 (e ^{ℽ x} ) + V_{R }/ 2 ( e
^{–ℽ x} )

– ( I_{R} Z_{C} ) / 2 (
e ^{–ℽ x} )

Rearranging terms

= V_{R }( e ^{ℽ x} + e ^{–ℽ
x} ) / 2 + ( I_{R} Z_{C} ) ( e ^{ℽ x} – e ^{–ℽ
x} ) / 2

Cos h^{ }ℽx
= ( e ^{ℽ x} + e ^{–ℽ x} ) / 2 and

Sin h^{ }ℽx
= ( e ^{ℽ x} – e ^{–ℽ x} ) / 2

Therefore

V = V_{R }Cos h^{ }ℽx + ( Z_{C}
) I_{R} Sin h^{ }ℽx

Sending end
voltage is obtained by putting x = l

V_{S} = V_{R
}Cos h^{ }ℽl + ( Z_{C} ) I_{R} Sin h^{ }ℽl

Now ℽl = √ ( yz )
l = √ ( yl )( zl ) = √ YZ

Where Y = yl =
Total admittance of the transmission line

Z = zl = Total impedance of the
transmission line

V_{S} = V_{R }Cos h^{ }√
YZ + ( Z_{C} ) I_{R} Sin h^{ }√ YZ

Compare this equation with V_{S} =
AV_{R} + BI_{R}_{}

A = Cos h^{ }√ YZ

B = Z_{C} Sin h^{ }√ YZ

**Current Equation**

Substitute value
of A1 and A2 in the equation ( 5 )

I = √ ( y /z ) {
A1 e^{√(yz) x} – ^{ }A2 e
^{-√(yz) x} }……. ( 5 )

I = 1 / Z_{C}
[{ V_{R }+ I_{R} Z_{C} } / 2 ]( e ^{ℽ x} ) – [{ V_{R }– I_{R} Z_{C}
} / 2 ]( e ^{–ℽ x} )

I = [{ V_{R}
/ Z_{C} + I_{R} } / 2 ]( e ^{ℽ x} ) – [{ V_{R}
/ Z_{C} _{ }– I_{R}
} / 2 ]( e ^{–ℽ x} )

= { ( V_{R} / Z_{C} ) / 2 }( e
^{ℽ x} ) – { ( V_{R} / Z_{C }) / 2 }( e ^{–ℽ x}
) +

{ ( I_{R} ) / 2
}( e ^{ℽ x} ) + {( I_{R} ) / 2 }( e ^{–ℽ x} )

= V_{R} / Z_{C} { ( e ^{ℽ
x} ) – ( e ^{–ℽ x} ) / 2 } +

I_{R} { ( e ^{ℽ
x} ) + ( e ^{–ℽ x} ) / 2 }

I = I_{R} Cos h^{ }ℽx
+ V_{R} / Z_{C} Sin h^{
}ℽx

Sending end voltage is obtained by putting x =
l

I_{S} = I_{R}
Cos h^{ }ℽl + V_{R} / Z_{C} Sin h^{ }ℽl

= V_{R} / Z_{C} Sin h^{ }ℽl
+ I_{R} Cos h^{ }ℽl

Again ℽl = √ ( yz
) l = √ ( yl )( zl ) = √ YZ

I_{S} = V_{R}
/ Z_{C} Sin h^{ }√ YZ + I_{R} Cos h^{ }√ YZ

Compare this equation with I_{S} =
CI_{R} + DV_{R}

C = 1 / Z_{C} Sin h^{ }√ YZ

D = Cos h^{ }√ YZ

**Derive AD – BC =
1**

**A = Cos h ^{ }√ YZ **

**B = Z _{C} Sin
h^{ }√ YZ **

**C = Sin h ^{ }√ YZ / Z_{C} **

**D = Cos h √ YZ **

Now

AD = ( Cos h^{
}√ YZ )( Cos h^{ }√ YZ ) = Cos h^{2} √ YZ

BC = ( _{ }Sin h^{ }√ YZ )( Sin h^{ }√
YZ ) = Sin h^{2} √ YZ

Therefore

**AD – BC = Cos h ^{2}
√ YZ – Sin h^{2} √ YZ = 1 **

**Points to be noted**

Cos h ( a + jb )
= { Cos h a × Cos h jb + Sin h a × Sin h jb }

Sin h ( a + jb )
= { Sin h a × Cos h jb + Cos h a × Sin h jb }

A = D = Cos h √
YZ

= 1 + ( YZ/2 ) + (Y^{2}Z^{2}
/ 24 ) + …….

B = Z_{C}
Sin h^{ }√ YZ

= Z ( 1
+ ( YZ/2 ) + (Y^{2}Z^{2} / 120 ) + …….)

C = ( 1 / Z_{C
}) Sin h^{ }√ YZ

= Y ( 1 + ( YZ/6
) + (Y^{2}Z^{2} / 120 ) + …….)

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