ABCD Parameters of Long Transmission Line

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

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 = √ ( r² + x² )

y = Admittance per unit length of line = √ ( g² + b² )

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

d²V / dx² = z ( dI / dx )

from equation ( 2 )

d²V / dx² = 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

Long Transmission Line: A, B, C and D Parameters

 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 )

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Long Transmission Line: 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

Long Transmission Line: 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 h² √ YZ

BC = (  Sin h √ YZ )( Sin h √ YZ ) = Sin h² √ YZ

Therefore

AD – BC = Cos h² √ YZ  – Sin h² √ YZ = 1

Long Transmission Line: Important Formula

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 ) + (Y²Z² / 24 ) + …….

B = Z_C Sin h √ YZ

   = Z ( 1 + ( YZ/2 ) + (Y²Z² / 120 ) + …….)

C = ( 1 / Z_C ) Sin h √ YZ

= Y ( 1 + ( YZ/6 ) + (Y²Z² / 120 ) + …….)