15/04/2022

ABCD Parameters of Long Transmission Line

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.


ABCD-parameters-of-long-transmission-line.png


Let

VS = Sending end voltage to neutral

VR = Receiving end voltage to neutral

IS = Sending end current

IR = 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

 At the receiving end side x = 0, V = VR and I = IR

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

VR = A1 + A2 ……….. ( 6 )

IR = √ ( y /z ) ( A1 –  A2 ) ….. ( 7 )

Characteristic constant ZC

ZC = √ ( z / y )……. ( 8 )

Propagation constant

ℽ = √ ( yz )……. ( 9 )

Relation between propagation constant and characteristic constant

ZC 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 = VR ………( 10 )

A1 –  A2 = IR / √ ( y /z )……… ( 11 )

Addition of equation ( 10 ) and ( 11 )

2A1 = VR + IR / √ ( y /z )

A1 = { VR + IR / √ ( y /z ) } / 2

As ZC = √ ( z / y )

A1 = { VR + IR ZC } / 2

Substitute value of A1 into equation ( 10 )

A1 + A2 = VR

{ VR + IR / ZC  } / 2 + A2 = VR

A2 = VR – { VR + IR ZC } / 2

A2 = VR / 2 – IR  ZC / 2

A2 = { VR – IR ZC } / 2

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

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Voltage Equation

V = A1 e√(yz) x + A2 e -√(yz) x 

V = [{ VR + IR ZC } / 2 ] ( e ℽ x ) + [ { VR – IR ZC } / 2 ]( e –ℽ x )

    = VR / 2 ( e ℽ x ) + ( IR ZC ) / 2 (e ℽ x ) + VR / 2 ( e –ℽ x )  

        – ( IR ZC ) / 2 ( e –ℽ x )

Rearranging terms

    = VR ( e ℽ x + e –ℽ x ) / 2 + ( IR ZC ) ( 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 = VR Cos h ℽx + ( ZC ) IR Sin h ℽx

Sending end voltage is obtained by putting x = l

VS = VR Cos h ℽl + ( ZC ) IR 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

VS = VR Cos h √ YZ + ( ZC ) IR Sin h √ YZ

Compare this equation with VS = AVR + BIR

A = Cos h √ YZ

B = ZC 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 / ZC [{ VR + IR ZC } / 2 ]( e ℽ x )  – [{ VR – IR ZC } / 2 ]( e –ℽ x )

I = [{ VR / ZC + IR } / 2 ]( e ℽ x ) – [{ VR / ZC  – IR } / 2 ]( e –ℽ x )

  = { ( VR / ZC ) / 2 }( e ℽ x ) – { ( VR / ZC ) / 2 }( e –ℽ x ) +

                        { ( IR ) / 2 }( e ℽ x ) + {( IR ) / 2 }( e –ℽ x )

   = VR / ZC { ( e ℽ x ) – ( e –ℽ x ) / 2 } +

                        IR { ( e ℽ x ) + ( e –ℽ x ) / 2 }

I = IR Cos h ℽx +  VR / ZC Sin h ℽx

 Sending end voltage is obtained by putting x = l

IS = IR Cos h ℽl + VR / ZC Sin h ℽl

   = VR / ZC Sin h ℽl + IR Cos h ℽl 

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

IS = VR / ZC Sin h √ YZ + IR Cos h √ YZ  

Compare this equation with IS = CIR + DVR

C = 1 / ZC Sin h √ YZ  

D = Cos h √ YZ

Derive AD – BC = 1

A = Cos h √ YZ

B = ZC Sin h √ YZ

C = Sin h √ YZ / ZC

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

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 ) + (Y2Z2 / 24 ) + …….

B = ZC Sin h √ YZ

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

C = ( 1 / ZC ) Sin h √ YZ

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

  

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