General Relation between voltage and current of transmission line: Derivation

We  know from the equivalent circuit of a transmission line (mentioned in a post before ) per unit length:

Series impedance, Z = R + jωL

Shunt admittance, Y = G + jωC

 

Let us consider a short section of transmission line PQ of length dx, at a distance x from the sending end A.

At   P,          voltage = V

                       Current = I

Then at Q,     voltage = V + dV

                            Current = I + dI 

Now we get for short section dx,

                          Series impedance = ( R + jωL )dx

                          Shunt admittance = ( G + jωC )dx

Therefore, Voltage drop from P to Q may be considered to be due to the current I flowing through the impedance  ( R + jωL )dx

                         V – ( V + dV ) = I ( R + jωL )dx

                   Or ,    - dV | dx   =    I ( R + jωL )   --------(i)

And the decrease in current from P to Q may be considered due to the voltage V applied to 
the shunt admittance   ( G + jωC )dx

                       I – ( I + dI ) = V ( G + jωC )dx

                   Or,  - dI | dx  = V  ( G + jωC ) ---------(ii)

Now differentiating (i) with respect to x, we get –

- d ²V | dx ²   =    ( R + jωL ) dI | dx

Or,    d ²V | dx ²   =    ( R + jωL )  ( - dI | dx )

                                  =  ( R + jωL )    ( G + jωC )V   [ from (ii) ]

So,  d ²V | dx ²   = P ²V   ------------(iii)

Where, P ² = ( R + jωL )    ( G + jωC )

P is propagation constant and a complex quantity may represent as

 P  = œ  + jß

Similarly from (ii) we get,  d ²I | dx ²   = P ²I   ----------(iv)

The equations (iii) and (iv) are standard linear differential equations and their solution in exponential form can be written as, 

V  =  A e ^-Px   + B e ^Px      ------------(v)

I   =   C e ^-Px  +  D e ^Px      ---------(vi)

Where A, B, C, D are arbitrary constants

              A, B having dimensions of voltage

             C, D having dimensions of current

So,     V = A e ^–œx  e ^–jßx   + B e ^œx  e ^jßx      

                          [ as  P  = œ  + jß ]

             I =  C e ^–œx  e ^–jßx    + D e ^œx  e ^jßx     

                        ---I term---         ---2^nd term ----

The first term in the equation of voltage and current are called incident waves which travels from source to load and suffer phase shift ß . The second term represents the reflected wave, which travels from load to source. The two are the general equation of voltage and current from sending end.

Now the complete general solution for transmission line, we consider the following  figure  in which load impedance is Z_R ( not equal to characteristic impedance).

  Now differentiating  (v) we get –

 

    dV | dx  =  - PA e ^-Px   + PB e ^Px  

     

   or,     I ( R + jωL )   = P (A e ^-Px  -  B e ^Px )             [from (i)]

or,   I  =   P (A e ^-Px  -  B e ^Px )  |  ( R + jωL )   

         

Since        cosh Px  =  ( e^Px  +  e ^–Px )| 2

                    sinh Px  =  ( e^Px  -  e ^–Px )| 2

therefore,  cosh Px   -  sinh Px    =   e  ^-Px 

                               cosh Px   +  sinh Px    =   e  ^Px

Now putting these values into (v) and (vii) –

V = A (cosh Px   -  sinh Px  ) + B (cosh Px   +  sinh Px ) 

V  =  coshPx ( A + B ) – sinhPx ( A – B )    --------(viii)

And   I =  {coshPx ( A - B ) – sinhPx ( A + B ) }  | Z_o   ------(ix)

Now condition at the input end is 

                  x = 0

                 V = V_s

                 I  =  I_s 

                coshPx = 1

               sinhPx = 0

Hence we get,    V_s  = A + B

                                 I_s = ( A  -  B ) | Z_o

Putting the values in (viii)  and (ix)  we get  – 

V  =  V_s  coshPx   –   I_s  Z_o  sinhPx 

I    =  I_s coshPx     -    sinhPx   V_s | Z_o

These  are the completely general equations of transmission line for voltage and current.

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