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Again, we will look at a transmission line circuit in Figure fig:FVTRLine to find the input impedance on a transmission line. Figure 1: Transmission Line connects generator and the load.

The equations for the voltage and current anywhere (any z) on a transmission line are

Using the equations from the previous section, we can replace the transmission line with its input impedance, Figure fig:FVTRLineEqCirc. Figure 2: Transmission Line connects generator and the load.

### Forward voltage phasor as a function of load impedance

From Figure fig:FVTRLineEqCirc, we can find the input voltage on a transmission line using the voltage divider.

Using Equation eq:FVitlfin, we can also find the input voltage. The input voltage equation at the generator $z=-l$ is:

Since these two equations represent the same input voltage we can make them equal.

Rearranging the equation, we find $\tilde {V}_0^+$.

### Forward voltage phasor as a function of input reflection coefficient

There is another way to find the input impedance as a function of the input reflection coefficient.

We write KVL for the circuit in Figure fig:FVTRLineEqCirc.

Using Equations eq:FVitlfin-eq:FVitlfin, we can also find the input voltage and current. Input voltage and current equation at the generator $z=-l$ are:

Substituting these two equations in Equation eq:FVKVL we get

We can re-write this equation as follows.

Using that $\Gamma _{in} = \Gamma _L e^{-2 j \beta l}$ is the input reflection coefficient, and multiplying through with $Z_0$.

Rearranging the equation, we get $\tilde {V}_0^+$

$\Gamma _{in} = \Gamma _L e^{-2 j \beta l}$ is the input reflection coefficient.

### Special case - forward voltage when the generator and transmission-line impedance are equal

Because the generator’s impedance is equal to the transmission line impedance, we will use the second equation. When $Z_g=Z_0$ we see that the denominator simplifies into $Z_0 (1+\Gamma _{in}) +Z_g (1-\Gamma _{in}) = Z_0 (1+\Gamma _{in}+1-\Gamma _{in})=2$, and we can further simplify the fraction to get the final value of $\tilde {V}_0^+=\frac {V_g}{2} e^{-j \beta l}$.