Transmission line power flow

Power on a Transmission line splits between forward (incident) and backward (reflected) waves, plus dissipation in any losses. For a lossless line, the analysis is clean:

Average incident power:

$P_{\text{av}}^i=\frac{|V_0^{+}{|}^2}{2Z_0}.$

Average reflected power:

$P_{\text{av}}^r=-|\Gamma {|}^2{P}_{\text{av}}^i=-|\Gamma {|}^2\frac{|V_0^{+}{|}^2}{2Z_0}.$

The minus sign indicates that reflected power flows toward the generator (opposite to the incident).

Net average power delivered to load:

$P_{\text{av}}=P_{\text{av}}^i+P_{\text{av}}^r=(1-|\Gamma {|}^2)\frac{|V_0^{+}{|}^2}{2Z_0}.$

Where $\Gamma$ is the Reflection coefficient.

Independence of position

On a lossless line, both $P_{\text{av}}^i$ and $P_{\text{av}}^r$ are independent of position $d$ along the line. The amplitudes $|V_0^{+}|$, $|V_0^{-}|$ are constants set by the source and load — they don’t depend on where along the line you measure.

This is a consequence of energy conservation in the absence of dissipation: power is neither lost nor gained as the wave traverses the line. What enters the input end equals what reaches the load plus what bounces back to the input.

(Voltage and current magnitudes do depend on position because of the standing wave interference, but the time-average power flow doesn’t — the interference doesn’t transport energy, it just redistributes the instantaneous power.)

Why the factor 1/2

For sinusoidal $v(t)=V_0\cos (\omega t)$ and $i(t)=I_0\cos (\omega t+\varphi )$:

$⟨p⟩=⟨vi⟩=\frac{V_0{I}_0}{2}\cos \varphi .$

The $\frac{1}{2}$ comes from $\overline{{\cos }^2}=\frac{1}{2}$. For a pure forward wave, $V=V_0^{+}{e}^{-j\beta z}$ and $I=(V_0^{+}/Z_0)e^{-j\beta z}$, in phase (for lossless $Z_0$ real), so $\cos \varphi =1$:

$P_{\text{av}}^i=\frac{|V_0^{+}{|}^2}{2Z_0}.$

Phasor formula

In phasor analysis, time-average power is

$P_{\text{av}}=\frac{1}{2}\text{Re}[\tilde{V}{\tilde{I}}^{\ast }].$

Where $\tilde{V},\tilde{I}$ are the phasors and ${}^{\ast }$ is complex conjugate. This works at any point on the line — substitute the position-dependent phasors and you get the local time-average power.

For a forward wave at position $d$ (lossless, $Z_0$ real):

$\tilde{V}=V_0^{+}{e}^{j\beta d},\tilde{I}=(V_0^{+}/Z_0)e^{j\beta d}.$

$P_{\text{av}}^i=\frac{1}{2}\text{Re}\left[V_0^{+}{e}^{j\beta d}\cdot \frac{V_0^{+\ast }}{Z_0}{e}^{-j\beta d}\right]=\frac{|V_0^{+}{|}^2}{2Z_0}.$

Phase factors cancel — independent of $d$. ✓

Worked example

A 1.05 GHz generator with $V_g=10\sin (\omega t+30°)$, $Z_g=10$ Ω, drives a 50 Ω lossless line of length 67 cm into a load $Z_L=100+j50$ Ω. Phase velocity is $0.7c$.

Compute:

• $\beta =\omega /u_p=2\pi (1.05\times 10^9)/(0.7\times 3\times 10^8)=31.4$ rad/m.
• $\beta l=31.4\cdot 0.67=21.04$ rad. Reduce mod $2\pi$: $21.04-3\cdot 2\pi =2.19$ rad.
• $\Gamma =(Z_L-Z_0)/(Z_L+Z_0)=(50+j50)/(150+j50)=0.45\angle 27°$.
• $Z_{\text{in}}=Z_0(Z_L+j{Z}_0\tan \beta l)/(Z_0+j{Z}_L\tan \beta l)$. Compute numerically.
• ${\tilde{V}}_i=V_g{Z}_{\text{in}}/(Z_g+Z_{\text{in}})$, recover $V_0^{+}$ from the line equations.
• $P_{\text{av}}^L=(1-|\Gamma {|}^2)|V_0^{+}{|}^2/(2Z_0)\approx (1-0.2)|V_0^{+}{|}^2/100$.

The mechanics are tedious; the structure is what matters. Power flow on a TL has incident, reflected, and delivered components; each is computable from the wave amplitudes.

In lossy lines

For a lossy line ($\alpha \ne 0$), the forward wave amplitude decays as $e^{-\alpha d}$ moving toward the load, so

$P_{\text{av}}^i(d)=\frac{|V_0^{+}{|}^2}{2Z_0}{e}^{-2\alpha d}.$

Power flow does depend on position — the difference is dissipated in $R^{\prime }$ and $G^{\prime }$ along the line. For a long lossy line, the total loss is the integral of dissipation along the length, equivalent to the input minus what arrives at the load.

In a “long-haul” telecom or power transmission line, this is the central engineering concern: minimizing $\alpha$ to deliver useful power at the far end.