# 2.9 Power Flow on a Lossless Transmission Line

Our discussion thus far has focused on the voltage and current attributes of waves propagating on a transmission line. Now we examine the flow of power carried by the incident and reflected waves. We begin by reintroducing Eqs. (2.63a) and (2.63b) with z=−d:

V˜(d)=V 0+(e j β d+Γ e−j β d),(2.98a)
I˜(d)=V 0+Z 0(e j β d−Γ e−j β d).(2.98b)

In these expressions, the first terms represent the incident-wave voltage and current, and the terms involving Γ represent the reflected-wave voltage and current. The time-domain expressions for the voltage and current at location d from the load are obtained by transforming Eq. (2.98) to the time domain:

υ(d,t)=ℜ e[V˜e j ω t]=ℜ e[|V 0+|e j ϕ+(e j β d+|Γ|e j θ r e−j β d)e j ω t]=|V 0+|[cos(ω t+β d+ϕ+)+|Γ|cos(ω t−β d+ϕ++θ r)],(2.99a)
i(d,t)=|V 0+|Z 0[cos(ω t+β d+ϕ+)−|Γ|cos(ω t−β d+ϕ++θ r)],(2.99b)

where we used the relations V 0+=|V 0+|e j ϕ+ and Γ=|Γ|e j θ r that were introduced earlier as Eqs. (2.31a) and (2.62), respectively.

# 2.9.1 Instantaneous Power

The instantaneous power carried by the transmission line is equal to the product of υ(d,t) and i(d,t):

P(d,t)=υ(d,t)i(d,t)=|V 0+|[cos(ω t+β d+ϕ+)+|Γ|cos(ω t−β d+ϕ++θ r)]×|V 0+|Z 0[cos(ω t+β d+ϕ+)−|Γ|cos(ω t−β d+ϕ++θ r)]=|V 0+|2 Z 0[cos 2(ω t+β d+ϕ+)−|Γ|2 cos 2(ω t−β d+ϕ++θ r)] (W).(2.100)

Per our earlier discussion in connection with Eq. (1.31), if the signs preceding ω t and β d in the argument of the cosine term are both positive or both negative, then the cosine term represents a wave traveling in the negative d direction. Since d points from the load to the generator, the first term in Eq. (2.100) represents the instantaneous incident power traveling towards the load. This is the power that would be delivered to the load in the absence of wave reflection (when Γ=0). Because β d is preceded by a minus sign in the argument of the cosine of the second term in Eq. (2.100), that term represents the instantaneous reflected power traveling in the +d direction—away from the load. Accordingly, we label these two power components

P i(d,t)=|V 0+|2 Z 0 cos 2(ω t+β d+ϕ+) (W),(2.101a)
P r(d,t)=−|Γ|2|V 0+|2 Z 0 cos 2(ω t−β d+ϕ++θ r) (W).(2.101b)

Using the trigonometric identity

cos 2 x=1 2(1+cos 2 x),

the expressions in Eq. (2.101) can be rewritten as

P i(d,t)=|V 0+|2 2 Z 0[1+cos(2 ω t+2 β d+2 ϕ+)],(2.102a)
P r(d,t)=−|Γ|2|V 0+|2 2 Z 0[1+cos(2 ω t−2 β d+2 ϕ++2 θ r)].(2.102b)

We note that in each case, the instantaneous power consists of a dc term (not varying in time) and an ac term that oscillates at an angular frequency of 2 ω.

# 2.9.2 Time-Average Power

From a practical standpoint, we usually are more interested in the time-average power flowing along the transmission line, P av(d), than in the instantaneous power P(d,t). To compute P av(d), we can use a time-domain approach or a computationally simpler phasor-domain approach. For completeness, we consider both.

# Time-Domain Approach

The time-average power is equal to the instantaneous power averaged over one time period T=1/f=2 π/ω. For the incident wave, its time-average power is

P av i(d)=1 T∫0 T P i(d,t)d t=ω 2 π∫0 2 π/ω P i(d,t)d t.(2.103)

Upon inserting Eq. (2.102a) into Eq. (2.103) and performing the integration, we obtain

which is identical with the dc term of P i(d,t) given by Eq. (2.102a). A similar treatment for the reflected wave gives

Note that the expressions for P av i and P av r are independent of d, which means that the time-average powers carried by the incident and reflected waves do not change as they travel along the transmission line. This is as expected, because the transmission line is lossless.

The net average power flowing towards (and then absorbed by) the load shown in Fig. 2-23 is

# Figure 2-23 The time-average power reflected by a load connected to a lossless transmission line is equal to the incident power multiplied by |Γ|2.

Figure 2 23 Full Alternative Text

# Phasor-Domain Approach

For any propagating wave with voltage and current phasors V˜ and I˜, a useful formula for computing the time-average power is

where I˜∗ is the complex conjugate of I˜. Application of this formula to Eqs. (2.98a) and (2.98b) gives

P av=1 2 ℜ e[ V 0+(e j β d+Γ e−j β d)V 0+∗Z 0(e−j β d−Γ∗e j β d) ]=1 2 ℜ e[ |V 0+|2 Z 0(1−|Γ|2+Γ e−j 2 β d−Γ∗e j 2 β d) ]=|V 0+|2 2 Z 0{ [1−|Γ|2]+ℜ e[|Γ|e−j(2 β d−θ r)−|Γ|e j(2 β d−θ r)] }=|V 0+|2 2 Z 0{ [1−|Γ|2]+|Γ|[cos(2 β d−θ r)−cos(2 β d−θ r)] }=|V 0+|2 2 Z 0[1−|Γ|2],(2.108)

which is identical to Eq. (2.106).

Concept Question 2-15: According to Eq. (2.102b), the instantaneous value of the reflected power depends on the phase of the reflection coefficient θ r, but the average reflected power given by Eq. (2.105) does not. Explain.

Concept Question 2-16: What is the average power delivered by a lossless transmission line to a reactive load?

Concept Question 2-17: What fraction of the incident power is delivered to a matched load?

Concept Question 2-18: Verify that

1 T∫0 T cos 2(2 π t T+β d+ϕ)d t=1 2,

regardless of the values of d and ϕ, so long as neither is a function of t.