# 2.7 Wave Impedance of the Lossless Line

The standing-wave patterns indicate that on a mismatched line the voltage and current magnitudes are oscillatory with position along the line and in phase opposition with each other. Hence, the voltage to current ratio, called the wave impedance Z(d), must vary with position also. Using Eqs. (2.63a) and (2.63b) with z=−d,

Z(d)=V˜(d)I˜(d)=V 0+[e j β d+Γ e−j β d]V 0+[e j β d−Γ e−j β d]Z 0=Z 0[ 1+Γ e−j 2 β d 1−Γ e−j 2 β d ]=Z 0[ 1+Γ d 1−Γ d ] (Ω),(2.74)

where we define

Γ d=Γ e−j 2 β d=|Γ|e j θ r e−j 2 β d=|Γ|e j(θ r−2 β d)(2.75)

as the phase-shifted voltage reflection coefficient, meaning that Γ d has the same magnitude as Γ, but its phase is shifted by 2 β d relative to that of Γ.

In the circuit of Fig. 2-17(a) at terminals B B′ at an arbitrary location d on the line, Z(d) is the wave impedance of the line when “looking” to the right, i.e., towards the load. Application of the equivalence principle allows us to replace the segment to the right of terminals B B′ with a lumped impedance of value Z(d), as depicted in Fig. 2-17(b). From the standpoint of the input circuit to the left of terminals B B′, the two circuit configurations are electrically identical.

# Figure 2-17 The segment to the right of terminals B B′ can be replaced with a discrete impedance equal to the wave impedance Z(d).

Figure 2 17 Full Alternative Text

Of particular interest in many transmission-line problems is the input impedance at the source end of the line where d=l, which is given by

Z in=Z(d=l)=Z 0 1+Γ l 1−Γ l(2.76)

with

Γ l=Γ e−j 2 β l=|Γ|e j(θ r−2 β l).(2.77)

By replacing Γ with Eq. (2.59) and using the relations

e j β l=cos β l+j sin β l,(2.78a)
e−j β l=cos β l−j sin β l,(2.78b)

Eq. (2.76) can be written in terms of the normalized load impedance z L as

From the standpoint of the generator circuit, the transmission line can be replaced with an impedance Z in, as shown in Fig. 2-18. The phasor voltage across Z in is given by

V˜i=I˜i Z in=V˜g Z in Z g+Z in.(2.80)

# Figure 2-18 At the generator end, the terminated transmission line can be replaced with the input impedance of the line Z in.

Figure 2 18 Full Alternative Text

Simultaneously, from the standpoint of the transmission line, the voltage across it at the input of the line is given by Eq. (2.63a) with z=−l:

V˜i=V˜(−l)=V 0+[e j β l+Γ e−j β l].(2.81)

Equating Eq. (2.80) to Eq. (2.81) and then solving for V 0+ leads to

This completes the solution of the transmission-line wave equations given by Eqs. (2.21) and (2.23) for the special case of a lossless transmission line. We started out with the general solutions given by Eq. (2.26), which included four unknown amplitudes, V 0+,V 0−,I 0+, and I 0−. We then determined that Z 0=V 0+/I 0+=−V 0−/I 0−, thereby reducing the unknowns to the two voltage amplitudes only. Upon applying the boundary condition at the load, we were able to relate V 0− to V 0+ through Γ, and finally, by applying the boundary condition at the source, we obtained an expression for V 0+.

# Example 2-7: Complete Solution for υ(d, t) and i(d, t)

A 1.05 GHz generator circuit with series impedance Z g=10 Ω and voltage source given by

υ g(t)=10 sin(ω t+30°)(V)

is connected to a load Z L=(100+j 50) Ω through a 50-Ω, 67-cm long lossless transmission line. The phase velocity of the line is 0.7 c, where c is the velocity of light in a vacuum. Find υ(d,t) and i(d,t) on the line.

Solution: From the relationship u p=λ f, we find the wavelength

λ=u p f=0.7×3×10 8 1.05×10 9=0.2 m,

and

β l=2 π λ l=2 π 0.2×0.67=6.7 π=0.7 π=126°,

where we have subtracted multiples of 2 π. The voltage reflection coefficient at the load is

Γ=Z L−Z 0 Z L+Z 0=(100+j 50)−50(100+j 50)+50=0.45 e j 26.6°.

With reference to Fig. 2-18, the input impedance of the line given by Eq. (2.76) is

Z in=Z 0(1+Γ l 1−Γ l)=Z 0(1+Γ e−j 2 β l 1−Γ e−j 2 β l)=50(1+0.45 e j 26.6°e−j 252°1−0.45 e j 26.6°e−j 252°)=(21.9+j 17.4) Ω.

Rewriting the expression for the generator voltage with the cosine reference, we have

υ g(t)=10 sin(ω t+30°)=10 cos(90°−ω t−30°)=10 cos(ω t−60°)=ℜ e[10 e−j 60°e j ω t]=ℜ e[V˜g e j ω t] (V).

Hence, the phasor voltage V˜g is given by

V˜g=10 e−j 60°=10−60° (V).

Application of Eq. (2.82) gives

V 0+=(V˜g Z in Z g+Z in)(1 e j β l+Γ e−j β l)=[ 10 e−j 60°(21.9+j 17.4)10+21.9+j 17.4 ]⋅(e j 126°+0.45 e j 26.6°e−j 126°)−1=10.2 e j 159° (V).

Using Eq. (2.63a) with z=−d, the phasor voltage on the line is

V˜(d)=V 0+(e j β d+Γ e−j β d)=10.2 e j 159°(e j β d+0.45 e j 26.6°e−j β d),

and the corresponding instantaneous voltage υ(d,t) is

υ(d,t)=ℜ e[V˜(d)e j ω t]=10.2 cos(ω t+β d+159°)+4.55 cos(ω t−β d+185.6°) (V).

Similarly, Eq. (2.63b) leads to

I˜(d)=0.20 e j 159°(e j β d−0.45 e j 26.6°e−j β d),i(d,t)=0.20 cos(ω t+β d+159°)+0.091 cos(ω t−β d+185.6°) (A).