# 2.4 Wave Propagation on a Transmission Line

The two first-order coupled equations (2.18a) and (2.18b) can be combined to give two second-order uncoupled wave equations: one for V˜(z) and another for I˜(z). The wave equation for V˜(z) is derived by first differentiating both sides of Eq. (2.18a) with respect to z, resulting in

−d 2 V˜(z)d z 2=(R′+j ω L′)d I˜(z)d z.(2.19)

Then, upon substituting Eq. (2.18b) for d I˜(z)/d z, Eq. (2.19) becomes

d 2 V˜(z)d z 2−(R′+j ω L′)(G′+j ω C′)V˜(z)=0,(2.20)

or

where

Application of the same steps to Eqs. (2.18a) and (2.18b) in reverse order leads to

The second-order differential equations (2.21) and (2.23) are called wave equations for V˜(z) and I˜(z), respectively, and γ is called the complex propagation constant of the transmission line. As such, γ consists of a real part α, called the attenuation constant of the line with units of Np/m, and an imaginary part β, called the phase constant of the line with units of rad/m. Thus,

γ=α+j β(2.24)

with

# 2.4.1 Phasor-Domain Solution

In Eqs. (2.25a) and (2.25b), we choose the square-root solutions that give positive values for α and β. For passive transmission lines, α is either zero or positive. Most transmission lines, and all those considered in this chapter, are of the passive type. The gain region of a laser is an example of an active transmission line with a negative α.

The wave equations (2.21) and (2.23) have traveling wave solutions of the form:

V˜(z)=V 0+e−γ z+V 0−e γ z(V),(2.26a)
I˜(z)=I 0+e−γ z+I 0−e γ z(A).(2.26b)

As shown later, the e−γ z term represents a wave propagating in the +z direction, while the e γ z term represents a wave propagating in the −z direction (Fig. 2-9). Verification that these are indeed valid solutions is easily accomplished by substituting the expressions given by Eqs. (2.26a) and (2.26b), as well as their second derivatives, into Eqs. (2.21) and (2.23).

# Figure 2-9 In general, a transmission line can support two traveling waves: an incident wave (with voltage and current amplitudes (V 0+, I 0+)) traveling along the +z direction (towards the load) and a reflected wave (with (V 0−, I 0−)) traveling along the −z direction (towards the source).

Figure 2 9 Full Alternative Text

In their present form, the solutions given by Eqs. (2.26a) and (2.26b) contain four unknowns: the wave amplitudes (V 0+,I 0+) of the +z propagating wave and (V 0−,I 0−) of the −z propagating wave. We can easily relate the current wave amplitudes, I 0+ and I 0−, to the voltage wave amplitudes, V 0+ and V 0−, by using Eq. (2.26a) in Eq. (2.18a) and then solving for the current I˜(z). The process leads to

I˜(z)=γ R′+j ω L′[V 0+e−γ z−V 0−e γ z].(2.27)

Comparison of each term with the corresponding term in Eq. (2.26b) leads us to conclude that

V 0+I 0+=Z 0=−V 0−I 0−,(2.28)

where

is called the characteristic impedance of the line.

It seems reasonable that the voltage-to-current ratios of the two waves V 0+/I 0+ and V 0−/I 0− are both related to the same quantity, namely Z 0, but it is not immediately obvious as to why one of the ratios is the negative of the other. The explanation, which is available in more detail in Chapter 7, is based on a directional rule that specifies the relationships between the directions of the electric and magnetic fields of a TEM wave and its direction of propagation. On a transmission line, the voltage is related to the electric field E and the current is related to the magnetic field H. To satisfy the directional rule, reversing the direction of propagation requires reversal of the direction (or polarity) of I relative to V. Hence, V 0−/I 0−=−V 0+/I 0+.

In terms of Z 0, Eq. (2.27) can be cast in the form

I˜(z)=V 0+Z 0 e−γ z−V 0−Z 0 e γ z.(2.30)

According to Eq. (2.29), the characteristic impedance Z 0 is determined by the angular frequency ω of the wave traveling along the line and the four line parameters (R′, L′, G′, and C′). These, in turn, are determined by the line geometry and its constitutive parameters. Consequently, the combination of Eqs. (2.26a) and (2.30) now contains only two unknowns, namely V 0+ and V 0−, as opposed to four.

# 2.4.2 Time-Domain Solution for υ(z, t)

In later sections, we apply boundary conditions at the source and load ends of the transmission line to obtain expressions for the remaining wave amplitudes V 0+ and V 0−. In general, each is a complex quantity characterized by a magnitude and a phase angle:

V 0+=|V 0+|e j ϕ+,(2.31a)

V 0−=|V 0−|e j ϕ−.(2.31b)

After substituting these definitions in Eq. (2.26a) and using Eq. (2.24) to decompose γ into its real and imaginary parts, we can convert back to the time domain to obtain an expression for υ(z,t), the instantaneous voltage on the line:

υ(z,t)=ℜ e(V˜(z)e j ω t)=ℜ e[ (V 0+e−γ z+V 0−e γ z)e j ω t ]=ℜ e[|V 0+|e j ϕ+e j ω t e−(α+j β)z+|V 0−|e j ϕ−e j ω t e(α+j β)z].

The final expression for υ(z,t) is then given by

From our review of waves in Section 1-4, we recognize the first term in Eq. (2.32) as a wave traveling in the +z direction (the coefficients of t and z have opposite signs) and the second term as a wave traveling in the −z direction (the coefficients of t and z are both positive). Both waves propagate with a phase velocity u p given by Eq. (1.30):

u p=f λ=ω β.(2.33)

Because the wave is guided by the transmission line, λ often is called the guide wavelength. The factor e−α z accounts for the attenuation of the +z propagating wave, and the factor e α z accounts for the attenuation of the −z propagating wave.

To gain a physical understanding of what that means, we later provide a thorough treatment of standing waves in Section 2-6.2.

# 2.4.3 Time-Domain Solution for i(z, t)

In the phasor domain, the expression for I˜(z) given by Eq. (2.30) is similar to that for V˜(z), except that the current amplitudes are divided by the characteristic impedance of the transmission line, Z 0, and the current wave traveling in the −z direction has a minus sign (second term in Eq. (2.30)). Since Z 0 is, in general, a complex quantity (see Eq. (2.29)), we need to express it as

Z 0=|Z 0|e j ϕ z.(2.34)

To obtain an expression for i(z,t), we insert the expressions given by Eqs. (2.31) and (2.34) into Eq. (2.30), multiply both terms by e j ω t, and then take the real part:

i(z,t)=ℜ e(I˜(z)e j ω t)=ℜ e [ |V 0+||Z 0|e j ϕ+e−j ϕ z e j ω t e−(α+j β z) −|V 0−||Z 0|e j ϕ−e−j ϕ z e j ω t e(α+j β z) ],

which yields

Comparison of the expression for i(z,t) with that given by Eq. (2.32) for υ(z,t) reveals that:

1. The amplitudes of the current waves are reduced by |Z 0| when compared with the amplitudes of the voltage waves.

2. The amplitude of the second term in Eq. (2.35)—corresponding to the current wave traveling in the negative z direction—has a minus sign.

3. The reference phase angle in the expression for i(z,t) has an extra component, namely −ϕ z, when compared with the reference phase angle of υ(z,t).

# Example 2-1: Air Line

An air line is a transmission line in which air separates the two conductors, which renders G′=0 because σ=0. In addition, assume that the conductors are made of a material with high conductivity so that R′≈0. For an air line with a characteristic impedance of 50 Ω and a phase constant of 20 rad/m at 700 MHz, find the line inductance L′ and the line capacitance C′.

Solution: The following quantities are given:

Z 0=50 Ω,β=20 rad/m,f=700 MHz=7×10 8 Hz.

With R′=G′=0, Eqs. (2.25b) and (2.29) reduce to

β=ℑ m[ (j ω L′)(j ω C′) ]=ℑ m(j ω L′C′)=ω L′C′,Z 0=j ω L′j ω C′=L′C′.

The ratio of β to Z 0 is

β Z 0=ω C′,

or

C′=β ω Z 0=20 2 π×7×10 8×50=9.09×10−11 (F/m)=90.9 (pF/m).

From Z 0=L′/C′, it follows that

L′=Z 0 2 C′=(50)2×90.9×10−12=2.27×10−7 (H/m)=227 (nH/m).