# 2.6 The Lossless Transmission Line: General Considerations

According to the preceding section, a transmission line is fully characterized by two fundamental parameters: its propagation constant γ and its characteristic impedance Z 0, both of which are specified by the angular frequency ω and the line parameters R′, L′, G′, and C′.

These conditions allow us to set R′=G′≈0 in Eq. (2.22), which yields

γ=α+j β=j ω L′C′,(2.44)

which in turn implies that

For the characteristic impedance, application of the lossless line conditions to Eq. (2.29) leads to

which now is a real number. Using the lossless line expression for β [Eq. (2.45b)], we obtain the following expressions for the guide wavelength λ and the phase velocity u p:

λ=2 π β=2 π ω L′C′,(2.47)
u p=ω β=1 L′C′.(2.48)

Upon using Eq. (2.10), Eqs. (2.45b) and (2.48) may be rewritten as

where μ and ε are, respectively, the magnetic permeability and electrical permittivity of the insulating material separating the conductors. Materials used for this purpose are usually characterized by a permeability μ 0=4 π×10−7 H/m (the permeability of free space). Also, the permittivity ε is often specified in terms of the relative permittivity ε r defined as

ε r=ε/ε 0,(2.51)

where ε 0=8.854×10−12 F/m≈(1/36 π)×10−9 F/m is the permittivity of free space (vacuum). Hence, Eq. (2.50) becomes

u p=1 μ 0 ε r ε 0=1 μ 0 ε 0⋅1 ε r=c ε r,(2.52)

where c=1/μ 0 ε 0=3×10 8 m/s is the velocity of light in free space. If the insulating material between the conductors is air, then ε r=1 and u p=c. In view of Eq. (2.51) and the relationship between λ and u p given by Eq. (2.33), the wavelength is given by

where λ 0=c/f is the wavelength in air corresponding to a frequency f. Note that, because both u p and λ depend on ε r, the choice of the type of insulating material used in a transmission line is dictated not only by its mechanical properties but by its electrical properties as well.

According to Eq. (2.52), if ε r of the insulating material is independent of f (which usually is the case for commonly used TEM lines), the same independence applies to u p.

This is an important feature to consider when digital data are transmitted in the form of pulses. A rectangular pulse or a series of pulses is composed of many Fourier components with different frequencies. If the phase velocity is the same for all frequency components (or at least for the dominant ones), then the pulse’s shape does not change as it travels down the line. In contrast, the shape of a pulse propagating in a dispersive medium becomes progressively distorted, and the pulse length increases (stretches out) as a function of the distance traveled in the medium (Fig. 2-3), thereby imposing a limitation on the maximum data rate (which is related to the length of the individual pulses and the spacing between adjacent pulses) that can be transmitted through the medium without loss of information.

Table 2-2 provides a list of the expressions for γ, Z 0, and u p for the general case of a lossy line and for several types of lossless lines. The expressions for the lossless lines are based on the equations for L′ and C′ given in Table 2-1.

# Table 2-2 Characteristic parameters of transmission lines.

Table 2 2 Full Alternative Text

# 2.6.1 Voltage Reflection Coefficient

With γ=j β for the lossless line, Eqs. (2.26a) and (2.30) for the total voltage and current become

V˜(z)=V 0+e−j β z+V 0−e j β z,(2.54a)

I˜(z)=V 0+Z 0 e−j β z−V 0−Z 0 e j β z.(2.54b)

These expressions contain two unknowns, V 0+ and V 0−. According to Section 1-7.2, an exponential factor of the form e−j β z is associated with a wave traveling in the positive z direction, from the source (sending end) to the load (receiving end). Accordingly, we refer to it as the incident wave with V 0+ as its voltage amplitude. Similarly, the term containing V 0−e j β z represents a reflected wave with voltage amplitude V 0− traveling along the negative z direction from the load to the source.

To determine V 0+ and V 0−, we need to consider the lossless transmission line in the context of the complete circuit, including a generator circuit at its input terminals and a load at its output terminals, as shown in Fig. 2-12. The line of length l is terminated in an arbitrary load impedance Z L.

At the sending end, at z=−l, the line is connected to a sinusoidal voltage source with phasor voltage V˜g and internal impedance Z g. Since z points from the generator to the load, positive values of z correspond to locations beyond the load; therefore, they are irrelevant to our circuit. In future sections, we will find it more convenient to work with a spatial dimension that also starts at the load but whose direction is opposite of z. We shall call it the distance from the load d and define it as d=−z, as shown in Fig. 2-12.

# Figure 2-12 Transmission line of length l connected on one end to a generator circuit and on the other end to a load Z L. The load is located at z=0, and the generator terminals are at z=−l. Coordinate d is defined as d=−z.

Figure 2 12 Full Alternative Text

The phasor voltage across the load, V˜L, and the phasor current through it, I˜L, are related by the load impedance Z L as

Z L=V˜L I˜L.(2.55)

The voltage V˜L is the total voltage on the line V˜(z) given by Eq. (2.54a), and I˜L is the total current I˜(z) given by Eq. (2.54b). Both are evaluated at z=0:

V˜L=V˜(z=0)=V 0++V 0−,(2.56a)

I˜L=I˜(z=0)=V 0+Z 0−V 0−Z 0.(2.56b)

Using these expressions in Eq. (2.55), we obtain

Z L=V 0++V 0−V 0+−V 0−Z 0.(2.57)

Solving for V 0− gives

V 0−=Z L−Z 0 Z L+Z 0 V 0+.(2.58)

From Eq. (2.58), it follows that

where

z L=Z L Z 0(2.60)

is the normalized load impedance. In many transmission line problems, we can streamline the necessary computation by normalizing all impedances in the circuit to the characteristic impedance Z 0. Normalized impedances are denoted by lowercase letters.

In view of Eq. (2.28), the ratio of the current amplitudes is

The reflection coefficient Γ is governed by a single parameter, the normalized load impedance z L. As indicated by Eq. (2.46), Z 0 of a lossless line is a real number. However, Z L is in general a complex quantity, as in the case of a series RL circuit, for example, for which Z L=R+j ω L. Hence, in general Γ also is complex and given by

where |Γ| is the magnitude of Γ and θ r is its phase angle. Note that |Γ|≤1.

On the other hand, when the load is an open circuit (Z L=∞), Γ=1 and V 0−=V 0+, and when it is a short circuit (Z L=0), Γ=−1 and V 0−=−V 0+ (Table 2-3).

# Table 2-3 Magnitude and phase of the reflection coefficient for various types of loads. The normalized load impedance z L=Z L/Z 0=(R+j X)/Z 0=r+j x, where r=R/Z 0 and x=X/Z 0 are the real and imaginary parts of z L, respectively.

Table 2 3 Full Alternative Text

# Example 2-3: Reflection Coefficient of a Series RC Load

A 100 Ω transmission line is connected to a load consisting of a 50 Ω resistor in series with a 10 pF capacitor. Find the reflection coefficient at the load for a 100 MHz signal.

Solution: The following quantities are given (Fig. 2-13):

R L=50 Ω,C L=10 pF=10−11 F,Z 0=100 Ω,f=100 MHz=10 8 Hz.

# Figure 2-13 RC load (Example 2-3).

Figure 2 13 Full Alternative Text

The normalized load impedance is

z L=Z L Z 0=R L−j/(ω C L)Z 0=1 100(50−j 1 2 π×10 8×10−11)=(0.5−j 1.59) Ω.

From Eq. (2.59), the voltage reflection coefficient is

Γ=z L−1 z L+1=0.5−j 1.59−1 0.5−j 1.59+1=−0.5−j 1.59 1.5−j 1.59=−1.67 e j 72.6°2.19 e−j 46.7°=−0.76 e j 119.3°.

This result may be converted into the form of Eq. (2.62) by replacing the minus sign with e−j 180°. Thus,

Γ=0.76 e j 119.3°e−j 180°=0.76 e−j 60.7°=0.76−60.7°,

or

|Γ|=0.76,θ r=−60.7°.

# Example 2-4: |Γ| for Purely Reactive Load

Show that |Γ|=1 for a lossless line connected to a purely reactive load.

Solution: The load impedance of a purely reactive load is Z L=j X L. From Eq. (2.59), the reflection coefficient is

Γ=Z L−Z 0 Z L+Z 0=j X L−Z 0 j X L+Z 0=−(Z 0−j X L)(Z 0+j X L)=−Z 0 2+X L 2 e−j θ Z 0 2+X L 2 e j θ=−e−j 2 θ=1 e j(π−2 θ),

where θ=tan−1 X L/Z 0. Hence, for Γ=|Γ|e j θ r,

|Γ|=1

and

θ r=π−2 θ=π−2 tan−1 X L Z 0.

# 2.6.2 Standing Waves

Using the relation V 0−=Γ V 0+ in Eqs. (2.54a) and (2.54b) yields

V˜(z)=V 0+(e−j β z+Γ e j β z),(2.63a)
I˜(z)=V 0+Z 0(e−j β z−Γ e j β z).(2.63b)

These expressions now contain only one (yet to be determined) unknown, V 0+. Before we proceed to solve for V 0+, however, let us examine the physical meaning underlying these expressions. We begin by deriving an expression for |V˜(z)|, which is the magnitude of V˜(z). Upon using Eq. (2.62) in Eq. (2.63a) and applying the relation |V˜(z)|=[V˜(z)V˜∗(z)]1/2, where V˜∗(z) is the complex conjugate of V˜(z), we have

|V˜(z)|={ [ V 0+(e−j β z+|Γ|e j θ r e j β z) ] ⋅[ (V 0+)∗(e j β z+|Γ|e−j θ r e−j β z) ] }1/2=|V 0+|[ 1+|Γ|2+|Γ|(e j(2 β z+θ r)+e−j(2 β z+θ r)) ]1/2=|V 0+|[ 1+|Γ|2+2|Γ|cos(2 β z+θ r) ]1/2,(2.64)

where we have used the identity

e j x+e−j x=2 cos x(2.65)

for any real quantity x. To express the magnitude of V˜ as a function of d instead of z, we replace z with −d on the right-hand side of Eq. (2.64):

|V˜(d)|=|V 0+|[1+|Γ|2+2|Γ|cos(2 β d−θ r)]1/2.(2.66)

By applying the same steps to Eq. (2.63b), a similar expression can be derived for |I˜(d)|, which is the magnitude of the current I˜(d):

|I˜(d)|=|V 0+|Z 0[1+|Γ|2−2|Γ|cos(2 β d−θ r)]1/2.(2.67)

The variations of |V˜(d)| and |I˜(d)| as a function of d, which is the position on the line relative to the load (at d=0), are illustrated in Fig. 2-14 for a line with |V 0+|=1 V, |Γ|=0.3, θ r=30°, and Z 0=50 Ω. The sinusoidal patterns are called standing waves and are caused by the interference of the two traveling waves. The maximum value of the standing-wave pattern of |V˜(d)| corresponds to the position on the line at which the incident and reflected waves are in-phase [2 β d−θ r=2 n π in Eq. (2.66)]. Therefore, they add constructively to give a value equal to (1+|Γ|)|V 0+|=1.3 V. The minimum value of |V˜(d)| occurs when the two waves interfere destructively, which occurs when the incident and reflected waves are in phase-opposition [2 β d−θ r=(2 n+1)π]. In this case, |V˜(d)|=(1−|Γ|)|V 0+|=0.7 V.

# Figure 2-14 Standing-wave pattern for (a) |V˜(d)| and (b) |I˜(d)| for a lossless transmission line of characteristic impedance Z 0=50 Ω, terminated in a load with a reflection coefficient Γ=0.3 e j 30°. The magnitude of the incident wave |V 0+|=1 V. The standing-wave ratio is S=|V˜|m a x/|V˜|min=1.3/0.7=1.86.

Figure 2 14 Full Alternative Text

The standing-wave pattern describes the spatial variation of the magnitude of V˜(d) as a function of d. If one were to observe the variation of the instantaneous voltage as a function of time at location d=d m a x in Fig. 2-14, that variation would be as cos ω t and would have an amplitude equal to 1.3 V [i.e., υ(t) would oscillate between −1.3 V and +1.3 V]. Similarly, the instantaneous voltage υ(d,t) at any location d will be sinusoidal with amplitude equal to |V˜(d)| at that d.

Close inspection of the voltage and current standing-wave patterns shown in Fig. 2-14 reveals that the two patterns are in phase opposition (when one is at a maximum, the other is at a minimum, and vice versa). This is a consequence of the fact that the third term in Eq. (2.66) is preceded by a plus sign, whereas the third term in Eq. (2.67) is preceded by a minus sign.

The standing-wave patterns shown in Fig. 2-14 are for a line with Γ=0.3 e j 30°. The peak-to-peak variation of the pattern (|V˜|min=(1−|Γ|)|V 0+| to |V˜|m a x=(1+|Γ|)|V 0+|) depends on |Γ|. For the special case of a matched line with Z L=Z 0, we have |Γ|=0 and |V˜(d)|=|V 0+| for all values of d, as shown in Fig. 2-15(a).

# Figure 2-15 Voltage standing-wave patterns for (a) a matched load, (b) a short-circuited line, and (c) an open-circuited line.

(a) ZL = Z0

Figure 2 15-a Full Alternative Text

(b) ZL = 0 (shortcircuit)

Figure 2 15-b Full Alternative Text

(c) ZL = ∞ (opencircuit)

Figure 2 15-c Full Alternative Text

The other end of the |Γ| scale, at |Γ|=1, corresponds to when the load is a short circuit (Γ=−1) or an open circuit (Γ=1). The standing-wave patterns for those two cases are shown in Figs. 2-15(b) and (c); both exhibit maxima of 2|V 0+| and minima equal to zero, but the two patterns are spatially shifted relative to each other by a distance of λ/4. A purely reactive load (capacitor or inductor) also satisfies the condition |Γ|=1, but θ r is generally neither zero nor 180° (Table 2-3). Exercise 2.9 examines the standing-wave pattern for a lossless line terminated in an inductor.

Now let us examine the maximum and minimum values of the voltage magnitude. From Eq. (2.66), |V˜(d)| is a maximum when the argument of the cosine function is equal to zero or a multiple of 2 π. Let us denote d m a x as the distance from the load at which |V˜(d)| is a maximum. It then follows that

|V˜(d)|=|V˜|m a x=|V 0+|[1+|Γ|],(2.68)

when

2 β d m a x−θ r=2 n π,(2.69)

with n=0 or a positive integer. Solving Eq. (2.69) for d m a x, we have

where we have used β=2 π/λ. The phase angle of the voltage reflection coefficient, θ r, is bounded between −π and π radians. If θ r≥0, the first voltage maximum occurs at d m a x=θ r λ/4 π, corresponding to n=0, but if θ r<0, the first physically meaningful maximum occurs at d m a x=(θ r λ/4 π)+λ/2, corresponding to n=1. Negative values of d m a x correspond to locations past the end of the line; therefore, they have no physical significance.

Similarly, the minima of |V˜(d)| occur at distances d min where the argument of the cosine function in Eq. (2.66) is equal to (2 n+1)π, which gives the result

|V˜|min=|V 0+|[1−|Γ|],when (2 β d min−θ r)=(2 n+1)π,(2.71)

with −π≤θ r≤π. The first minimum corresponds to n=0. The spacing between a maximum d m a x and the adjacent minimum d min is λ/4. Hence, the first minimum occurs at

The ratio of |V˜|m a x to |V˜|min is called the voltage standing-wave ratio S, which from Eqs. (2.68) and (2.71) is given by

This quantity is often referred to by its acronym, VSWR or the shorter acronym SWR, and it provides a measure of the mismatch between the load and the transmission line:

Concept Question 2-5: The attenuation constant α represents ohmic losses. In view of the model given in Fig. 2-6(c), what should R′ and G′ be in order to have no losses? Verify your expectation through the expression for α given by Eq. (2.25a).

Concept Question 2-6: How is the wavelength λ of the wave traveling on the transmission line related to the free-space wavelength λ 0?

Concept Question 2-7: When is a load matched to a transmission line? Why is it important?

Concept Question 2-8: What is a standing-wave pattern? Why is its period λ/2 and not λ?

Concept Question 2-9: What is the separation between the location of a voltage maximum and the adjacent current maximum on the line?

# Example 2-5: Standing-Wave Ratio

A 50 Ω transmission line is terminated in a load with Z L=(100+j 50) Ω. Find the voltage reflection coefficient and the voltage standing-wave ratio.

Solution: From Eq. (2.59), Γ is given by

Γ=z L−1 z L+1=(2+j 1)−1(2+j 1)+1=1+j 1 3+j 1.

Converting the numerator and denominator to polar form yields

Γ=1.414 e j 45°3.162 e j 18.4°=0.45 e j 26.6°.

Using the definition for S given by Eq. (2.73), we have

S=1+|Γ|1−|Γ|=1+0.45 1−0.45=2.6.

# Example 2-6: Measuring ZL

A slotted-line probe is an instrument used to measure the unknown impedance of a load, Z L. A coaxial slotted line contains a narrow longitudinal slit in the outer conductor. A small probe inserted in the slit can be used to sample the magnitude of the electric field and, hence, the magnitude |V˜(d)| of the voltage on the line (Fig. 2-16). By moving the probe along the length of the slotted line, it is possible to measure |V˜|m a x and |V˜|min and the distances from the load at which they occur. Use of Eq. (2.73), namely S=|V˜|m a x/|V˜|min, provides the voltage standing-wave ratio S. Measurements with a Z=50 Ω slotted line terminated in an unknown load impedance determined that S=3. The distance between successive voltage minima was found to be 30 cm, and the first voltage minimum was located at 12 cm from the load. Determine the load impedance Z L.

# Figure 2-16 Slotted coaxial line (Example 2-6).

Figure 2 16 Full Alternative Text

Solution: The following quantities are given:

Z 0=50 Ω,S=3,d min=12 cm.

Since the distance between successive voltage minima is λ/2,

λ=2×0.3=0.6 m,

and

β=2 π λ=2 π 0.6=10 π 3(rad/m).

From Eq. (2.73), solving for |Γ| in terms of S gives

|Γ|=S−1 S+1=3−1 3+1=0.5.

Next, we use the condition given by Eq. (2.71) to find θ r:

2 β d min−θ r=π, n=0 (first minimum),

which gives

θ r=2 β d min−π=2×10 π 3×0.12−π=−0.2 π (rad)=−36°.

Hence,

Γ=|Γ|e j θ r=0.5 e−j 36°=0.405−j 0.294.

Solving Eq. (2.59) for Z L, we have

Z L=Z 0 1+Γ 1−Γ=50 1+0.405−j 0.294 1−0.405+j 0.294=(85−j 67) Ω.