# 2.12 Transients on Transmission Lines

Thus far, our treatment of wave propagation on transmission lines has focused on the analysis of single-frequency, time-harmonic signals under steady-state conditions. The impedance-matching and Smith chart techniques we developed, while useful for a wide range of applications, are inappropriate for dealing with digital or wideband signals that exist in digital chips, circuits, and computer networks. For such signals, we need to examine the transient transmission line response instead.

Let us start by considering the case of a single rectangular pulse of amplitude V 0 and duration τ, as shown in Fig. 2-41(a). The amplitude of the pulse is zero prior to t=0, V 0 over the interval 0≤t≤τ, and zero afterwards. The pulse can be described mathematically as the sum of two unit step functions:

V(t)=V 1(t)+V 2(t)=V 0 u(t)−V 0 u(t−τ),(2.143)

where the unit step function u(x) is

u(x)=1 for x>0,0 for x<0.(2.144)

# Figure 2-41 A rectangular pulse V(t) of duration τ can be represented as the sum of two step functions of opposite polarities displaced by τ relative to each other.

Figure 2 41 Full Alternative Text

The first component, V 1(t)=V 0 u(t), represents a dc voltage of amplitude V 0 that is switched on at t=0 and then retains that value indefinitely. The second component, V 2(t)=−V 0 u(t−τ), represents a dc voltage of amplitude −V 0 that is switched on at t=τ and remains that way indefinitely. As can be seen from Fig. 2-41(b), the sum V 1(t)+V 2(t) is equal to V 0 for 0<t<τ and equal to zero for t<0 and t>τ. This representation of a pulse in terms of two step functions allows us to analyze the transient behavior of the pulse on a transmission line as the superposition of two dc signals. Hence, if we can develop basic tools for describing the transient behavior of a single step function, we can apply the same tools for each of the two components of the pulse and then add the results to obtain the response to V(t).

# 2.12.1 Transient Response to a Step Function

The circuit shown in Fig. 2-42(a) consists of a generator, composed of a dc voltage source V g and a series resistance R g, connected to a lossless transmission line of length l and characteristic impedance Z 0. The line is terminated in a purely resistive load R L at z=l.

# Figure 2-42 At t=0+, immediately after closing the switch in the circuit in part (a), the circuit can be represented by the equivalent circuit in part (b). This is because from that instant and until a reflection is received back from the load the generator circuit “sees” only an impedance Z 0.

(a) Transmission-line circuit

Figure 2 14-a Full Alternative Text

(b) Equivalent circuit at t = 0+

Figure 2 14-b Full Alternative Text

The switch between the generator circuit and the transmission line is closed at t=0. The instant the switch is closed, the transmission line appears to the generator circuit as a load with impedance Z 0. This is because, in the absence of a signal on the line, the input impedance of the line is unaffected by the load impedance R L. The circuit representing the initial condition is shown in Fig. 2-42(b). The initial current I 1+ and corresponding initial voltage V 1+ at the sending end of the transmission line are given by

I 1+=V g R g+Z 0,(2.145a)
V 1+=I 1+Z 0=V g Z 0 R g+Z 0.(2.145b)

The combination of V 1+ and I 1+ constitutes a wave that travels along the line with velocity u p=1/μ ε immediately after the switch is closed. The plus-sign superscript denotes the fact that the wave is traveling in the +z direction. The transient response of the wave is shown in Fig. 2-43 at each of three instances in time for a circuit with R g=4 Z 0 and R L=2 Z 0. The first response is at time t 1=T/2, where T=l/u p is the time it takes the wave to travel the full length of the line. By time t 1, the wave has traveled halfway down the line; consequently, the voltage on the first half of the line is equal to V 1+, while the voltage on the second half is still zero (Fig. 2-43(a)). At t=T, the wave reaches the load at z=l, and because R L≠Z 0, the mismatch generates a reflected wave with amplitude

V 1−=Γ L V 1+,(2.146)

where

Γ L=R L−Z 0 R L+Z 0(2.147)

is the reflection coefficient of the load. For the specific case illustrated in Fig. 2-43, R L=2 Z 0, which leads to Γ L=1/3. After this first reflection, the voltage on the line consists of the sum of two waves: the initial wave V 1+ and the reflected wave V 1−. The voltage on the transmission line at t 2=3 T/2 is shown in Fig. 2-43(b); V(z,3 T/2) equals V 1+ on the first half of the line (0≤z<l/2) and (V 1++V 1−) on the second half (l/2≤z≤l).

# Figure 2-43 Voltage and current distributions on a lossless transmission line at t=T/2, t=3 T/2, and t=5 T/2 due to a unit step voltage applied to a circuit with R g=4 Z 0 and R L=2 Z 0. The corresponding reflection coefficients are Γ L=1/3 and Γ g=3/5.

(a) V(z) at t = T/2

Figure 2 43-a Full Alternative Text

(b) V(z) at t = 3T/2

Figure 2 43-b Full Alternative Text

(c) V(z) at t = 5T/2

Figure 2 43-c Full Alternative Text

(d) I(z) at t = T/2

Figure 2 43-d Full Alternative Text

(e) I(z) at t = 3T/2

Figure 2 43-e Full Alternative Text

(f) I(z) at t = 5T/2

Figure 2 43-f Full Alternative Text

At t=2 T, the reflected wave V 1− arrives at the sending end of the line. If R g≠Z 0, the mismatch at the sending end generates a reflection at z=0 in the form of a wave with voltage amplitude V 2+ given by

V 2+=Γ g V 1−=Γ g Γ L V 1+,(2.148)

where

Γ g=R g−Z 0 R g+Z 0(2.149)

is the reflection coefficient of the generator resistance R g. For R g=4 Z 0, we have Γ g=0.6. As time progresses after t=2 T, the wave V 2+ travels down the line toward the load and adds to the previously established voltage on the line. Hence, at t=5 T/2, the total voltage on the first half of the line is

V(z,5 T/2)=V 1++V 1−+V 2+=(1+Γ L+Γ L Γ g)V 1+ (0≤z<l/2),(2.150)

while on the second half of the line the voltage is only

V(z,5 T/2)=V 1++V 1−=(1+Γ L)V 1+ (l/2≤z≤l).(2.151)

The voltage distribution is shown in Fig. 2-43(c).

So far, we have examined the transient response of the voltage wave V(z,t). The associated transient response of the current I(z,t) is shown in Figs. 2-43(d)–(f). The current behaves similarly to the voltage V(z,t), except for one important difference. Whereas at either end of the line the reflected voltage is related to the incident voltage by the reflection coefficient at that end, the reflected current is related to the incident current by the negative of the reflection coefficient. This property of wave reflection is expressed by Eq. (2.61). Accordingly,

I 1−=−Γ L I 1+,(2.152a)
I 2+=−Γ g I 1−=Γ g Γ L I 1+,(2.152b)

and so on.

The final value of V is given by

V∞=V 1+​+​V 1−​+​V 2+​+​V 2−​+​V 3+​+​V 3−​+⋯=V 1+[1​+​Γ L​+​Γ L Γ g​+​Γ L 2 Γ g​+​Γ L 2 Γ g 2​+​Γ L 3 Γ g 2​+⋯]=V 1+[(1​+​Γ L)(1​+​Γ L Γ g​+​Γ L 2 Γ g 2​+⋯)]=V 1+(1​+​Γ L)[1+x+x 2+⋯],(2.153)

where x=Γ L Γ g. The series inside the square bracket is the geometric series of the function

1 1−x=1+x+x 2+⋯ for |x|<1.(2.154)

Hence, Eq. (2.153) can be rewritten in the compact form

V∞=V 1+1+Γ L 1−Γ L Γ g.(2.155)

Upon replacing V 1+, Γ L, and Γ g with Eqs. (2.145b), (2.147), and (2.149) and simplifying the resulting expression, we obtain

The voltage V∞ is called the steady-state voltage on the line, and its expression is exactly what we should expect on the basis of dc analysis of the circuit in Fig. 2-43(a), where we treat the transmission line as simply a connecting wire between the generator circuit and the load. The corresponding steady-state current is

# 2.12.2 Bounce Diagrams

Keeping track of the voltage and current waves as they bounce back and forth on the line is a rather tedious process. The bounce diagram is a graphical presentation that allows us to accomplish the same goal but with relative ease. The horizontal axes in Figs. 2-44(a) and (b) represent the position along the transmission line, while the vertical axes denote time. Figures 2-44(a) and (b) pertain to V(z,t) and I(z,t), respectively. The bounce diagram in Fig. 2-44(a) consists of a zigzag line indicating the progress of the voltage wave on the line. The incident wave V 1+ starts at z=t=0 and travels in the +z direction until it reaches the load at z=l at time t=T. At the very top of the bounce diagram, the reflection coefficients are indicated by Γ=Γ g at the generator end and by Γ=Γ L at the load end. At the end of the first straight-line segment of the zigzag line, a second line is drawn to represent the reflected voltage wave V 1−=Γ L V 1+. The amplitude of each new straight-line segment equals the product of the amplitude of the preceding straight-line segment and the reflection coefficient at that end of the line. The bounce diagram for the current I(z,t) in Fig. 2-44(b) adheres to the same principle except for the reversal of the signs of Γ L and Γ g at the top of the bounce diagram.

# Figure 2-44 Bounce diagrams for (a) voltage and (b) current. In (c), the voltage variation with time at z=l/4 for a circuit with Γ g=3/5 and Γ L=1/3 is deduced from the vertical dashed line at l/4 in (a).

(a) Voltage bounce diagram

Figure 2 44-a Full Alternative Text

(b) Current bounce diagram

Figure 2 44-b Full Alternative Text

(c) Voltage versus time at z = l/4

Figure 2 44-c Full Alternative Text

Using the bounce diagram, the total voltage (or current) at any point z 1 and time t 1 can be determined by first drawing a vertical line through point z 1, and then adding the voltages (or currents) of all the zigzag segments intersected by that line between t=0 and t=t 1. To find the voltage at z=l/4 and T=4 T, for example, we draw a dashed vertical line in Fig. 2-44(a) through z=l/4, and we extend it from t=0 to t=4 T. The dashed line intersects four line segments. The total voltage at z=l/4 and t=4 T therefore is

V(l/4,4 T)=V 1++Γ L V 1++Γ g Γ L V 1++Γ g Γ L 2 V 1+=V 1+(1+Γ L+Γ g Γ L+Γ g Γ L 2).

The time variation of V(z,t) at a specific location z can be obtained by plotting the values of V(z,t) along the (dashed) vertical line passing through z. Figure 2-44(c) shows the variation of V as a function of time at z=l/4 for a circuit with Γ g=3/5 and Γ L=1/3.

# Example 2-16: Pulse Propagation

The transmission-line circuit of Fig. 2-45(a) is excited by a rectangular pulse of duration τ=1 ns that starts at t=0. Establish the waveform of the voltage response at the load given that the pulse amplitude is 5 V, the phase velocity is c, and the length of the line is 0.6 m.

# Figure 2-45 Example 2-16.

(a) Pulse circuit

Figure 2 45-a Full Alternative Text

(b) Bounce diagram

Figure 2 45-b Full Alternative Text

(c) Voltage waveform at the load

Figure 2 45-c Full Alternative Text

Solution: The one-way propagation time is

T=l c=0.6 3×10 8=2 ns.

The reflection coefficients at the load and the sending end are

Γ L=R L−Z 0 R L+Z 0=150−50 150+50=0.5,Γ g=R g−Z 0 R g+Z 0=12.5−50 12.5+50=−0.6.

By Eq. (2.143), the pulse is treated as the sum of two step functions: one that starts at t=0 with an amplitude V 10=5 V and a second one that starts at t=1 ns with an amplitude V 20=−5 V. Except for the time delay of 1 ns and the sign reversal of all voltage values, the two step functions generate identical bounce diagrams, as shown in Fig. 2-45(b). For the first step function, the initial voltage is given by

V 1+=V 01 Z 0 R g+Z 0=5×50 12.5+50=4 V.

Using the information displayed in the bounce diagram, it is straightforward to generate the voltage response shown in Fig. 2-45(c). Note that at t=2 ns, the voltage at the load consists of both V 1+=4 V, and its reflection V 1−=Γ L V 1+=2 V. Thus, the total voltage at the load at t=2 ns is 6 V. A similar process occurs at 3 ns due to the negative step function, which generates −6 V. Consequently, the sum of the two step-function contributions adds up to zero and stays that way until t=6 ns.

# Example 2-17: Time-Domain Reflectometer

A time-domain reflectometer (TDR) is an instrument used to locate faults on a transmission line. Consider, for example, a (formerly matched) long underground or undersea cable that gets damaged at some distance d from the sending end of the line. The damage may alter the electrical properties or the shape of the cable, causing it to exhibit at the fault location an effective resistance R L f. A TDR sends a step voltage down the line, and by observing the voltage at the sending end as a function of time, it is possible to determine the location of the fault and its severity.

If the voltage waveform shown in Fig. 2-46(a) is seen on an oscilloscope connected to the input of a 75 Ω matched transmission line, determine (a) the generator voltage, (b) the location of the fault, and (c) the fault shunt resistance. The line’s insulating material is Teflon with ε r=2.1.

# Figure 2-46 Time-domain reflectometer of Example 2-17.

(a) Observed voltage at the sending end

Figure 2 46-a Full Alternative Text

(b) The fault at z = d is represented by a fault resistance Rf

Figure 2 46-b Full Alternative Text

Solution: (a) Since the line is properly matched, R g=R L=Z 0. In Fig. 2-46(b), the fault located a distance d from the sending end is represented by a shunt resistance R f. For a matched line, Eq. (2.145b) gives

V 1+=V g Z 0 R g+Z 0=V g Z 0 2 Z 0=V g 2.

According to Fig. 2-46(a), V 1+=6 V. Hence,

V g=2 V 1+=12 V.

(b) The propagation velocity on the line is

u p=c ε r=3×10 8 2.1=2.07×10 8 m/s.

For a fault at a distance d, the round-trip time delay of the echo is

Δ t=2 d u p.

From Fig. 2-46(a), Δ t=12 μ s. Hence,

d=Δ t 2 u p=12×10−6 2×2.07×10 8=1,242 m.

(c) The drop in level of V(0,t) shown in Fig. 2-46(a) represents V 1−. Thus,

V 1−=Γ f V 1+=−3 V,

or

Γ f=−3 6=−0.5,

where Γ f is the reflection coefficient due to the effective fault resistance R L f that appears at z=d.

From Eq. (2.59),

Γ f=R L f−Z 0 R L f+Z 0,

which leads to R L f=25 Ω. This fault load is composed of the parallel combination of the fault shunt resistance R f and the characteristic impedance Z 0 of the line to the right of the fault:

1 R L f=1 R f+1 Z 0,

so the shunt resistance must be 37.5 Ω.

# Example 2-18: Pulse Transmission

When excited by a pulse source of amplitude 1 V and duration 5 μ s, a transmission line—with unknown characteristic impedance Z 0 and unknown load resistance R L—exhibits the voltage response shown in Fig. 2-47(b) when observed at the middle of the line. The line relative permittivity is ε r=2.25. Determine (a) the length of the line, (b) Z 0, and (c) R L.

# Figure 2-47 Circuit and voltage response of Example 2-18.

(a) Transmission line

Figure 2 47-a Full Alternative Text

(b) Observed voltage at midpoint of the line

Figure 2 47-b Full Alternative Text

Solution: (a) The phase velocity is

u p=c ε r=3×10 8 2.25=2×10 8 m/s.

It takes 10 μ s for the initial part of the pulse to reach the middle of the line, so it would take it T=20 μ s to reach the end of the line. Hence,

l=u p T=2×10 8×20×10−6=4000 m.

(b) From Eq. (2.145b),

V 1+=V g Z 0 R g+Z 0.

In the present case, V g=1 V, R g=100 Ω, and the waveform in Fig. 2-47(b) indicates that V 1+=0.5 V. Solving for Z 0 leads to

Z 0=100 Ω.

(c) The pulse that appears between 30 μ s and 35 μ s is due to reflection by the load and its amplitude V 1−=0.3 V. Hence,

Γ L=V 1−V 1+=0.3 0.5=0.6.

Also,

Γ L=R L−Z 0 R L+Z 0.

Using Γ L=0.6 and Z 0=100 Ω leads to

R L=400 Ω.

Concept Question 2-27: What is transient analysis used for?

Concept Question 2-28: The transient analysis presented in this section was for a step voltage. How does one use it for analyzing the response to a pulse?

Concept Question 2-29: What is the difference between the bounce diagram for voltage and the bounce diagram for current?