In chapter 6 we deal problems mostly with loads present linear resistive
characteristic. However, in many applications, we should consider nonlinear load.
Load-line technique is a graphical technique analyzing the voltage and current along a
transmission line connected with a nonlinear load. Our demo is based on the
following example.
Consider a transmission-line system terminated by a passive nonlinear element
having the I-V relationship of VL  50I L I L , and at the source end is connected with
a resistor of 200Ω and a DC voltage of 50V. The characteristic impedance of the
transmission-line is Z0=50Ω, and T=1μs. To obtain the time variation of the voltages
VS and VL at the source and at the load ends, respectively, following the closure of the
switch S at t=0, using the load-line technique.
At t=0+ at z=0:
50  200 I S  VS ........................(1)
VS  V 
IS  I  
V VS

................(2)
Z0
50
V+ and I+ are the voltage and current, respectively, of the (+) wave set up immediately
after closure of the switch. We can solve VS and IS graphically by constructing the
straight lines representing them. (See Fig. 6.48 in page 406 of the textbook) The point
of intersection is denoted by A, which gives the values of VS and IS.
When the (+) wave reaches the load end z=l at t=T, a (-) wave is set up.
At t=T+ at z=l:
VL  50 I L I L ..............................................................(3)
VL  V   V 
V   V  2V   VL
IL  I  I 

........................(4)
Z0
50


V- and I- are the (-) wave voltage and current, respectively. The solution of VL and IL
is given by the intersection of the intersection of the curve representing equation (3)
and the straight line of slope -1/50 corresponding to equation (4). The intersection is
denoted by B.
When the (-) wave reaches the source end z=0 at t=2T, it sets up a reflection,
denote this to be the (-+) wave.
At t=2T+ at z=0:
50  200I S  VS .......................................................................(5)
VS  V   V   V  
I S  I   I   I  
V   V   V   2V   VS

...............(6)
Z0
50
V-+ and I-+ are the (-+) wave voltage and current, respectively. Noting from equation
(5) that for VS=V++V-, IS=(V+-V-)/50, we see that equation (6) represents a straight
line of slope 1/50 passing through B. Thus, the solution of (5) and (6) is given by
point C in Fig. 6.48.
Continuing analyze the circuit in this manner, we observe that the solution
consists of obtaining the points of intersection on the source and load V-I
characteristics, that is, equation (1) and equation (3), by drawing successively straight
lines of slope 1/Z0 and -1/Z0, beginning at the origin and with each straight line
originating at the previous point of intersection, as shown in Fig. 6.48. To be more
specific, solutions of VS and IS are obtained by drawing the intersection of the straight
lines of slope 1/Z0 and equation (1), whereas that of VL and IL are obtained by
drawing the intersection of the straight lines of slope -1/ Z0 and equation (3). The
location of the straight line is determined by the previous point of intersection.
Now that we have the time-variations of VS, IS, VL, and IL. The voltage and
current of an arbitrary point on the transmission-line at time t can be obtained. To
illustrate the method, consider the following examples.
When t=1.5T at z=0.6L, the voltage and current should be equal to that at the
load end when t=1T+. However, at the same time, at z=0.4L, the (-) wave has not yet
traveled to this point, so the voltage and current is equal to that at the source end when
t=0+.