5.7: Bounce Diagrams

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Now this new \(V_{2}^{+}\) will head back towards the load and...hmm...things are going to get kind of messy and complicated. Fortunately for us, transmission line engineers came up with a scheme for keeping track of all of the waves bouncing back and forth on the line. The scheme is called a bounce diagram. A bounce diagram consists of a horizontal distance line, which represents distance along the transmission line, and a vertical time axis, which represents time since the battery was first connected to the line. Just to keep things conceptually clear, we usually first start out by showing the line, the battery, the load and a switch, S, which is used to connect the source to the line. It doesn't hurt to make a little sketch like Figure \(\PageIndex{1}\), and write down the length of the line, \(Z_{0}\) and \(v_{p}\), along with the source and load resistances. Now we draw the bounce diagram, which is shown in Figure \(\PageIndex{2}\).

Two horizontal wires are connected at their left ends by a voltage source V_s with the positive end facing up, and at their right ends by a load resistor R_L. A resistor R_s is located in the top left corner, and a switch is located just to the right of this resistor. The horizontal distance between the load resistor and the right end of the switch is L. For this distance, impedance value Z_0 is the square-rooted fraction of distributed L over distributed C and voltage value v_p is one divided by the square-rooted product of distributed L and distributed C.

Empty bounce diagram consisting of a horizontal line representing an x-axis with 0 at the left end and L at the right end, with a vertical axis of positive time extending upwards from each end. The expression for Gamma_vs, the difference between R_s and Z_0 divided by their sum, is written by x=0. The expression for Gamma_vL, the difference between R_L and Z_0 divided by their sum, is written by x=L.

Normally, you would not put the formula for \(\Gamma_{vs}\) and \(\Gamma_{vL}\) by \(x=0\) and \(x=L\) in the diagram, but rather their values. This will become clear when we do an example. The next thing we do is calculate \(V_{1}^{+}\) and draw a straight line on the bounce diagram (nominally at a slope of \(\frac{1}{v_{p}}\)) which will represent the initial signal going down the line. We mark a \(\tau = \frac{L}{v_{p}}\) on the vertical axis to show how long it takes for the wave to reach the end of the line, as shown in Figure \(\PageIndex{3}\).

A line is drawn on the diagram from Figure 2 above, stretching from the bottom left corner to the point tau on the right time axis. The line is labeled as V1+, equal to the product of Z_0 and V_s divided by the sum of Z_0 and R_s.

Once the initial wave hits the load, a second, reflected wave \(V_{1}^{-} = \Gamma_{vL} V_{1}^{+}\) is sent back the other way so we add it to the bounce diagram. This is shown in Figure \(\PageIndex{4}\). Since all of the waves move with the same phase velocity, we should be careful to draw all of the lines with the same slope. Note that the time when the reflected wave hits the generator end is a total round trip time of \(2 \tau\). (This simple concept is one which students often forget come test time, so be forewarned!)

The diagram from Figure 3 above is shown with the addition of a new diagonal line labeled V1-, which slants up and to the left from the point t = tau on the right time axis to the point t = 2 tau on the left time axis.

We saw that the next thing that happens is that another wave is reflected from the generator, so we add that to the bounce diagram as well. This is shown in Figure \(\PageIndex{5}\).

The diagram from Figure 4 above with the addition of a new diagonal line V2+, slanting up from the point t = 2 tau on the left t-axis to the point t = 3 tau on the right t-axis. The lines for V1+ and V2+ are parallel. — Figure \(\PageIndex{5}\): The third wave

Finally, one last wave, as we are almost bounced right off the diagram, as shown in Figure \(\PageIndex{6}\)!

The diagram from Figure 5 above is shown with the addition of a diagonal line V2-, slanting up and to the left and parallel to the V1- line. The left end of the V2- line rises past the region visible on the diagram.

OK, so we've got a bounce diagram, so what? Having the diagram is only part of the solution. We still have to see what good they are. Let's do a numerical example, as it is maybe a little more illustrative, and certainly will be easier to write out than all these ratios all the time. We will just pick some typical numbers, and then work out the answers. Let's let \(V_{S} = 40 \mathrm{~V}\), \(R_{S} = 150 \mathrm{~\Omega}\), \(Z_{0} = 50 \mathrm{~\Omega}\), and \(R_{L} = 16.7 \mathrm{~\Omega}\). The line will be 100 meters long, and \(v_{p} = 2 \times 10^{8}\ \frac{\mathrm{m}}{\mathrm{s}}\). The circuit is labeled with these numerical values in Figure \(\PageIndex{7}\).

The circuit from Figure 1 above is labeled with a voltage source of 40 V, a source resistance of 150 Ohms, a load resistance of 16.7 Ohms, and a line distance of 100 meters between the switch and the load resistor. The line has impedance 50 Ohms and a v_p of 2 times 10 to the 8th power, in meters per second.

First we calculate the reflection coefficients \[\begin{array}{l} \Gamma_{vL} &= & \dfrac{R_{L} - Z_{0}}{R_{L} + Z_{0}} \\[4pt] &= & \dfrac{16.7 - 50}{16.7 + 50} \\[4pt] &= & -0.50 \end{array}\] and \[\begin{array}{l} \Gamma_{vS} &=& \dfrac{R_{S} - Z_{0}}{R_{S} + Z_{0}} \\[4pt] &=& \dfrac{150-50}{150+50} \\ &=& 0.50 \end{array}\]

The initial voltage signal \(V_{1}^{+}\) is \[\begin{array}{l} V_{1}^{+} &=& \dfrac{50}{50 + 150} \cdot 40 \\[4pt] &=& 10(V) \end{array}\]

and the propagation time is \[\begin{array}{l} \tau &=& \dfrac{L}{v_{p}} \\[4pt] &=& \dfrac{100 \mathrm{~m}}{\left(2 \times 10^{8}\right) \frac{\mathrm{m}}{\mathrm{s}}} \\[4pt] &=& 0.5 \ \mu \mathrm{s} \end{array}\]

So we draw the bounce diagrams seen in Figure \(\PageIndex{8}\).

Bounce diagram from Figure 6 above with tau labeled as 0.5 microseconds, Gamma_vs labeled as 0.5 and occurring at x=0, and Gamma_vL labeled as -0.5 and occurring at x=100. The four voltage lines in the diagram are labeled from bottom to top with values of 10 V, -5 V, -2.5 V, and 1.25 V. A vertical dotted line is drawn halfway between the x-endpoints.

Now, here's how we use a bounce diagram, once we have it. Suppose we want to know what \(V(t)\), the voltage as a function of time, would look like halfway down the line. We draw a vertical line at the place we are interested in (the dotted line in Figure \(\PageIndex{8}\)) and then just go up along the line, adding voltage to whatever we had before whenever we cross one of the "bouncing" signal lines. Thus for the line as shown we would have for \(V(t)\) what we see in Figure \(\PageIndex{9}\).

Graph of voltage in units of Volts at location x=50 meters, as a function of time t in microseconds. The graph takes the form of four straight horizontal segments with instantaneous transitions: V = 10 for x between 0.25 and 0.75, V = 5 for x between 0.75 and 1.25, V = 2.5 for x between 1.25 and 1.75, and V = 3.75 for x greater than 1.75.

For the first \(0.25 \ \mu \mathrm{s}\) we have no voltage, because \(V_{1}^{+}\) has not reached the halfway point yet. The voltage then jumps to \(+10 \mathrm{~V}\) when \(V_{1}^{+}\) comes by. It stays like that until the \(V_{1}^{-}\) of \(-5 \mathrm{~V}\) comes by \(0.5 \ \mu \mathrm{s}\) later. The voltage then remains constant at \(5 \mathrm{~V}\) until the \(V_{2}^{+}\) of \(-2.5 \mathrm{~V}\) comes along to drop the total voltage down to only 2.5 volts. When \(V_{2}^{-}\) comes along, it has been switched back to a positive voltage wave by the negative load reflection coefficient, and so now the voltage jumps back up to \(3.75 \mathrm{~V}\). It will keep oscillating back and forth until it finally settles down to some asymptotic value.

What will that asymptotic value be? One approach is to write down the following equation: \[V(x, \infty) = V_{1}^{+} \left(1 + \Gamma_{L} + \Gamma_{L} \Gamma_{S} + \Gamma_{L}{ }^{2} \Gamma_{S} + \ldots\right)\]

which we can re-write as \[V_{1}^{+} \left(1 + \Gamma_{L} \Gamma_{S} + \left(\Gamma_{L}\Gamma_{S}\right)^{2} + \ldots\right) + \Gamma_{L} V_{1}^{+} \left(1 + \Gamma_{L} \Gamma_{S} + \left(\Gamma_{L}\Gamma_{S}\right)^{2} + \ldots\right)\]

Now, remembering the infinite sum relationship \[\sum_{n=0}^{\infty} x^{n} = \frac{1}{1 - x}\]

for \(|x| < 1\) (which is always the case for a reflection coefficient). We can substitute Equation \(\PageIndex{7}\) for the terms inside the parentheses in Equation \(\PageIndex{6}\) and we get \[\begin{array}{l} V(x, \infty) &= V_{1}^{+} \left(\dfrac{1}{1 - \Gamma_{L} \Gamma_{S}} + \dfrac{\Gamma_{L}}{1 - \Gamma_{L} \Gamma_{S}}\right) \\[4pt] &= V_{1}^{+} \dfrac{1 + \Gamma_{L}}{1 - \Gamma_{L}\Gamma_{S}} \end{array}\]

We will leave it as an exercise to the reader to show that if we substitute Equations \(5.6.9\) [link], \(5.6.14\) [link], and finally \(5.6.17\) [link] into Equation \(\PageIndex{8}\) we will eventually get: \[V(x, \infty) = \frac{R_{L}}{R_{L} + R_{S}} V_{S}\]

Look back at Figure \(\PageIndex{1}\) and see if Equation \(\PageIndex{9}\) makes any sense. It should. If we wait long enough, it is reasonable to expect that any "transmission line" effects should go away, and we would be back to the same situation we would have if the line was just some wire connecting the source to the load. In this case, the load resistor and the source resistor would form a voltage divider, and we would expect the voltage across the load to be determined by the voltage divider equation. That's all Equation \(\PageIndex{9}\) is saying!

What do we do if we want, say, the voltage across the load with time? To do this we move up the right-hand side of the bounce diagram, and count voltage waves as we move across them. We start out at zero, of course, and do not see anything until we get to \(0.5 \ \mu \mathrm{s}\). Then we cross the \(10 \mathrm{~V}\) \(V_{1}^{+}\) wave and we cross the \(-5 \mathrm{~V}\) \(V_{1}^{-}\) wave at the same time, so the voltage only goes up to \(+5 \mathrm{~V}\). Likewise, another \(1 \ \mu \mathrm{s}\) later, we cross both the \(-2.5 \mathrm{~V}\) \(V_{2}^{+}\) wave and the \(+1.25 \mathrm{~V}\) \(V_{2}^{-}\) wave, and so the voltage ends up at the \(3.75 \mathrm{~V}\) position as in Figure \(\PageIndex{10}\).

Graph of V, in units of Volts, across the load as a function of time in units of microseconds. The value of V is 5 for t = 0.5 to 1.5, and V is 3.75 for values of t greater than 1.5, with instantaneous transitions.

We can also use the bounce diagram to find the voltage as a function of position, for some fixed time \(t_{0}\) as in Figure \(\PageIndex{11}\).

Bounce diagram with Gamma_vS = 0.5, Gamma_vL = -0.5, and x = 100 meters. The 10V wave connects the bottom left corner to t = 0.5 microseconds on the right time axis, and the -5V wave connects the upper end of the first line to t = 1.0 microseconds on the left time axis. A horizontal dotted line is drawn across the graph at t = 0.75 microseconds.

To do this, we draw a horizontal line at the time we are interested in, say \(0.75 \ \mu \mathrm{s}\). Now, for each position \(x\), we go from the bottom of the diagram, up to the horizontal line, adding up voltage as we go. Thus for the example: we get what we see in Figure \(\PageIndex{12}\). For the first half of the line, we cross the \(+10 \mathrm{~V}\) \(V_{1}^{+}\), but that's it. For the second half of the line we cross both the \(+10 \mathrm{~V}\) line as well as the \(-5 \mathrm{~V}\) \(V_{1}^{-}\) wave, and so the voltage drops down to \(5 \mathrm{~V}\).

Graph of V in units of Volts at time 0.75 microseconds, as a function of x in units of meters. The graph takes the value of V = 10 for x = 0 to 50 and the value of V = 5 for x = 50 to 100, with instantaneous transitions.

Of particular interest to many of you will be the way in which a pulse moves down a line and is reflected etc. This is also quite easy to do with a reflection diagram, if we simply break the pulse into two waves, one which has a positive swing at \(t=0\) and another which is a negative going wave at \(t = \tau_{p}\), where \(\tau_{p}\) is the pulse width of the pulse being generated. The way we do this is suggested in Figure \(\PageIndex{13}\). We replace the pulse generator with two battery/switch combinations. The first circuit is just like we have seen so far, with a battery equal to the open circuit pulse height of the generator, and a switch which closes at \(t = 0\). The second circuit has a battery with an amplitude of minus the pulse height, and a switch which closes at \(t = \tau_{p}\), the pulse width of the pulse itself.

A pulse is simulated with a combination of two circuits: a voltage source of V_p connected to a resistor R_s and a switch that closes at t=0, and a voltage source of -V_p connected to a resistor R_s and a switch that closes at t = tau_p. — Figure \(\PageIndex{13}\): Simulating a pulse with two batteries and two switches

By superposition, we can just add these two generators, one after the other, and see how the pulse goes down the line. Suppose \(V_{p}\) is 10 volts, \(\tau_{p} = 0.25 \ \mu \mathrm{s}\), \(R_{S} = 50 \mathrm{~\Omega}\), \(Z_{0} = 50 \mathrm{~\Omega}\), and \(R_{L} = 25 \mathrm{~\Omega}\). With the numbers, we find that \(V_{1}^{+} = 25 \mathrm{~V}\). \(\Gamma_{vL} = \frac{-1}{3}\), and \(\Gamma_{vS} = 0\). Let's assume that the propagation time on the line is still \(0.5 \ \mu \mathrm{s}\) to get from one end of the line to the other.

We draw the bounce diagram, and launch two waves, one which leaves at \(t=0\), has an amplitude of \(V_{1}^{+} = 5 \mathrm{~V}\). The second wave leaves at a time \(\tau_{p}\) later, and has an amplitude of \(-5 \mathrm{~V}\).

Bounce diagram with Gamma_vS = 0, Gamma_vL = -0.33, and an x-axis of 0 to 100 meters. A 5V line connects the bottom left corner to the point t=0.5 microseconds on the right time axis. A -1.67V line connects the upper end of the +5V line to the point t = 1.0 microseconds on the left time axis. A -5V line parallel to the 5V line starts at the point tau_p above the origin, and a +1.67V line parallel to the -1.67V starts at the upper end of this line.

Now when we want to see what the voltage as a function of time looks like, we again draw a line up the middle, and add voltages as we cross them. Here we see, again, no voltage until we cross the first wave at \(0.25 \ \mu \mathrm{~s}\), which pops us up to \(+5 \mathrm{~V}\). At a time \(0.25 \ \mu \mathrm{~s}\) later, however, the \(-5 \mathrm{~V}\) wave comes along, and we go back down to zero. At \(t = 0.75 \ \mu \mathrm{~s}\), the reflected \(-1.67 \mathrm{~V}\) pulse comes along, and so we see that. Since the source is matched to the line, \(\Gamma_{vS} = 0\) and so this is the end of the story, as shown in Figure \(\PageIndex{15}\).

Graph of V at 50 meters, in units of volts, as a function of t in units of microseconds. For t=0.25 to 0.5, V is 5. For t=0.5 to 0.75, V is 0. For t=0.75 to 1.0, V is -1.67. All transitions are instantaneous.

You can get somewhat more interesting waveforms if you go someplace where the two pulses at least partially overlap. Let's look at say, \(x = 87.5 \mathrm{~m}\). Figure \(\PageIndex{16}\) is the bounce diagram.

Bounce diagram with x running from 0 to 100 meters, Gamma_vs = 0, and Gamma_vL = -0.33. A diagonal +5V line slants from the lower left corner to the value of 0.5 microseconds on the right time axis, and a -5V line parallel to this starts at the point tau_p on the left time axis. A diagonal -1.67V line slants from the right endpoint of the +5V line to the value of 1.0 microseconds on the left time axis. A +1.67 line runs parallel to this, starting at the right endpoint of the -5V line. A vertical dotted line is drawn through the graph at the point x=87.5 m.

And Figure \(\PageIndex{17}\) is the voltage waveform we get.

Graph of V at x=87.5 m, in terms of t. At some t greater than 0, V goes from 0 to 5 V and stays there for some time before dropping to 3.33 V and remaining constant for a time. Then it drops to -1.67 V and remains constant for a time before returning to 0. All transitions are instantaneous.

This time the \(1.67 \mathrm{~V}\) pulse gets to us before the \(+5 \mathrm{~V}\) pulse has completely passed, and so we drop from \(5 \mathrm{~V}\) to \(3.33 \mathrm{~V}\). Then, when the \(-5 \mathrm{~V}\) wave goes by, we drop down to \(-1.67 \mathrm{~V}\) for a little while, until the \(+1.67 \mathrm{~V}\) wave comes along to bring us back to zero.