I have developed a mental image of the transmission line as an energy field that can sustain the resistive current flow from the source as the wave moves down the line, and even as it reflects. I would still like to understand how the transition from a resistive load occurs....the following quote from my old college textbook is almost metaphysical...
"until the driver becomes aware of the impedance mismatch at the end of the line the line looks resistive"
Thanks for your interest in High-Speed Digital Design, Good question!
The behavior of a transmission line is defined by a set of mathematical state-variable equations. As is usually the case with mathematical formulations, the choice of state representation greatly affects the utility of the final solution. The two most popular representations of the "state" of a transmission line are the voltage/current model, and the forward/reverse wave model. Both will give the same result. Both models involve two functions which each vary with position and time.
(a) The voltage/current model expresses the current flowing through the line, and the voltage appearing on it, at every position.
(b) The forward/reverse wave model expresses the intensity of wave propagation in the forward direction, and the intensity of wave propagation in the reverse direction, at every position.
When we are thinking about the input impedance of a transmission line, the voltage/current model is the most useful. From it we may directly derive the voltage/current ratio (that is, the input impedance) at the input to the line.
When we are thinking about a fast edge propagating down the line, it is convenient to use the forward/ reverse model. In the first instant, before the wave hits the end of the line, we have a forward wave (no reverse wave), and that simplifies the mathematics.
Either formulation can lead to the correct answer.
(**Extra for experts: In a more general waveguide, the medium may support many other modes of propagation, some of which bang back and forth across the cavity in complex 3-D patterns. A complete description of the cavity involves evaluation of the intensity of each mode. In a multi-mode fiber-optic cable, for example, there may be hundreds, or thousands, of propagation modes. In a transmission line there are only two ways to go: there and back. The transmission line mathematics therefore simplify into this two-mode formulation.)
To answer your question, I am going to start with the forward/reverse model, and then switch gears to the other model to get the final answer.
Starting with formulation (b), we imagine a driver pushing a signal into the line. The signal travels down the line toward the load. Until the reflected wave returns to the driver, we have one and only one wave on the line.
Now we will use a tricky property of the forward/ reverse model. When we have a wave traveling in one direction only, the ratio of voltage to current (i.e., the wave impedance) is constant, and equal to the characteristic impedance of the transmission line. This is the property that, in effect, defines the term "characteristic impedance".
As long as there is only one wave on the line, this property holds true. That is, until a reflected wave returns from the far end of the line, the driver, sees a constant V/I ratio. It appears to be driving a purely resistive load, with a resistance equal to the characteristic impedance of the line.
At some point the outgoing wave encounters the load, reflects, and returns to the source. When the reflected wave returns to the driver, the returning wave superimposes itself onto the outgoing wave. Here's where we switch representations to see what is going on at the driver. The equations for changing representations look like this:
V(x,t) = Fwd(x,t) + Rev(x,t)
I(x,t) = [Fwd(x,t) - Rev(x,t)]/Z0
At the driver, when we superimpose the waves, the voltages add, but the currents subtract (because the waves are flowing in opposite directions).
As you can see, when a returning wave hits the driver, the non-zero term Rev(0,t) is going to change both voltage and current values at position zero. In other words, it's going to mess up the V/I ratio at the driver. The input impedance of the transmission line/load structure may no longer look resistive, or it may look resistive with a different value. Anything could happen. Past this point in time, the input impedance of the transmission line/load structure is a function of both line and load.
Generally speaking, if the far-end load has an impedance greater than Z0, the initial effect (when the reflected wave hits the driver) is to raise the effective input impedance of the line. If, on the the other hand, the far-end load has an impedance smaller than Z0, the initial effect (when the reflected wave hits the driver) is to lower the effective input impedance of the line.
These examples are easier to work out if you imagine the line delay being much longer than the 10-90 rise time of the source. For shorter lines, the same principles apply, but the waveforms all mush together in a way that makes it difficult to see what's happening. For very short lines, the effect of the load is felt almost immediately, before the rising edge even gets to its 10value.
Dr. Howard Johnson