Perky little Miss Flynn, my fifth-grade physical-education teacher, used to admonish her flock: "Never go in for half-measures; always try your hardest." I like to test rules, so as life progressed, I tried drinking as much as possible and driving as fast as my car would go. I discovered that her rule did not apply to those situations. It doesn't apply to digital engineering, either. I know a circuit where half-measures are not only acceptable, but necessary (Figure 1).
When you close switch A, a half-sized step emanates from the source. It propagates across the pc-board trace (the thick blue line) toward B. When the half-step slams into B, it finds no load to absorb any power, so the half-step bounces back, returning to the source. At the source, resistor R1terminates the reflected half-step. After that time, the outbound and reflected currents perfectly cancel at all points, bringing the structure to a stable, fully charged condition. The whole scenario takes precisely one round-trip time.
Do you suppose the outbound step must be exactly half-sized, or will any other size of waveform do the trick? If you change the source-termination resistor R1 to a value other than Z0, will the circuit work in a new way?
I say no. The proof involves the principle of conservation of energy—something that Miss Flynn may not appreciate, but in the world of physics, a good energy-balance equation easily sums up the operation of a complicated system without bogging you down in details.
Begin by calculating the total energy supplied to the transmission line. With the circuit in a state of perfect rest, assume the source conveys a step voltage into the transmission line with amplitude V i. The characteristic impedance of a pc-board trace, in the range of frequencies and geometries that high-speed digital work uses, nearly approximates a pure resistance. Until the first reflection returns from the far end, the source thinks it's driving a resistive load having a value of Z0. The power, P i, delivered to the transmission line under these conditions equals Vi2/Z0. Over time equal to twice the line delay, or 2T, the total energy, Ei, that the source delivers equals:
E i=2TVi 2/Z0
At the end of one full round-trip cycle, if the line achieves stasis fully charged to voltage VT at all points, with zero current, then all of the energy supplied during the period 2T must be stored in the distributed capacitance, C, of the transmission line. This stored energy equals (1/2)CVT2. The tricky part of this argument is knowing how to calculate the capacitance, C, which is just the line delay, T, divided by its characteristic impedance, Z0 (Reference 1). Making that substitution for C produces an expression for the total energy stored at the conclusion of a successful cycle:
ES=(1/2)(T/Z0)VT 2=(1/2)TVT 2/Z0
Equating the stored energy, ES, to the supplied energy, Ei, reveals that the initial step amplitude, Vi, must equal half of VT. A larger initial step amplitude supplies too much energy; the signal rattles about, slowly dissipating. A smaller step supplies too little; the driver takes multiple round trips to finish the job. Perfect operation of this circuit requires a precise half-measure.
Johnson, Howard, and Martin Graham, High-Speed Digital Design: A Handbook of Black Magic, Prentice Hall, April 1993, pg 149, equation 4.40.