## Half Measures

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
R_{1}terminates 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 R_{1} to a value other than Z_{0}, 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 Z_{0}. The power, P _{i}, delivered to the transmission line under these conditions
equals V_{i}^{2}/Z_{0}. Over time equal to twice the
line delay, or 2T, the total energy, E_{i}, that the
source delivers equals:

E _{i}=2TV_{i} ^{2}/Z_{0}

At the end of one full round-trip cycle, if the line achieves stasis fully
charged to voltage V_{T} 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)CV_{T}^{2}. 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, Z_{0} (Reference 1). Making that
substitution for C produces an expression for the total energy stored at
the conclusion of a successful cycle:

E_{S}=(1/2)(T/Z_{0})V_{T} ^{2}=(1/2)TV_{T} ^{2}/Z_{0}

Equating the stored energy, E_{S}, to the supplied energy, E_{i}, reveals that the initial step amplitude, V_{i}, must equal half of V_{T}. 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*.

### Reference

**[1]**Johnson, Howard, and Martin Graham, High-Speed Digital Design: A Handbook of Black Magic, Prentice Hall, April 1993, pg 149, equation 4.40.