The horizontal axis represents various physical positions along the transmission line from the source position (at the far left) to the load (at the right). The vertical dimension represents the flow of time, beginning at t ₀, when the driver first impresses onto the line a rising step edge of amplitude A.

As the step edge interacts with various obstacles along the way, each encounter spawns a new reflected signal. The time-space diagram tracks the magnitude of all the reflection products. Each arrow is labeled according to the attenuation factors (reflection coefficients) it encounters. The four reflection coefficients ₁ to ₄ are schematically defined at the top of the figure. Assume for this simple example that all four coefficients ₁ to ₄ are small, meaning that the line is well-terminated at both ends (₁ and ₄) and that the obstacle in the middle, whatever it is, generates only mild reflections ( ₂ and ₃ ). In general, the amplitude of any step passing through obstacle n is multiplied by a factor (1+ n). For simplicity, the figure leaves out these (1+ n) terms under the assumption that, in this discussion, n is small so (1+ n) must be reasonably close to one.

The first thing you should notice about the diagram is that all the first-order products (green arrows), having bounced one time, are heading from right to left. None of these products reach the endpoint. Only second-order products (blue arrows) and higher order even-numbered products can reach the endpoint. Because each reflection attenuates the signal, the higher order products are very small. In Figure 1 the higher order products appear in gray, denoting that they are too small to worry about.

The next thing to notice is that each of the second-order products has been attenuated by two small coefficients. For example, both ₄ and ₃ attenuate the product arriving at time t₂. Both ₂ and ₁ attenuate the product arriving at time t ₃. In both cases, the surviving signal has been double-attenuated. That's the beauty of a double-end-terminated net. All second-order reflection products have been attenuated twice. It hardly matters what kind of obstacle lies in the middle; the terminators always get a chance to damp out the second-order-reflected products.

Contrast that behavior with what would happen on a plain end-terminated line. In that case, the magnitude of the coefficient ₁ would equal almost unity. (A powerful, low-impedance driver creates a reflection coefficient at the source of approximately –1.) The second-order term at time t₃ would then loom much larger.

Similarly, on a plain source-terminated line, the reflection coefficient ₄ would be practically +1, enlarging the second-order term at time t₂.

The both-ends termination atenuates all second-order reflection products, improving signal quality over any single termination. Mathematically, reducing the magnitude of both ₄ and ₁ renders your design impervious to variations in ₂ and ₃.

Of course, the big disadvantage of the both-ends termination is the half-amplitude received signal. The driver (whose source impedance matches the characteristic impedance of the transmission line) produces only a half-sized step. This half-sized step remains half-sized at the end-terminated endpoint. It takes an especially sensitive receiver to work with a both-ends-terminated transmission line.

The both-ends termination is an excellent choice for very high-speed serial links in which you anticipate encountering connectors, vias, or other impedance discontinuities in the middle of the line and for which you can afford a super-sensitive receiver.

All Publications by Dr. Howard Johnson except as noted.
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