The answers to the Circuit Cellar 264 (July) Engineering Quotient are now available. The problems and answers are listed below, along with a schematic.
Problem 1a—Is it possible to transmit on-off (DC) signals between two pieces of equipment in both directions simultaneously on the same wire, in much the same way that telephones do for audio?
Source: D. Tweed, CC264
Answer 1a—Why not? Hybrids work just as well at DC as they do for audio; you just need a receiver with balanced inputs, like an RS-422 buffer:
All resistors are the same value (e.g., 4,700 Ω) and the transmit driver needs to be a voltage source (low impedance).
If the transmitter switches between, say, 0 V and 5 V, the opposite receiver will see a voltage differential of 0 V and 2.5 V, respectively, while the local receiver will just see 0V.
For long lines, you’ll probably want to use lower resistances and you’ll want to limit the slew rate of the transmitter so that the receiver doesn’t produce glitches on the transitions of the local transmitter.
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If the RS-422 receiver is replaced with an op-amp differential amplifier with a gain of 2, then any analog voltage transmitted by one end will be reproduced at the other end.
Problem 1b—But doesn’t a true hybrid use transformers, or at least some tricky transformer simulation with op amps to ensure the transmitted signal does not appear on the receive port?
Answer 1b—No. A hybrid is just a bridge circuit, with one arm of the bridge replaced by the line and the termination at the far end. The transmit signal is applied to two opposite corners of the bridge and the receive signal is taken from the other two corners.
In order to provide the Tx/Rx isolation, the bridge must be balanced, which in the example above, means that the lower resistor on each side must match the impedance of the line/far end combination. For DC and short lines, a simple resistor suffices. At audio frequencies and with the long unshielded twisted pairs used in telephony, a more complex matching impedance is required.
Transformers are used only because it’s the easiest way (and the only passive way) to get a balanced drive and/or receive signal — the transmit driver and receiver cannot share a ground. In order to mass produce phones that were dirt cheap, yet simple and reliable, the phone company figured out how to use a multi-winding transformer to provide the both the isolation and the balanced/unbalanced conversion in both directions, usually with a single resistor and capacitor to provide the line matching. As noted, modern electronic phones use active electronics to achieve the same things.
As always, the theory is simple, but the practical implementations can get complicated.
Problem 2a—The conventional way to calculate the magnitude (length) of a vector is to take the square root of the sum of the squares of its components. On small processors, this can be somewhat difficult (especially the square root operation), and various approximations are used instead.
One approximation that works surprisingly well for 2-D vectors and complex numbers is to take the absolute values of the two components, compare them, then add 1/3 of the smaller to the larger.
What is the maximum error using this method?
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Answer 2a—If we restrict the discussion to unit vectors at various angles A, the x component is cos(A) and the y component is sin(A), and the correct magnitude is 1.
Furthermore, let’s concentrate on angles between 0 and 45° — then we know that both cos(A) and sin(A) are positive and that cos(A) > sin(A). (The absolute value and compare operations provide the symmetry that covers the rest of the unit circle.) The approximation then gives the result
Magnitude = cos(A) + sin(A)/3
Graphing this shows that this is most negative (0.943) at 45° and most positive (1.054) at approximately 18.4° (the actual angle is given by atan(1/3) —can you show why?). The peak error is therefore –5.7%, +5.4%.
Problem 2b—Is there a similar formula that gives even better results?
Answer 2b—Yes. One more multiplication operation gives a result that has less than 4% error:
Magnitude = 0.960433 × max(|x|, |y|) + 0.397826 × min(|x|, |y|)
This function is most negative at 0° and 45°, and most positive at 22.5°. The error is ± 3.96%. This form is well-suited to DSPs that have multiply-accumulate units. The two constants can be expressed as 62943/65536 and 26072/65536, respectively.
Contributor: David Tweed
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