These are the answers to the EQ questions that appeared in Circuit Cellar 260 (March 2012).

Problem 1—In an RS-232 interface, why is the idle or “mark” level a negative voltage?

Answer 1—RS-232 was developed in the days when people were connecting electromechanical teletypes to telephone lines with modems, and in fact, at that time the design of any equipment connected to a phone line was tightly controlled by the phone company.

The reason RS-232 uses a negative voltage for its idle state is the same reason the phone lines themselves use a negative voltage relative to ground for power—copper wires in long cables potentially exposed to moisture are significantly less likely to corrode if they have a negative DC bias on them.

Problem 2—Similarly, why does the “mark” level correspond to a logical “high” level on the TTL side of the interface?

Answer 2—Again, back in the days when RS-232 was developed, the primary logic families in use were DTL and TTL. Both of these technologies draw significantly less power when a signal is in the high state than in the low state, so the high state is preferred for the inactive state of any signal.

Problem 3—What does the following C function compute? You may assume the input argument is a positive integer. Answer 3—Remember the algorithm that computes integer square roots by subtracting successive odd numbers from the input value? This function extends that concept to computing integer cube roots.

The reason this works is that taking the differences between successive values is the discrete equivalent of taking a derivative in the continuous world. The derivative of a cubic curve is a quadratic, and the derivative of a quadratic is a straight line. To generate a “straight line” in the discrete world, you just add a constant to a variable.

When you look at successive squares—0, 1, 4, 9, 16, 25, etc.—the differences are 1, 3, 5, 7, 9, etc. This is why the algorithm that subtracts odd numbers works for computing square roots.

When you look at the successive cubes—0, 1, 8, 27, 64, 125, etc.—the first set of differences is the sequence 1, 7, 19, 37, 61, etc. This doesn’t look very useful until you take the differences between those numbers, which are: 6, 12, 18, 24, etc., which is obviously another straight line.

Problem 4—Suppose you are given some calibration constants for a sensor in the form of four-digit hexadecimal (16-bit) integers, and you are told that the format of these numbers is “7 integer bits and 9 fractional bits.” How would you go about converting these constants to floating-point so that you could, for example, work with them in a spreadsheet?

Answer 4—The direct way to convert numbers in an arbitrary fixed-point representation to the equivalent floating-point value is to figure out what the representation for “1.000” would be in the fixed-point notation and then divide the given numbers by that constant.

In this case, with 7 integer bits and 9 fraction bits, “1.000” would be represented as binary 0000001.000000000, or 0x0200. So, if you are given a constant of, say, 0x5453, just divide it by 0x0200 to find out that it represents the value “42.162.”

Contributed by David Tweed (eq at circuitcellar.com)