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# Just Sine Me in, Scotty

The latest question in our Certification Corner series

A Question of Peak Performance Question from the Oct. 17, 2012 issue
(Exam level: CRBE)

What is the mathematical relationship of the peak voltage and the RMS voltage value of a uniform triangular waveform?

A. The peak value and the RMS value in volts are numerically equal
B. The peak value is the sine value and the RMS value is the cosine value of the square root of 12 times Vp to p in volts
C. The RMS voltage value is the peak voltage value divided by the square root of 3
D.

where n is the number of cycles considered (normally based on a nominal 1 second or the waveform frequency) and k is Boozeman’s constant (after Victor Boozeman) normally rounded to 1.57 or π/2
E. RMSE = Vp times the square root t divided by T where T is the duration of the waveform (nominally ¼ of the cycle) and T is the total count of cycles or the frequency of the waveform

SBE certification is the emblem of professionalism in broadcast engineering. To help you get in the certification exam-taking frame of mind, Radio World Engineering Extra poses a typical question in every issue. Although similar in style and content to the exam questions, these are not from past exams nor will they be on future exams in this exact form.

Fig. 1: RMS Value of Uniform Triangle Wave
In our attempt to kindle interest and involvement in the Society of Broadcast Engineers’ certification process, we have been moving methodically through a selection of topics highlighting the grades and specialty certificates offered in the program. We hope this has provided our readers with the confidence that one can be successful and stimulated the desire to obtain certification.

This issue’s column is a return to basics, designed to focus and strengthen your grasp of fundamentals. Mastery of them is a notable element of the SBE CRBE exam.

VOLTAGE MEASUREMENT

Of the two voltage universes that we have at our disposal, alternating current (AC) is undoubtedly the dominant. Direct current (DC) is sort of a stepchild and is used mostly in the broadcast realm for device power and control.

The measurement of DC essentially is just one parameter, an instantaneous peak voltage, plus or minus, normally from the reference point of zero volts or ground.

In contrast, AC waveforms can only be described by mathematical analysis. It usually requires trigonometry for pure sine waves and functional derivatives for more complex, non-sinusoidal waveforms. For discussion purposes, let’s focus on pure sine waves for a moment and leave the complex waveforms for a CPBE question.

In the meantime, we’ll look around nearby for a pure sine wave to study. Let’s put our fingers in the outlet on the wall for a 60 Hz sine wave (just kidding). Fig. 1 is a drawing of a single cycle of a pure sine wave such as you would find on the power line with related proportional voltage levels. Since this wave is symmetrical (equal +/– potential) around a zero voltage axis, we could measure its peak value in one polarity only, or its peak-to-peak value.

Another way to consider it would be its average voltage. If we took the average of the entire envelope, that would be zero as the positive and negative polarities cancel out. So let’s take the average of half a cycle. A statistical iteration of a nearly infinite number of evenly phased spaced (say 1 degree of advancement) would give us an average value of 63.7 percent of the peak instantaneous value.

Already you can see that AC voltage measurements are somewhat more complex than DC, and for most of us to keep it simple, we perform our measurements with a meter that does the math for us. What most meters are actually measuring is this average value.

POWER EQUIVALENCE

But is there a numerical value that would be more useful and universal to us than this average value? A common thread between AC and DC is equivalent power.

Humans receive most of their information visually, so let’s use a visual analogy. Power up a suitable incandescent light bulb with DC and achieve a reference luminosity of, say, 10 candelas. If that bulb requires 100 volts DC to reach this point, then substituting an AC voltage we discover that a peak value of 141.4 volts would be needed to reach that same 10 candela level. The average AC value turns out to be 90 volts (141.4 x 0.637). This is problematic because if we are measuring equivalent levels of power (luminosity), we would expect to measure equivalent voltages.

To rationalize the numbers at 100 volts between DC and AC, we create a new value of 100 volts AC which, by convention, we call the “effective value” or 0.707 times the peak value. As previously mentioned, although your trusty Simpson 260 might be actually registering average voltage, the AC scales are calibrated in this effective value.

Now we can work backwards, and in doing so, we identify that an infinite number of voltage samples of the sine wave added up to the 0.707 relationship indicates the RMS value of the sine wave. The RMS value is not quite the same as the average value; it is the square root of the mean of the squares of the sample values, sometimes called the quadratic mean.

OK, SO THIS WAS A HARD ONE

In previous discussions of exam-taking strategies, one caution we urged repeatedly is to always read the question and identify what is actually being asked. In this case, our question is asking for the RMS value of a triangular wave. If we were to repeat the RMS calculation on a uniform triangular wave, we would find that effective value would be 0.577 as a multiplier or approximately 1.73 as a divisor (actual value is the square root of 3). To solve this exactly requires the solution to something called a Taylor series equation. A brute force method of calculating the value is to take an estimate of the value using known samples at even intervals. For example, if we choose nine values (see Fig. 1 again) we get that the RMS value for a triangle wave equals approximately:

By increasing the number of samples, we would find that the value converges to 0.577 (try this and see if you are curious). Fig. 1 shows how the above values are derived.

This reduced effective value makes logical sense as there is less area “under the curve” for a triangle wave when compared with a pure sine wave.

So answer (c) was the most correct.

If you’d like to explore sine and triangular waveforms more, one of the better explanation explorations from an infinite number of Internet locations is a paper by Kenneth B. Cartwright; we’ve posted the link to it at radioworld.com/links.

Once again, “Schaum’s Outline of Electric Circuits” and Dr. Frederick Terman’s multitude of textbooks thoroughly cover the subject.

CLASSIC BUC BOGUS

Our two excellent editors (and managers) let me have some fun cooking up these bogus answers. Your humble author really got out of control on this issue; all answers but the correct entry were really bogus.

Answer (a) is wrong as peak and RMS are very different values. Answers (b) and (e) read well but are gobbledygook.

Answer (d) is a super bogus entry made up of pieces and parts from the “equation” section of my Microsoft Word program. Doesn’t this equation have that official and scholarly patina?It looks devastating, but means nothing.

The next SBE certification exams will be given at NAB on April 9. Closing date for signing up for the exams is March 22. If you are interested and ready to take the exams, we strongly suggest that you sign up soonest as the next following exams are scheduled for June in the local chapters.

Remember a dream is just a dream … a goal is a dream with a plan and a deadline. So get yourself and your confreres motivated to become certified or advance a grade today.

Charles “Buc” Fitch, P.E., CPBE, AMD, is a frequent contributor to Radio World. Missed some Certification Corners or want to review them for your next exam? See the “Certification” tab under Columns at radioworld.com.

You Impede Me, Coaxially Question for next time
(Exam level: CPBE)

What factors determine the characteristic impedance of coaxial transmission line?

a. Diameters of the inner and outer conductors.
b. The speed of light divided by the wavelength of use, all over the square root of the unit ratio areas of the surface of the inner and outer conductors, if you want to be really exact.
c. The square root of the individual diameters of the inner and outer conductors multiplied together, divided by the constant K, which is the velocity factor of the manufactured material.
d. 0.707 times the equivalent impedance of a waveguide of comparable area as the outer conductor; this is an important relationship for circular to rectangular waveguide converters.
e. The fine impedance value is factory-set by the density and spacing of ceramic spacers.

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