We Got It Wrong Last Time
     

What’s the Ratio, Kenneth?
Question posed in the June 9 issue
(Exam level: CBRE)

At 100 percent AM modulation, what is the ratio of peak antenna current to unmodulated current?

a. 16 to 1
b. 8 to 1
c. 4 to 1
d. 2 to 1
e. 1 to 1

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

I am blessed with an abundance of good, caring friends. No person deserves to have so many people who care about one’s feelings and are so supportive.

No sooner did the June 9 issue of RW Engineering Extra hit the mailboxes of my buddies than I started getting calls and e-mails, each telling me gently that the correct answer to the latest question was not among the choices. How could this happen?

Shown is a thermocouple ammeter on a meter plug inserted into the output jack as it would be to make a tower current measurement. Photo courtesy Kintronic
The explanation involves a description of how this column is written.

As I come upon an interesting issue or field a question from a client that has a distinct, compact and illustrative answer, I make a note of it for possible use here.

Among many references I use to double-check answers or cite for further reading are such tomes as the ARRL “ARRL Handbook for Radio Communications” and the classic McGraw-Hill tech series, as well as older license exam Q&A prep books. In the latter I have found many “compact questions” to develop into Certification Corner articles.

RWEE editor Michael LeClair and I run through possible choices, attempting to balance the mix of certification grade and subject areas. We also try to identify just plain interesting discussion or theme lines to help you prepare and get pumped up for the exam with an enjoyable and intriguing dissertation. Hopefully this helps the reader, and us, to learn something useful in a fun way.

Among our choices in preparing the June column was a question from an old license prep manual in my father’s library, “Radio Operating Q & A” by Arthur R. Nilson and J.L. Hornung. On Page 317 of the second printing of the eighth edition, published in 1946, was the above question on antenna current, with the answer: 2 to 1. That is, Nilson and Hornung tell us the 100 percent modulated current rises to twice the unmodulated current flowing into the antenna. Nilson and Hornung’s answer is wrong.

WHY?

But this discussion isn’t getting us to the right answer; so back to the question.

Any power transfer calculation has three factors: The resistance of the load (into which the power is sent), the voltage impressed on the load and the current of the supplied power. In this question we’re focused mainly on the current.

Consider a typical 1,000 watt AM station with an atypical 50 ohm pure resistance antenna. With just an unmodulated carrier, the current will be 4.47 amps.

If you have a line current meter on the output of your 1,000 watt AM transmitter, you will recognize that number as the line current of 1,000 watts into your 50 ohm transmission line with unmodulated carrier.

The direct formula for antenna current (where “m” is the modulation factor and 1 = 100 percent, 0.5 = 50 percent etc.) is:

How did we get there? Basic AM theory tells us that at 100 percent modulation, the power in the sidebands created will be 50 percent of the carrier power, or 500 watts in this case. Substituting 1,500 into the original formula we obtain a line current of 5.47 amps.

5.47 amps/ 4.72 amps = 1.225

So the correct answer is really 1.225 to 1 — which was not one of the selections.

Please accept my apology if this has caused you weeks of perplexity.

By the way, don’t go running out to the ammeter in your ATU and expect to see the meter peaking 1.225 times higher when your mod monitor is showing peaks at 100 percent. Most of these ammeters are thermocouple types and so have a very high hysteresis and damping factor. A little twitch up is about all you’ll see unless you introduce steady tones.

SOMETIMES THERE’S AN ERROR

Incidentally, this question was legendary for the number of times and various forms in which it appeared on the FCC Second (in the context of maritime AM) and First Class examinations — so much so that Robert Shrader addresses the subject area in his industry standard, “Electronic Communication.”

Where might the question error have popped up?

As my good buddy Cris Alexander notes, the ratio between the sum of the voltages of the carrier plus the sidebands at 100 percent (vis-à-vis modulated to unmodulated) is 2 to 1. The question may have been corrupted by Nilson and Hornung, or two questions fused.

We should mention that with the volume of test material that must be prepared for each SBE exam cycle, sometime and someday there will be some question errors. (Remember that there is no test question list for SBE exams, as there is for FCC exams; pretty much every SBE exam is a new document.)

Answer all SBE exam questions as best you can; but if you feel that a question was written in a confusing or inconclusive manner or that the correct answer wasn’t listed, point this out when you submit your exam papers to the proctor. Give that person a written short explanation along with your correct answer. Only with everyone’s help will the exam quality stay high and improve further.

If you have been studying for a ham ticket, you will have seen updates that several questions were removed from the FCC amateur question list, for the reasons noted above. A convenient place to check for questions removed from the pool is www.arrl.org/withdrawn-questions.

It’s tough to write effective exams. Our praise and admiration should be given to all those who work on the SBE certification exams.

The deadline for signing up for the next cycle of SBE certification exams is Dec. 31, 2010 for exams given at the Local Chapters between Feb. 4–14, 2011. If you’re a veteran, you may be reimbursed for the Certification Exam fee. Contact Megan Clappe at SBE headquarters for details: mclappe@sbe.org.

Missed some Certification Corners or want to review them for your next exam? Find back columns at rwonline.com/section/certification-corner.

REFERENCES:
“Schaum’s Outline of Electronic Communication,” by Lloyd Temes, McGraw-Hill, 1979, pages 1–27

“Radio Engineering,” by Frederick Terman, McGraw-Hill, 1937; section on modulation

“Electronic Communication,” by Robert Shrader, fourth edition, McGraw-Hill, 1980, page 398

“Radio Engineering Handbook,” edited by Keith Henney, fifth edition, McGraw-Hill, 1959, Beverly Dudley section

… AND I’M HERE TO HELP
Question for next time
(Exam level: CBRE)

It’s a little after 8 p.m. and your five-station cluster has settled into night automation operation. As usual, you are the last in the building. But now there are two people at the front door, holding up federal FCC IDs to the security camera. They request to inspect your station(s). Are they entitled to inspect at this late hour?

a. No, it’s a constitutional right to have an attorney present when questioned. They can come back when your attorney is present.
b. No, it’s after business hours, and meaningful management presence is absent
c. No, you’re off the clock
d. No, tours are only given by appointment
e. Yes

Rating People: 16   Average Rating:     
Comment List:

Let's murk (opposite of clear) things up a bit - Using the 1000 watts into 50 ohms example Voltage = square root of Power x Impedance 223.6v = square root of 1000w x 50 ohms Maximum voltage at 100% modulation is twice as no modulation 447.2v = 223.6 x 2 Peak voltage = R.M.S. voltage x 1.414 632.4v = 447.2v x 1.414 Current = Voltage divided by Impedance 12.65a = 632.4v divided by 50 ohms Current ratio = Peak Current divided by Average Current 2.83 to 1 = 12.65a divided by 4.47a Just playing with my calculator.
By Ralph on 10/18/2010
Probably buried in the formulas in the article is the answer to my question, but I'm just not that conversant with the theory or the math. With 100% modulated double-sideband AM I have forgotten what the ratio of peak power to unmodulated power is; 2 to 1?, 3 to 1?, 4 to 1? For example, if I had carrier output of 750 watts (such as a well-tuned class C amplifier would have generated with the old FCC amateur rules of 1 KW DC input to the final)would my transmission line/antenna need to be rated for 3 KW peak power, or perhaps some lesser amount. Oh - an so I wouldn't have to ask again once my memory fades, is there a convenient formula to compute the answer?
By Charles Frodsham on 11/25/2010
Charles, I will dig into this. To summarize the outstanding questions: With 100% modulated double-sideband AM, what is the ratio of peak power to unmodulated power? If one has carrier output of 750 watts, would the transmission line/antenna need to be rated for 3 kW peak power? or perhaps some lesser amount? Is there a convenient formula to compute the answer? What is the peak current/voltage on the line? What are the ratings of some pedestrian coax cables? I'll be back.
By Buc Fitch on 11/29/2010
No, the *peak* current with modulation is still two times carrier level. The 22.5 percent rule applies to antenna current *indicated* on an average-reading antenna ammeter.
By Anonymous on 10/21/2010

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