3
\$\begingroup\$

The question is as follows:

An interfering signal from a transmitter is found to have a frequency of 57 MHz (TV Channel 2 is 54 - 60 MHz). This signal could be the:

And the options are as follows:

A) crystal oscillator operating on its fundamental
B) second harmonic of a 10 metre transmission
C) seventh harmonic of an 80 metre transmission
D) third harmonic of a 15 metre transmission

Now in my mind options B and D are the same frequency as the second harmonic of a 10 meter transmission is 30 MHz*2=60 MHz and the third harmonic of a 15 meter transmission should be 20 MHz*3=60 MHz. I don’t see how one answer could be more correct than the other.

\$\endgroup\$
9
  • 3
    \$\begingroup\$ I'm voting to close this question as off-topic because it's not a general EE question, it might be better off at ham.stackexchange.com. \$\endgroup\$
    – Marcus Müller
    Commented Nov 8, 2018 at 16:56
  • \$\begingroup\$ "harmonics" are sometimes counted differently; in some contexts (not really in this, but whatever, ham radio exam questions are full of incorrect and / or vague formulations in many countries of this earth), only the odd harmonics count; in some I could imagine you'd only count the even harmonics. I don't know – maybe this is one of these cases. \$\endgroup\$
    – Marcus Müller
    Commented Nov 8, 2018 at 17:08
  • \$\begingroup\$ Also, let me do the quick math: $$ f=\frac c\lambda = \frac{3\cdot 10^8 \frac{\text m}{\text s}}{15\,\text m }= 20\,\text{MHz,} $$ so no matter which harmonics you count, 57 MHz is not a multiple of 20 MHz and B and D are both equally wrong. \$\endgroup\$
    – Marcus Müller
    Commented Nov 8, 2018 at 17:13
  • 1
    \$\begingroup\$ I fully agree this is a stupid question, but if your frequency control is not reliably with 3 MHz (capitalization is important, as M = mega = 1 million and m = milli = 1/1000) of 57 MHz, that's a frequency error of more than 5%: If your oscillator is that unreliable, no matter what you do with it, turn it of, and repair or replace it. No discussions: that is a broken oscillator. \$\endgroup\$
    – Marcus Müller
    Commented Nov 8, 2018 at 17:18
  • 3
    \$\begingroup\$ The 15 metre band is 21 - 21.45 MHz, so the third harmonic will be above 63 MHz - too high. 10 Metres is 28 - 29 MHz, so the second harmonic will be 56 - 58 MHz - looks like B is the right answer. (where did the 20 Hz and 30 Hz come from?) \$\endgroup\$
    – Peter Bennett
    Commented Nov 8, 2018 at 18:47

1 Answer 1

Reset to default
9
\$\begingroup\$

The 15 metre band is 21 - 21.45 MHz, so the third harmonic will be above 63 MHz - too high. 10 Metres is 28 - 29.7 MHz, so the second harmonic will be 56 - 59.4MHz - looks like B is the right answer.

You have to look at the defined band frequency limits, rather than the frequency calculated from the band name.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ This is why this should be on the ham site! \$\endgroup\$
    – mike65535
    Commented Nov 8, 2018 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.