# Forum

## Wavelength "modulation"

22 posts by 8 authors in: Forums > CWNA - Enterprise Wi-Fi Admin
Last Post: December 14, 2011:
• First of all, please excuse me if this is not the right place to post this comment. This said...
On CWNA's Study Guide, question 11 at page 24 asks about "[i]wave properties which can be modulated to encode data[/i]". There are 4 possible answers, Amplitude, Frequency, Phase and Wavelength.
On one hand, existing modulations are ASK, FSK and PSK. [i]WSK[/i] does nor exist.
OK then.
However, we all know that [b]c = Lf[/b], where [b]c[/b] is the speed of light in the medium, [b]L[/b] is wavelength and [b]f[/b] is frequency.
Since it would be pretty hard to modulate [b]c[/b] with our current earthling technology, I think that Wavelength is [i]implicitly[/i] modulated when using FSK. In other words, Wavelength can be modulated indeed.

That's why I reckon that all ABCD answers are correct.
Please correct me if I am wrong.

• Well we know AM and FM as Amplitude and Frequency modulation types. I also know of phase change modulation.

I see what you are saying. If FM modulation is used then wavelength is changing also.
Wikipedia - [quote]For periodic waves, frequency has an inverse relationship to the concept of wavelength[/quote]

• There you go!
"[i]For periodic waves, frequency has an inverse relationship to the concept of wavelength[/i]" can be written as:

[b]f = c/L[/b] which is the very same formula I posted earlier.

• I agree that when using Frequency modulation that changing wavelengths seems to be a result.

It also seems to me that to determine if a frequency modulation has taken place, the question would be "has the wavelength changed"? Frequency changes are a result of a change in wavelengths. One would not "Change the Frequency". One is actually changing the wavelength and the frequency changes as a result.

Just my opinion but I am not an expert.... yet. :-)

• from a nitpicky, argument for the sake of argument point of view, frequency modulation is wavelength modulation. the thing is, you are modulating about .4 percent of a wavelength with FSK

In practice wavelength is used to describe the nature of broad swaths of RF that behave in similar ways, and frequency is what is modulated with FSK. in ham radio, 420 - 450 mhz is known as the 70 CM band.

when the frequency is changed via FSK, it is changed the same amount whether channel 1 or channel 14 is used. if you measured this change in wavelength, there would be different changes for each channel, expressed as a tiny percentage

which is more realistic ( using numbers selected at random to demonstrate a point, not 802.11 specific numbers )

Channel Center Frequency + and - 100 mhz

or

.0024 - .0026 wavelength ( Channel 1 )
.0023995 - .0025995 ( Channel 2 )

so if you just have to be right in a smirking teenager who thinks he is cool because he can prove that adults aren't always right kind of way, yes.

• By (Deleted User)

Yep, I agree. While making edits to the existing book, I submitted this comment to the authors a few weeks ago. Will be updated in next version.

• Wavelength and frequency are inherently related through the previously mentioned formula, involving lamda ( wavelength ), f ( frequency ) and c ( velocity of propagation through the medium involved ) . The latter does not always correspond to the speed of light, in radio communications. For example, consider the system with which TACAMO command aircraft use to communicate with nuclear submarines which are submerged at the time ( i.e. not on surface utilizing other modes of communication ).

http://en.wikipedia.org/wiki/TACAMO
http://en.wikipedia.org/wiki/Extremely_low_frequency

A few things to note from the video:

The frequency required to penetrate salt water is extremely low. Due to the fact that the frequency is so low, a very long antenna is required ( several miles long in this case ). Note that they try to get the antenna as vertical as possible. This is due to the fact that if we used a horizontal wire ( simply trailing behind the aircraft ), most of the electric field vector would be literally short circuited by the ?horizontal? sea water.
When we propagate a radio wave in an absolute vacuum ( even outer space is not considered a perfect vacuum due to space dust etc ), the signal will propagate at the speed of light. We normally see 3 times 10 to the power 8 being used as the speed of light. Even that is an approximation ( rounded off ), as the true speed of light is a little less than that:

http://en.wikipedia.org/wiki/Speed_of_light

When the signal from the TACAMO aircraft enters the salt water, refraction takes place, just as it does when we have a Wi-Fi signal. This bending of the wavefront occurs when we pass from a medium of lower density to one of higher density and vice versa ( Snell?s Law ).

When the signal hits the water, the velocity changes. The frequency remains the same however. From our formula of v = f times lamda, if the velocity has reduced, and the frequency has remained the same, the wavelength must reduce.
For our purposes in Wi-Fi, we can assume that in free space ( ingnoring the small effect due to atmospheric gases, dust and water vapor etc ) the velocity remains the same. Thus, if the frequency of the signal increases, the wavelength ?automatically? decreases. I spoke today with a friend of mine who is a physics professor about this issue. He told me the following interesting thing:

"Notice that the eye responds to frequency - because the chemical reactions in the retina depend on the energy of the incoming photons, which is proportional to their frequency - so that's why colors under water stay the same; otherwise a red object above the surface would look blue to a diver"

When pioneer radio engineers ( Armstrong etc ) began working on AM, FM etc, they had to decide whether to name the new modulation systems frequency modulation or wavelength modulation systems. As with everything to do with radio, there is very little new under the sun ( in other words, nearly every ?invention? in radio utilizes some work that has been done before in a related field ). Oscillators had been used to generate signals. In electronics, we say ( simply by historical convention ) that we change the frequency of the signal. Wavelength is changed as well, but convention dictated that frequency was the term used. Radio engineers basically used the same convention as electronics engineers and hence used the term frequency modulation. If the original developers of electronic oscillators had decided to use the term ?wavelength? instead, then radio engineers would probably have followed suit.

http://en.wikipedia.org/wiki/Edwin_Howard_Armstrong

In radio engineering, frequency shifts can occur, when the transmitter/receive pair are moving relative to one another. This occurs with satellite communication systems operating in either geostationary orbit or low-earth orbit. In low earth orbit, the satellites appear to move overhead from horizon to horizon, while in geostationary orbit, the satellite actually moves in a figure of eight type pattern due to the fact that the earth is not a perfect sphere, gravitational effects from the moon etc. This change in frequency is known as the doppler effect. It doesn?t usually concern us much in Wi-Fi except for some effect with the proposed use of Wi-Fi for vehicles exchanging data with roadside public safety equipment etc.

In summary, convention dictates that we use the term Frequency Modulation as opposed to wavelength modulation. In order to help reduce confusion among beginners, I?d recommend that the expression ?Wavelength Modulation? simply be dropped from the list of possible answers.

A while back, someone e-mailed me about FSK. I thought I?d just put a few technical notes down about a misconception about FSK, for those who are interested:

Imagine that we have an oscillator running at 1 MHz ( one million cycles per second ). Now imagine that we modulate that signal ( the carrier signal ) with a 1 KHz signal ( one thousand cycles per second ), using FSK. Many diagrams will show the output of the modulator as containing:

1. The original carrier frequency at 1 MHz

2. The upper sideband at 1.001 MHz

3. The lower sideband at 0.999 MHz.

These diagrams seem to suggest that we have simply taken two frequencies ( 1.001 MHz and 0.999 MHz ) and ?plopped them down? on the frequency spectrum.

In reality, things are more complex than that. We can look on FSK as a subset of Frequency Modulation. With frequency modulation, we utilize Bessel Functions to calculate the practical occupied bandwidth of an FM signal. This applies to FSK as well. Bessell functions will give us a number of values representative of the ouput of the modulator.

http://en.wikipedia.org/wiki/Bessel_function

In order for circuit designers to know how much bandwidth an FSK signal will occupy ( usually the double-sided Nyquist Bandwidth ), they need to know the maximum possible bit rate entering the modulator. The maximum fundamental frequency ( from Fourier analysis ) will occur when we have an alternating pattern of ones and zeroes. From this, we can determine:

When we have mark and space frequencies ( corresponding to one?s and zeroes ) on either side of the ?at rest? carrier frequency, the difference between those two frequencies does not represent the required system bandwidth as many documents and books wrongly tell us.

When an oscillator changes frequency from it?s ?at rest state? to a frequency above or below that rest value, a large range of other frequencies are produced by the modulation process. We now need to consider the modulation index of the modulator ( given by the frequency deviation divided by the modulating frequency ). From this value, we can use a Bessel Function chart and find the number of significant side frequencies that are produced by the modulator. We can then plot these values on either side of the resting frequency and use the difference between those values to give us the double sided Nyquist Bandwidth.

When we have a singer at a radio station, the FM modulator ( unlike digital TV, analog FM will be with us for a long, long time, although digital FM will slowly pick up pace ) produces a very large number of frequencies. The FCC allocates a certain amount of bandwidth to each FM station.

Usually 0.2 MHz ( 200 KHz ) of bandwidth is used ( including guard bands, just as we use in Wi-Fi ). However, the frequency range ( prior to filtering ) of the modulated singer?s voice is much higher than this. Experimentally, it was found that not all the frequencies need be transmitted for the music to be heard and ?enjoyed?. The larger the frequency bandwidth of the channel, the greater the ?fidelity? to the original music ( hence the term Hi-Fi ). However, the greater the bandwidth, the less channels can be ?fitted? in the FM spectrum of 88 to 108 MHz ( approx ). As with all radio engineering, a compromise had to be achieved between fidelity of voice/music reproduction and the number of channels that can be fitted. Eaxctly the same considerations take place with our cellphones and conventional landlines.

Dave

• Many thanks for this great explanation, Dave!
Especially the point about Doppler effect -very clear!

On the other hand, I did not mean to bother anyone. If so, please accept my apologies.

• Hi jamrb1963, welcome to the forums.

Don't take the "smirking teenager" comment to heart, I am sure it wasn't meant as dicouragement. I would class that as challenging and inquisative, it's what drives things forward. Questioning is good.

Dave1234 is one of the cleverest guys I know when it comes to this technology, the forum is lucky to have him and I have never seen a technical question that stumps him. Anything that gets an answer from Dave is a benefit to all users, so don't feel bad about asking questions - in fact, keep asking, as that's how you learn. It's OK to make mistakes.

Not that you did in this case - you were asking a fair question, so here's my input.

You question was really with regard to the practice test rather than how real world how-things-work.

Look at the big picture in that the practice tests are aimed at helping you to prepare for the exam. The exam is based on the exam objectives http://www.cwnp.com/exams/pw0104_objectives.pdf.

If you cross reference this to the study guide, you will see that you are expected to know about modulation, and the types of common modulation used, those being amplitude, frequency and phase.

That's what you need to know for the exam. The key terms and what they relate to. We don't use wavelength modulation, it's not referred to as a technology, and it's not used as a frame of reference, which is clear from looking at the study guide.

In the real world, there is not much between the physical effect of changeing wavelength vs changing frequency when it comes to the effect on a wave. Same result, different terminology.

But, think again, this is an exam prep question for the PW0-104 exam, so where you could argue your point as technically correct, it's not actually relevant to what the exam wants to test your understanding of. The fact that you do know slightly more is great, but don't let it make you lose sight of the simpler answers.

There are a lot of questions to make you think, in practice tests and in the exam, and I do encourage you to keep posting your questions. A lot of readers gain very good help by reading posts, but are sometimes nervous to ask, so by doing so you are helping the wider community.

Neil Mac
CWNT

• Thanks so much, Neil!
I will try my best to both pass the CWNA exam :-) and help others whenever I see a chance to.

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