Forum

http://www.wiley.com/WileyCDA/WileyTitle/productCd-0470438908,descCd-ERRATA.html

Thanks for posting the link to the CWNA Study Guide Errata. It currently lists the two biggest errors: the fact that two files needed for the book's lab exercises were left off the CD.

A spreadsheet called LinkBudget.xls is needed for Exercise 3.6

A jpeg image called floorplan.jpg is needed for Exercise 16.4

Both files can now be downloaded from the URL provided. We have found some other minor errors (spelling, punctuation, etc) and they will be addressed in future versions of the Errata.

There is a sentence In Page 203 on CWNP material (PWO-104): "the lower the ratio,the less resistant the signal is to interference and greater the data rate will be."

I think that maybe wrong.I think should be:

The higher the ratio,should be the less resistant the signal is to interference and greater the data rate will be?

Is that right?

KevinZhu Escribi?3:

There is a sentence In Page 203 on CWNP material ???¡¥????????PWO-104???¡¥?????¡é?€?¡ã: ???¡é?¡é?????¡­?¡°the lower the ratio,the less resistant the signal is to interference and greater the data rate will be.???¡é?¡é???????? I think that maybe wrong.I think should be: The higher the ratio,should be the less resistant the signal is to interference and greater the data rate will be?

Well, a ratio is a comparison between two (2) quantities. Usually ratios are comparing apples and oranges. In this case, bits transmitted vs. the bits encoded. I just don't like the way the sentence is worded either way and I think we will be removing the sentence in the next edition.

I brought up the same question in another post but got no response so far, so I put it here for discussion.

We know that the coding rate is the ratio between data bits and the whole transmitted bits. For instance, if the data bits per OFDM symbol are 36, and the coded bits per OFDM symbol are 48, then the coding rate will be 36:48, i.e. 3:4.

Two conclusions might be made:

(1) The higher the coding rate, the more data bits, accordingly the less redundancy bits. As a result, the less resistant the signal is to interference.

(2) The lower the coding rate, the less data bits, accordingly the more redundancy bits. As a result, the more resistant the signal is to interference.

If what I thought above is correct, and if the "ratio" does mean the "coding rate," then the sentence "the lower the ratio, the less resistant the signal is to interference" could be a bit confusing. Why not "the lower the ratio, the more resistant the signal is to interference"?

Ive been having a problem with a review question that I hope someone can clarify.

Chapter 4 - RF Signal and Antenna Concepts

Q11. In order to establish a 4-mile point-to-point bridge link in the 2.4GHz ISM band, what factors should be taken under consideration? (Choose all that apply.)

A. Fresnel zone with 40 percent or less blockage
B. Earth bulge calculations
C. Minimum of 16dBi of passive gain
D. Proper choice of semi directional antenna
E. Proper choice of highly directional antenna.

I always choose A & E, yet the answers to the review question says A & D. Yet in the chapter it states that semi directional antennas are used for short to medium range links upto 1 mile for planar antennas and up to 2 miles for yagi antennas.

Which is right?

Hi,

On Chapter 3, page 71 while talking about Decibel we have:
[quote]... Now let's go back and calculate the bels from the access point to laptop2 example by using logarithms. Remember that bels are used to calculate the ratio between two powers. So let's refer to the power of the access point as P_AP and the power of [b]laptop1 as P_L1[/b]...[/quote]

I think the text should read:
[quote]...and the power of [b]laptop2 as P_L2[/b]...[/quote]
to be consistent with the previous sentence.

The following formula also needs to be corrected to represent laptop2:
[quote]... this example would be y=log10(P_AP/[b]P_L1[/b])...[/quote]

_should read_

[quote]... this example would be y=log10(P_AP/[b]P_L2[/b])...[/quote]

And actually when in the text the authors decide to replace P_L1 with the power level they end up using P_L2 power level instead as you can read in:

[quote]... If you plug in the power values, the formula becomes y = log10 (100/1), or y = log10 (100)...[/quote]

Can you please review and confirm if I'm right or missing something?

Thanks,