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In the Errata list Page 53 and 54.
To: A signal is transmitted at 100mW. After passing through a coaxial cable, it is measured with a signal strengh of 10mW. What is the loss of the cable in DB?
....
Comment:
The bolding should be on the dB answer. since the question asks for dB loss not mW. The mW loss of the cable is obviously 100mW - 10mW = 90mW.

Question:
I have no problem to understand --
100mW / 10 = 10mW
so the loss of the cable in dB is 10db.
In comment says, when we talk
about The mW loss of the cable is obviously.
100mW - 10mW = 90mW.
yes, it looks like obviously, it also tell me that
10dB is equivalent to 90mW.
Could some one tell me how to convert 90mW to 10dB????

Phelan999,
I understand what you are saying, i think you need to think of this question laid out like this:

100mW - Xdb = 10mW

Then solve for X. In this case, the question is still asking for X, which happens to be the the loss of the cable in db's.

Thanks for the reply, but you are not get my point.
I understant how to get 10dB, this is the answer
for loss of cable in dB.
I also can see the loss of calbe in mW as comment said, 100mW - 10mW =90mW
But my question is I can not understand why 10dB and 90mW is not equivalent.
Would you please point to me why the two loss of cabe values are not matching to each other, which concept I have been missed?

phelan99,

Would you agree that the difference between 1000mW and 100mW in dB is 10dB? Yet 900mW does not equal 90mW.

milli Watts are absolute units and decibels are relative units.

See WLANstan's reply to Shivany's post "When is a person ready for the CWNA?" under the CWNP Program forum.

Regards,

moe

Oops..no injury was intended to either party. I meant to address you as phelan999 and Shivanry. I apologize for the errors.

Thanks for the help! I am not worry about the typo of my name. I only want to understand the topic.
Any help is appreciated.

Thanks, Moe, for pointing out the difference between RELATIVE and ABSOLUTE measurements.

If one were to pump 10 Watts into the cable instead of 100mW, the output would be 1 Watt, so the difference is 9 Watts - but the cable loss is still 10 dB - the RATIO of the input power to output power.

You could solve this problem corectly a bunch of different ways. What usually works for me is converting power to dBm, because most loss/gain is listed in dB. Then all I have to do is add and subtract the values. But use what ever method you find easy to understand and use (that gets you the correct results).

Hi:

relative? absolute?

20 mW is absolutely twenty units of power.
20 dB is absolutely a pure number equal to 100.

"I was run over by a herd of 20 dB elephants" is the same thing as saying "I was run over by a herd of 100 elephants."

"The power has been reduced to 1/100th of its previous strength" is the same thing as saying "The power has been reduced by (a factor of) 20 dB."

I think of an RF power gain or loss first and foremost as a multiplication or division process. I don't think of it as an addition or subtraction process (or as relative versus absolute things). Ok, the factors are converted to powers of ten and then arithmetically added. Underneath it is still multiplication.

The CWNAv3 study guide RF math section goes the other way and is awfully confusing (sorry). Fortunately answering the practice exam and exam questions correctly does not require becoming comfortable with the study guide approach. Learn your RF math elsewhere than this study guide.

I hope this helps. Thanks. /criss