WiMAX RF Engineer Certification Material Chapter 2, Review RF Fundamental

After my first self learning modul on RF WiMax Network, this is the second modul that I’ve promised you on my first post. Well.. let’s we start..

Power Levels calculation in dBm, Watts, and ìV/m

• dB stands for Decibels, it’s a measurement unit to measure power ratio (P2/P1, where P2 is the power being measured, and P1 is the reference). This measurement unit is also valid for voltage ratio (V2/V1).
• dBm stands for dB for reference 1 mW (Audio, 600 Ohm) or 1 mV (Television, 75 Ohm) or 1 mW (Radio Frequency, 50 Ohm), the last one is what we aim to learn… right? :p
• We can describe both expressions above in formula, as:

• about ìV/m, until now I couldn’t  find any good reference on this matter. Please share to me if you have one.
• I have several links that maybe useful for us to understand this concept deeper :
  Decibel watts, dB conversion, Conversion (these links is also my resource in writing this post )

Bandwidth channel, and WiMax adaptive modulation and coding calculation, using nyquist and shannon observations.

Well… for everyone have ever heard or read about shannon and nyquist observation theorem, you may go to next paragraph. ^_^

Nyquist theorem(quoted from here):

A theorem stating that when an analogue waveform is digitised, only the frequencies in the waveform below half the sampling frequency will be recorded. In order to reconstruct (interpolate) a signal from a sequence of samples, sufficient samples must be recorded to capture the peaks and troughs of the original waveform. If a waveform is sampled at less than twice its frequency the reconstructed waveform will effectively contribute only noise. This phenomenon is called “aliasing” (the high frequencies are “under an alias”). This is why the best digital audio is sampled at 44,000 Hz – twice the average upper limit of human hearing. The Nyquist Theorem is not specific to digitised signals (represented by discrete amplitude levels) but applies to any sampled signal (represented by discrete time values), not just sound.

where  is the highest frequency at which the signal can have nonzero energy.

C is channel capacity -maximum rate of information carried-; B is the Bandwidth of the channel in hertz; S is the signal power; and N is the noise power. We can conclude that the maximum rate of information carried depends on bandwidth, signal level, and the noise level.

with equation above, we can calculate Bandwidth -read: capacity- of signal with certain Signal to Noise Ratio.  For the complete equation you can follow the link above.

Oke, now we will have a discussion about Bandiwdth Channel and WiMAX adaptive modulation and coding equation.

My interpretation about Bandwidth Channel here is about question: how much data will a channel carry in one second? -channel capacity-

So, from reading shannon equation above there are logical relationships between Bandwidth and Quality S/N -Signal to Noise Ratio- in determining Capacity.

1. The greater Bandwidth occupied, then bigger Capacity will produced.
2. Bigger Capacity, needs bigger S/N (quality) level. This means bigger Power is needed.
3. From 1st and 2nd expression above, we may conclude that there’s a trade off between Bandwidth and Power due to Capacity issue.

as we know that, Eb= S/C -> S=Eb*C and No=N/B -> N=No*B so Eb/No=(Eb*C)/(No*B), the shannon equation might be written as:

the relationship above gives:

the expressions above lead us to 2 conditions on telecommunication system:

1. Bandwidth limited System
2. Power limited System

Picture is taken from: gaussianwaves.blogspot.com, if you wanna read the complete article you can folow this link.

Please note, for M-ary Adaptive Modulation System, Es/No=(Eb/No)*k -> where k equals to code_rate*log(base2)(M), assume code rate = 1-> Es/No=SNR*(Tsymbol/Tsampling), so we can rewrite relation above as:

(Eb/No)*k = SNR*(Tsymbol/Tsampling)

Eb/No = SNR*(Tsymbol/Tsampling)/k

Eb/No(dB)= SNR (dB) + 10 log(base10) (Tsymbol/Tsampling) – log(base2)k …..1)

And if Encoding is added on the system, then we should calculate the code rate and the coding gain.

Encoding aims to make the channel robust from error,  therefore required Eb/No increases as effect of coding gain for certain required BER.

which lead to equation …1)

Eb/No(dB)= SNR (dB) + 10 log(base10) (Tsymbol/Tsampling) – log(base2)k

where k equals to code_rate*log(base2)(M).

Maybe we need a real calculation example, for further comprehension.

Example: is there anyone have? hehehe

well.. I’ll try to find it first… salam.