A4-Signal Strength and Capacity

=⌘ A4-Signal Strength and Capacity = Main focus in the previous lectures was on propagation effects. We will first repeat the main conclusions from last lecture on electromagnetic signals, and then introduce the capacity of a system based on Shannon's theorem.

New literature:
 * J. Noll, K. Baltzersen, A. Meiling, F. Paint, K. Passoja, B. H. Pedersen, M. Pettersen, S. Svaet, F. Aanvik, G. O. Lauritzen. '3rd generation access network considerations'. selected pages from FoU R 3/99, Jan 1999
 * H. Holma, A. Toskala (eds.), "WCDMA for UMTS", John Wiley & sons, Oct 2000, selected pages

Comments


Figure: Illustrating reduction of capacity in network A (top) and blinding of phones in cell (B)

More detailed discussions on these effects can be found in the literature indicated above.

=⌘ Signal/noise ratio = $$ \mathrm{SNR} = {P_\mathrm{signal} \over P_\mathrm{noise}} $$

$$ \mathrm{SNR (dB)} = 10 \log_{10} \left ( {P_\mathrm{signal} \over P_\mathrm{noise}} \right ) $$,

where P is average power

[source: Wikipedia]
 * why talking about noise?
 * dB, $$\mbox{dB}_m,\ \mbox{dB}_a $$
 * near-far problem

=⌘ Shannon Theorem =
 * The fundamental theorem of information theory, or just Shannon's theorem, was first presented by Claude Shannon in 1948.
 * Given a noisy channel with channel capacity C and information transmitted at a rate R, then if R < C there exist codes that allow the probability of error at the receiver to be made arbitrarily small. This means that theoretically, it is possible to transmit information nearly without error at any rate below a limiting rate, C.


 * See [[File:LarsLundheim-Telektronikk2002.pdf]]: The channel capacity of a band-limited information transmission channel with additive white, Gaussian noise. This capacity is given by an expression often known as &#8220;Shannon&#8217;s formula&#8221;: $$ C = W\ \mathrm{log}_2(1+P/N) $$ [bits/s]

with W as system bandwidth, and $$ P/N = \frac{P}{N_0 W} $$ in case of interference free environment, otherwise $$ N_0 W + N_\mathrm{interference} $$, where $$N_0 = k_B T_K$$ with $$k_B$$ as Boltzmann constant and $$T_K$$ as temperature in Kelvin.

=⌘ Shannon - formula=

$$ C = W\ \mathrm{log}_2(1+P/N) $$ [bits/s]

Exercises
 * calculate capacity for W= 200 kHz, 3.8 MHz, 26 MHz, (all cases P/N = 0 dB, 10 dB, 20 dB)
 * If the SNR is 20 dB, and the bandwidth available is 4 kHz, what is the capacity of the channel?
 * If it is required to transmit at 50 kbit/s, and a bandwidth of 1 MHz is used, what is the minimum S/N required for the transmission?

[source: Wikipedia, Telektronikk 2002]

Comments


Figure: Calculation of Shannon capacity for GSM (GPRS, EDGE), UMTS (packet data, HSDPA) and 802.11b



Figure: Log_10 function and related power. The power expressed in dB is 10 times the log_10 of the normalised power.

There are also the abbreviations
 * $$dB_m$$ stands for power with respect to 1 mW. How much is 0 dB_m and 10 dB_m?
 * $$dB_a$$ Power of a sound (or music).

=⌘ Cell capacity in UMTS =

UMTS has already good spectrum efficiency with respect to Shannon.
 * Modulation schemes like QPSK and 16-QAM are applied to achieve higher bandwidth.
 * Higher modulation schemes need a higher signal to noise ration, Why?

⌘Multiple-Input, Multiple-Output: MIMO


Using multiple paths (wave propagation) in a MIMO system to enhance the bandwidth of the system.

Comments
The dominating factor in capacity is the bandwidth. If you double the bandwidth being used at a given frequency, you double the capacity. However, bandwidth is limited, both due to spectrum licensing and due to electronics. Typical electronic amplifiers have a linearity of 10%, meaning: An amplifier operating at 1.8 GHz will have 180 MHz bandwidth. If you want to achieve more bandwidth, you need to invest in more expensive receivers and transmitters, pushing the costs for the hardware.

Wifi at 2.4 GHz is a typical example using cheap hardware. The ISM band ranges from 2400 to 2485 MHz, and a receiver or transmitter needs only to linearity of 85/2400 = 3.5%

The MIMO principle allows that every antenna on the transmit side can communicate to all other antennas on the receiver side. Example: a 3 x 2 MIMO system will have 3 transmit ($$T_1, T_2, T_3$$) and 2 receive ($$R_1, R_2$$) antenna. In that case, $$T_1$$ will have waves travelling to both $$R_1$$ and $$R_2$$. The more these channels T_1 -> R_1 and T_1 -> R_2 are independent, the more information can be transferred.

The effect on capacity is given by the number of spatial streams, being the minimum of the number of antennas on transmit and receive side. In our case of a 3 x 2 MIMO systems we will have an increase of $$ min{N,M} = min {2, 3} = 2 $$.

For further readings, I recomment: http://mwrf.com/markets/understanding-benefits-mimo-technology

⌘MIMO laptop


Figure: A MIMO equipped laptop (Source:Valenzuela, BLAST project)

=⌘ Range versus SNR =

Lessions learned
Let's start What have we learned?
 * antenna characteristics and gain
 * what happens if I double the frequency (900 - 1800 - 2400 MHz)?
 * minimum GSM receiver sensitivity
 * other questions related to radio?

Range


[Source Valenzuela, BLAST project]

why is there no relation to frequency?