# Radio propagation equation (A4,B1)

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Radio propagation equation (A4,B1)

 Course UNIK4700, UNIK9700 Radio propagation for mobile and wireless communications 2017/09/12 1315-1600h by Josef Noll The objective of this lecture is to explain the principles of radio propagation. What will we learn today Modes of Propagation Free Space Propagation Transmit Power EIRP Permittivity Permeability Attenuation Reflection, Scattering, Diffraction References: Propagation, Free space attenuation, propagation equation, transmit power, Attenuation, Permittivity, Permeability, EIRP, Reflection, Diffraction, Scattering, Shannon

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Forelesning fra Scheme@ifi.uio.no - IP: 10.64.127.141,

# Test yourself, answer these questions

• Convert to dB, dBm: $P_r = P_t \cdot G_t \cdot G_r \cdot L_{free}$
• What is the exact difference between permeability and permittivity?
• What is Relative permittivity? and Relative Permeability?
• Define Propagation constant

# left over from 5Sep2017

1. ) Video recording is now in place, please check on the UNIK4700 for the format of the .wmv recordings, or select from https://plus.google.com/communities/107897964181233758835
2. ) Regarding the assignments: Please check 19Sep and 26Sep for the topics to be presented

# focus 2016

• Structured presentation of the topics of the course, see Media:UNIK4700-L1-Overview.pdf
• Application case, e.g. Tanzania, Country in Africa
• Assignments and work for 2nd presentation

Focus 2015

• Presentation of Slobodan on "Propagation equation"
• Presentation of Marshed on "Radio channels in WPAN"
• also: Shannon, Capacity
• define 2nd presentation

Title
UNIK4700/UNIK9700 Signal and Capacity
Author
Josef Noll,
Footer
Radio propagation equation (A4,B1)
Subfooter
UNIK4700/UNIK9700

# ⌘ UNIK 4700: Radio & Mobility

3rd lecture block

• Free space propagation equation, what is the power level at a receive antenna, see Media:PropagationConstant.pdf for a more detailed explanation of the terms being used here
• Capacity of mobile and wireless propagation,

# ⌘ Signal Strength and Capacity

Building .... Networks
History, Now and Future
History
Pioneers: Maxwell, Hertz,...
1G, 2G,... 5G networks
Frequencies and Standards
Future Challenges
A-Basics of Communication
Electromagnetic Signals
Digital communication: Signal/Noise Ratio
Signal strength and Capacity: Shannon
B-Antennas and Propagation
Free Space Propagation
Antennas, Gain, Radiation Pattern
Multipath Propagation, Reflection, Diffraction
Attenuation, Scattering
Interference and Fading (Rayleigh, Rician, …)
Mobile Communication dependencies
C-Propagation models
Environments (indoor, outdoor to indoor, vehicular)
Outdoor (Lee, Okumura, Hata, COST231 models)
Indoor (One-slope, multiwall, linear attenuation)
D-System Comparison
Proximity: RFID, NFC
Short Range: ZigBee, Bluetooth, ANT+,...
WLAN/Wifi/802.11...
Mobile: GSM, UMTS, IMT-A (WiMAX, LTE)
E-Mobility
Mobile Network mobility
IP mobility
F-Network Building
5G and Future Networks
5G Heterogeneous Networks
Basic Internet
Video Distribution Networks
Coverage simulations
Coverage simulations
Traffic simulations
Network Capacity simulations
Building .... Networks

# ⌘ 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

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

• why talking about noise?
• dB, $\mbox{dB}_m,\ \mbox{dB}_a$
• near-far problem

[source: Wikipedia]

# ⌘ 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 “Shannon’s formula”: $C = W\ \mathrm{log}_2(1+P/N)$ [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]

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.

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

 $R_\mathrm{max}=\mathrm{log}_2(1 + SNR)$ [Source:Valenzuela, BLAST project]

### 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?

## Relation of Bit Error Rate and SNR

 The Bit Error Rate (BER) describes the statistics of bits being "false" due to propagation effects. BER is typically described in BER = 10-3. Typical values for reasonable communications are: voice: BER < 10-3 data: BER < 10-6

# ⌘ Free Space Propagation

Building .... Networks
History, Now and Future
History
Pioneers: Maxwell, Hertz,...
1G, 2G,... 5G networks
Frequencies and Standards
Future Challenges
A-Basics of Communication
Electromagnetic Signals
Digital communication: Signal/Noise Ratio
Signal strength and Capacity: Shannon
B-Antennas and Propagation
Free Space Propagation
Antennas, Gain, Radiation Pattern
Multipath Propagation, Reflection, Diffraction
Attenuation, Scattering
Interference and Fading (Rayleigh, Rician, …)
Mobile Communication dependencies
C-Propagation models
Environments (indoor, outdoor to indoor, vehicular)
Outdoor (Lee, Okumura, Hata, COST231 models)
Indoor (One-slope, multiwall, linear attenuation)
D-System Comparison
Proximity: RFID, NFC
Short Range: ZigBee, Bluetooth, ANT+,...
WLAN/Wifi/802.11...
Mobile: GSM, UMTS, IMT-A (WiMAX, LTE)
E-Mobility
Mobile Network mobility
IP mobility
F-Network Building
5G and Future Networks
5G Heterogeneous Networks
Basic Internet
Video Distribution Networks
Coverage simulations
Coverage simulations
Traffic simulations
Network Capacity simulations
Building .... Networks

# ⌘ Maxwell's Equation in a source free environment

Source free environment and free space:

$\nabla \cdot \vec E = 0 \qquad \qquad \qquad \ \ (1)$

$\nabla \times \vec E = -\frac{\partial}{\partial t} \vec B \qquad \qquad (2)$

$\nabla \cdot \vec B = 0 \qquad \qquad \qquad \ \ (3)$

$\nabla \times \vec B = \mu_0 \epsilon_0 \frac{\partial}{\partial t} \vec E \qquad \ \ \ (4)$

where div is a scalar function
$\mbox{div}\,\vec v = {\partial v_x \over \partial x} + {\partial v_y \over \partial y} + {\partial v_z \over \partial z} = \nabla \cdot \vec v$
and curl is a vector function
$\mbox{curl}\;\vec v = \left( {\partial v_z \over \partial y} - {\partial v_y \over \partial z} \right) \mathbf{i} + \left( {\partial v_x \over \partial z} - {\partial v_z \over \partial x} \right) \mathbf{j} + \left( {\partial v_y \over \partial x} - {\partial v_x \over \partial y} \right) \mathbf{k} = \nabla \times \vec v$

[Source: Wikipedia]

# ⌘ Wave equation

Taking the curl of Maxwell's equation

$\nabla \times \nabla \times \vec E = -\frac{\partial } {\partial t} \nabla \times \vec B = -\mu_0 \varepsilon_0 \frac{\partial^2 \vec E } {\partial t^2}$

$\nabla \times \nabla \times \vec B = \mu_0 \varepsilon_0 \frac{\partial } {\partial t} \nabla \times \vec E = -\mu_o \varepsilon_o \frac{\partial^2 \vec B}{\partial t^2}$

yields the wave equation:
${\partial^2 \vec E \over \partial t^2} \ - \ {c_0}^2 \cdot \nabla^2 \vec E \ \ = \ \ 0$

${\partial^2 \vec B \over \partial t^2} \ - \ {c_0}^2 \cdot \nabla^2 \vec B \ \ = \ \ 0$

with $c_0 = { 1 \over \sqrt{ \mu_0 \varepsilon_0 } } = 2.99792458 \times 10^8$ m/s

[Source: Wikipedia]

# ⌘ Homogeneous electromagnetic wave

A single frequency electro (E)-magnetic (B) wave is described by

$\vec E(\vec r) = E_0 e^{j(\omega t - \vec k \cdot \vec r) }$,

$\vec B(\vec r) = B_0 e^{j(\omega t - \vec k \cdot \vec r) }$,

[Source: Wikipedia]

where

• $\vec r = (x, y, z)$ and $\vec k = (k_x, k_y, k_z)$ so?
• $j \,$ is the imaginary unit
• $\omega = 2 \pi f \,$ is the angular frequency, [rad/s]
• $f \,$ is the frequency [1/s]
• $e^{j \omega t} = \cos(\omega t) + j \sin(\omega t)$ is Euler's formula

with the group velocity (free space = speed of light) $c = { c_0 \over n } = { 1 \over \sqrt{ \mu \varepsilon } }$ and
the refraction index $n = \sqrt{ \mu \varepsilon \over \mu_0 \varepsilon_0$

• What is the difference between a static and a dynamic field
• Develop the relations for a plain wave
• Assume a plane wave: $E_x, H_y$. Show that $\frac{E_x}{H_y}=Z_0 = \sqrt{\mu_0/\varepsilon_0}$

The differences between a plane wave, a cylindrical wave and a spherical wave is defined through the surface of wave. Assume propagation in z-direction

• a wave having no variation in x,y direction is called a plane wave, and is represented as e.g. $\vec E(\vec r) = E_x e^{j(\omega t - k_z \cdot z) }$,
• a wave looking like a "cylinder" is called a cylindrical wave. Such a wave is typically generated by a line source, or the diffraction at the edge of a roof
• a wave looking like a "ball" is called a spherical wave. Such a wave is typically generated by a point source, e.g. a short antenna. If you move long away from the wave, then the spherical wave will become a plane wave. We talk about the far-field of an antenna, when the wave is treated like a plane wave. This is typically expressed as a phase variation being less than 90 degrees: $\varphi < 90\deg$

Q: More about wave equation in physics

A: See the good examples at http://www.physicsclassroom.com/class/waves/Lesson-2/The-Wave-Equation

Q: I still don't understand the propagation equation

A: The propagation equation describes how a wave propagates. It addresses the change from an electrical field indicating a magnetic field and vice versa. See the video from Bucknell University: https://www.youtube.com/watch?v=xkG86pwaOH0

## ⌘ Free space propagation

Power received in an area in a distance R from transmitter:

• area of a sphere is $A_s = 4*\pi*R^2$
• power transmitted from isotropic antenna is $P_t$
• antenna area of receiver is $A_r = \lamda^2/4\pi$
• power received in A_r = P_r

$P_r = P_t * A_r / A_s = P_r = P_t * A_r / (4*\pi*R^2)$

thus

$P_r = P_t \ G_t\ G_r\ \left (\frac{\lambda}{4\pi r} \right )^2\cdot$

• convert into dB
• provide examples for f = 10 MHz, 1 GHz, 100 GHz
• discuss influences on radiation pattern

Develop the propagation equation, see (http://www.antenna-theory.com/basics/friis.php)

How much is 0 dB_m and 10 dB_m?

• Convert dBm to mW is: mW = 10^(x/10), x = number of dBm
• Convert mW to dBm is: dBm = 10*log10(y), y = number of mW

So you get:

• 0 dBm = 10^(0/10) = 1 mW
• 10 dBm = 10^(10/10) = 10 mW

Free space attenuation $L = 92,4 + 20 \log(d \mathrm{[km]}) + 20 \log(f \mathrm{[/GHz]})$