radio frequency communication systems, antenna theory and microwave devices

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TEL 213/05 Telecommunication Principle

Tutorial 3: RADIO FREQUENCY COMMUNICATION SYSTEMS, ANTENNA THEORY AND MICROWAVE

DEVICES

Semester January 2012



Radio Frequency bands showing name, frequency range and wavelength



Electromagnetic Spectrum



Example

• Calculate the wavelength of a 3MHz transmission signal



Resistance / Impedance

• Transmission line equivalent circuit



Example



Standing Waves

• If a resistive load equal to the characteristic impedance of a line is connected at the end of the line, the signal is

absorbed by the load and power is dissipated as heat.



Basic Roadmap of Unit 2



Digital Versus Analog Communication

Digital Communication Analogue CommunicationMore immune to noise as signal can be regenerated if it is below the threshold.

Less immune to noise

Has error detection and correction techniques

No error detection and correction

Compatible with time division multiplexing

Compatible with frequency division multiplexing

Smaller ICs possible with greater processing capability

Smaller ICs not possible

Can be processed using digital signal processing techniques including signal manipulation

Signal manipulation not possible with DSP

More bandwidth required Less bandwidth requiredMore complex and requires more circuitry

Less complex with less required circuitry



Sampling

• The challenge is always to represent analog signals in digital form to ease transmission.

• To do this, sampling of analog data needs to be done.

• Sampling is a process of approximation (estimation) of an analog quantity.

• After sampling, mathematical modeling can be done to represent the signal in digital form.

• Sampling must be done at regular intervals and must cover most of the data (at least twice the bandwidth frequency) to have an accurate depiction of the whole data.

• This concept is called the Nyquist-Shannon Sampling Theory

• The phenomenon called “aliasing” (misrepresentation) will happen if not enough samples are taken to represent the whole population.

Sampling Frequency = 2 * Bandwidth



Sampling an Analog Signal



Example

• If a 12 bits A/D converter were to be used, how many voltage increments are there?

• 12 bits would produce 212 or 4096 voltage increments.



Example

• Calculate the minimum voltage step increment for a 10 bit A/D converter assuming that the input voltage is from 0V to 6V.

• Number of voltage levels =2 to the power of 10=1024 voltage levels

• Number of Increments = 1024-1=1023

• Minimum voltage step increment or maximum amount of error=



Example

• An information signal to be transmitted digitally is a rectangular wave with a period of . It is given that the wave will be adequately passed if the bandwidth includes the fourth harmonic. Calculate the signal frequency, the frequency of the fourth harmonic and the minimum sampling frequency (Nyquist rate).



Noise and S/N

• Noise is an electronic signal that is a mixture of many random frequencies at different amplitudes that gets added to a radio or information signal as it is transmitted from one place to another as it is processed

• The signal-noise (S/N) ratio is also called SNR, and is an indication of the relative strengths of the signal and noise in a communication system. The stronger the signal and the weaker the noise, the higher the S/N ratio



Formula for S/N calculation



Example



Bit Error Ratio in digital Communication System

• Bit error ratio (BER) is defined as the possibility of a bit being received in error in a digital communication system



Example



Solution – ERF Table



Bit Error Rate (BER)



Example

• Find the BER of a 100kbits/s assuming unipolar transmission. The SNR is given as 1.2dB.



Data Transmission

• All data need to be converted to ASCII code first: http://www.ascii-code.com/



Value “M” being transmitted serially

M = 010011101(first 0 is not taken into account, and total is 8 bits). t is the time between each bit, known as bit interval. Bit time is the total time taken and can be expressed in bps



Example 1

What is the bit time at 230.4kbps?

st 34.410*34.4230400

1 6



No PAM

• Compute the bit rate that it will take to transmit a decimal number '201' using a bit interval of 1 microseconds using no modulation (serial transmission)



With PAM

• Repeat the transmission with 2 bits of PAM transmission and calculate the bit rate of this transmission.



PCM

• Digitizing = converting analogue signals to digital signals.• Pulse code modulation (PCM) is commonly used

The resulting 4 bit PAM of a signal is found to be 10,9,8,11. Draw the resultant PCM.

Solution10=10109=10018=100011=1011

10 9



FSK

• Frequency-shift keying (FSK) uses two sine wave frequencies are used to represent binary 0s and 1s

Binary Signal

FSK Signal



Problem with FSK



Binary Phase Shift Keying



DPSK



Line Encoding



Example

• Transmit the word “Data Com” in the transmission line and calculate the LRC/BCC and parity of VRC (odd)



Solution – Step 1

• D = 01000100

• A = 01000001

• T = 01010100

• A = 01000001

• <space> = 00100000

• C = 01000011

• O = 01001111

• M = 01001101



Solution – Step 2

A (input) B (input) Q (even parity) (odd parity)0 0 0 10 1 1 01 0 1 01 1 0 1

Q

Character D A T A C O M LRC or

BCC(LSB) 0 1 0 1 0 1 1 1 1

0 0 0 0 0 1 1 0 01 0 1 0 0 0 1 1 00 0 0 0 0 0 1 1 00 0 1 0 0 0 0 0 10 0 0 0 1 0 0 0 1

(MSB) 1 1 1 1 0 1 1 1 1Parity of VRC

(odd)1 1 0 1 0 0 0 1 0



XOR Addition

•When the number of 1's is ODD, the result of the XOR operation is '1'.

•When the number of 1's is EVEN (or none present), the result of the XOR operation is '0'.



Detecting Errors

Error Bit



Hamming Detection (FEC) Example

• The data word is 01101010. Use Hamming FEC Method to transmit this data across the transmission line



Step 1 – Hamming – find the value of n

8 bit data word to be transmitted

Experimented Value (2)

False

8 bit data word to be transmitted

Experimented Value (4)

True (therefore n=4)



Step 2: Initial Hamming Table

• Total bits required = 8+4=12 bits

• Insert the 4 required Hamming bits between the transmitted data:

12 11 10 9 8 7 6 5 4 3 2 1

H 0 1 H 1 0 H 1 0 H 1 0



Step 3:

• Determine which bits are already ‘1’12 11 10 9 8 7 6 5 4 3 2 1

H 0 1 H 1 0 H 1 0 H 1 0

Position 2 = 0010

Position 5 = 0101

Position 8 = 1000

Position 10 = 1010



Step 4: XOR Bits that are ‘1’



Step 5: Insert Final XOR into Hamming Table

12 11 10 9 8 7 6 5 4 3 2 1H 0 1 H 1 0 H 1 0 H 1 012 11 10 9 8 7 6 5 4 3 2 10 0 1 1 1 0 0 1 0 1 1 0

Final Transmitted Data



Hartley’s Law

C=2B

whereby C = channel capacity expressed in bits per second and B is the channel bandwidth

whereby the S/N is the signal to noise ratio in power.

)1)(2log( 10 N

SBC

With noise



Example – Shanon-Hartley’s Law

Calculate the maximum channel capacity of a voice-graded telephone line with a bandwidth of 3100hz and a S/N of 30dB.

10003log)10

30(log

)10/log(

log10

11

P

dBantiP

PdB

bpsC

N

SBC

31000)10(3100

1097.9)3(32.31001log32.31001log

1001log3100)10001(log3100)1(log

102

222

32

5log

56200

31000

)3100(2

31000

2log

2

2

N

antiN

B

CN

32 channels of multilayer encoding is required for a maximum channel capacityOf 31kbps with S/N of 30dB



How a pulse propagate though a transmission line



How a pulse propagate though a transmission line

• Assume that the length of the line and other characteristic incur a time delay of 500ns.

• Therefore 500ns after the switch is closed, an output pulse will occur at the end of the line.

• At this time, the voltage across the output capacitance C4 is equal to 5V or half of the supply voltage.

• The instant that the output capacitance charges to its final value of 5V, all current flow in the line ceases, causing any magnetic field around the inductors to collapse. The energy stored in the magnetic field L4 is equal to the energy stored in the output capacitance C4. Therefore, a voltage of 5V is induced into the inductor.

• The polarity of this voltage will be in such a direction that it adds to the charge already in the capacitor. Thus, the capacitor will charge to two times the applied 5V, or 10V.

•

• A similar effect then takes place in L3. The magnetic field across L3 collapses, doubling the voltage charge on C3. Next, the magnetic field around L2 collapses; charging C2 to 10V.

• The same effect occurs in L1 and C1.

• Once the signal reaches the right end of the line, a reverse charging effect takes place on the capacitors from right to left. The effect is as though the signal were moving from the output to the input. This moving charge from right to left is the reflection, or reflected wave, and the input wave from the generator to the end of the line is the incident wave.



Incident and Reflected Wave



Possibilitites

• There are 3 possibilities that can happen when a load is placed on the transmission line:– Matched Line – no standing waves / reflection occur– Shorted Line – All wave reflected back to generator. Standing

waves present and magnitude depends on length of the line (if half wavelength, reflected is exactly 180 degrees out of phase with incident wave, causing both to cancel out – no standing waves).

– Open Line – All wave radiated. All energy reflected. Standing waves are maximum.



Case 1: Matched Line



Case 2: Shorted Lines



Case 3: Open Lines



What actually happens when a line is mismatched?



Formulas for Standing Wave



Example

• Calculate the SWR if a 75 ohm antenna load is connected to a 50 ohm transmission line.

Note: If Matching occurs, SWR=1. If open circuit, SWR=infinity as all signal is reflected



Smith Chart



Resistance Circles



Resistance and Reactance Circles



Example – Smith Chart Plotting

• An antenna is connected to a 53.5Ω transmission line. And the load is 40Ω. Find the normalized impedance and plot this on the Smith Chart.



Solution



Example – Smith Chart Plotting

• An antenna is connected to a 53.5-j20Ω transmission line. And the load is 40+j30 Ω. Find the normalized impedance and plot this on the Smith Chart.



Solution



Smith Chart External Scales



Example

• The operating frequency for a 6 meter coaxial cable with a characteristic impedance of 53.5 ohm is 140MHz. The load is resistive, with a resistance of 93 ohm. What is the impedance seen by the transmitter?

The impedance variations along a line repeat for every half wavelength, and therefore for every full wavelength, for the purposes of calculation, only 0.3 is needed. The Smith Chart is normalized to the characteristic impedance of the cable which is 53.5 ohm:



Step 1:

• Find SWR=1.74 from the bottom scale. Once this is found, draw a vertical line across the Smith Chart. Then find the prime center of the graph on the real axis. From the prime center, draw a circle with your geometry set so that the perimeter of the circle would intersect the SWR=1.74 straight line drawn earlier.



Step 2:



Step 3:

• Plot the wavelength, 0.3. Since you are interested to find the impedance seen by the generator, find the scale on the chart that says "Wavelengths toward generator". Plot this measurement into the chart. Draw a radius line from the prime center towards the perimeter of the wavelength plot so that this line intersects. Find the intersection between the SWR circle and the radius line.



Step 4:



Step 5:

• The actual impedance seen by the transmitter is: .



Antenna theory- Fleming’s Left Hand Rule is applied

Fleming’s Left Hand



When there is a load…



Efficient radiation



Signal Loss – free space loss (FSL)

Important to understand how this equation is derived!



Example

• A radio wave is transmitted from point A and point B. It is given that the distance between point A and point B is 5km and the frequency of transmission is 6MHz. Assuming free-space loss only; calculate the loss experienced by the signal in dB.



Thank you!

radio frequency communication systems, antenna theory and microwave devices

Engineering

peoples university tel

signal frequency

transmission signal

information signal

electronic signal

signalnoise sn ratio

bandwidth frequency

digital communication