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 Communication
More 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 required
More 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

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

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 µ34.410*34.4
230400
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.
Solution
10=1010
9=1001
8=1000
11=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 1
0 1 1 0
1 0 1 0
1 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 0
1 0 1 0 0 0 1 1 0
0 0 0 0 0 0 1 1 0
0 0 1 0 0 0 0 0 1
0 0 0 0 1 0 0 0 1
(MSB) 1 1 1 1 0 1 1 1 1
Parity 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 1
H 0 1 H 1 0 H 1 0 H 1 0
12 11 10 9 8 7 6 5 4 3 2 1
0 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
S
BC +=
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
S
BC
31000)10(3100
1097.9)3(32.31001log32.31001log
1001log3100)10001(log3100)1(log
102
222
==
≈===
=+=+=
32
5log
5
6200
31000
)3100(2
31000
2
log
2
2
=
=
====
N
antiN
B
C
N
32 channels of multilayer encoding is required for a maximum channel capacity
Of 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. I
f 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.

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

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