Tutorial Transmission 2011

Tutorial Wireless Transmission

Copyright Dr K.Sandrasegaran

Page 1 15/12/2011

A number of questions in this tutorial are based on Stallings Wireless Communications and Networks

Chapter 5 which is given in UTS Online.

Tutorial 1 (Frequency Bands)

Q. Complete the following table Band Half-wave Antenna

sizes

Propagation Mechanism Applications

VLF =

Very Low

Frequency

3 kHz 30

kHz

50km-5km Limited Human speech, Music

LF =

Low Frequency

MF =

Medium Frequency

HF =

High Frequency

VHF =

Very High

Frequency

UHF =

Ultra High

Frequency

SHF =

Super High

Frequency

EHF =

Extra High

Frequency

Q. What are the benefits and challenges that arise as we move to higher frequencies in wireless

communication?

Q. Why is there a need to regulate wireless transmission?

Tutorial 1 (Signals)

Q. Distinguish between the terms spectrum, bandwidth, and data rates.

Q. Define the following: Fundamental Frequency, DC Component, Bandwidth, Carrier Frequency, and Period.

Q. a) What are the x and y axes in the following domains?

X axis Y axis

Amplitude domain

Frequency domain

Phase domain

b) Draw a sinusoidal signal in the amplitude, frequency and phase domain.

c) Explain how each domain is used by an engineer in the field?

Tutorial 1 (Antennas)

Q. Briefly define the following terms: isotropic radiator, radiation pattern, omni-directional antenna,

sectorized antenna, switched diversity , diversity combining , dipole , smart antenna.

Q. Draw the top-view of the radiation pattern of an omnidirectional antenna, directional antenna, and

sectorized antenna? For the directional antenna, indicate the main lobe, side lobes, max transmit power, beam

width and 3dB half power of the antenna.

Q. The gain of antenna is defined the equation:

Define all of the terms in the equation. 2

2

2

44

c

AfAG ee

==



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Q) Find the optimum wavelength and frequency for a half-wave dipole of length 10m. Ans=15 MHz

Q) The audio power of the human voice is concentrated at about 300 Hz. Antennas of the appropriate size for

this frequency are impracticably large, and hence the need to modulate to a higher (carrier) frequency for

which the antenna size is smaller.

a. What is the length of an half wavelength antenna for sending radio at 300 Hz? Ans=500 km b. With a half-wave antenna of length 1 m, what carrier frequency would we use? Ans= 15OMHz

Q) If a source of electromagnetic energy is placed at the focus of the

paraboloid antenna, and if the paraboloid is a reflecting surface, then the

wave will bounce back in lines parallel to the axis of the paraboloid. To

demonstrate this, consider the parabola y2 =2px shown in Figure 5.16. Let

P(x 1,y1) be a point on the parabola, and PF be the line from P to the focus.

Construct the line L through P parallel to the x-axis and the line M tangent

to the parabola at P. The angle between L and M is and the angle

between PF and M is . The angle is the angle at which a ray from F strikes the parabola at P. Because the angle of incidence equals the angle

of reflection, the ray reflected from P must be at an angle to M. Thus, if we can show that = , we have demonstrated that rays reflected from

the parabola starting at F will be parallel to the x-axis.

a) First show that tan( )= (p/y1) Hint: Recall from trigonometry that the slope of a line is equal to the tangent of

the angle the line makes with the positive x direction. Also recall that the slope of the line tangent to a curve at

a given point is equal to the derivative of the curve at that point.

b) Now show that tan( )= (p/y1), which demonstrates that = .Hint: Recall from trigonometry that the

formula for the tangent of the difference between two angles l and 2 is tan( 2 - l) =(tan 2 - tan l) / (1 + tan 2 * tan 1).

Q) For each of the antenna type listed below, what is the effective area and gain at a wavelength of 30 cm? Repeat

for a wavelength of 3 mm. Assume that the actual area for the horn and parabolic antennas is '.

Tutorial 1 (Signal Propagation)



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Q) It is often more convenient to express distance in km rather than m and frequency in MHz rather than Hz.

Rewrite the path loss equation using these dimensions: LdB = 20log(f) + 20log(d) 147.56dB

Q) Assume that two antennas are half-wave dipoles and each has a directive gain of 3 dB. If the transmitted power

is 1W and the two antennas are 10 km apart, what is the received power? Assume that the antennas are aligned so

that the directive gain numbers are correct and that the frequency used is 100 MHz. Ans=-86.4 dBW

Q) Suppose a transmitter produces 50 W of power.

a. Express the transmit power in units of dBm and dBW. Ans= 17dBW, 47dBm b. If the transmitter's power is applied to a unity gain antenna with a 900-MHz carrier frequency, what is the

received power in dBm at a free space distance of 100 m? Ans= -24.52 dBm

c. Repeat (b) for a distance of 10 km. Ans= -64.52 dBm d. Repeat (c) but assume a receiver antenna gain of 2. Ans= -61.52 dBm

Q) A microwave transmitter has an output of 0.1 W at 2 GHz. Assume that this transmitter is used in a microwave

communication system where the transmitting and receiving antennas are parabolas, each 1.2 m in diameter.

a. What is the gain of each antenna in dBs? Ans= 25.46 dB

b. What is the EIRP (effective radiated power) of the transmitted signal? Ans=15.46 dBW

c. If the receiving antenna is located 24 km from the transmitting antenna over a free space path, find the

available signal power out of the receiving antenna in dBm. Ans= -55.14 dBm

Q.) [Haykin]

a) What is receiver sensitivity? What factors affect the receiver sensitivity?

b) A commercial mobile receiver for data transmission has a sensitivity of -90 dBm. Assuming a 100-milliwatt

transmitter and free-space path loss between the transmitting and receiving isotropic antennas, what is the radius

of the service area of this receiver at a transmission frequency of 800 MHz? Ans= 9.2 km

Q) If there are no intervening obstacles, the optical line of sight can be expressed as d = 3.57 h ,where d is the

distance between an antenna and the horizon in kilometres and h is the antenna height in meters. Using a value for

the earth's radius of 6370 km, derive this equation. Hint: Assume that the antenna is perpendicular to the earth's

surface, and note that the line from the top of the antenna to the horizon forms a tangent to the earth's surface at

the horizon. Draw a picture showing the antenna, line of sight, and earth's radius to help visualize the problem.

Q) Determine the height of a TV antenna that must be able to reach customers up to 80 km away. Ans= 378m

Q. Fresnel Zone: Draw a diagram to illustrate the concept of Fresnel Zones. As a general rule of thumb, radio

planning engineers must keep the first Fresnel Zone free of obstructions in order to obtain transmission under free

space conditions. The radius of the first Fresnel Zone is given by the equation R1 = sqrt ( d1 d2/( d1 + d2) )

Given that the separation of the transmit and receiver is 500m and the height of the antennas are 10m, compute R1 at points that are intervals of 100m from the transmitter. What is the maximum height of an obstacle at each point?

Transmitter

Distance (d1)

100m 200m 300m 400m 500m

R1

Obstruction

Height



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Tutorial 1 Transmission Impairments

Q. Briefly define the terms attenuation, fading, multipath propagation, absorption, ISI, shadowing,

Q) What is the thermal noise and SNR of a 10 kHz channel carrying 1000 watts of power operating at 50C?

Answer: -163.5 dBW, 193.5 dB.

Q. If the received signal level for a particular digital system is 151 dBW and the receiver system effective noise

temperature is 1500 K, what is Eb/No for a link transmitting 2400 bps?

Answer 12 dBW

Tutorial 1 (Simple Link Planning) Q UTS Kuringai and Broadway are to be linked by a microwave communication link for transmission of video lectures with

the following specification:

Radio carrier frequency = 1 GHz

Channel bandwidth = 10 MHz

Length of the link = 20 km

Velocity of propagation = 3 x 108 ms

-1

Maximum permissible transmit power = 100 W (output of

transmit antenna) Equivalent channel noise temperature = 17 deg Celsius

Modulation technique : QPSK

Assume unit gain antennas.

a) Compute the wavelength of the transmission. 0 .3m

b) Compute the transmit power in dBW and dBm. 20 dBW = 50dBm

c) Compute the free space or path loss between transmitter and receiver in dB. 118.5 dB

d) Compute the received signal power level in dBW. - 98.5 dBW

e) Compute the channel thermal noise power in dBW. -134 dBW

f) Compute the signal to noise ratio at the receiver and prove that it is 35.5 dB. Express this as a power ratio. 35.5 dB3548.13

g) Compute the maximum data rate that can be achieved using the Nyquist capacity criteria. 40 Mbps

M= 4

h) Compute the maximum data rate that can be achieved using the Shannon criteria. 118 Mbps

i) What is the maximum data rate that can be transmitted on this microwave link? 40 Mbps

Can the maximum data rate on this link be improved. If so how ?

Q. Repeat the above question for a satellite link has the following specifications:

Operating Frequency = 5 GHz

Channel bandwidth = 100 MHz

Distance from Earth = 38500km

Transmit Power = 8 Watt

Transmit Antenna Gain = 12 dBi

Receiver Antenna = 50m parabolic

Receiver antenna efficiency = 79%

Equivalent Receive Noise Temp = 85 Kelvin

Velocity of propagation = 3 x 108 ms-1

Modulation technique : 8-QAM

Tutorial 1 (Mutiplexing)

Tutorial 1 (Modulation)

Please fill the following items into the appropriate place in the following table: AM, FM, PM, ASK, FSK, PSK,



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FDM, TDM, CDM, DSSS, FHSS

multiplexing techniques

digital modulation techniques

analog modulation techniques

spread spectrum techniques

Draw phasor diagrams for the following systems and state the number of bits per symbol in each diagram. [8

Marks]

BPSK

QPSK (UMTS) 8-PSK (EDGE) 16-QAM (HSPDA)

Q. Identify the following modulation schemes using the figure below and write short notes on it:

Q. The Figure below shows how the MSK signal is generated. Sketch the MSK output for 1001101



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Tutorial Transmission 2011