RAE3601/201/0/2016

Tutorial letter 201/0/2016

Radio Engineering III (Theory)

RAE3601

Year Module

Department of Electrical and Mining Engineering

IMPORTANT INFORMATION:

This tutorial letter contains important information about your module.

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ASSIGNMENT 2.

QUESTION 1: INTRODUCTORY TOPICS.

1.1. Which three components make up or comprise an electronic communication system.

(3)

Answer: All electronic communication systems have a transmitter, a communication channel or

medium, and a receiver.

1.2. Describe the term Noise? (2)

Answer: Noise is the general term applied to any phenomenon that degrades or interferes with the

transmitted information.

1.3. Define what baseband transmission and broadband transmission mean? (2)

Answer: In a communication system, baseband information signals can be sent directly and

unmodified over the medium or can be used to modulate a carrier for transmission over the medium.

Putting the original voice, video, or digital signals directly into the medium is referred to as baseband

transmission.

1.4. What are the two disadvantages of digital communication? (2)

Answer: The most important is the bandwidth size required by a digital signal. With binary

techniques, the bandwidth of a signal can be two or more times greater than it would be with analog

methods. Also, digital communication circuits are usually more complex than analog circuits.

1.5. The code 10011101is being transmitted over a digital line. Draw the diagrams of a serially and

parallel transmission of the code, given that the least significant bit (LSB) is on the right and the most

significant bit (MSB) is on left. (4)

Answer:

Parallel data transmission of the code 10011101

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1.6. What is the name given to undesirable interference that is added to a signal being transmitted?

Answer: Noise

1.7. Name the technique used to extract multiple intelligence signals that have been transmitted

simultaneously over a single communication channel.

Answer: Demultiplexing

1.8. Briefly explain the difference between quantization and aliasing.

Answer: Aliasing is when the signal is sampled below the required sampling frequency. Aliasing

causes a new signal near the original to be created. This signal has a frequency of fs-fm. When the

sampled signal is eventually converted back to analogue by a D/A converter, the output will be the

alias fs-fm, not the original signal fm.

On the other hand, quantization is the conversion of a sampled signal, which is discrete in time but

continuous in value, into a signal which is discrete in value. One of the basic choices in quantization

is the number of discrete quantization levels to use. The fundamental trade-off in this choice is the

resulting signal quality versus the amount of data needed to represent each sample.

1.9. Complete the table below by providing any three examples of the types of

communication below.

Type of communication Examples

Simplex Broadcasting e.g. television.

Half duplex Walkie talkies

Full duplex Telephone such as cellular

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QUESTION 2: NOISE

2.1. Two random signal voltage generators are connected in series. The two voltage generators have

maximal rms amplitudes e1 = 1V and e2 = 10V respectively.

2.1.1. Derive the expression for the rms voltage amplitude at the terminals of this two source

network.

2.1.2. Comment on the result if the specific numerical values in this example are used.

Answer: 2.1.1. The following is the answer

2.1.2. The above result clearly illustrate that within a network that contains more than one source,

main contributor to the noise voltage is the source with the highest amplitude.

2.2 Resistors R1=20 kΩ and R2=50 kΩ are at room temperature T = 290K. For a given bandwidth of

BW = 20 kHz find the thermal noise voltage for the three cases shown in Figures 1 below:

2.2.1. Each resistor separately.

2.2.2. Their serial combination.

2.2.3. Their parallel combination

Figure 1: Resistor combinations.

Answers :

2.2.1. Each of the two stand-alone separate resistors generates thermal noise as follows:

2.2.2. Equivalent resistance Rs of the two resistors in series is:

2.2.3. Equivalent resistance Rp of the two resistors in parallel is

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Question 2.3. An amplifier with the input signal power of 5 × 10−6 W, noise input power 1 × 10-6 W,

has output signal power of 50 × 10−3 W and the output noise power 40×10-3 W.

2.3.1. Calculate the noise factor F.

2.3.2. The nose figure NF of this amplifier.

Answer : 2.3.1. By definition we find:

2.3.2. NF=10log (4) = 6dB

QUESTION 3: AMPLITUDE MODULATION TRANSMISSION

3.1. What is the main disadvantage of DSB and SSB signals?

Answer: The main disadvantage of DSB and SSB signals is that they are harder to recover, or

demodulate, at the receiver. Demodulation depends upon the carrier being present. If the carrier is not

present, then it must be regenerated at the receiver and reinserted into the signal.

3.2. Name the circuit that causes one signal to modulate another, and give the names of the two

signals applied to this circuit.

Answer: The modulator and the two signals applied to it are the intelligence and the carrier.

3.3. The antenna current of an AM transmitter is 12 A when unmodulated but increases to 13 A when

modulated. Calculate the percent modulation.

Answer:

M=0.59=59%

3.4. A 1500 kHz carrier and kHz intelligence signal are combined in a non-linear device. List all the

frequency components produced. (4)

Answer:

(a) A dc level,

(b) Components of each of the two original frequencies

(c) Components of the sum and difference frequencies of the two original frequencies.

(d) Harmonics of the two original frequencies.

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3.5. A standard AM broadcast station is allowed to transmit modulating frequencies up to 5 kHz. If

the AM station is transmitting on a frequency of 980 kHz calculate:

3.5.1. The maximum and minimum upper and lower sidebands. (2)

3.5.2. The total bandwidth occupied by the AM station.

Solution:3.5.1

fUSB=980+5=985

fLSB=980 -5=975

3.5.2. total bandwidth occupied is given by

fUSB- fLSB=985-975= 10kHZ

3.6. An antenna has an impedance of 40 V. An unmodulated AM signal produces a current of 4.8 A.

The modulation is 90 percent. Calculate:

3.6.1. The carrier power. (2)

3.6.2. The total power. (2)

3.6.3. The sideband power.

Answer:

3.7. A transmitter experiences an antenna current change from 4.8 A unmodulated to 5.1 A.

What is the percentage of modulation?

Answer:

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3.8. Which type of balanced modulator gives the greatest carrier suppression?

Answer: The balanced modulator that gives the greatest carrier suppression is the IC balanced

modulator

3.9. Which four signals and frequencies appear at the output of a low-level diode modulator?

Answer:

(a) A dc level,

(b) Components of each of the two original frequencies

(c) Components of the sum and difference frequencies of the two original frequencies.

(d) Harmonics of the two original frequencies.

What is the most critical component value in a diode detector circuit? Explain.

Answer: Semiconductor diodes have a certain turn-on voltage. Accordingly the voltage has to reach a

certain level before the diode is able to operate reasonably efficiently.

4.1. List at least five categories in which FM is used?

Answer: FM radio, VCR, Television sound, Satellite TV, Personal computer sound card, Narrowband

amateur radio, narrowband public service channels.

4.2. What determines the frequency deviation rate, or how many times per second the carrier

frequency deviates above and below its center frequency.

Answer: The amount of change in carrier frequency produced by the modulating signal is known as

the frequency deviation fd. Maximum frequency deviation occurs at the maximum amplitude of the

modulating signal. The frequency of the modulating signal determines the frequency deviation rate,

or how many times per second the carrier frequency deviates above and below its center frequency. If

the modulating signal is a 500-Hz sine wave, the carrier frequency shifts above and below the center

frequency 500 times per second.

4.3. What is the deviation ratio of TV sound if the maximum deviation is 25 kHz and the maximum

modulating frequency is 15 kHz?

Answer:

4.4 The total signal power of FM transmitter is Pt= 100W while modulation index is mf =

2.0.

4.4.1. Calculate power levels contained in all frequency components.

4.4.2. If frequency of the used modulation signal is fm = 1.0 kHz estimate the FM signal

bandwidth requirement.

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Answer :

4.5. Discuss preemphasis and deemphasis as applied to FM transmission.

Answer: Most of the content of a modulating signal, particularly voice, is at low frequencies.

In voice communication systems, the bandwidth of the signal is limited to about 3 kHz, which

permits acceptable intelligibility. To overcome this problem, most FM systems use a technique

known as pre-emphasis that helps offset high-frequency noise interference. At the transmitter, the

modulating signal is passed through a simple network that amplifies the high- frequency components

more than the low-frequency components. The pre-emphasis circuit increases the energy content of

the higher frequency signals so that they become stronger than the high-frequency noise components.

This improves the signal-to-noise ratio and increases intelligibility and

fidelity. To return the frequency response to its normal, “l at” level, a de-emphasis circuit, a simple

low-pass filter with a time constant of 75 μs, is used at the receiver.

The combined effect of preemphasis and deemphasis is to increase the signal-to-noise ratio for the

high- frequency components during transmission so that they will be stronger and not masked by

noise.

4.6. An FM signal, FMs= 2000sin[2πx108t+(2πx104)cos (πx104t)]t is applied to a 50Ω

antenna. Determine:

4.6.1. The carrier frequency.

4.6.2. The transmitted power.

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4.6.3. The modulation index.

4.7.4. The intelligence frequency.

4.6.4. The bandwidth using Carson’s rule or Bessel table.

4.6.5. Power in the largest and smallest sidebands (use Bessel table provided).

Answer:

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ASSIGNMENT 3

QUESTION 1: FREQUENCY RECEIVERS.

1.1. Discuss the difference between the limiter and discriminators as applied in FM receivers.

Answer: In FM receivers, one or more of the IF amplifier stages are used as a limiter, to remove any

amplitude variations on the FM signal before the signal is applied to the demodulator. Typically,

limiters are simply conventional class A IF amplifiers. Circuits used to recover the original

modulating signal from an FM transmission are called demodulators, detectors, or discriminators.

1.2. A certain receiver provides a voltage of 200,000 (106dB) prior to its limiter. The limiter

quietening voltage is 200mV. Determine the receiver’s sensitivity? (2)

Answer: To reach the quietening, the input must be:

1.3. Draw a diagram of a phase locked loop (PLL) and explain how it works? (8)

Answer:

A phase locked Loop

A phase-locked loop (PLL) is a frequency- or phase-sensitive feedback control circuit used in

frequency demodulation, frequency synthesizers, and various filtering and signal detection

applications. All phase-locked loops have the three basic elements, 1. A phase detector is used to

compare the FM input, sometimes referred to as the reference signal, to the output of a VCO.

2. The VCO frequency is varied by the dc output voltage from a low-pass filter.

3. The low-pass filter smoothes the output of the phase detector into a control voltage that varies the

frequency of the VCO.

The primary job of the phase detector is to compare the two input signals and generate an output

signal that, when filtered, will control the VCO. If there is a phase or frequency difference between

the FM input and VCO signals, the phase detector output varies in proportion to the difference.

1.4. Complete the frequency spectrum of an FM signal as shown below showing the values of the

lower and upper sidebands as shown in the diagram

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Figure 3. frequency spectrum of an FM signal

Answer:

1.5. Suppose you are driving along the Jan smuts avenue listening to 5FM radio. Suddenly you hear

another radio station overriding the 5FM radio. Explain the phenomenon that just happened?

(3)

Answer: The phenomenon is called Capture effect. It happens when the stringer signal locks on to the

weaker signal. As a matter of fact the receiver can be alternating between the two stations.

1.6. A PLL is setup so that its VCO free runs at 10MHz. The VCO does not change frequency until

the input is within 50kHz of 10Mhz. After that condition, the VCO follows the input to ±200kHz of

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10Mhz before the VCO starts to free run again. Determine the lock and the capture ranges of the

PLL.

Answer: 300kHz and 40kHz

1.7. Describe the process of quadrature detection.

Answer: Quadrature detection uses a phase-shift circuit to produce a phase shift of 90° at the

unmodulated carrier frequency. The most commonly used phase-shift arrangement is shown the

figure above. The frequency-modulated signal is applied through a very small capacitor (C1) to the

parallel-tuned circuit, which is adjusted to resonate at the center carrier frequency. At resonance, the

tuned circuit appears as a high value of pure resistance. The small capacitor has a very high reactance

compared to the tuned circuit impedance. Thus, the output across the tuned circuit at the carrier

frequency is very close to 90° and leads the input. When frequency modulation occurs, the carrier

frequency deviates above and below the resonant frequency of the tuned circuit, resulting in an

increasing or a decreasing amount of phase shift between the input and the output.

QUESTION 2: DIGITAL COMMUNICATIONS

2.1. List four (4) uses of pulse code modulation (PCM).

Answer: PCM is utilized to basically convert analog network signals to digital network signals. PCM

finds use in space communications, Computers, compact discs, digital telephony.

2.2 Discuss flat top sampling and natural sampling?

Answer: Natural Sampling is a practical method of sampling in which pulse have finite width equal

to τ. Sampling is done in accordance with the carrier signal which is digital in nature

while Flat top sampling is like natural sampling i.e; practical in nature. In comparison to natural

sampling flat top sampling can be easily obtained. In this sampling techniques, the top of the samples

remains constant and is equal to the instantaneous value of the message signal x(t) at the start of

sampling process. Sample and hold circuit are used in this type of sampling.

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2.3. What is quantization?

Answer: Quantization is the process of converting a continuous range of values into a finite range of

discreet values. This is a function of analog-to-digital converters, which create a series of digital

values to represent the original analog signal.

2.4. An audio is band-limited to 15kHz. What is the minimum sample frequency if this signal is to be

digitized?

Answer: 30KHz

2.5. What is quantization noise?

Answer: This is the noise or error that is introduced when the analog signal is sampled or converted

into a digital format. It ca be overcome by oversampling.

QUESTION 3: TRANSMISSION LINES 3.1. Define velocity of propagation. (2) Answer: velocity of propagation (VoP) is the ratio of the velocity of propagation of a

signal in a transmission line to its velocity in free space. However, this ration is called velocity factor (VF)

3.2. Define near end cross talk and attenuation relative to testing twisted pair

cable. (4) Answer: Twisted-pair cable specifications also include attenuation and near-end

cross talk. Attenuation means the amount by which the cable attenuates the signal. The longer the cable, the greater the amount of loss in the cable and the smaller the output. The other key specification is near-end cross talk (NEXT). Cross talk refers to the signal that is transferred from one twisted pair in a cable to another by way of capacitive and inductive coupling. Near-end cross talk is the signal appearing at the input to the receiving end of the cable.

3.3. Draw an equivalent of a transmission line and explain the physical

significance of each element.

Answer:

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When the length of a transmission line is longer than several wavelengths at the signal frequency, the two parallel conductors of the transmission line appear as a complex impedance. The wires exhibit considerable series inductance whose reactance is significant cant at high frequencies. In series with this inductance is the resistance of the wire or braid making up the conductors, which includes inherent ohmic resistance plus any resistance due to skin effect. Furthermore, the parallel conductors form a distributed capacitance with the insulation, which acts as the dielectric.

3.4. Explain how an unbalanced line differs from balanced line Answer: Transmission lines can be balanced or unbalanced. A balanced line is one

in which neither wire is connected to ground. Instead, the signal on each wire is referenced to ground. The same current flows in each wire with respect to ground, although the direction of current in one wire is 180° out of phase with the current in the other wire. In an unbalanced line, one conductor is connected to ground.

3.5. Determine the wavelength (λ) of a 100 MHz signal in free space and while

through an RG-8A/U coaxial cable whose velocity of propagation is .

3.6. Discuss the two types of losses found in transmission lines?

Answer: The attenuation is the amount of power lost per 100m of cable expressed in decibels at 100 MHz. Attenuation is directly proportional to cable length and increases with frequency. If the load at the end of a line is an open circuit or a short circuit or has an impedance other than the characteristic impedance of the line, the signal is not fully absorbed by the load. When a line is not terminated properly, some of the energy is reflected from the end of the line and actually moves back up the line, toward the generator. This creates what is known as a standing wave.

3.7. A citizen band transmitter operating at 27MHz with a 4W output is connected through a 10m of RG-8A/U cable to an antenna that has an input resistance of 300Ω. Determine the:

3.7.1 The reflection coefficient

3.7.2 The electrical length of the cable in wavelengths (λ)

3.7.3. The VSWR

Answer: =0.71

λ=7.67

VSWR=6

3.7. Derive the equation for the time required for energy to propagate through a transmission line? (See page 518 of the recommended book. Modern electronic communication 9th edition)

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QUESTION 4: ANTENNAS

4.1. Two λ/2 dipoles are separated by 50km. they are aligned for optimum reception. The transmitter feeds it antenna with 10W at 144MHz. Calculate the power received.

Answer: ( See page 615 of modern electronic communication 9th

edition) 4.2. Define a collinear array and driven array? Answer: collinear antenna array is an array of dipole antennas mounted in such a manner that the corresponding elements of each antenna are parallel and collinear, that is they are located along a common line or axis. A driven array antenna in which each individual element is connected to the transmitter or receiver through a phase shifter controlled by a computer. 4.3. An antenna has a gain of 14 dB. It is fed by an RG-8/U transmission line 25

m long whose attenuation is 3 dB/10 m at 220 MHz. The transmitter output is 50 W. Calculate: 4.3.1 The transmission line loss 4.3.2 The effective radiated power

Answer:

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4.4. Define radiation resistance and its significance.

Answer: The antenna that is radiating electromagnetic energy appears to the generator as an ideally resistive electrical load so that the applied power is consumed as radiated energy. The resistive component is called the antenna radiation resistance. This resistance does not dissipate power in the form of heat, as in electronic circuits. Instead, the power is dissipated as radiated electromagnetic energy. When the radiation resistance of the antenna matches the characteristic impedance of the transmission line, the SWR is minimum and maximum power reaches the antenna.