Principles of Communications Lecture FINAL

Principles of CommunicationsBSECE IV - 2

PRINCIPLES OF COMMUNICATIONS

Communication – it is the process of sending and receiving messages or information from one location to another via some communication link.

Components of a Communication System

1. Transmitter – it is the source of information. It processes the information so as to make it suitable for transmission and subsequent reception. It performs encoding and modulation.

2. Transmission Channel – provides a means of transporting signal between transmitter and receiver.

3. Receiver – a device that accepts the transmitted signals from the transmission medium and then converts back to their original form. It performs decoding and demodulation.

Brief Historical Background

1820 –Danish physicist H. C. Oersted showed that an electric current produces magnetic field.

1831 –British physicist Michael Faraday discovered that the magnet in motion can generate electricity.

1837 –Samuel Morse invented the Telegraph. 1864 –James Clark Maxwell predicted the Theory of Electromagnetism

which became the theoretical basis of radio. 1876 –Alexander Graham Bell invented the Telephone. 1887 –Heinrich Rudolf Hertz demonstrated the effect of radio in space. He

showed that radio waves can be reflected, refracted, diffracted, etc. 1893 –Nicola Tesla outlined the basic principle of radio transmission and

reception. He saw the possible use of radio waves in long distance wireless communication.

1894 –Guigliermo Marconi invented the Marconi Antenna. He provided the f irst complete system of wireless communication.

1906 –Lee de Forest invented the triode vacuum tubes which provide the first tonn of practical electronic amplification and really opened door for

wireless communication. 1920 –Commercial radio broadcasting began in Pennsylvania. 1931 –Edwin Howard Armstrong patented FM. 1935 –Monophonic FM commercial broadcasting began. 1941 –1945 – Popular use of TV Broadcasting. 1948 –Bell lab scientists Shockley, Brattain, and Bardeen invented the

Transistor. 1951 –TV Broadcasting reached Philippines shore. 1953 –Color TV system was adopted. 1957 –Russia launches the first world satellite called Sputnik. 1958 –Kilbey and Noyce develop the first IC. NASA launches their first

satellite. 1961 –Stereo Broadcasting in FM band was adopted.

CommunicationREASONS FOR MODULATION To maintain the frequency of the signal To prevent any unwanted energy to interfere with the signal information To reduce the required antenna length For multiplexing To maintain equipment limitation

METHODS OF ANALOG MODULATION Amplitude modulation Angle modulation

Frequency modulation Phase modulation

CommunicationLIMITATION OF COMMUNICATION SYSTEM Noise (If noise level becomes too high, information is lost) Bandwidth of frequency allocated for the transmitted signal

BASIC CONCEPT OF COMMUNICATIONModulation

process of putting low-frequency information (audio signal) onto high frequency carrier (radio signal) for transmission. Transmission takes place at the high frequency which has been modulated to carry low frequency information.

Demodulation process of separating the low-frequency information signal and high

frequency carrier.

Radio Spectrum Frequency Wavelength

ELF(Extremely Low Frequency) 30 Hz to 300 Hz Megametric Waves

VF(Voice Frequency) 300 Hz to 3 KHz Hectokilometric Waves

VLF(Very Low Frequency) 3 KHz to 30 KHz Myriametric Waves

LF(Low Frequency) 30 KHz to 300 KHz Kilometric Waves

MF(Medium Frequency) 300 KHz to 3 MHz Hectometric Waves

HF(High Frequency) 3 MHz to 30 MHz Decametric Waves

VHF(Very High Frequency) 30 MHz to 300 MHz Metric Waves

UHF(Ultra High Frequency) 300 MHz to 3 GHz Decimetric Waves

SHF(Super High Frequency) 3 GHz to 30 GHz Millimetric Waves

EHF(Extremely High Frequency) 30 GHz to 300 GHz Centimetric Waves

Infrared Light 300 GHz to 3 THz Decimillimetric Waves

Infrared Light 3 THz to 30 THz Centimillimetric Waves

Infrared Light 30 THz to 300 THz Micrometric Waves

Visible Light 300THz to 3 PHz Decimicrometric Waves

Ultraviolet Light 3 PHz to 30 PHz

X-rays 30 PHz to 300 PHz

Gamma Rays 300 PHz to 3 EHz

Cosmic Rays 3 EHz to 30 EHz

DECIBELS

• The range of powers, voltages and currents encountered in radio engineering is too wide to be expressed on linear scale. Consequently, logarithmic scale based on the decibel (dB, one tenth if a Bel ) is used. The decibel does not specify a magnitude of a power, voltage or current but in ratio between two values of them. Gains and losses in circuits or radio paths are expressed in decibels.

Decibels and the Logarithmic Scale

• It is a unit that describes a ratio. It is not an absolute unit but rather indicates the relation between two powers. Symbol used to indicate the gain of the loss in the system.

Decibels

• Gain = 10 log (output / input) = 10 log (P2 /P1)

• Loss = 10 log (input / output) = 10 log (P1/P2)

– Note: Doubling the power means 3 dB gains; likewise, halving the power means a 3 dB loss

Formulas

• it is a logarithmic value that shows the difference between the measured value and the laid down standard value.

Absolute Level

• Resistance: R= 600 Ω (average value if a VF telephone signal line)

• Power: P= 1 mW (power produced in a mic, if a “A” is strongly spoken)

• Voltage: V= 0.775 V

• Current: I= 1.29 mA

Standard Values (At the Zero Relative Level Point)

• Absolute Power Level = 10 log Pm/1 mW

• Absolute Voltage Level =20 log Vm/0.775 V

• Absolute Current Level=20 log Im/1.29 mA– Note: Across a resistance of 600Ω, the absolute levels of voltage,

current, and power have the same value, if the laid down standard values are used. In the field, the absolute power level and absolute voltage level are used for telecom path measurements

Standard Values (At the Zero Relative Level Point)With the standard resistance @ 600Ω

• Absolute Voltage Level=10 log (Pm/1mW) + 10 log (600/Rm)

Calculation of absolute voltage level with different resistance

• Absolute Current Level=10 log (Pm/1 mW) + 10 log (Rm/600)

Calculation of absolute current level with different resistance

• 10 log (Pm/1mW) = 10 log (Vm/0.775 V) + 10 log (600/Rm)

Conversion from the power level to the voltage level and vice versa

• 10 log (Pm/1mW) = 10 log (Vm/0.775 V) + 10 log (Rm/ 600)

Conversion from the power level to the current level and vice versa

dBm• dBm – absolute power level referred to 1mW. The term

dbm was originally used for telephone and audio work and, when used in that context, implies an impedance of 600 Ω, the nominal impedance of a telephone line. When it is desired to define a relative transmission level in a circuit, dBT is preferred.

http://en.wikipedia.org/wiki/Omega

Equations:

• dBm = 10 log ( Pm / 1mW)• dBm dB = dBm

• dBm1 dB = 10dB1/10 10dbm2/10

• Note: 0 dBm = 1 mW

dBv• dBv – absolute voltage level, referred to 0.775 V. dBv is

used in audio work when the impedance is not 600 Ω and no specific impedance is implied.

• dBv = 10 log (Vm / 0.775 V)


dBi

• dBi – absolute current level, referred to 1.29 mA at 600 Ω.

• dBi = 10 log (Im / 1.29 mA)


dBv s and dBvps

• dBv s – absolute voltage level in the sound channel, referred to 0.775 V

• dBvps – absolute noise voltage level in the sound channel, referred to 0.775 V and CCIT weighted

dBv

– decibels relative to 1 volt.

dBv = 20 log (Vm ⁄ 1V)

dBrn

– a weighted circuit noise power unit in dB referenced to 1pW (-90 dBm, 1000 Hz)which is 0 dBrn. This is the noise unit of 144 weighting network where the old 144 telephone handset was the devise used.

dBrn = 10 log (Nm ⁄ 1x10 ˆ-12W)

dBrnc– is weighted noise power in dBrn, measured by a noise

measuring set ‘C- message’ weighting.

Pure 1kHz test tone :dBrnc=dBm+ 90F1A weigthted noise:dBrnc= dBa+ 6601A/F1A weighted: dBrnc= dBa+ 6

- The dBrn 30kHz flat noise measurements are noise readings taken with a filter that has a flat response from 20Hz to 30kHz. It was found to be typically 1.5 dB than dBrnc readings for equal noise power level.

dBrnC0

Is noise measure in dBrnC also referred to zero transmission level point.

dBrnC0= dBrnC - TLPdB

pWp picowatt of noise power, psopometrically weighted. I pWp

= 800 Hz tone at -90 dBm

This unit is used in psopometric noise weighting, which assumes a perfect receiver thus the weighing curve corresponds to the frequency response of the human ear only.

dBrn = 10 log pWp *Note: -90 dBm = 1x10-12 W = 0 dBrnC

dBmp

A unit of noise power in dBm, measured with psophometric weighting.

dBmp = 10 log pWp / 10-3

dBmp = 10 log pWp - 90= dBa – 84

= dBm – 2.5 (for flat noise 300-3400 Hz)

• dBr-means dB ‘relative level’. Used to define transmission level at various points in a circuit or system referred to the zero relative level point.

dBr=dBm-dBm0

• dBm0 – dBm referred to or measured at, a point of zero transmission level

dBm0=dBm-dBr

• dBm0p – the abbreviation for absolute noise power referred to or measured at a point zero relative transmission level , psophometrically weighted.

dBm0p=dBmp-dBr

dBa

• Stands for dB adjusted.• It is an expression of the relative loudness of sounds in

air as perceived by the human ear.• This is a weighted circuit noise power referred to -85

dBm, which is 0 dBa or 10-11.5

• It is measured with a noise meter at the receiving end.

The meter is calibrated on a 1000 Hz tone such that 1mW (0 dBm) gives a reading of +85 dBm.

If 1mW is spread over the band 300-3400 Hz as random white noise, the meter will read as -82 dBm or 0 dBa.

The main advantages with dBA are:

adapted to the human ear response to sound possible to measure with low cost instruments

Pure 1 kHz test tone : dBa=dBm+85

A 3 kHz band of random noise :

dBa=dBm+82

601A/F1A weighted : dBa=dBm+77

dBa0

Is referred to as circuit noise power in dBa at a point of zero relative transmission level (0 dB).

It is preferable to convert circuit noise measurement values from dBa to dBa0 as this makes it unnecessary to know or state the relative transmission level at the point of measurement.

dBa0= dBa + TLPdB

dBv0

Is defined as the absolute voltage level, and also referred to the relative level.

dBv0 = dBv + dBr

dBd – used for expressing the gain of an antenna referred to

a dipole.

dBi – used for expressing the gain of an antenna referred to

an isotropic radiator.

THE NEPER

A transmission unit used in number of Northern European countries.

Np = ½ loge (P2/P1) Note : 1 Np = 8.686dB

NOISEANALYSIS

NOISE ANALYSIS

• Noise in communication system originates both in the channel and in the communication equipment. Noise consist of undesired signals and inhibit communications. It cannot be avoided completely, but its effects can be reduced by various means, such as reducing signal bandwidth, increasing the transmitter power, and using low noise amplifiers for weak signals.

NOISE ANALYSIS

• Noise- any unwanted form of energy tending to interfere with the easy reception and reproduction of wanted signals.

• Distortion- any deviation in the signal caused by the imperfect response of the system to the desired signal.

• Interference-is any contamination by external signals from human sources, other transmitter, power lines, switching circuits and others.

EFFECTS OF NOISE

1. Limits the performance of every systems.2. It affects the sensitivity of the system.3. It limits the range of the system for a given transmitter

power.4. It forces a reduction in the bandwidth of the system.

SOURCES OF NOISE

A. External noise- noise that is generated outside the receiver or circuit.

1. Atmospheric noise- noise that come from natural disturbances occurring in the atmosphere. Often called static electricity. Ex. Lightning charges(>20MHz not significant)

2. Extraterrestrial/Deep space noise- noise from the outer space(8 to 1.5GHz). This energy components are absorbed by the earth’s atmosphere before they can reach the atmosphere.

• Solar noise- generated directly from the sun’s heat.• Cosmic/Black-body noise- noise created by the stars.

3. Man-made/Industrial noise- the most troublesome form of noise that is usually produced by mankind. Ex. Power lines, fluorescent lights, electric motors power generating and switching equipments, spark producing mechanism etc.

B. Internal noise- noise introduced by the receiver itself.1. Thermal/Johnson/Brownian/White/Gaussian noise- noise

generates due to the rapid and random movements of electrons, atoms, molecules inside a resistive component due to thermal agitation.

Pn=kTBWPno=kT

Where,Pn= noise powerPno= noise power density @ 1Hz

T= temperature, kelvink= Boltzman’s constant (1.38 x 10-23 J/k)

So that,Vn= √4kTBWRL

In= √4kTBWG

NOISE DUE TO SEVERAL SOURCES

• Noisy resistor’s in series, VnT=√4kTBWRTseries InT=√4kTBWGTseries

VnT=√V2n1+ V2

n2+….+V2nn

• Noisy resistor’s in parallel VnT=√4kTBWRTparallelInT=√4kTBWGTparallel

InT=√I2n1+I2

n2+….I2nn

Shot noise/ transistor noise• Caused by random variations in the arrival of

electrons or holes at the output electrode of an amplifying device and appearing as a randomly varying noise current superimposed on the output.

• Shot noise is due to the corpuscular nature of transport. In 1918, WALTER SCHOTTKY discovered shot noise in tubes and developed Schottky's theorem. Shot noise is always associated with direct current flow.

Shot noise/ transistor noise•

Transit time noise• Noise due to the time taken by the electrons to

travel from the emitter to the collector of the transistor

• Its greatest effect is a higher frequencies particularly in the microwave region. It is otherwise known as ‘HIGH-FREQUENCY NOISE’

Flicker noise• Noise appearing at frequencies below 1 kHz• Directly proportional to emitter current & junction

temperature and inversely proportional to frequency

• Negligible at about above 500 Hz• Known as LOW FREQUENCY NOISE, EXCESS

NOISE, MODULATION NOISE or PINK NOISE

Miscellaneous noise

Harmonic distortion• Occurs when unwanted harmonics of a signal are

produced through non linear amplification• Harmonics are integer multiples of the original

input signal • Total harmonic distortion is hardly perceptible to

the human ear• Every component adds some level of distortion,

but most distortion is insignificant

Harmonic distortion•

Intermodulation distortion• Occurs when unwanted sum and difference

frequencies are produced• Caused by non-linear behavior of the signal

processing being used• Creates additional signals at frequencies that are

the sum and difference frequencies of the original frequencies and at multiples of those sum and difference frequencies.

Frequency Spectrum of intermodulation distortion in a radio-frequency signal passed through the linear broad-band amplifier.

Impulse noise• Characterized by high amplitude peaks of

short duration in the total noise spectrum• Consists of sudden bursts of irregularly

shaped pulses that generally last between a few microseconds and several milliseconds

Partition noise

• Occurs whenever current has to divide between two or more electrodes and results to random fluctuations in the process

Burst noise• Low frequency noise found in transistors• It appears as a series of burst of two or more

levels of discrete voltage or current levels, as high as several hundred microvolts, at random and unpredictable times

• Also called as ‘POPCORN NOISE’

Burst Noise

Avalanche noise• Large noise spikes present in the avalanche current due

to the oscillation that results in the avalanching action

Reactance noise effects

Reactance noise effects•

Signal to Noise Ratio

Relative measure of the desired signal power to the output noise power. Identifies the noise content at a specific point but not useful in relating how much additional noise is injected to the circuit. The higher the value, the better the system is.

S / N = Ps / Pn

S / N = 10 Log ( Ps / Pn ) or S / N = 20 Log ( Vs / Vn )

Noise Figure / Noise Factor/ Noise Ration

Usually used to specify exactly how noisy a device is. It is a figure of merit, indicating how much a component, stage, or series of stages degrades the signal to noise ratio of a system. The noise figure of a totally noiseless device is 1 or 0 dB. The higher the noise figure, the worse the signal to noise ration at the output.

NR = Input ( S / N ) / Output ( S / N )

FdB = 10 log ( S / N )1 / ( S / N )0 = 10 log NR

Noise due to several amplifier in cascade,

Req = R1 + ( R2 / A12 ) + ( R3 / A1

2

A22 )

Req = R1 + Req

F = 1 + ( Req / Ra )

FTotal = F1 + ( F2 – 1 ) / A1 + (F3 – 1 ) / A1A2 + . . . + ( Fn – 1 ) / A1A2 . . . An-1

Noise Figure in terms of equivalent noise resistance,

When two or more amplifiers are cascaded ( Friiss’ Formula )

FdB = 10 Log FTotal

NOISE TEMPERATURE

Equivalent noise temperature referred at the input,

F1 = 1 (Te / To ) or Te = To ( F – 1 )

Ttotal = Te1 + Te2 / A1 + Te3 / A1A2 + . . . + Teu / A1A2 + . . . + An – 1

Employed extensively for antennas and low-noise amplifier especially with microwave receivers and their associated antenna system. It allows easy calculation of noise power since the equivalent noise temperature ( Te ) of microwave antennas and their coupling networks are then simply additive. It is convenient to use since microwave antenna and receiver manufacturers usually provide Te information for their equipment.

F = Noise Power, dBTe = Equivalent Noise ResistanceT0 = Reference Temperature, 17^C = 17 C + 273 C = 290

AMPLITUDEMODULATION

Amplitude Modulation

• A method of analog modulation in which the amplitude of the carrier signal is made proportional with the instantaneous amplitude of the modulating signal. The frequency of the carrier is kept constant

Amplitude Modulation

• Formulas: VAM( t ) = ( Vc + em ) Sin ( wc t ) VAM = Vc Sin ( wct ) – [ ( mVc/2 ) Cos ( wc + wm ) t ] + [ (mVc/2) Cos ( wc – wm ) t ]

Kinds of AM WAVE1. Undermodulation (i.e. Em<Ec)2. 100% Modulation (i.e. Em=Ec)3. Over Modulation (i.e. Em>Ec)

FREQUENCY SPECTRUM OF THE AM SIGNAL

4. Carrier Frequency , fc5. Upper sideband frequency , fusb6. Lower sideband frequency, flsb

Modulation Index- defined as the ration of the amplitude of the modulating signal to the amplitude of the carrier signal

POWER COMPONENT OF AN AM SIGNAL

1. Carrier Power, Pc2. Upper sideband power, Pusb3. Lower sideband power, Plsb

POWER CALCULATIONS:

Modulation by Several Signals

•

Current Calculation

•

Voltage Calculationwhere:

Vt = total voltageVc = voltage of the carrier signal

where:It = total voltageIc = voltage of the carrier signal

Efficiency of Modulation

* Efficiency is 33.33% for 100% modulation

Bandwidth for Simultaneous Modulation

Total Bandwidth = 2 x fm x nWhere :

fm = frequency of the modulating signaln = the number of channels

Types of Devices that generate AM

• AM Generator – is a device that generates low power amplitude modulation. It gives high importance to simplicity rather than efficiency.

• AM Transmitter – is a device that generates high power amplitude modulation. Contrary to AM Generator it focuses mainly on higher efficiency requirements.

Transmitter Requirements:

1. It must generate signal with the right type of modulation2. It must have sufficient power3. It must have the right carrier frequency4. It must have reasonable efficiency5. The output signal must be coupled to the antenna.

AM Modulator Circuits

• 1.) Low-level modulation – used predominantly for low-power, low capacity systems. The modulation takes place prior to the output element of the final stage of the transmitter. Less modulating signal is required to achieve high percentage of modulation.

AM Modulator Circuits

• 2.) High-level modulation - Modulation takes place in the final element of the final stage where the carrier signal is at its maximum amplitude; - Requires much higher amplitude modulating signal power to achieve a reasonable percentage modulation

Advantages/Disadvantages of AM

1. AM station requires narrower channel than FM.2. Transmitting and receiving sections are much

simpler and cheaper.3. AM area of reception is wider.4. Limited deviation is required since it is not

possible to exceed 100% modulation without causing severe distortion.

5. In AM, most of the transmitted power is in the carrier, which does not contain any information.

Single Sideband Modulation (SSBSC)

The signal takes up fewer spectrums. This allow twice as many signals to be transmitted.Improvement in S/N ratio that can be achieved by reducing the bandwidth.Typical SSB transmitters are designed to handle only average power level.In SSB, the transmitter output is expressed in terms of Peak Envelope Power. (PEP)

PEP=(Vp)2/2RL

ADVANTAGES DISADVANTAGES

• Less Bandwidth is required.

• Power conservation. Power for the sidebands is increased.

• Reduced noise interference due to reduced bandwidth.

• Selective fading. Sidebands will not experience transmission impairments.

• The generation and reception of SSB signals is complicated.

• The SSB transmitter and receiver need to have an excellent frequency stability.

• Requires accurate, complex and expensive tuning circuits

Power Saving

•

METHODS OF GENERATING SSBSC

• Filter Method – for perfect SSB signal, it is necessary to pass one sideband completely and reject the other. This makes the sideband filter as the most complex component.

Types: Crystal lattice – commonly used in single sideband systems and

has high quality factor which may range up to 50, 000.

Ceramic – made from lead zirconate-titanate which exhibits the piezoelectric effect and operates similarly to a crystal but does not have a high Q-factor. Less expensive, smaller and more rugged. Q-factor is up to 2000.

Mechanical – a mechanically resonant transducer, which receives electrical energy converts it to mechanical vibrations and then back to electrical energy as output. More rugged, larger, heavier and, therefore, impractical for mobile communication. Q-factor ranges up to 10,000.

Surface acoustic wave – SAW use acoustic energy rather than

electromechanical energy to provide excellent performance for precise bandpass filtering. But SAWs have extremely high insertion loss which cannot be used to filter low-level signals. Often used in TV and radar. Applicable to high frequencies

Balanced Modulator

• Is a device used to suppress the carrier signal. It resembles the conventional push-pull amplifier in circuitry but not in operation.

• Sometimes called balanced lattice modulator• It has two inputs : a single frequency carrier and the

modulating signal, which may be a single frequency or a complex waveform.

• For it to operate properly the amplitude of the carrier must be greater than that of the modulating signal.

• Phase-shift Method – the undesired sideband is cancelled in the output of the modulator. Designed specifically to avoid the serious complexity on filters and their inherent disadvantages while still providing the desired result. It makes use of two balanced modulators and two phase-shifts network.

• Third Method/ Weaver Method – very complex method and not often used commercially.

Applications of Single Sideband Modulation

1. Mobile system, in which the weight and low power consumption must be low

2. Point-to-point communications3. Land, air and maritime mobile

communications4. Radio Navigation and Amateur radio5. Military communications6. Television7. Telemetry

Forms of Amplitude Modulation

1. A3E – DSBFC, is probably the most often used form of amplitude modulation. It is sometimes called conventional AM. In AM DSBFC, all frequencies in the modulated signal are transmitted.

2. F3E – FM, is widely used for a variety of radio communications applications. It is a method of angle modulation in which the amplitude of the carrier is kept constant, while its frequency and rate of change are varied by the modulating signal

3. G3E – PM, is a type of electronic modulation in which the phase of a carrier wave is varied in order to transmit the information contained in the signal.

4. H3E – SSBFC, reducing the carrier by a small amount in order to maintain some degree of compatibility with older AM receivers.

5. R3E – SSBRC, a small amount of carrier signal is transmitted in order that receivers with the necessary circuitry can synchronize with the transmitted signal.

6. J3E - SSBSC transmission generally attempts to reduce the (amplitude) level of the carrier signal to as close as possible to zero.

7. A3C – AM Facsimile, amplitude modulation facsimile; the main carrier is modulated either directly or by a frequency modulated subcarrier;

8. F3C – FM Facsimile, modulation frequency facsimile, by direct modulation of carrier frequency;

9. A1A – Telegraphy by ON-OFF Keying, without modulation by an audio frequency.

10.C3F – television, in amplitude modulation, with vestigial sideband;

11.H8E – ISB, modulation in which the radio-frequency carrier is reduced or eliminated and two channels of information are transmitted, one on an upper and one on a lower sideband

12.F1B – frequency modulation telegraphy with automatic reception, without using a modulating subcarrier;

ITU SIGNAL CLASSIFICATION

1. First Symbol (letter) – type of modulation of the main carrier.

2. Second Symbol (number) – number of signals modulating the main carrier.

3. Third Symbol (letter) – type of information to be transmitted.

4. Fourth Symbol (letter) – details of the signal5. Fifth Symbol (letter) – nature of multiplexing

ANGLE MODULATION

- Is a type of analog modulation in which the sinusoidal reference signal is varied in accordance with a modulating signal.

FREQUENCY MODULATION (F3E)

- It is a method of angle modulation in which the amplitude of the carrier is kept constant, while its frequency and rate of change are varied by the modulated signal. (i.e. instantaneous frequency of the carrier is made to vary by an amount proportional to the modulating signal amplitude).

eFM (t) = Ecsin[wct + mfsinwmt]

CARRIER SWING

- The total variation in frequency from lowest to highest. It is also defined as the difference between the maximum positive and negative deviation of the carrier.

C.S. = 2δ

FREQUENCY DEVIATION (δ)-amount of oscillation frequency increase or decrease around the carrier frequency(i.e, frequency of the modulating signal determines the rate of frequency deviation).

MODULATION INDEX- it is defined as the ratio of actual frequency deviation to the modulating frequency.

DEVIATION RATIO- worst case modulation index, in which the maximum permitted frequency deviation and the maximum permitted audio frequency are used.

PERCENT MODULATION- it refers of the actual frequency deviation to the maximum allowable frequency deviation.

FREQUENCY DEVIATION (δ)-amount of oscillation frequency increase or decrease around the carrier frequency(i.e, frequency of the modulating signal determines the rate of frequency deviation).

MODULATION INDEX- it is defined as the ratio of actual frequency deviation to the modulating frequency.

PERCENT MODULATION- it refers of the actual frequency deviation to the maximum allowable frequency deviation.

DEVIATION RATIO- worst case modulation index, in which the maximum permitted frequency deviation and the maximum permitted audio frequency are used.

Disadvantages of FM1. Reception is limited to line-of-sight.

2. A much wider channel is required or excessive use of spectrum space.

3. More complex and costly circuits particularly for modulation and demodulation.

Methods of Generating FM1. Direct Method

a.Reactance Modulator – a reactance tube is connected to the tank circuit of an oscillator.

b.Varactor Diode Modulator – an FM generator utilizing a voltage variable capacitor which when reversed biased will vary its junction capacitance thereby producing direct FM of an oscillator.

c. Transistor Reactance Modulator

d. Linear IC Direct FM Modulator2. Indirect Method (i.e. Armstrong Modulator) – FM is only produced as long as the place shift is being varied. Armstrong modulator uses a crystal oscillator for frequency stability. The modulating signal is fed to an integrator, the output of which is also fed to a balanced modulator (PM) together with a local oscillator signal that is shifted 90 degrees in phase.

1. Direct MethodMethods of Generating FM

Phase Modulation (G3E) – An angular modulation in which the phase of the carrier is varied in accordance with the instantaneous amplitude of the modulating signal.

VPM (t) = Ec Coswct + mpCoswmt

Phase Modulation

δ = ∆ØVm

Phase ModulationPhase Deviation – Is the amount of phase shifts that occur when it is acted on by a modulating signal (i.e. In PM, the instantaneous frequency deviation is directly proportional to the first derivative or slope of the modulating signal).

Modulation Index – the modulation index is proportional to the amplitude of the modulating signal, independent of its frequency.

In PM, the modulation index is proportional to the amplitude of the modulating signal, independent of its frequency.

In PM, phase deviation is proportional to both the amplitude of the modulating signal and modulating frequency.

Radio Receivers

1. AM Radio Receiver – device designed to receive an AM signal between 526.5 to 1705 kHz band with a channel assignment spaced at 9kHz.

2. FM Radio receiver – device designed to received a frequency modulation wave between 88 to 108 MHz band with channel assignment speed to 200 kHz.

Essential Functions of a Receiver

1. Reception2. Selection3. Detection4. Reproduction

END OF LECTURE

SAMPLEPROBLEMS

Problems:

1. Covert the following thermal noise power to dBm : 0.001 µW and 1 pW.

0.001 µW to dBm :

Solution:

dBm = 10 log ( 0.001 µW / 1x10 -3 )

dBm = -60 dBm

1 pW to dBm :

dBm = 10 log ( 1 pW / 1x10 -3 )

dBm = -90dBm

Problems:

2. Covert the following thermal noise to watts: -150dBm and -174 dBm. A network with +7 dBm and +11 dBm input has an insertion loss of 3 dB. What is the ouput in dBm?

Solution:-150 dBm to Watts:

-150 dBm = 10 log ( P / 1x10 -3 )

P = 1 x 10 -18 Watts

-174 dBm to Watts:

-174 dBm = 10 log ( P / 1x10 -3 )

P = 3.98 x 10 -18 Watts

Convert 7 dBm and 11 dBm to Watts:

7 dBm = 10 log ( P / 1x10 -3 )

7 dBm = 5.01 mW

11 dBm = 10 log ( P / 1x10 -3 )

11 dBm = 12.59 mW

Pin = 5.01 mW + 12.59 mW ;

Pin = 17.6 mW

dBm = 10 log ( 17.6 mW / 1x10 -3 )

dBm = 12.46 dBm

dBm = 12.46 dBm - 3dB = 9.46 dBm

9.46 dBm = 10 log ( P / 1x10 -3 )

P = 8.83 mW

Problems:

3. At the input to the receiver of a standard telephone channel, the noise power is 50 µW and the signal power is 20mW. Calculate the Shannon limit for the capacity of the above channel under these conditions and when the signal power is halved.

Solution:S / N = ( 20 mW / 50 µW )

S / N = 400

Shannon Limit: ( 3.32 ) ( BW ) Log10 [1 + ( S / N )]

Shannon Limit: 26.791 kbps

When Signal Power is halved; S = 10mW

S / N = 200Shannon Limit: ( 3.32 ) ( BW ) Log10 [1 + ( S / N )]


Problems:

4. A 2kHz channel has a signal-to-noise ratio of 24 dB. Calculate the maximum capacity of this channel. Assuming constant transmitting power, calculate the maximum capacity when the channel bandwidth is halved and reduced to a quarter of the original value.

Solution:Shannon Limit: ( 3.32 ) ( BW ) Log10 [1 + ( S / N )]

Convert 24 dB to Signal to Noise Ratio:24 dB = 10 Log ( S / N )

S / N = 251.19Shannon Limit: ( 3.32 ) ( BW ) Log10 [1 + ( 251.19)] Shannon Limit: 15.947 kbpsWhen Channel Bandwidth is halved; BW = 1khz


When Channel Bandwidth is reduced to a quarter; BW = 500 Hz


Problems:

5. Calculate the capacity of a standard telephone channel with a signal-to-noise ratio of 32 dB.Solution:

Shannon Limit: ( 3.32 ) ( BW ) Log10 [1 + ( S / N )]

Convert 32 dB to Signal to Noise Ratio:

32 dB = 10 Log ( S / N )S / N = 1584.89Shannon Limit: ( 3.32 ) ( 3100 ) Log10 [1 + ( 1584.89)]


7. Calculate the thermal noise power in both watts and dBm for the bandwidth of 100 kHz and temperature of 100°C.Solution:N=KTBWN= (1.38 x 10-23)(100+ 273K)(100000)N=5.15 x 10-16NdBm= 10 log(5.15 x 10-16/ 1 x 10-3)NdBm= -122.88 dB

8. Determine the bandwidth necessary to produce 8 x 10-17 watts of thermal noise power at a temperature of 17°C over a 1 MHz bandwidth.Solution:N= KTBW8 x 10-17 = (1.38 x 10-23)(17 + 273K)BWBW= 19990.05 Hz or 20 kHz

10. A diode noise generator is required to produce 10µVof noise in a receiver with an input impedance of 75Ω, resistive and a noise power bandwidth of 200kHz. What must be the current through the diode be?

Given: Vn=10µV, R=75Ω, BW=200kHz Req’d: Id

Sol’n: In= In=√2qeIdBWId=

11. Given an AM amplifier which operates at 20˚C over a 10kHz bandwidth has a 220Ω input resistor. Find the input noise voltage.

Given: T=20˚C + 273= 293K, BW= 10khzR=220Ω

Req’d: Vn

Sol’n: Vn=√4kTBWR=√4(k)(293K)(10kHz)(220Ω) Vn= 188.68nV

12. Calculate the noise voltage for a 1kΩ resistor at 17˚C tuned by an LC circuit with a bandwidth of 1MHz.

Given: R=1kΩ, T=17˚C + 273= 290K, BW=1Mhz

Req’d: Vn

Sol’n: Vn=√4kTBWR=√4(k)(290K)(1MHz)(1kΩ)Vn= 4µV

13. Calculate the noise voltage at the input of a TV rf amplifier using a device that has 200Ω equivalent noise resistance and a 300Ω input resistance. The bandwidth of the amplifier is 6MHz and the temperature is 17˚C.

Given: R1=300Ω, R2=200Ω, BW=6MHz, T= 17˚C + 273= 290K

Req’d: V1

Sol’n: V2=√4kTBWR2=√4(k)(290K)(6MHz)(200Ω) V2=4.383µVV1=

14 . Calculate the noise voltage at the input of a RF simplifier using a device that has a 200 Ω equivalent noise resistance and a 300 Ω input resistance. The bandwidth of the amplifier is 6 MHz and the temperature is 27 ᵒC.

Given: Required:Rn = 200 Ω Vi

Ri = 300 ΩBW = 6 MHzT = 27 ᵒC

16. A noise source has two resistors in series at two different temperature: R1 = 100 Ω at 300 K and R2 = 200 Ω at 400 K. Find the total noise voltage and noise power produced at the load RL = 300 Ω, over a bandwidth of 100 kHz.

Given: Required:R1 = 100 Ω at 300 K VT, VL

R2 = 200 Ω at 400 KRL = 300 ΩBW = 100 kHz

18. A parallel tuned circuit having Q of 20 is resonated to a 200 MHz with a 10 pF capacitor. If this circuit is maintained at 17oC, what noise voltage will a wideband voltmeter measure when placed across it?

19. An amplifier has a noise voltage of 4.0µV and is operating over a frequency range having a maximum frequency of 460kHz. What is the lowest frequency of operation does the amplifier have if the input resistor is 200kΩ and the ambient temperature is 17oC.

20. A diode noise generator is required to produce 10µV of noise in a receiver with an input impedance of 75Ω, resistive, and a noise power bandwidth of 200kHz. What must the current through the diode be?

Number 22

• Calculate the S/N ratio in dB for a receiver output of 4 volts signal and 0.48 noise voltage.

Solution and Answer for No.22

Number 23

• Determine the equivalent noise temperature for a noise figure of 10 dB.

Number 24

• Determine the overall noise factor and noise figure for the three cascaded amplifiers with the ff. parameters: A1=3 dB; A2=13 dB; A3=10 dB; NF1=10 dB; NF2=6 dB; NF3=10 dB.

Number 25

• Determine the noise figure for an amplifier with an input signal to noise ratio of 30 dB and output signal to noise ratio of 24 dB.

28. The noise produced by a resistor is to be amplified by a noiseless amplifier having a voltage gain of 75 and a bandwidth of 100kHz. A sensitive meter at the output reads 240µV. Assuming operation at 37oC, calculate the resistors resistance if the bandwidth is cut to 25kHz, determine also the expected output meter reading.

29. A voice transmission occupies a channel 30 kHz wide. Suppose a spread spectrum system is used to increase its bandwidth to 10 MHz If the signal has a total signal power of -110 dBm at the receiver input and the system noise temperature referred to the same point is 300K. Calculate the signal-to-noise ratio for both system.

30. Determine the noise current for a vacuum tube diode with a forward bias of 1mA over a 100kHz bandwidth. Determine also the equivalent noise voltage of the diode.

Given: Ib = 1mA ; BW = 100kHz

Req’d: In = ? ; Vn= ?

Sol’n:

In

31. A transistor amplifier has a measured signal to noise power of 10 at its input and 5 at its output. Calculate the transistor’s noise figure in decibel.

Given:

32. Suppose the noise power at the input to a receiver is 1nW in the bandwidth of interest. What would be the required signal power for a signal to noise ratio of 25dB?

33. For a three cascaded amplifier stages, each with noise figure of 3dB and power gains of 10dB, determine the total noise figure in decibel.

34. For a three cascaded amplifier stages, each with noise figure of 3dB and power gains of 10dB, determine the total noise figure in decibel.

Given: A1 = A2 = A3 = 10dB = 10 Req’d: NFTNF1 = NF2 = NF3 = 3dBF1 = F2 = F3 = 2

Sol’n: NFT = 10log FT; FT = F1 + (F2-1)/A1 + (F3-1)/A1A2

NFT = 10 log [2+ (2-1)/10 + (2-1)/100]NFT = 3.24 dB

35. A microwave antenna (Teq=25K) is coupled through a network (Teq=30K) to a microwave receiver with Teq=60K referred to its input. Calculate the noise power at its input for a 2 MHz bandwidth. Determine the receiver’s noise figure.

Given: Teq1 = 25K, Teq2 = 30K, Teq3 = 60K, BW = 2MHz Req’d: N, NFSol’n: TeqT = Teq1 + Teq2 + Teq3 = 25K + 30K + 60K

N = kTB = (1.38x10-23J/K)(2x106Hz)(115K)N = 3.17 x 10-15 W

F = 1 + Teqreceiver/To = 1 + 60K/290K = 1.2069NF = 10 log F = 10 log 1.2069NF = 0.82 dB

36. A satellite receiving system includes a dish antenna Teq=35K connected via coupling network Teq=40K to a microwave receiver Teq=52K referred to its input. Determine the receiver’s noise figure.

Given: Teq = 52K (receiver) Req’d: NF (receiver)Sol’n: F = 1 + Teq/To = 1 + 52K/290K = 1.18

NF = 10 log F = 10 log 1.18 NF = 0.72 dB

37. Calculate the noise power at the input of a microwave receiver with an equivalent noise temperature of 45K. It is fed from an antenna with a 35K equivalent noise temperature and operates over a 5MHz bandwidth.

Given: Teq(input) = 45K, Teq(antenna) = 35K, B = 5MHz Req’d: NSol’n: Teq = Teq(input) + Teq(antenna) = 45+35 Teq= 80K

N= kTB = (1.38x10-23J/K)(5x106Hz)(80K)N= 5.52 x 10-15 W

38. An amplifier consists of two stages. Stage one has a gain of 12 dB and a noise figure of 2 dB. Stage two has a gain of 20 dB and a noise figure of 5 dB. Calculate the noise figure in decibel, and the equivalent noise temperature, in Kelvin, for the amplifier.

Given: NF1 = 2 dB, F1 = 1.58; NF2 = 5 dB, F2 = 3.16A1 = 12 dB = 15. 85, A2 = 20 dB = 100, To = 290 K

Req’d: NFT, Te

Sol’n: FT = F1 + (F2 – 1)/ A1 = 1.58 + (3.16 – 1)/15.85 FT = 1.72NFT = 10 log FT = 10 log 1.72 NFT = 2.36 dB

Te = To (FT – 1) = 290K (1.72 - 1)Te = 208.8 K

Problem 39

• The first stage of a two stage amplifier has a voltage gain of 10, a 600Ω resistor, a 1,600Ω equivalent noise resistance and a 27 kΩ output resistance. For the 2nd stage the values are 25, 81 kΩ, 10 kΩ and 1 MΩ respectively. Calculate the equivalent input noise resistance of this two stage amplifier.







Given: Av1 = 10, R1 = 600Ω, NR1 = 1,600Ω, Ro1 = 27 kΩAv2 = 25, R2 = 81 kΩ, NR2 = 10 kΩ, Ro2 = 1 MΩ

Required: NRi = ?

Solution:R1T = R1 + NR1

R1T = 600Ω + 1,600ΩR1T = 2,200 Ω R2 = (Ro1*R2)/ (Ro1+R2) + NR2

R2 = (27 Ω* 81Ω)/(27Ω + 81Ω) + 10ΩR2 = 30.2 k ΩR3 = 1 MΩNRi = R1 + (R 2/A1

2) + R3/(A12A2

2)NRi = 2.2 kΩ + (30.2 kΩ/ 102 ) + ( 1 MΩ/ 102*252)NRi = 2,518 Ω





















Problem 40

• The RF amplifier of a receiver has an input resistance of 1,000 Ω and an equivalent shot-noise resistance of 200 Ω, a gain of 25 and a load resistance of 125 kΩ, given that the bandwidth is 1 MHz and the temperature is 20 0C. Calculate the equivalent noise voltage at the input of this amplifier with an impedance of 75 Ω, calculate the noise figure.





Given: Ri = 1,000 Ω Rn = 200Ω, RL = 125 kΩA1 = 25, BW = 1 MHz, T = 20 oC, Ra = 75 Ω

Required: Vn = ?NF = ?

Solution:Req = R1 + Rn + (RL/A1

2)Req = 1,000 Ω + 200 Ω + (125,000 Ω / 252)Req = 1,400 ΩVn = (4RKTB)1/2

Vn = (4 x 1,400 x 293 x 1,000,000)1/2 Vn = 4.758 x 10-6 VR’eq = Req – R1

R’eq =1,400 Ω -1,000 ΩR’eq = 400 ΩNF = 10 log ( 1 + R’eq/ Ra )NF = 10 log ( 1 + 400/75 )

NF = 8.02 dB












Problem 41

• A 3 stage amplifier is to have an overall noise temperature no greater than 70K. The overall gain of the amplifier is to be at least 45 dB. The amplifier is to be built by adding a low noise first stage to an existing two stage amplifier that has the gains and noise figure of 20 dB and 15 dB and 3 dB and 6 dB respectively. What is the minimum gain that the 1st stage have? Using the gain calculated, what is the noise figure that the first stage have?

Given: AT = 45, A L = 2, A3 =3.98Required: A1 = ? , NF1 = ?Solution:

NF1 = 3 dB = 2NF2 = 6 dB = 3.98AT = A1 AL A3

A1 = 10 log [ (10AT/10/10A3/10 x10AL/10)]A1 = 10 log [ 104.5/(102x101.5)]

A1 = 10 dB

Teq = To ( F – 1 )70 = 290 ( F – 1 )F = 1.241F1 = 1.1261NF1 = 10 log ( F1 )NF1 = 10 log ( 1.1261 )

NF1 = 0.515 dB

42. A three stage amplifier is to have than overall noise temperature no greater than 70K. The overall gain of the amplifier is to be at least 45 dB. The amplifier is to be built by adding a low noise first stage to an existing two-stage amplifier that has the gains and noise figures of 20 dB & 15 dB and 3 dB & 6 dB respectively. What is the minimum gain that the first stage has?

Given: Teq = 70 K, At = 45 dB, A2 = 20 dB, A3 = 15 dBNF2 = 3 dB, NF3 = 6 dB

Req’d: A1, NF1

Sol’n: At = A1 + A2 + A3

A1 = At – A2 – A3 = 45 – 20 – 15 A1 = 10 dB

NF2 = 3 dB, F2 = 2 FT = F1 + (F2-1)/(A1)+(F3-1)/(A1A2)NF3 = 6 dB, F3 = 4 36/19 = F1 + (2-1)/(10)+(4-1)/(10)(20)FT = 1 + (70/290) = (36/29) F1 = 1.126, NF1 = 0.515

43. A three stage amplifier system has a 3 dB bandwidth of 200 kHz determined by an LC circuit at its input, and operates at 22°C. The first stage has a power gain of 14 dB and a noise figure of 3 dB. The 2nd and 3rd stages are identical with power gain of 20 dB and noise figure of 8 dB. The output load is 300 ohms. The input noise is generated by a 10kΩ resistor. Calculate a). the noise voltage and power at the input and the output of the system assuming ideal noise loss amplifier; b).the overall noise figure for the system; c). the actual output noise voltage and power of a TV receiver having a bandwidth of 7 MHz and operating temperature of 27 °C consists of an amplifier.

Given: At = 3 dB, A1 = 14 dB, A2 = A3 = 20 dB, BW = 200 kHzNF1 = 3 dB, NF2 = NF3 = 8 dB, Ri = 10000 ohms, Ro = 300 ohmsT = 22 + 273 = 293 K

Req’d: Vni, Vno, NF, Vnt, Pnt

Sol’n:Vni = √[(4)(Ri)(k)(T)(BW)]Vni = √[(4)(10000)(1.38×10-23 )(293)(200000)]Vni = 5.7×10-6 V

Vno = √[(4)(Ro)(k)(T)(BW)]Vno = √[(4)(300)(1.38×10-23 )(293)(200000)]Vno = 0.99×10-6 V

44. A three stage amplifier has the stages with the following specifications: Stage 1 has a power gain of 10 and a noise figure of 2. For stages 2 and 3, the values are 25 & 4 and 30 & 5 respectively. Calculate the power gain, noise figure and noise temperature for the entire amplifier assuming matched condition.

Given: Stage 1: A1 = 10, F1 =2 Req’d: AT, NFT, TeqStage 2: A2 = 25, F2 = 4Stage 3: A3 = 30, F3 = 5

Sol’n: AT = A1 A2 A3 = (10)(25)(30) AT = 7500FT = F1 + (F2-1)/A1 + (F3-1)/A1A2

FT = 2 + (4-1)/10 + (5-1)/25 = 2.316NFT = 10log FT = 10log 2.316 FT = 3.65dBTeq = To (FT-1) = 290K (2.316-1) Teq = 381.64K

45. The four stages of an amplifier have gains and noise figures of 12 dB, 15 dB, 20 dB & 17 db and 2 dB, 4 dB, 6 dB & 7dB respectively. Calculate the overall noise figure in decibel.

Given: Stage 1: A1 = 12 dB = 15.85 NF1 = 2 dB, F1 = 1.58Stage 2: A2 = 15 dB = 31.62 NF2 = 4 dB, F2 = 2.51Stage 3: A3 = 20 dB = 100 NF3 = 6 dB, F3 = 3.98Stage 4: A4 = 17 dB = 50.12 NF4 = 7 dB, F4 = 5.01

Req’d: NFT

Sol’n: FT = F1 + (F2-1)/A1 + (F3-1)/A1A2 + (F4-1)/A1A2A3

FT = 1.58 + (2.51-1)/15.85 + (3.98-1)/(15.85)(31.62) + (5.01-1)/(15.85)(31.62)(100)FT = 1.68NFT = 10 log FT = 10 log 1.68NFT = 2.25 dB

46. The front end of a TV receiver having a bandwidth of 7MHz and operating at a temperature of 27 degrees Celsius consists of an amplifier having a gain of 15 followed by a mixer whose gain is 20. The amplifier has a 300 Ω input resistance and noise equivalent resistance of 500 Ω. For the mixer, the values are 2.2kΩ and 13.5kΩ respectively. And the load resistance is 470kΩ. Calculate the equivalent resistance of the cascaded amplifier for the TV receiver.

Given: A1 = 15, Ri1 = 300 Ω, Req1 = 500Ω BW = 7MHz, RL = 470k ΩA2 = 20, Ri2 = 2.2kΩ, Req2 = 13.5kΩ T = 27 C = 300K

Req’d: ReqT

Sol’n: R1 = Ri1 + Req1 = 300 Ω + 500Ω = 800ΩR2 = Ri2 + Req2 = 2.2kΩ + 13.5kΩ = 15.7kΩR3 = RL = 470kΩReqT = R1 + R2/A1

2 + R3/A12A2

2

ReqT = 800Ω + 15.7kΩ/(15)2 + 470kΩ/(15)2(20)2

ReqT = 875Ω

PROBLEMSPART 2

1. In order to modulate a signal of 98% given that the peak amplitude of the unmodulated carrier voltage is 37V, what must be the value of the modulated carrier voltage?Given: M = 98% Vc = 37VRequired: Vm

Solution:m = ; m = ; m = 0.98

= mVc= (0.98)(37V) = 36.26V

2. The carrier maximum value is 600mV and the carrier minimum value is 247mV. Calculate the percentage modulation, modulation index.Given: Vmax= 600mVVmin= 247mVRequired: m, M and Vm

Solution:m == m = 0.4168M = x 100% =X 100%M = 41.6765%

3. The total AM signal power is 2800W. The carrier power is 2000W, what is the power in each sideband and the percentage modulation.Given: PT= 2800W

PC=2000W Required:

a.) Pusb

b.) Plsb c.) M

Solution:PT = PC ( 1+ m2/2);

2800W = 2000W (1+ m2/2) m = 0.8944 x 100%

M = 89.44%

Pusb = Plsb = m2Pc / 4 =

Pusb = Plsb = 40W

4. An AM broadcast station has an unmodulated transmitter power of 50kW and uses an antenna with a 10Ω effective resistance. Calculate the transmitter carrier amplitude.Given: PT= 50kW andR= 10ΩRequired: VC

Solution:P = ; 50kW = ; V = 707.1068 Vrms

Vp = Vrms ; Vp = 1kV

5. The antenna current produced by an unmodulated carrier is 2.4A into an antenna with 75Ω resistance. When we apply amplitude modulation the antenna current rises to 2.7A. Compute for M and m. Also determine the power of the carrier.

Given: IC = 2.4A IT = 2.7A R = 75ΩRequired: M, m and Pc

Solution:

(2.7A)m = 0.7289 M = 72.89%

Pc = Ic2R

= (2.4A)2(75Ω)Pc = 432 W

Current Calculation

𝑉 𝑡=𝑉𝑐√1+𝑚22

𝐼 𝑡=𝐼𝑐 √1+𝑚2

2where:

It = total voltageIc = voltage of the carrier signal

6. For an AM DSBFC modulator with a carrier frequency of 100kHz and a maximum modulating signal of 5kHz, determine:

a) Frequency limits for the upper and lower sidebandsb) Bandwidthc) Upper and lower frequencies produced by a

modulating signal of 3kHz.

Given: fc= 100kHz ; fm(max)= 5kHz Required: a.) LSB & USB

b.) Bc.) fusf & flsf

Solution:a.) LSB = [fc - fm(max)] to fc

= [100kHz – 5kHz] to 100kHz LSB = 95kHz to 100kHza.) USB = fc to [fc + fm(max)]

= 100kHz to [100kHz + 5kHz] USB = 100kHz to 105kHz

b.) B = 2fm(max)

= 2(5kHz ) B = 10kHZc.) given that fm=3kHz

fusf = fc + fm

= 100kHz + 3kHzfusf = 103kHzflsf = fc – fm

= 100kHz - 3kHzflsf = 97kHz

7. For an AM waveform with a maximum peak voltage of 18Vp and a minimum peak voltage of 2Vp, compute for:

a) Upper and lower peak amplitude side frequenciesb) Peak amplitude of the unmodulated carrierc) Peak change in the amplitude of the enveloped) Modulation indexe) Percent modulation

Given: Vmax= 18VpVmin= 2Vp

Required: a.) Eusf & Elsf

b.) Ec

c.) Em

d.) me.) M

Solution:a.) Eusf = Elsf =

= Eusf = Elsf = 4V

b.) Ec = = Ec = 10V

Solution:c.) Em = = = Ec = 8V

d.) m = = m = 0.8

e.) M = x 100% = x 100% M = 80%

8. For an AM DSBFC transmitter with an unmodulated carrier power Pc = 100W that is modulated simultaneously by three modulating signals with coefficients of modulation: 0.2, 0.4, 0.5. Determine:

a) Total coefficient of modulationb) Upper and lower sideband powerc) Transmitted power

Given: Pc = 100W m1 = 0.2 m2 = 0.4 m3 = 0.5

Required : a.) mt

b.) Psbt

c.) Pt

Solution:a.) mt=

= mt = 0.67

b.) Psbt = = Psbt =

c.) Pt = Pc ( 1 = 100W ( 1 + )Pt = 122.45W

9. For the maximum positive envelope voltage of 12V and a minimum positive envelope amplitude of 4V. Determine the modulation coefficient and percent modulation.

Given: Vmax = 12V ; Vmin = 4V

Required: m & M

Solution:m =

= m = 0.5

M = x 100% = X 100%M = 50%

10. What is the maximum modulating signal frequency that can be used with AM DSBFC system with a 20kHz bandwidth?Given: B = 20kHzRequired: fm(max)

Solution:B = 2fm(max)

fm(max) = B/2 = 20kHz / 2

fm(max) = 10kHz

11. For an AM DSBFC wave with an unmodulated carrier voltage of 18Vp and load resistance of 72Ω and a modulation coefficient of 0.6 Determine:

a) Unmodulated carrier powerb) Total sideband powerc) Upper and lower sideband powerd) Total transmitted power.

Given:Vc = 18Vp RL= 72ΩRequired:

a) Pc

b) Psb

c) Pusb & Plsb

d) Pt

Solution:a.) Pc = Vc

2 / 2R ; = (18V)2 / 2(72Ω)Pc = 2.25W

b.) Psb = m2Pc / 2 = (0.6)2(2.25W) / 2 Psb = 0.405 W

c.) Pusb = Plsb = m2Pc / 4 = (0.6)2(2.25W) / 4

Pusb = Plsb = 0.2025 Wd.) Pt = Pc ( 1+ m2/2)

= Pc ( 1+ m2/2)= (2.25W) (1+ 0.62/2)

Pt = 2.655W

12.An antenna has an impedance of 40Ω. An unmodulated AM signal produces a current of 6.2A. The modulation is 85 percent. Determine:a) the Current Powerb) the Total Powerc) the Sideband Power

Given: R=40Ω ; I=6.2A ; m=0.85 since M=85%Req’d: Pc , PT , PSB

Sol’n:

13. An AM signal has a 12W carrier and 1.5W in each sideband. What is the percent modulation of this AM signal?Given:

Req’d : M = ?

Sol’n:

14.How much power appears in one sideband of an AM signal of 5kW transmitter modulated by 80%?

Given:

Req’d:

Sol’n:

15.An AM signal with 100 percent modulation has an upper sideband power of 32W. What is the carrier power?

Given: since 100% modulation, m=1 ; Req’d: Sol’n:

then, at m=1,

substituting at the first equation,

16. An AM wave displayed on an oscillator has values of Vmax = 4.8. If it has a modulation coefficient of 0.4, determine Vmin.Given: Vmax = 4.8 ; m=0.4Req’d: Vmin

Sol’n:

Amplitude Modulation Problems

17.Suppose that an AM signal, the Vmaxp-p is 7.8 and Vminp-p is 2.8.Determine the Percent Modulation.

Given:Vmaxp-p = 7.8Vminp-p = 2.8

Required:Percent Modulation, (M)

Working Solution:M = [ (Vmax - Vmin) / (Vmax + Vmin) ] x 100*Since given values are peak to peak, and we only need the peak values, let Vmax = 3.9 and Vmin be equal to 1.4

M = [ (3.9 – 1.4) / (3.9 + 1.4) ] x 100

M = 47.17 %


18.An AM DSBFC has a peak modulated carrier voltage of 18 Volts and a load resistance of 27Ω, and a modulation index of 1. Solve for the Power Carrier.

Given:Vc = 18 V

RL = 27 Ω

Required:Power Carrier, (Pc)

Working Solution:

Pc = [ (Vc)2 / (2)(RL) ]

Pc = [ (18)2 / (2)(27) ]

Pc = 6 watts

m = 1


19.A 500 kW carrier is to be modulated to 78%. Determine total transmitted power.

Given:Pc = 500 kW

M = 78%

Required:Transmitted Power, (Pt)

Working Solution:

Pt = [Pc][ 1 + ((m)2 / (2)) ]

Pt = [500kW][ 1 + ((0.78)2 / (2)) ]

Pt = 652.1kW

*Since given is M = 78%, derive the modulation index from M, which is 0.78


20.An AM DSBFC transmitter with an unmodulated carrier power of 100W that is simultaneously modulated by three modulating signals with coefficients of modulation m = 0.80, m = 0.65, and m = 0.12. Determine the total coefficient of modulation.

Given:m1 = 0.80

m2= 0.65

Required:Total coefficient of modulation, (mt)

Working Solution:

mt = (m1 + m2 + m3)1/2

mt = 1.08

m3 = 0.12

mt = (0.80 + 0.65 + 0.12)1/2

PROBLEMSPART 3

1. Determine the percent modulation of an FM wave with a frequency deviation of 15 kHz for (a.) FM broadcast, (b.) for TV broadcast.

Solution:a.) M=x100(15kHz/75kHz)*100= 20%b.) M=x100(15kHz/25kHz)*100= 60%

2. A given FM transmitter is modulated with a single sinusoid. The output for no modulation is 100 watts into a 50-ohm resistive load. The peak frequency deviation of the transmitter is carefully increased from zero until the first sideband amplitude in he output is 0. under these conditions determine (a.) the average power at the carrier frequency, (b.) the average power in all the remaining sidebands, (c.) the average power in the second-order sidebands.

Given: J1=0, J0=-0.4 @β=3.8 R=50 ohms, Pt=100 watts

Req’d: Pc(ave), PSB(ave),PSB @2nd-order SBSol’n: Pc=(Jo

2 (3.8)/ Jo2 (0))(100W)=16W

PSB= Pt- Pc=100W-16W=84WPc=2(Jo

2 (3.8)/ Jo2 (0))(100W)=34W

3. Determine the frequency deviation and carrier swing required to provide 80% modulation in the FM broadcast band. Repeat this for an FM signal serving as the audio portion of a TV broadcast.

Solution:a.) FM broadcast band:M=

80= x 100actual freq. deviation= (80/100)*75=60 kHzCS=2xfreq.deviation=2x60kHz=120 kHzb.) Sound in TV broadcast:80=Freq.deviation= (80/100)*25=20kHzCS= 2xfreq.deviation=40 kHz

4. Determine the permissible range in modulation index for commercial FM that has a 50 to 20kHz modulating frequency.Given: fm1=50Hz; fm2=20kHz; δmax=75kHz (for commercial FM)Required: range of modulation index (m)Solution:mf= mf1= to mf2=

3.75 Hz to 1.5kHz

4. If an 18MHz band were to be considered for use with the same standards that apply to the 88-108MHz FM broadcast band. How many FM stations could be accommodated?Given:

6. An FM signal has a deviation of 10kHz and is modulated by a sine wave with a frequency of 5kHz. The carrier frequency is 150MHz and the signal has a total power of 12.5W, operating into an impedance of 50Ω. Calculate: modulation index and the bandwidth.

Given: fm=5kHz; δ=10kHz; fc=150MHz; PT=12.5W; RL=50Ω

Required: a. mb. B

Solution:a. mf = ; = mf = 2

b. F

B = 2 x fm x number of sig. sidebands = 2 (5kHz)(4)

B = 40kHz

There are 4 significant sidebands according to Bessel’s function for this

signal

7. For the FM modulator with a modulation index of 1, a modulating signal vm(t)=8sin(21000t) and an unmodulated carrier vm(t)=15sin(2540t) and the load resistance is 85Ω. Determine :a) the unmodulated carrier power b) the power in the angle modulated wave

Given: from vm(t)=8sin(21000t); Vm=8V and fm=1kHz from vm(t)=15sin(2540t) ; VC=15V and fC=540Hz RL=85Ω;

Required: a. PC

b. PT

Solution:a.PC = = PC = 1.32W

b. PT = (J02 + 2J1

2 + 2J22 + 2J3

2 + 2J42 + 2J5

2 ) PC

= [0.222 + 2(0.58)2 + 2(0.35)2 + 2(0.13)2 +2(0.03)2] (1.32W)

PT = 1.3224 W

ADDITIONALPROBLEMS!

SampleProblems:1. What is the bandwidth required for an FM signal in which

the modulating frequency is 10 kHz and the maximum deviation is 50 kHz?

Approximate BW Formula = 2 ( δ + fm )Approximate BW Formula = 2 ( 50kHz + 10kHz )

Approximate BW Formula = 120kHz

Narrowband FM = 2fm

Narrowband FM = 2 (10 kHz)

Narrowband FM = 20 kHz

Wideband FM = 2δWideband FM = 2 (50 kHz)

Wideband FM = 100 kHz

2. Determine the frequency deviation and the carrier swing of an FM signal which is the audio portion of a TV signal and has a percent modulation of 85%?

Sample Problems:

C.S. = 2δ

Percent Modulation = [ (δactual) / (δmax) ] x 100

C.S. = 127.500kHz

Percent Modulation = [ (δactual) / (75kHz) ] x 100

C.S. = 2(63.750kHz)

Frequency Deviation (δ) = 63.750kHz

3. Determine the relative total power of the carrier and side frequency when the modulation index is 0.25 for a 10kW FM transmitter.

Sample Problems:

Pt = 9.892kW

For a modulation index of 0.25:

Modulation Index Carrierm J0 J1

0.25 0.98 0.12

Pt = (J02

+ 2J12)(P)

Pt = [(0.98)2 + 2(0.12)2][10kW]

4. Determine the permissible range in modulation index for commercial FM that has a 50 to 20kHz modulating frequency.

Sample Problems:

mf1 = (δ / fm)

mf2 ≤ X ≤ mf1

Permissible range in modulation index = 3.75 ≤ X ≤ 1500

mf1 = (75kHz / 50Hz)mf1 = 1500

mf2 = (δ / fm)mf2 = (75kHz / 20kHz)mf2 = 3.75

5. If an 18MHz band were to be considered for use with the same standards that apply to a 88 – 108MHz FM broadcast band, how many FM stations could be accommodated?

Sample Problems:

Number of stations = 18MHz / 200kHzNumber of stations = 90 Stations

For a standard 88 – 108MHz FM broadcast:

108MHz – 88MHz = 20MHzNumber of stations = 20MHz / 200kHzNumber of stations = 100 Stations

For the given condition:

PROBLEMSPART 4

Phase Modulation Problems

1. A phase modulator has a sensitivity of 2 rad/V. Calculate the phase deviation produced with a sine wave input of 5 volts peaks.

Given:K = 2 rad/V

Vm = 5 Volts

Required:Phase deviation, ( σ )

Working Solution:

σ = (K)(Vm)

σ = (2 rad/V)(5 Volts)

σ = 10 rad

2. A phase modulator has a phase deviation of 24 rad. If it has a sensitivity of 5 rad/Volts, What is the peak voltage of the phase modulator?Given:σ = 24 rad

K = 5 rad/Volts

Required:Peak Voltage, (Vm)

Working Solution:


σ = (K)(Vm)Vm = σ / K

Vm = 4.8Vm = ( 24 rad ) / ( 5 rad/ Volts )

3. A PM modulator has a phase deviation sensitivity of 2.5 radian / V and a modulating signal of Vm(t) = 2Cos(4000π)t. Calculate the peak phase deviation.

Given:K = 2.5 radian/ V

Vm(t) = 2Cos(4000π)t

Required:Peak Phase Deviation, ( σ )

Working Solution:

Σ = 5 rad


σ = (K)(Vm)

σ = (2.5 radian/V) (2)

LAST PROBLEMS

1. A broadband radio transmitter radiates 10kW of carrier power when the percentage modulation is 75%. How much is the transmitted power in DSBFC and in SSBSC?

Given: Pc=10kW;M=75%

Required: PT in DSBFC and SSBSC

Solution:For DBSFC : PT=PC () ; PC = 10kW()PT= 12.81kWFor SSBSC:PT=PC() ; PC =10kW()PC = 1.41kW

The carrier power of a signal that undergone the process of modulation by suppressing the carrier wave is 9kW. The modulated wave is vm(t)=8sin(21000t) and the unmodulated wave is vm(t)=10sin(2200t). What is the modulating index, carrier power and the bandwidth of the signal if the maximum modulating frequency is 5kHz?Given: from vm(t)=8sin(2850t); Vm=8V &fm=1kHz from vm(t)=10sin(2200t); Vc=10V & fc=200Hz

PT=9kW

Required: m, PCand B if fm=5kHz

Solution:a. m=; m=;m=0.8

b. PT=PC(); 9kW=PC() PC=56.25kW

c. B=2fm; B=2(5kHz); B=10kHz

3. A DSBFC AM transmission of 3kW is fully modulated. Calculate the power transmitted, if it is transmitted as a single sideband signal.

Given: PT=3kW; m=1 (problem indicates that the signal is

fully modulated)

Required: PC

Solution:a. For DBSFC : PT=PC () ; 3kW= PC ()Pc= 2kWIf it is transmitted as a single sideband: PT=PC();PT=2kW();PT=500W

Submitted by:LIBUNAO, Ralph ThomasMAGALLON, Mark JosephMAGLALANG, Jess BrianMARIÑO, AlvinMENDOZA, MalvinMORAL, Vincent MillerROCAVERTE, DacelyROYO, KristelSALCEDA, Jane MarielleSALUD, Kenneth KarlSORREDA, Neil Adrian

BS-ECE IV-2

Submitted to:Dr. Carlos Sison

Principles of Communications Lecture FINAL

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