1 ANGLE MODULATION CHAPTER 3 Part 2 (cont’d) Transmission / modulation Reception / demodulation...

ANGLE MODULATIONCHAPTER 3

Part 2 (cont’d)

Transmission / modulation

Reception / demodulation

EKT 231 : COMMUNICATION SYSTEMEKT 231 : COMMUNICATION SYSTEM

CHAPTER 3 : ANGLE MODULATIONCHAPTER 3 : ANGLE MODULATION

• Angle modulated waveforms generally classified as :– Low modulation index – Medium modulation index– High modulation index

• Low –index FM systems sometimes called as Narrow Band FM or NBFM

Generation of FM

• Two major FM generation:i) Direct method:

i) Straight forward, requires a VCO whose oscillation frequency has linear dependence on applied voltage.

ii) Advantage: large frequency deviationiii) Disadvantage: the carrier frequency tends to

drift and must be stabilized.iv) example circuit:

i) Reactance modulatorii) Varactor diode

Generation of FM (cont’d)

ii) Indirect method: i. Frequency-up conversion.

ii. Two ways:

a. Heterodyne method

b. Multiplication method

iii. One most popular indirect method is the Armstrong modulator

Simple direct FM modulator

TomasiElectronic Communications Systems, 5e

Accousticalmechanical energy

Cm varied causes the resonant frequency varied

Varactor diode direct FM modulator

TomasiElectronic Communications Systems, 5e

Deviate the frequency via capacitances changes

JFET reactance modulator

Called as reactance modulator – JFET (active device)

The modulating signal varies with the reactant of Q causes corresponding change in the resonant frequency

Armstrong modulator

IntegratorBalanced modulator

Down converter

Frequency multiplier (x n)

Crystal oscillatorPhase shifter

Vc(t)fc

Vm(t)fm

FM Detection/Demodulation• FM demodulation

– is a process of getting back or regenerate the original modulating signal from the modulated FM signal.

– It can be achieved by converting the frequency deviation of FM signal to the variation of equivalent voltage.

– The demodulator will produce an output where its instantaneous amplitude is proportional to the instantaneous frequency of the input FM signal.

FM detection (cont’d)

• To detect an FM signal, it is necessary to have a circuit whose output voltage varies linearly with the frequency of the input signal.

• The most commonly used demodulator is the PLL demodulator. Can be use to detect either NBFM or WBFM.

PLL Demodulator

Phase detector

Low pass filter

Amplifier

FM input

• The phase detector produces an average output voltage that is linear function of the phase difference between the two input signals. This low frequency component is selected by LPF.

PLL Demodulator

• After amplification, part of the signal is fed back through VCO where it results in frequency modulation of the VCO frequency. When the loop is in lock, the VCO frequency follows or tracks the incoming frequency.

• Let instantaneous freq of FM Input,

fi(t)=fc +k1vm(t),

and the VCO output frequency,

f VCO(t)=f0 + k2Vc(t);

f0 is the free running frequency.

• For the VCO frequency to track the instantaneous incoming frequency,

fvco = fi

PLL Demodulator f0 + k2Vc(t)= fc +k1vm(t)

• If VCO can be tuned so that fc=f0, then

• Where Vc(t) is also taken as the output voltage, which therefore is the demodulated output

)()( 10 tvkfftV mcc

)()( 1 tvktV mc

Comparison AM and FM• Its the SNR can be increased without increasing transmitted power

about 25dB higher than in AM

• Certain forms of interference at the receiver are more easily to suppressed, as FM receiver has a limiter which eliminates the amplitude variations and fluctuations.

• The modulation process can take place at a low level power stage in the transmitter, thus a low modulating power is needed.

• Power content is constant and fixed, and there is no waste of power transmitted

• There are guard bands in FM systems allocated by the standardization body, which can reduce interference between the adjacent channels.

Application of FM

• used by most of the field VHF portable, mobile and base radios in exploration use today.

• It is preferred because of its immunity to noise or interference and at the frequencies used the antennas are of a reasonable size.

Summary of angle modulation

Summary (cont’d)

• Bandwidth:

a) Actual minimum bandwidth from Bessel table:

b) Approximate minimum bandwidth using Carson’s rule:

)(2 mfnB

)(2 mffB

Summary (cont’d)

• Multitone modulation (equation in general):

21 mmci KvKv

....cos2cos2 2211 tftfci

......sinsin 22

Summary (cont’d)

..].........sinsinsin[

]sinsinsin[

tmtmtV

Summary (cont’d)-Comparison NBFM&WBFM

ANGLE MODULATION

Part 3•Advantages

•Disadvantages

Advantages• Wideband FM gives significant improvement in the SNR at the

output of the RX which proportional to the square of modulation index.

• Angle modulation is resistant to propagation-induced selective fading since amplitude variations are unimportant and are removed at the receiver using a limiting circuit.

• Angle modulation is very effective in rejecting interference. (minimizes the effect of noise).

• Angle modulation allows the use of more efficient transmitter power in information.

• Angle modulation is capable of handing a greater dynamic range of modulating signal without distortion than AM.

Disadvantages

• Angle modulation requires a transmission bandwidth much larger than the message signal bandwidth.

• The capture effect where the wanted signal may be captured by an unwanted signal or noise voltage.

• Angle modulation requires more complex and inexpensive circuits than AM.

END OF ANGLE MODULATION

Revision (Decibel)

• Decibel compress the wide range of signals into a smaller group of numbers and make measurements and type of analysis of system performance more convenient.

Absolute power ratioPower gain

Absolute voltage ratioVoltage gain

AM DSB-FCAM DSB-SCAM SSB-FCAM SSB-SC

CHAPTER 3 : ANGLE MODULATIONCHAPTER 3 : ANGLE MODULATIONRevisionRevision

Where P1 = power level 1 (watts) P2 = power level 2 (watts) i.e the reference power

• If two powers expressed in same units, the ratio can be expressed as:

dB=10 log(10) (P1/P2)

In electronic circuits, to measure a power gain

or loss Ap(dB)=10log(10) (Pout/Pin)

Where Ap(dB) = Power gain (dB)

Pout/Pin = absolute power gain (unitless)

Av(dB)=20log(10) (Eout/Ein)

Where Av(dB) = Power gain (dB)

Eout = output voltage (volts)

Ein = output voltage (volts)

Eout/Ein = Absolute voltage gain (unitless)

Voltage gain in dB is expressed as

A dBm is a unit of measurement used to indicate the ratio of power level with respect to a fixed reference level (1mW)

OR…..“decibel relative to 1 milliwatt”

dBm = 10log(P/1mW)

Where P is the power in watts

1mW is the reference voltage

Similarly…..

A dBW is a unit of measurement used to indicate the ratio of power level with respect to a fixed reference level (1W)

OR…..“decibel relative to 1 watt”

dBW = 10log(P/1W)

Where P is the power in watts

1W is the reference voltage

Q1) Express the power values as dBW (a) 1W

(b) 0.73W

(c) 950W

Answer :

dBW=10log(P/1W);

0dBW,-1.37dBW, 29.8dBW.

Q2) Express 200W as dBW and as dBm

Answer :

dBm=10log(P/1mW);

dBW=10log(P/1W);

23dB,53dBm

Q3) Convert the absolute power ratio of 200 to a power gain in dB.

Answer :Ap(dB) =10log(10) (Pout/Pin) Ap(dB) =10log(10)(200) = 23dB

Q4) Convert the power gain of Ap = 23dB

to an absolute power ratio.

Answer :23dB =10log(10) (Pout/Pin) (Pout/Pin) = 200

Q5) A signal is attenuated from 5V to 0.1V. What is the corresponding value

in dB?

Answer :

Av(dB) = 20log(0.1/5)=-34dB. OR 34dB loss

Q6) A 100-W signal goes through a circuit with a 30-dB loss. What is the final power value?

Answer :

Ap(dB) = 10 log(P/100)=-30dB, P=0.1W

Q7) For an AM DSBFC wave with a peak unmodulated carrier voltage Vc = 10 Vp,frequency of 100kHz, a load resistor, RL = 10 , frequency of modulating signal of 10kHz and m = 1.0, Determine

i) Powers of the carrier and the upper and lower sidebands.

ii) Total power of the modulated wave.

iii) Bandwidth of the transmitted wave.

iv) Draw the power and frequency spectrum.

Answer the following questions based on information from Q7

Q8) For the same given values, determine questions (ii)-(iv) for a AM DSB-SC, AM SSB-FC and AM SSB-SC systems.

Q9) Determine also the percentage of power saved in each of the system design.

Solution Q7 : DSBFCi)

iii) Bandwidth = 2(fmmax) = 2(10kHz) = 20kHz

PPccct

5.7)5(4

clsbusb

5)10(2

Solution Q7 :For DSB-SC

iii) Bandwidth = 2(fmmax) = 2(10kHz) = 20kHz

5.2)5(4

fc 110kHz90kHz

Solution Q8 : For SSB-FC

iii) Bandwidth=fmmax=10kHz

25.6)5(4

100kHz 110kHzfc-fm

Solution Q8 : For SSB-SC

iii) Bandwidth=fmmax=10kHz

P ct 25.1)5(4

fc110kHzfc-fm

Solution Q10 :For DSB-SC

%67.16%)100(5.7

25.125.65.7

WPower

WWWPower

%67.66%)100(5.7

55.25.7

WPower

WWWPower

Solution Q10 :For SSB-FC

%33.83%)100(5.7

25.625.15.7

WPower

WWWPower

Solution Q10 :For SSB-SC

• An angle modulated signal with carrier frequency as described by the following equation

• Determine(i) The power of the modulated signal (let R=1Ω) ans:50W

(ii) Prove that (iii) The frequency deviation,Δf ans:12.387kHz

(iv) The deviation ratio, DR ans:12.387Hz

(v) The phase deviation, m ans:15 rad

(vi) Estimate the bandwidth of the modulated signal. ans:26.774kHz

5102 Xc

)2sin103sin5cos(10)( tkktttm c

)2cos203cos15 tkkktkci

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1 ANGLE MODULATION CHAPTER 3 Part 2 (cont’d) Transmission / modulation Reception / demodulation...

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