te4201-communication electronics 1 10. fm generation and transmission angle modulation angle...

1TE4201-Communication Electronics

10 . FM Generation and Transmission

Angle modulation

FM analysis, Bessel function

Bandwidth of FM waves

Noise suppression

Direct & Indirect FM generation


FM WaveFM Wave

Modulating intelligence waveModulating intelligence wave

Difference betweenDifference betweenAM WaveAM Wave

andandFM WaveFM Wave

Peak frequency of Peak frequency of FM waveFM wave

Peak amplitude of Peak amplitude of AM waveAM wave

Peak of intelligence wavePeak of intelligence wave


where :where :e= instantaneous voltagee= instantaneous voltageA= peak value of the carrier waveA= peak value of the carrier wavecc= carrier angular velocity (2= carrier angular velocity (2ffc c ))mmpp= maximum phase shift caused by intelligence signal amplitude= maximum phase shift caused by intelligence signal amplitudeii= modulating(intelligence signal angular velocity (2= modulating(intelligence signal angular velocity (2ffi i ))

FM AnalysisFM Analysis

tsinmtsinAe ifc FMFM tsinmtsinAe ipc PMPM

= maximum frequency deviation (shift ) caused by intelligence = maximum frequency deviation (shift ) caused by intelligence signal amplitudesignal amplitudemmff = FM = FM modulation indexmodulation index = ratio of maximum freq. deviation of = ratio of maximum freq. deviation of the carrier to the intelligence frequencythe carrier to the intelligence frequency

if f

m

FM analysis, Bessel functionFM analysis, Bessel function


Simple FM Generator

capacitormicrophone

soundwaves Oscillator

FM radiowaves

C + C L )CC(L21fL

C L LC21fO

C - C L )CC(L21fH

FM wave varies from fL to fH with fO at the center (when there is no sound waves)

Sound wave will cause capacitor microphone to change it’s capacitance by C


FM EquationFM Equation tsinmtsinAe ifc FMFM

if f

m wherewhere

sine of sinesine of sine

Bassel functionBassel function solution of the FM equation is solution of the FM equation is

componentf2sidebandondsect2cost2cos)m(J

componentfsidebandfirsttcostcos)m(Jcomponentcarriertcos)m(J

componentwaveFM)t(f

t3cost3cos)m(Jt2cost2cos)m(J

tcostcos)m(Jtcos)m(J)t(f

iicicf2

iicicf1

cf0

c

icicf3

icicf2

icicf1cf0c

...............!3n!3

2m

!2n!22

m

!1n!12

m

!n1n

2m

)m(Jwhere

6f

4f

2f

ffN


Solution of Bassel function (chart)Solution of Bassel function (chart)


Solution of Bassel function (graph)Solution of Bassel function (graph)

2f2fii

ffiiffCC

2f2fii


BWBW

4f4fi1i1

if

if fmthen

fmcesin

If If is fixed the mis fixed the mff x f x fii is constant . For small f is constant . For small fii , m , mf f will be large, and more side frequencies will be large, and more side frequencies

BWBW

10f10fi2i2

mmff = = 0.50.5

mmff = = 2.52.5

If maximum deviation of the FM wave is fixed at If maximum deviation of the FM wave is fixed at 20kHz.20kHz.Determine the Determine the BW required for an intelligence of fi = 10kHz.BW required for an intelligence of fi = 10kHz.

ExampleExample

mmf f = = ffii20kHz/10kHz = 2 …..from table pair of important sideband is 4pairs 20kHz/10kHz = 2 …..from table pair of important sideband is 4pairs (J(J00 to J to J44) or 8 side frequencies. BW required for an intelligence of f) or 8 side frequencies. BW required for an intelligence of f ii = 10kHz is = 10kHz is 8x10kHz = 8x10kHz = 80kHz80kHz

If maximum deviation of the FM wave is fixed at If maximum deviation of the FM wave is fixed at 20kHz.20kHz.Determine the Determine the BW required if the intelligence is changed to fi = 5kHz.BW required if the intelligence is changed to fi = 5kHz.

ExampleExample

mmf f = = ffii20kHz/5kHz = 4 …..from table pair of important sideband is 7pairs (J20kHz/5kHz = 4 …..from table pair of important sideband is 7pairs (J00 to Jto J77) or 14side frequencies. BW required for an intelligence of f) or 14side frequencies. BW required for an intelligence of f ii = 5kHz is = 5kHz is 14x5kHz = 14x5kHz = 70kHz70kHz

Bandwidth of FM wavesBandwidth of FM waves


Carson’s ruleCarson’s rule

BW = 2 (BW = 2 (maxmax + f + fii maxmax ) )

Approximate BW prediction of a FM wave can be made Approximate BW prediction of a FM wave can be made by Carson’s rule……. by Carson’s rule…….

If maximum deviation of the FM wave is fixed at If maximum deviation of the FM wave is fixed at 20kHz.20kHz.Determine the BW Determine the BW required for an intelligence signal having a higher cutoff frequency at fi required for an intelligence signal having a higher cutoff frequency at fi maxmax = = 10kHz.10kHz.

ExampleExample

BW = 2 (BW = 2 (maxmax + f + fii maxmax ) = 2 (20+ 10 )kHz = 60kHz ) = 2 (20+ 10 )kHz = 60kHzNote from important side frequency of Bessel function is BW= 80kHz.Note from important side frequency of Bessel function is BW= 80kHz. ( If neglecting sidebands having an amplitude less than 10% of maximum ( If neglecting sidebands having an amplitude less than 10% of maximum value of fundamental amplitude component (unmodulated carrier Jvalue of fundamental amplitude component (unmodulated carrier J0 0 = 1) = 1) then there will be 6 side frequencies (Jthen there will be 6 side frequencies (J00 to J to J3 3 ) as J) as J44 = 0.03 can be neglected. = 0.03 can be neglected. Then BW = 6x10kHz = 60kHz)Then BW = 6x10kHz = 60kHz)


Instantaneous voltage of the FM wave is e (t) = 2000 sin ( 2Instantaneous voltage of the FM wave is e (t) = 2000 sin ( 2 x10 x108 8 t+ 2 sin 2t+ 2 sin 2 x10x1044t )t )

It is applied to an antenna having an impedance of 50It is applied to an antenna having an impedance of 50. Determine:. Determine:(a)(a) Carrier frequency fCarrier frequency fCC

(b)(b) Transmitted power PTransmitted power PTT

(c)(c) mmf f , (d) f, (d) fii (e) BW (by two methods)(e) BW (by two methods)(f)(f) Power in the Power in the largestlargest and the and the smallest smallest sidebands predicted by the table of sidebands predicted by the table of

Bessel function .Bessel function .

ExampleExample

(a)(a) Carrier frequency fCarrier frequency fCC = 10 = 1088Hz =100MHzHz =100MHz(b) Transmitted power P(b) Transmitted power PTT = (E = (E rmsrms ) )22/ R = (2000/1.414 )/ R = (2000/1.414 )22/ 50/ 50 = 40kW = 40kW

(c) (c) mmf f = 2 (from = 2 (from 22 sin sin x10 x1044t ) t )

(d)(d) Information frequency (from 2 Information frequency (from 2 sin sin x10 x1044tt ) f ) fii = (10 = (1044/2)kHz = /2)kHz = 5kHz5kHz(e)(e) (Carson’s rule) (Carson’s rule) = m= mff x f x fii =2x5k=10kHz =2x5k=10kHz then BW = 2 (then BW = 2 (maxmax + f + fii maxmax ) = 2 (10+ 10 )kHz = 40kHz ) = 2 (10+ 10 )kHz = 40kHz

Bessel function) table, 4 pair (or 8 side frequencies, JBessel function) table, 4 pair (or 8 side frequencies, J0 0 to Jto J44 ) then BW = ) then BW = 8x5kHz = 40kHz8x5kHz = 40kHz

(f)(f) (Bessel function) table (Bessel function) table largest SB2largest SB2 is 58% of unmodulated carrier J is 58% of unmodulated carrier J0 0 = 1, = 1, then Pthen PSB2SB2 = (0.58E = (0.58E rmsrms ) )22/ R = (0.58x2000/1.414 )/ R = (0.58x2000/1.414 )22/ 50/ 50 = 13.5kW = 13.5kW

smallest SB4smallest SB4 is 3% of unmodulated carrier J is 3% of unmodulated carrier J0 0 = 1,then P= 1,then PSB4SB4 = (0.03E = (0.03E rmsrms ) )22/ R = (0.03x2000/1.414 )/ R = (0.03x2000/1.414 )22/ 50/ 50 = 0.036kW=36W = 0.036kW=36W


Broadcast FM Broadcast FM (entertainment)(entertainment)

200 kHz200 kHzChannel Channel widthwidth

Carrier1Carrier1

Maximum Maximum deviation deviation = = 75kHz75kHz


Carrier2Carrier2


100MHz100MHz


2x25kHz 2x25kHz guard guard bandband

0.1MHz0.1MHz 0.1MHz0.1MHz100.1MHz100.1MHz

100.2MHz100.2MHzCarrier accuracy Carrier accuracy 2kHz2kHz

+75kHz+75kHz -75kHz-75kHz

100.4MHz100.4MHz


= 75kHz = m= 75kHz = mff x f x fii = fixed to a maximum value. = fixed to a maximum value.Therefore mTherefore mff decreases if f decreases if fii increases increases (at high intelligent frequency no. of sideband is less because of small m(at high intelligent frequency no. of sideband is less because of small m f f ) )


Narrowband FM Narrowband FM (communications)(communications)

Narrowband FM is widely used in communications such as police, aircraft, Narrowband FM is widely used in communications such as police, aircraft, taxicabs, whether service, and private industry networks. Often voice taxicabs, whether service, and private industry networks. Often voice transmission whose highest frequency allowed is ftransmission whose highest frequency allowed is f ii = 3kHz. = 3kHz.

Narrowband FM has a (FCC) BW allocation of 10kHz to 30kHz. Narrowband FM has a (FCC) BW allocation of 10kHz to 30kHz.

Narrowband FM has a modulation index mNarrowband FM has a modulation index m f f (or a deviation ratio) of (or a deviation ratio) of (10kHz/3kHz)=3.3 to (30kHz/3kHz)= 10 compared to (75kHz/15kHz) = 5 in (10kHz/3kHz)=3.3 to (30kHz/3kHz)= 10 compared to (75kHz/15kHz) = 5 in Broadcast FMBroadcast FM

In FM, the waveform amplitude never varies just frequency.In FM, the waveform amplitude never varies just frequency. Therefore the Therefore the transmitted power must remain constant regardless of level of modulation. It is transmitted power must remain constant regardless of level of modulation. It is thus seen that whatever the energy is contained in the side frequencies has thus seen that whatever the energy is contained in the side frequencies has been obtained from the carrier.been obtained from the carrier. No additional energy is added during the No additional energy is added during the modulation process. In addition amplitude of the carrier in FM depends upon modulation process. In addition amplitude of the carrier in FM depends upon intelligence signalintelligence signal


(a)(a) Determine the permissible range in maximum modulation index for Determine the permissible range in maximum modulation index for commercial FM that has 30Hz to 15kHz modulating frequencies.commercial FM that has 30Hz to 15kHz modulating frequencies.

(b)(b) Repeat for a narrowband system that allows a maximum deviation of Repeat for a narrowband system that allows a maximum deviation of 10kHz and 100Hz to 3kHz modulating frequencies.10kHz and 100Hz to 3kHz modulating frequencies.

ExampleExample

5kHz15kHz75

fmto2500

Hz30kHz75

fm)a(

i(min)f

i(max)f

3.3kHz3kHz10

fmto100

Hz100kHz10

fm)b(

i(min)f

i(max)f

ExampleExample

Determine the Determine the relative relative total power of the carrier and side frequencies when mtotal power of the carrier and side frequencies when m f f = = 0.25 for a 10kW FM transmission0.25 for a 10kW FM transmission

kW10kW892.9144.0604.9PPPkW144.0kW10)12.0(PandkW604.9kW10)98.0(P

,powerngtransmittikW10forand12.0J,98.0J,25.0mat

sideandcarriertotal

2sideband

2carrier

10f


FM limiter and

detector

AM limiter and

detector

External noise

Noise at AM detector output

Noise absent at AM detector output

The most important advantage of FM over AM is the superior noise The most important advantage of FM over AM is the superior noise characteristics. Static noise is rarely heard on FM although it is quite common in characteristics. Static noise is rarely heard on FM although it is quite common in AM.AM.In FM intelligence is not carried by amplitude changes but instead by frequency In FM intelligence is not carried by amplitude changes but instead by frequency changes. The spikes of external noise picked up during transmission are clipped changes. The spikes of external noise picked up during transmission are clipped off by a off by a limiter circuitlimiter circuit and through detector circuits insensitive to amplitude and through detector circuits insensitive to amplitude changes.changes.Unfortunately the noise spike still causes an Unfortunately the noise spike still causes an undesired phase shiftundesired phase shift of the FM of the FM signal, and this frequency shift cannot be removed.signal, and this frequency shift cannot be removed.

Noise suppressionNoise suppression


If FM receiver is tuned to 96MHz, the receiver provide gain only for frequencies If FM receiver is tuned to 96MHz, the receiver provide gain only for frequencies near 96MHz. The noise also should be around this frequency since all other near 96MHz. The noise also should be around this frequency since all other frequencies are greatly attenuated.frequencies are greatly attenuated.

FM noise analysisFM noise analysis

Adding the noise to the desired signals will give a resultant signal with a different Adding the noise to the desired signals will give a resultant signal with a different phase angle then the desired FM signal alone. Since the noise can cause phase phase angle then the desired FM signal alone. Since the noise can cause phase modulation, it indirectly causes an undesired FM. Value of frequency deviation modulation, it indirectly causes an undesired FM. Value of frequency deviation caused by phase modulation is : caused by phase modulation is : = = x fx fii

WhereWhere = frequency deviation, = frequency deviation, = phase shift (radians), f= phase shift (radians), f ii = intelligence = intelligence frequencyfrequencyConsider the noise signal to be one-half of the amplitude of the desired signal. Consider the noise signal to be one-half of the amplitude of the desired signal. S/N = 2 intolerable in AM. But in FM it is not so bad as shown by the calculations S/N = 2 intolerable in AM. But in FM it is not so bad as shown by the calculations below. below. Since the noise “N” and signal “S” are at different frequencies (but in the same Since the noise “N” and signal “S” are at different frequencies (but in the same range, as dictated by receiver tuned circuits) the noise is a rotating vector with range, as dictated by receiver tuned circuits) the noise is a rotating vector with “S” as reference. The resultant phase “ “S” as reference. The resultant phase “ ” is maximum when “N” is ” is maximum when “N” is perpendicular to resultant vector “R”. Then perpendicular to resultant vector “R”. Then = sin= sin-1-1 (N/S) = sin (N/S) = sin-1-1 (1/2) = 30deg (1/2) = 30deg = 0.52 rad. and maximum possible frequency deviation is = 0.52 rad. and maximum possible frequency deviation is = = x fx fi i = 0.52 x = 0.52 x 15kHz = 7.5kHz15kHz = 7.5kHz

RR

NN

SS


ExampleExample

Determine the worst case output S/N for a broadcast FM program that has a maximum signal frequency of 5kHz. The input is S/N = 2

2)inFM(NS30

kHz5.2kHz75)outFM(

NS

2.5kHz 5kHz x 0.52 f x adr 0.52 30deg (1/2) sin (N/S) sin i-1-1

ExampleExample

Determine the worst case output S/N for a narrowband FM communication that has a maximum deviation of = 10kHz and maximum signal frequency of 3kHz. The input is S/N = 3

3)inFM(NS10

kHz1kHz10)outFM(

NS

kHz 1 3kHz x 0.34 f x adr 0.34 19.5deg (1/3) sin (N/S) sin i-1-1

In standard broadcast,maximum deviation is 75kHz at full intelligence signal In standard broadcast,maximum deviation is 75kHz at full intelligence signal amplitude. Now the deviation from noise is 7.5kHz then S/N of the amplitude. Now the deviation from noise is 7.5kHz then S/N of the FMFM detector detector will be (75kHz/7.5kHz) = will be (75kHz/7.5kHz) = 10 = S/N10 = S/N which is better than which is better than AMAM whose whose S/N = 2S/N = 2

FM S/N is better than AM S/NFM S/N is better than AM S/N


• Reversed bias Reversed bias capacitance of the capacitance of the Varactor Diode DVaractor Diode D22 is in parallel with is in parallel with

tuned capacitor Ctuned capacitor C11 . . C of DC of D22 change due change due

to change of to change of reverse bias by the reverse bias by the modulating signal. modulating signal.

Therefore output Therefore output freq. will change freq. will change

due to change of C due to change of C of Dof D22

Reactance Reactance ModulatorModulator

Crosby Crosby ModulatorModulator

VCO VCO ModulatorModulator

Varicap ModulatorVaricap Modulator

Varactor diode DVaractor diode D22

Varying reverse Varying reverse bias of Varactor bias of Varactor changes the changes the Varactor Varactor capacitancecapacitance

FM

modulatingsignal

FM outFM out

Modulating Modulating intelligence intelligence signalsignal

Carrier fCarrier fCC oscillatoroscillator

LL11CC11

Direct FM generation


Reactance Modulator (equivalent capacitance)Reactance Modulator (equivalent capacitance)

i1 iD

R

C

e Ceq= CRgmeg

cg

c11g jXR

ReejXR

eibutRie

c

mgmd jXR

Regegi

Rg

jXg1

RegjXRe

ieZ

m

c

mm

c

d

meqeqmm

c

c

CRgCThenCj1

CRg1j

RgjXZThen

XRbetochosenisRIf

• When modulating signal is connected to Gate of FET, VWhen modulating signal is connected to Gate of FET, VGSGS will change. This will change. This will make gwill make gmm to change. Then C to change. Then Ceqeq = CRg = CRgmm will change. will change.• If terminal “e” is connected to the LC tuned circuit of the carrier frequency If terminal “e” is connected to the LC tuned circuit of the carrier frequency oscillator, then carrier frequency will change producing FM wave.oscillator, then carrier frequency will change producing FM wave.

P

GS0mm V

V1gg


i1 iDR

C

eLeq=(CR/ gm)eg

Reactance Modulator (equivalent inductance)Reactance Modulator (equivalent inductance)

C

Cg

C1C1g jXR

)jX(eejXRei)jX(ie

C

mCgmd jXR

)eg(jXegi

mmmC

C

C

mCd gCRj

g1

gjXjXR

jXR)eg(jX

eiez

meqeq

mmm gCRLWhereLj

gCRjzthen

gCRj

g1If

Example Example

Show that the FET reactance Show that the FET reactance circuit shown will produce an circuit shown will produce an equivalent reactance of equivalent reactance of LLeqeq =(CR/g =(CR/gmm))


Reactance Reactance FM FM

modulatormodulator

i1 iD

R

C

RFC

C0 L0

C0

VDD

modulatingsignal

FET reactance circuit will provide changing equivalent FET reactance circuit will provide changing equivalent inductance proportional to modulating signal amplitude. As it is inductance proportional to modulating signal amplitude. As it is in parallel with Lin parallel with L00CC00 tuned oscillator (for carrier frequency), the tuned oscillator (for carrier frequency), the carrier frequency will deviate according to dL produced by the carrier frequency will deviate according to dL produced by the FET reactance circuit, creating an FM wave.FET reactance circuit, creating an FM wave.

FM

modulatingsignal

FET Reactance ModulatorFET Reactance Modulator


BJT Reactance ModulatorBJT Reactance Modulator

R

C

C0 L0

C0

VCC

modulatingsignal

Masteroscillator

FMoutput

BJ TRactancemodulator

R6

R5R2

R3R4

C3

• When terminal When terminal ““CC00=C=Ceqeq” is connected ” is connected to the Lto the L00CC00 tuned circuit tuned circuit of the carrier frequency of the carrier frequency oscillator, then carrier oscillator, then carrier frequency will change frequency will change resulting an FM wave.resulting an FM wave.

• When modulating signal is connected When modulating signal is connected to Base of BJT, Ito Base of BJT, IBB will change. This will will change. This will make make iibb to change. If current to change. If current generator of BJT having an equivalent generator of BJT having an equivalent in such a way that in such a way that iibb = g = gmmeegg where where ggmm=(=(iibb )/e )/eg g = = rre e =1/r=1/ree , then C , then Ceqeq = = CRgCRgmm = CR/r = CR/re e will change. Note that rwill change. Note that ree (=26mV/I(=26mV/ICC) will change with I) will change with ICC (= (= IIBB))

FM

modulatingsignal


Linear IC ,Voltage Controlled Oscillator (VCO) FM generation)Linear IC ,Voltage Controlled Oscillator (VCO) FM generation)

A voltage controlled oscillator (VCO) produces an output A voltage controlled oscillator (VCO) produces an output frequency that is directly proportional to a control voltage frequency that is directly proportional to a control voltage level. The circuitry necessary to produce such an oscillation level. The circuitry necessary to produce such an oscillation with a high degree of linearity between control voltage and with a high degree of linearity between control voltage and frequency was formerly productive on a discrete component frequency was formerly productive on a discrete component bias. But now low cost monolithic LIC VCO are available, they bias. But now low cost monolithic LIC VCO are available, they make FM generation extremely simple.make FM generation extremely simple.

Voltage Controlled Oscillator


VVV

43Note

VCRVV2f C

11

C0

The figure shown is a The figure shown is a 566VCO566VCO IC. It provide a high-quality FM generator with the IC. It provide a high-quality FM generator with the modulating voltage applied at VC terminal. The FM output can be taken as a modulating voltage applied at VC terminal. The FM output can be taken as a square or a triangular waveform from the IC. Feeding of any of these two outputs square or a triangular waveform from the IC. Feeding of any of these two outputs into an into an LC tank circuitLC tank circuit resonant at the center frequency of the VCO ( = carrier resonant at the center frequency of the VCO ( = carrier frequency of FM) subsequently provide standard FM signal by frequency of FM) subsequently provide standard FM signal by flywheelflywheel effect. effect.

Current

source

Schmitt trigger

Buffer amplifier

Buffer amplifier

VC

C1

R1

modulating input

V+

77

33

44

8866

55

11

566 VCO566 VCO


CROSBY ModulatorCROSBY Modulator

The mixer shown has the 90MHz carrier and 88MHz crystal oscillator signal as The mixer shown has the 90MHz carrier and 88MHz crystal oscillator signal as inputs. The mixer output only accepts the difference component of 2MHz. Which inputs. The mixer output only accepts the difference component of 2MHz. Which is fed to the discriminator. A Discriminator is the opposite of a VCO, in that it is fed to the discriminator. A Discriminator is the opposite of a VCO, in that it provides a dc level output based upon the frequency input. Discriminator output provides a dc level output based upon the frequency input. Discriminator output is zero dc if it has an input of exactly 2MHz which occurs when the transmitter is is zero dc if it has an input of exactly 2MHz which occurs when the transmitter is at precisely 90MHz. Any carrier drift up or down causes the discriminator output at precisely 90MHz. Any carrier drift up or down causes the discriminator output to go positive or negative, resulting in the appropriate primary oscillator to go positive or negative, resulting in the appropriate primary oscillator readjustment.readjustment.

ffCC =90 MHz =90 MHz f = 75 kHz f = 75 kHz (max)(max)

ffCC =30 MHz =30 MHz f = 25 kHz f = 25 kHz (max)(max)

90 MHz90 MHz88 MHz88 MHz

2 MHz2 MHz

ffCC =5 MHz =5 MHz f = 4.167 f = 4.167 kHz (max)kHz (max)

Reactance modulator

Primary oscillator

5MHz

Frequency Tripler and Doublers

Frequency Tripler

Power Amplifier

Frequency Converter

(Mixer)

To Antenn

a

Discriminator

Frequency Tripler and Doublers

Crystal oscillator

14.67 MHz

Audio input

Frequency Frequency stabilizatiostabilizatio

n circuitn circuit


Frequency Doublers Frequency Doublers

High-distortion High-distortion amplifier (class C)amplifier (class C)



Tuned to Tuned to 10MHz 10MHz LC2

1fC

Frequency Multipliers (doublers,triplers, etc.) are high distortion Frequency Multipliers (doublers,triplers, etc.) are high distortion amplifier which will produce high harmonic content at the output. Then amplifier which will produce high harmonic content at the output. Then the output is tuned to the desired frequency multiple of the input the output is tuned to the desired frequency multiple of the input frequency.frequency.In the above example shown above, a 5MHz carrier having a frequency In the above example shown above, a 5MHz carrier having a frequency deviation of deviation of f=4.167kHz enters the class C high distortion amplifier. f=4.167kHz enters the class C high distortion amplifier. Tank LC circuit of the amplifier output is tuned to 10MHz.Tank LC circuit of the amplifier output is tuned to 10MHz.Note that Note that f=4.167kHz is also doubled to 8.334kHz. The reason is as f=4.167kHz is also doubled to 8.334kHz. The reason is as follows: Suppose a 1 MHz deviating +0.1MHz enters the amplifier. The follows: Suppose a 1 MHz deviating +0.1MHz enters the amplifier. The output will double of output will double of 10.1 x2 = 20.210.1 x2 = 20.2 = a carrier of 2MHz and a = a carrier of 2MHz and a double double deviationdeviation of 0.2MHz. of 0.2MHz.


ffCC =5 MHz =5 MHz f = 0.1MHz f = 0.1MHz MixerMixer ffCC =10 MHz =10 MHz

f = 0.1MHzf = 0.1MHz

Tuned to Tuned to 10MHz 10MHz LC2

1fC

OscillatorOscillatorffoo =5 MHz =5 MHz

It is very important to note that in the case of a frequency converter It is very important to note that in the case of a frequency converter only the center frequency is changed, not the side frequency. only the center frequency is changed, not the side frequency. When 5.1MHz is mixed with 5MHz oscillator, the output will be When 5.1MHz is mixed with 5MHz oscillator, the output will be 5.1+5=10.1MHz5.1+5=10.1MHz meaning that the deviated frequency meaning that the deviated frequency remains the sameremains the same after the frequency conversion.after the frequency conversion.

frequency frequency converterconverter

Frequency ConvertersFrequency Converters


Wideband Armstrong FMWideband Armstrong FM

Preemphasis

90degphase

shif ter

Balancedmodulator

400kHzCrystal

OscillatorRF

amplifierx81 fi rst

Freq.multipliers

Mixer

x64 finalFreq.

multipliers

RFPower

amplifier

Oscillator

+

Wideband Armstrong FM system

f C =90.1MHzf d =75kHz

f C =1.41MHzf d =1.172Hz

f 0=33.81MHzf C =32.4MHzf d =1.172Hz

f C =400kHzf d =14.47Hzf C =400kHz

Audioinput

FMoutput

• MixerMixer is responsible for achieving the final carrier from the basic crystal is responsible for achieving the final carrier from the basic crystal oscillator frequencyoscillator frequency • Frequency multipliersFrequency multipliers are responsible for achieving the final deviation are responsible for achieving the final deviation frequency from the small deviation at the balanced modulatorfrequency from the small deviation at the balanced modulator

Indirect FM generation


Design of mixer freq. and multiplying factor of freq. multiplierDesign of mixer freq. and multiplying factor of freq. multiplier

1. Given: final= 88MHz ,1. Given: final= 88MHz , 75kHz, final multiplier= x64, before first xlier is 75kHz, final multiplier= x64, before first xlier is carrier fcarrier fCC = 400k and f = 400k and fdd=14.47Hz=14.47Hz

2. then before multiplying f2. then before multiplying fCC =(88MHz/64) =1.375MHz, f =(88MHz/64) =1.375MHz, fdd = (75kHz/64) = (75kHz/64) =1.172kHz=1.172kHz

3. before first xlier is f3. before first xlier is fdd=14.47Hz=14.47Hzthen first xlier is (1.172k/14.47)=81then first xlier is (1.172k/14.47)=81

4. before first xlier carrier f4. before first xlier carrier fCC =400k =400k then after first xlier 400kx81=32.4MHz, then fthen after first xlier 400kx81=32.4MHz, then f00 =32.4+1.375=33.81MHz =32.4+1.375=33.81MHz


HW: HW: A Wideband Armstrong FM system is to be transmitted at 88MHz. with maximum A Wideband Armstrong FM system is to be transmitted at 88MHz. with maximum deviation of deviation of 75kHz. A 400kHz crystal oscillator is used with a small deviation of 75kHz. A 400kHz crystal oscillator is used with a small deviation of 15Hz.at balanced modulator output. The final multiplier has a multiplication 15Hz.at balanced modulator output. The final multiplier has a multiplication factor of 64. factor of 64. (a) Design the freq. of Oscillator and the freq. multiplying factor of the multiplier (a) Design the freq. of Oscillator and the freq. multiplying factor of the multiplier before the mixer.before the mixer.Note that a mixer will not change the deviation while the carrier is changed and Note that a mixer will not change the deviation while the carrier is changed and that a multiplier will change both carrier and deviation freq.that a multiplier will change both carrier and deviation freq.(b) If the crystal available is 410kHz, what changes will you make to the above (b) If the crystal available is 410kHz, what changes will you make to the above FM system?FM system?

te4201-communication electronics 1 10. fm generation and transmission angle modulation angle...

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