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General Objective:o Analyze the various communication systems

used in digital modulation.

Specific Objectives:o Interpret the concept, types and the different

processes used in digital modulation.o Analyze ASK, FSK, PSK and QAM.o Compare the error performance of different

digital modulation systems.

General Objective:o Analyze the various communication systems

used in digital modulation.

Specific Objectives:o Interpret the concept, types and the different

processes used in digital modulation.o Analyze ASK, FSK, PSK and QAM.o Compare the error performance of different

digital modulation systems.

We have discussed some methods ofprocessing information involved in digitalcommunications.oDigital to digital encoding techniqueso These included UNIPOLAR, POLAR and BIPOLAR

techniques.oAnalog to digital conversiono These included PAM, PWM, PDM and PCM

All of these methods are done to ensureproper transmission of digital informationover a medium.

We have discussed some methods ofprocessing information involved in digitalcommunications.oDigital to digital encoding techniqueso These included UNIPOLAR, POLAR and BIPOLAR

techniques.oAnalog to digital conversiono These included PAM, PWM, PDM and PCM

All of these methods are done to ensureproper transmission of digital informationover a medium.

We will now investigate the methods ofconverting digital information to analog(digital to analog modulation).

oThese techniques include Amplitude shift keying (ASK) Frequency shift keying (FSK) Phase shift keying (PSK) Quadrature amplitude modulation (QAM) And other digital modulation techniques associated

with these.

We will now investigate the methods ofconverting digital information to analog(digital to analog modulation).

oThese techniques include Amplitude shift keying (ASK) Frequency shift keying (FSK) Phase shift keying (PSK) Quadrature amplitude modulation (QAM) And other digital modulation techniques associated

with these.

Digital Modulation

o transmittal of digitally modulated analog signals(carriers) between two or more points in acommunication system

Before proceeding with digital modulationtechniques, let us first compare bit rate andbaud rate.

o Bit rate refers to the rate of change of adigital information signal.

o Baud rate refers to the rate of change of asignal AFTER encoding and modulation haveoccurred.

Before proceeding with digital modulationtechniques, let us first compare bit rate andbaud rate.

o Bit rate refers to the rate of change of adigital information signal.

o Baud rate refers to the rate of change of asignal AFTER encoding and modulation haveoccurred.

o Bit rate is the number of bits transmitted perunit time. This is also called data rate. Theunit for bit rate is bits per second (bps)

o Baud rate is the number of symbolstransmitted per unit time. The other term forbaud rate is symbol rate. The unit for baudrate is baud.

o Bit rate is the number of bits transmitted perunit time. This is also called data rate. Theunit for bit rate is bits per second (bps)

o Baud rate is the number of symbolstransmitted per unit time. The other term forbaud rate is symbol rate. The unit for baudrate is baud.

Although the two quantities are different,they are closely related.

The two are related by the equationBaud = fb/log2 M = fb/N

where M = the coding level used. Thequantity log2 M = N is the number of bitsused in coding a level.

Although the two quantities are different,they are closely related.

The two are related by the equationBaud = fb/log2 M = fb/N

where M = the coding level used. Thequantity log2 M = N is the number of bitsused in coding a level.

1. Determine the baud rate for a 10kbps binarysignal to pass undistorted.

2. For the same 10kbps signal, determine thebaud rate if 2 bits are used to represent alevel.

3. In higher level PSK systems, such as 8PSK,three bits are used to represent a phasechange. If the baud rate is 15 baud, what isthe bit rate produced by this modulationsystem?

1. Determine the baud rate for a 10kbps binarysignal to pass undistorted.

2. For the same 10kbps signal, determine thebaud rate if 2 bits are used to represent alevel.

3. In higher level PSK systems, such as 8PSK,three bits are used to represent a phasechange. If the baud rate is 15 baud, what isthe bit rate produced by this modulationsystem?

The simplest digital modulation technique. A binary signal modulates the amplitude of

the analog carrier. For ASK, the analog output is either ON or

OFF (hence the term ON-OFF keying) For ASK, bit rate = baud rate = minimum

Nyquist bandwidth

The simplest digital modulation technique. A binary signal modulates the amplitude of

the analog carrier. For ASK, the analog output is either ON or

OFF (hence the term ON-OFF keying) For ASK, bit rate = baud rate = minimum

Nyquist bandwidth

Where:

Vask(t) – amplitude shift keying waveVm(t) – digital information signals (Volts)A/2 – unmodulated carrier amplitude (volts)A/2 – unmodulated carrier amplitude (volts)ωc – analog carrier radian frequency

(radians per second, 2πfct)Example:

Determine the baud and minimum bandwidth necessaryto pass a 10kbps binary signal using amplitude shift keying.

o Another simple, low performance type digitalmodulation system.

o In FSK, the binary signal varies the frequencyof the analog carrier in accordance with itsstate.

o For this system, two frequencies are required,the mark(1) frequency and the space(0)frequency.

o Another simple, low performance type digitalmodulation system.

o In FSK, the binary signal varies the frequencyof the analog carrier in accordance with itsstate.

o For this system, two frequencies are required,the mark(1) frequency and the space(0)frequency.

Where:Vfsk(t) – binary FSK waveformVfsk(t) – binary FSK waveform

fc – analog carrier center frequency (Hz)Vc – peak analog carrier amplitude (volts)

Δf – peak change (shift) in the analog carrier frequency (Hz)

Vm(t) – binary input signal (volts)

Frequency Deviationthe difference between either the mark or space frequency,and the center frequency or half the difference between themark and space frequencies

Δf – f requency deviation (hertz)Where:

| fm – fs | – absolute difference betweenthe mark and space frequencies (hertz)

Δf – f requency deviation (hertz)

fm – mark frequency, (fc + Δf)fs – space frequency, (fc – Δf)fc – carrier frequency

o For FSK, bit rate = baud rate

o The bandwidth is approximated by twice the sumof bit rate and frequency deviation.

Determine (a) the peak frequency deviation (b) minimumBW and (c) baud for a binary FSK with a mark frequency of49kHZ, a space frequency of 53kHz and an inputbit rate of 2kbps.

Example:

fa – highest fundamental frequency of the binary inputsignal (Hz)

Where:

Using Bessel Functions

fb – input bit rate (bps)

h = FM modulation index called the h-factor in FSK

fb – input bit rate (bps)

;unitless or

Using a Bessel table, determine the minimum bandwidthfor a binary FSK with a mark frequency of 49kHZ, a spacefrequency of 53kHz and an input bit rate of 2kbps.

Example:

Using a Bessel table, determine the minimum bandwidthfor a binary FSK with a mark frequency of 49kHZ, a spacefrequency of 53kHz and an input bit rate of 2kbps.

FSK Transmitter

Δf = Vm(t)k1

k1 – deviation sensitivity (Hz/V)

Δf = Vm(t)k1

Δf – peak freqeucny deviation (Hz)

Vm(t) – peak binary modulating signal voltage (V)

Where:

FSK ReceiverNon-coherent FSK demodulator (Envelope/Not PLL Carrier)

PowerSplitter

Envelopedetector

Coherent FSK demodulator (Synchronous/PLL Carrier)

PowerSplitter

FSK ReceiverPLL-FSK demodulator

Phasecomparator Amp

VoltageControlledoscillator

the most common circuit used fordemodulating binary FSK signal

o has a poorer performance than PSK or QAMo it is seldom used for high-performance digital radio

systemo its use is restricted to low-performance, low-cost,

asynchronous data modems for data communicationsover analog, voice band telephone lines

o a form of continuous phase frequency shiftkeying (CPFSK) with mark and spacefrequencies synchronized with input binaryrate separated by ½ of bit rate

o it requires synchronizing circuits and ismore expensive

o a form of continuous phase frequency shiftkeying (CPFSK) with mark and spacefrequencies synchronized with input binaryrate separated by ½ of bit rate

o it requires synchronizing circuits and ismore expensive

PSK is a form of M-ary modulation where achange in phase can be represented by manybits.

o The analog carrier’s phase is varied by thedigital signal.

PSK is a form of M-ary modulation where achange in phase can be represented by manybits.

o The analog carrier’s phase is varied by thedigital signal.

o For BPSK, N = 1, M = 2. Therefore, bit rate =baud, BW = fb

o two output phases are possible for a singlecarrier frequency (1 & 0)

o as the input signal changes state, the phaseof the output carrier shifts between 2angles that are 180̊

o For BPSK, N = 1, M = 2. Therefore, bit rate =baud, BW = fb

o two output phases are possible for a singlecarrier frequency (1 & 0)

o as the input signal changes state, the phaseof the output carrier shifts between 2angles that are 180̊

BalancedModulator

Binary Input Output PhaseLogic 0 180ºLogic 1 0º

Truth Table

PhasorDiagram

ConstellationDiagram

sinωct-sinωct

cosωct

-cosωct

cosωct

-cosωct

0º180º

0º180º Logic 1Logic 1Logic 0

Logic 0

PhasorDiagram

waveform

Example:

For a BPSK modulator with a carrier frequencyof 70 MHz, and an input bit rate of 10Mbps,determine the maximum and minimum upperand lower side frequencies, draw the outputspectrum, determine the minimum Nyquistbandwidth and calculate the baud.

Example:

For a BPSK modulator with a carrier frequencyof 70 MHz, and an input bit rate of 10Mbps,determine the maximum and minimum upperand lower side frequencies, draw the outputspectrum, determine the minimum Nyquistbandwidth and calculate the baud.

BPSK Receiver

Coherent Carrier Recovery – detectsand regenerates a carrier signal that isboth frequency and phase coherentwith the original transmit carrier

o angle modulated, constant amplitude digitalmodulation, M-ary encoding technique where N =2and M = 4

o four output phases are possible for a single carrierfrequency

o binary input data are combined into groups called‘dibit’

o the rate of change of the output baud is one-half ofthe input bit rate

Quaternary Phase-Shift Keying(Quadrature PSK)

o angle modulated, constant amplitude digitalmodulation, M-ary encoding technique where N =2and M = 4

o four output phases are possible for a single carrierfrequency

o binary input data are combined into groups called‘dibit’

o the rate of change of the output baud is one-half ofthe input bit rate

Quaternary Phase-Shift Keying

o With QPSK, bandwidth compression isrealized.oQPSK requires half the bandwidth of BPSK.

o For the same data rate with BPSK, QPSK hashalf the baud rate of BPSK (recall baud = bitrate/N; N = 1 for BPSK, 2 for QPSK).

o With QPSK, bandwidth compression isrealized.oQPSK requires half the bandwidth of BPSK.

o For the same data rate with BPSK, QPSK hashalf the baud rate of BPSK (recall baud = bitrate/N; N = 1 for BPSK, 2 for QPSK).

QPSKTransmitter

cosωct- sinωct cosωct+ sinωct

sinωct-sinωct

cosωct

sinωct-sinωct

cosωct

Q I1 0

Q I0 0

Q I1 1

Q I0 1

Binary Input QPSKOutputPhaseQ I

0 0 -135º0 1 -45º1 0 +135º1 1 +45º

-cosωct- sinωct -cosωct+ sinωct

-cosωct

Truth table

PhasorDiagram

Q I0 0 Q I

Circle

45°1/4π

135°3/4π

’00’

’01’

Circle

225°5/4π

315°7/4π

’10’

’11’

’00’

’01’

0 0 10 01

’10’

’11’

’00’

’01’

0 0 10 01

’10’

’11’

2fbBW 2

fbBaud

For a QPSK with an input data rate equal to10Mbps and a carrier frequency of 70MHz,determine the minimum bandwidth and thebaud.

Example:

For a QPSK modulator with an input bit rate of10Mbps and a carrier frequency of 70 MHz,determine the minimum Nyquist bandwidthand calculate the baud.

Example:

For a QPSK modulator with an input bit rate of10Mbps and a carrier frequency of 70 MHz,determine the minimum Nyquist bandwidthand calculate the baud.

Receivebinary data

o with 8-PSK, three bits are encoded, formingtribits and producing eight different outputphases

o with 8-PSK, n=3, M=8o thus bit rate is three times the baud rate

Receivebinary data

o with 8-PSK, three bits are encoded, formingtribits and producing eight different outputphases

o with 8-PSK, n=3, M=8o thus bit rate is three times the baud rate

Receivebinary data

o Q and I channel determine the polarityo Logic 1 – (+V)o Logic 0 – (-V)

o C and C’ channel determines the magnitudeo Logic 1 – (1.307 V)o Logic 0 – (0.541 V)

Receivebinary data

o Q and I channel determine the polarityo Logic 1 – (+V)o Logic 0 – (-V)

o C and C’ channel determines the magnitudeo Logic 1 – (1.307 V)o Logic 0 – (0.541 V)

BinaryInput

8-PSKOutputPhaseQ I C

0 0 0 -112.5º0 0 1 -157.5º0 1 0 -57.5º0 1 1 -22.5º

Receivebinary data

0 1 1 -22.5º1 0 0 +112.5º1 0 1 +157.5º1 1 0 +157.5º1 1 1 +22.5º

Truth table

Diagram

3fbBW 3

fbBaud

Receivebinary data

For an 8 PSK modulator, with an input datarate of 10Mbps and a carrier frequency of70MHz, determine the minimum bandwidthrequired and the baud rate.

Example:

For an 8-PSK modulator with an input bit rateof 10Mbps and a carrier frequency of 70 MHz,determine the minimum Nyquist bandwidthand calculate the baud.

Example:

For an 8-PSK modulator with an input bit rateof 10Mbps and a carrier frequency of 70 MHz,determine the minimum Nyquist bandwidthand calculate the baud.

o with 16-PSK, four bits are encoded, formingquadbits and producing 16 possible outputphases

o n = 4, M = 16o minimum bandwidth and baud is one-fourth the

bit rate Receivebinary data

4fbBaud

o with 16-PSK, four bits are encoded, formingquadbits and producing 16 possible outputphases

o n = 4, M = 16o minimum bandwidth and baud is one-fourth the

bit rate

BitCode

Phase BitCode

00000001001000110100010101100111

11.25º33.75º56.25º78.75º

101.25º123.75º146.25º168.75º

10001001101010111100110111101111

191.25º213.75º236.25º258.75º281.25º303.75º326.25º348.75º

Receivebinary data

00000001001000110100010101100111

11.25º33.75º56.25º78.75º

101.25º123.75º146.25º168.75º

10001001101010111100110111101111

191.25º213.75º236.25º258.75º281.25º303.75º326.25º348.75º

Truth table

PhasorDiagram

o QAM is a form of digital information similarto PSK except that the digital information iscontained in both the amplitude and thephase of the transmitted carrier.

o With QAM, amplitude and phase shiftkeying are combined in such a way that thepositions of the signaling elements on theconstellation diagrams are optimized toachieve the greatest distance betweenelements.

Receivebinary data

o QAM is a form of digital information similarto PSK except that the digital information iscontained in both the amplitude and thephase of the transmitted carrier.

o With QAM, amplitude and phase shiftkeying are combined in such a way that thepositions of the signaling elements on theconstellation diagrams are optimized toachieve the greatest distance betweenelements.

o 8 QAM is an M-ary encoding technique whereM = 8

o Unlike 8 PSK, the output of 8 QAM is not aconstant amplitude signal.

o The bandwidth required for 8 QAM is thesame as that of 8 PSK, and therefore theyhave the same baud rate as 8 PSK.

Receivebinary data

o 8 QAM is an M-ary encoding technique whereM = 8

o Unlike 8 PSK, the output of 8 QAM is not aconstant amplitude signal.

o The bandwidth required for 8 QAM is thesame as that of 8 PSK, and therefore theyhave the same baud rate as 8 PSK.

3fbBW 3

fbBaud

Receivebinary data

BinaryInput

8-QAMoutput

Q00001111

I00110011

C01010101

Amplitude0.765V1.848V0.765V1.848V0.765V1.848V0.765V1.848V

Phase-135°-135°-45°-45°

+135°+135°+45°+45°

o Q and I channeldetermine the polarityo Logic 1 – (+V)o Logic 0 – (-V)

o C channel determinesthe magnitudeo Logic 1 – (1.848 V)o Logic 0 – (0.765 V)

Truth Table

Q00001111

I00110011

C01010101

Amplitude0.765V1.848V0.765V1.848V0.765V1.848V0.765V1.848V

Phase-135°-135°-45°-45°

+135°+135°+45°+45°

o C channel determinesthe magnitudeo Logic 1 – (1.848 V)o Logic 0 – (0.765 V)

sinωct-sinωct

cosωctQ I C1 0 1

Q I C1 0 0

Q I C0 0 0

Q I C1 1 1

Q I C1 1 0

Q I C0 1 0

sinωct-sinωct

cosωct

sinωct-sinωct

-cosωctQ I C0 0 1

Q I C0 0 0

Q I C0 1 1

Q I C0 1 0

PhasorDiagram

sinωct-sinωct

-cosωct

0 1 00 0 0

0 1 10 0 1

o 16 QAM is an M-ary encoding techniquewhere M = 16

o The bandwidth required for 16-QAM is thesame as that of 16-PSK, and therefore theyhave the same baud rate as 16-PSK. Receive

binary data

o 16 QAM is an M-ary encoding techniquewhere M = 16

o The bandwidth required for 16-QAM is thesame as that of 16-PSK, and therefore theyhave the same baud rate as 16-PSK.

4fbBaud

Receivebinary data

o Q’ and I’ channeldetermines themagnitudeo Logic 1 – (0.821 V)o Logic 0 – (0.22 V)

BinaryInput 16-QAM output

Q Q” I I”

0000000011111111

0000111100001111

0011001100110011

0101010101010101

0.311V0.850V0.311V0.850V0.850V1.161V0.850V1.161V0.311V0.850V0.311V0.850V0.850V1.161V0.850V1.161V

-135°-165°-45°-15°

-105°-135°-75°-45°135°165°45°15°

105°135°75°45°

Truth Table

o Q’ and I’ channeldetermines themagnitudeo Logic 1 – (0.821 V)o Logic 0 – (0.22 V)

0000000011111111

0000111100001111

0011001100110011

0101010101010101

0.311V0.850V0.311V0.850V0.850V1.161V0.850V1.161V0.311V0.850V0.311V0.850V0.850V1.161V0.850V1.161V

-135°-165°-45°-15°

-105°-135°-75°-45°135°165°45°15°

105°135°75°45°

sinωct-sinωct

cosωct

sinωct-sinωct

cosωct

1 1 1 11 1 1 0

1 0 1 11 0 1 0

1 1 0 01 1 0 1

1 0 0 01 0 0 1

sinωct-sinωct

-cosωctPhasor

Diagram

sinωct-sinωct

-cosωctConstellation

Diagram

0 0 1 10 0 1 0

0 1 1 10 1 1 0

0 0 0 00 0 0 1

0 1 0 00 1 0 1

Example:

For a 16-QAM modulator with an input bit rateof 10Mbps and a carrier frequency of 70 MHz,determine the minimum Nyquist bandwidthand calculate the baud.

Example:

For a 16-QAM modulator with an input bit rateof 10Mbps and a carrier frequency of 70 MHz,determine the minimum Nyquist bandwidthand calculate the baud.

o ratio of the transmission bit rate to theminimum bandwidth required for a particularmodulation scheme

o it is generally normalized to 1Hz bandwidthand thus, indicates the number of bits that canbe propagated through a medium for each ofhertz of bandwidth

o ratio of the transmission bit rate to theminimum bandwidth required for a particularmodulation scheme

o it is generally normalized to 1Hz bandwidthand thus, indicates the number of bits that canbe propagated through a medium for each ofhertz of bandwidth

(Hz)bandwidthminimum

(bps)ratebitontransmissiEfficiencyBW Bits/cycle

Modulation EncodingScheme

OutputsPossible

MinimumBandwidth

Baud BWEfficiency

ASK Single Bit 2 fb fb 1

FSK Single Bit 2 fb fb 1

BPSK Single Bit 2 fb fb 1BPSK Single Bit 2 fb fb 1QPSK Dibits 4 fb/2 fb/2 28-PSK Tribits 8 fb/3 fb/3 38-QAM Tribits 8 fb/3 fb/3 316-PSK Quadbits 16 fb/4 fb/4 416-QAM Quadbits 16 fb/4 fb/4 432PSK Five Bits 32 fb/5 fb/5 5

64-QAM Six Bits 64 fb/6 fb/6 6

Example:

For an 8-PSK system, operating with aninformation bit rate of 24 kbps, determine (a)baud (b) minimum bandwidth and (c)bandwidth efficiency.

Example:

For an 8-PSK system, operating with aninformation bit rate of 24 kbps, determine (a)baud (b) minimum bandwidth and (c)bandwidth efficiency.

o it is the process of extracting a phasecoherent reference carrier from a receiversignal

o sometimes called “phase referencing”o three methods ( squaring loop, costas loop

and remodulator)

o it is the process of extracting a phasecoherent reference carrier from a receiversignal

o sometimes called “phase referencing”o three methods ( squaring loop, costas loop

and remodulator)

o as with any digital system, digital radiorequires precise timing or clocksynchronization between the transmit and thereceive circuitry

o because of this, it is necessary to regenerateclocks at the receiver that are synchronous withthose at the transmitter

o as with any digital system, digital radiorequires precise timing or clocksynchronization between the transmit and thereceive circuitry

o because of this, it is necessary to regenerateclocks at the receiver that are synchronous withthose at the transmitter

o A digital modulation where the binary inputinformation is contained in the differencebetween two successive signaling elementsrather than the absolute phase

o The difference in phase of the two signalingelement determines the logic condition ofthe data.

o A digital modulation where the binary inputinformation is contained in the differencebetween two successive signaling elementsrather than the absolute phase

o The difference in phase of the two signalingelement determines the logic condition ofthe data.

DBPSKTransmitter

DBPSKReceiver

1 1 1 1 1 1

1 1 1 1 11

00 0 00

0 0 0 0 00

0°0° 0° 0° 0° 0°0° 180° 180° 180°180° 180°Referencebit 0°0° 0° 0° 0° 0°0° 180° 180° 180°180° 180°

timing diagram (tx)

Referencebit

0°0° 0° 0° 0° 0°0° 180° 180° 180°180° 180°180°Reference

phase1 1 1 1 1 1 100 0 00Recovered bit stream

DBPSKInput phase

timing sequence (rx)

o The bit error performance for variousmultiphase digital modulation systems isdirectly related to the distance between pointson a signal state-space (constellation) diagram.

o Probability of error P(e) is a theoretical(mathematical) expectation of the bit error ratefor a given system.

o Bit error rate (BER) is an empirical (historical)record of actual bit error performance.

o Probability of error is a function of the carrier-to-noise power ratio, or more specifically, theaverage energy per bit-to-noise density ratio,and the number of possible encodingconditions used.

o The bit error performance for variousmultiphase digital modulation systems isdirectly related to the distance between pointson a signal state-space (constellation) diagram.

o Probability of error P(e) is a theoretical(mathematical) expectation of the bit error ratefor a given system.

o Bit error rate (BER) is an empirical (historical)record of actual bit error performance.

o Probability of error is a function of the carrier-to-noise power ratio, or more specifically, theaverage energy per bit-to-noise density ratio,and the number of possible encodingconditions used.

o the ratio of the average carrier power (thecombined power of the carrier and itsassociated sidebands) to the thermal noisepower

0.001)C(log10)C( watts

)(kTBP wattsN Where:o N – thermal noise power (watts)o k – Boltzman’s constanto T – temperature (K)o B – bandwidth

0.001KTBlog10)N(dBm

ratiounitlessKTB

Mathematically, the carrier-to-noise ratio is:

Noise in dB

ratiounitlessKTB

)N(-)C(NClog10(dB)

dBmdBm

Stated in dB

(J/bit)CTE bb

o the energy of a single bit of information

Where:o Eb – energy of a single bit (Joules per bit)o Tb – time of a single bit (seconds)o C – carrier power (watts)

bdBJb Elog10)(E

Where:o Eb – energy of a single bit (Joules per bit)o Tb – time of a single bit (seconds)o C – carrier power (watts)

Stated in dBJ

because Tb = 1/ fb, where fb is the bit ratein bits per second, Eb can be written as:

(J/bit)fCE

Stated in dBJ

(J/bit)fCE

flog10-Clog10fClog10)(E

o the thermal noise power normalized to a 1-Hz bandwidth

Hz) /(WBNN 0

Where:o No – noise power density (watts per Hz)o N – thermal noise power (watts)o B – Bandwidth (Hz)

Stated in dBm

Blog10-)N(

Blog10-0.001

Nlog10)(N

Blog10-)N(

Blog10-0.001

Nlog10)(N

Hz) /(WKTB

Tlog100.001

Klog10)(N dBm0

o the ratio of the energy of a single bit tonoise power present in 1 Hz of bandwidth

B /Nf /C

oo EEb/N/N0 is used to compare two or more digitalmodulation system that use differenttransmission rates (bit rates), modulationschemes or encoding techniques.

stated in dB

Nlog10-Elog10fBlog10

NClog10(dB)

Nlog10-Elog10fBlog10

NClog10(dB)

o The minimum C/N power ratio for QAM systems is less thanthat required for comparable PSK systems.

o The higher the level of encoding, the higher the minimumC/N ratio.

Example

For a QPSK system and the given parameters,determine:

o Carrier power in dBmo Noise power in dBmo Noise power density in dBmo Energy per bit in dBJo C/N in dBo Eb/N0 ratio

Pc = 10-12 W fb = 60kbpsPN = 1.2 x 10-14 W B = 120 kHz

For a QPSK system and the given parameters,determine:

o Carrier power in dBmo Noise power in dBmo Noise power density in dBmo Energy per bit in dBJo C/N in dBo Eb/N0 ratio

Pc = 10-12 W fb = 60kbpsPN = 1.2 x 10-14 W B = 120 kHz

o The bit error performance for the various multiphase digitalmodulation systems is directly related to the phase distancebetween points on signal state space diagram.

o For example, for BPSK signal state space diagram, the twosignal points (logic 1 ad logic 0) have maximum separation(d) for a given power level (D).

o In essence, one BPSK signal state is the exact negative of theother

o The bit error performance for the various multiphase digitalmodulation systems is directly related to the phase distancebetween points on signal state space diagram.

o For example, for BPSK signal state space diagram, the twosignal points (logic 1 ad logic 0) have maximum separation(d) for a given power level (D).

o In essence, one BPSK signal state is the exact negative of theother

o If a noise Vector (VN) is combined with the signal vector (VS),it is effectively shifts the phase of the signaling vector (VSE) toalpha degrees.

o If the phase shifts exceeds 90°, the signal element is shiftedbeyond the threshold point into the error region

o For PSK systems, the threshold point (TP) is givenby:

o The error performance of higher PSK systems can becompared to the BPSK (which is referred to as theantipodal performance) simply by determining therelative decrease in error distance between points on asignal state-space diagram. For PSK systems, thegeneral formula for the maximum distance betweensignaling points is:

180sin x(2(d)DistanceError

Where:o d – error distanceo M – number of phaseso D – peak signal amplitude

o From these equations we can see that

oThe higher levels of modulation require agreater Eb/N0 to reduce the effect of noiseinterference

oThe higher the level of modulation, thesmaller the angular separation betweensignal points and the smaller the errordistance.

o From these equations we can see that

oThe higher levels of modulation require agreater Eb/N0 to reduce the effect of noiseinterference

oThe higher the level of modulation, thesmaller the angular separation betweensignal points and the smaller the errordistance.

o The general expression for the bit errorprobability of an M-phase PSK is:

erf ( z )Mlog1

Where erf = error function

MlogMπ

Determine the minimum bandwidth requiredto achieve a P(e) of 10-7 for an 8PSK systemoperating at 10Mbps with a carrier-to-noisepower ratio of 11.7 dB.

o For a large number of signal points (i.e. M>4)QAM outperforms PSK. This is because thedistance between signaling points in a PSKsystem is smaller than the distance betweenadjacent signaling points for a QAM system.

Where:o d – error distanceo L – number of levels on each axiso D – peak signal amplitude

o The general expression for the bit errorprobability of an L-level QAM is:

zerfL1L

Where erfc(z) = complimentary error function

Example

Which system requires the higher EEb/N/N0 ratiofor a probability of error of 10-6, a fourlevel QAM system or an 8-PSK system?

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