Chapter 5 Digital Bandpass Modulation and …silage/Chapter5SVU.pdfEE4512 Analog and Digital...

EE4512 Analog and Digital Communications Chapter 5

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• Binary Amplitude Shift KeyingBinary Amplitude Shift Keying

•• Pages 212Pages 212--219219

•• The analytical signal for binary amplitude shift keying The analytical signal for binary amplitude shift keying (BASK) is:(BASK) is:

ssBASKBASK(t(t) =) = ssbasebandbaseband(t(t) sin (2) sin (2ππ ffCC t)t) (S&M Eq. 5.1)(S&M Eq. 5.1)

The signal The signal ssbasebandbaseband(t(t) can be ) can be any twoany two shapes over a bit time shapes over a bit time TTbb butbut it is usually a rectangular signal of amplitude 0 for a it is usually a rectangular signal of amplitude 0 for a binary 0 and amplitude A for binary 1. Then BASK is also binary 0 and amplitude A for binary 1. Then BASK is also known as known as onon--off keyingoff keying (OOK).(OOK).

•• The binary amplitude shift keying (BASK) signal can be The binary amplitude shift keying (BASK) signal can be simulated in simulated in SystemVueSystemVue..

ssBASKBASK(t(t) = ) = ssbasebandbaseband(t(t) sin 2) sin 2ππ ffCC t (S&M Eq. 5.1)t (S&M Eq. 5.1)

Baseband binary PAM signal 0 Baseband binary PAM signal 0 →→ 1 V, r1 V, rbb = 1 kb/sec= 1 kb/sec

Sinusoidal carrier fSinusoidal carrier fCC = 20 kHz, A= 20 kHz, Acc = 5 V= 5 V

MultiplierBASK signalBASK signal

•• A BASK signal is a baseband binary PAM signal multiplied A BASK signal is a baseband binary PAM signal multiplied by a carrier (S&M Figure 5by a carrier (S&M Figure 5--3). 3).

Unmodulated sinusoidal carrier

Baseband binary PAM signal

BASK signal

•• The unipolar binary PAM signal can be decomposed into a The unipolar binary PAM signal can be decomposed into a polar PAM signal and DC level (S&M Figure 5polar PAM signal and DC level (S&M Figure 5--4). 4).

Unipolar binary PAM signal

Polar binary PAM signal

DC level

0 → 1 V

± 0.5 V

•• The spectrum of the BASK signal is (S&M Eq. 5.2):The spectrum of the BASK signal is (S&M Eq. 5.2):

SSBASKBASK(f(f) = ) = FF( ( ssASKASK(t(t) )) ) == FF( ( ssbasebandbaseband(t(t) sin (2) sin (2ππ ffC C t) )t) )SSBASKBASK(f(f) = 1/2 j) = 1/2 j (Sbaseband(f – fC) + Sbaseband(f + fC) )

The analytical signal for the baseband binary PAM signal is:The analytical signal for the baseband binary PAM signal is:

ssbasebandbaseband(t(t) = ) = ssPAMPAM(t(t) + A/2 ) + A/2 (S&M Eq. 5.3)(S&M Eq. 5.3)SSbasebandbaseband(f(f) = ) = SSPAMPAM(f(f) + A/2 ) + A/2 δδ(f) (f) (S&M Eq. 5.4)(S&M Eq. 5.4)

Therefore by substitution (S&M Eq. 5.5):Therefore by substitution (S&M Eq. 5.5):

SSBASKBASK(f(f) = 1/ 2j ( ) = 1/ 2j ( SSPAMPAM(f(f –– ffCC) + A/2 ) + A/2 δδ(f (f –– ffCC) ) –– SSPAMPAM(f(f + f+ fCC) ) –– A/2 A/2 δδ(f + f(f + fCC) )) )

•• TheThe bibi--sided sided power spectral density PSD of the BASK power spectral density PSD of the BASK signal is (S&M Eq. 5.7):signal is (S&M Eq. 5.7):

GGBASKBASK(f(f) = 1/4 ) = 1/4 GGPAMPAM(f(f –– ffCC) + 1/4 ) + 1/4 GGPAMPAM(f(f + f+ fCC))+ A+ A22/16 /16 δδ(f (f –– ffCC) + A) + A22/16 /16 δδ(f + f(f + fCC))

For a rectangular polar PAM signal (For a rectangular polar PAM signal (±± A):A):

GGPAMPAM(f(f) = (A/2)) = (A/2)2 2 / r/ rb b sincsinc22 ((ππ f / rf / rbb)) (S&M Eq. 5.8)(S&M Eq. 5.8)

•• The The singlesingle--sidedsided power spectral density PSD of the BASK power spectral density PSD of the BASK signal is:signal is:

GGBASKBASK(f(f) = 1/2 ) = 1/2 GGPAMPAM(f(f + f+ fCC) + A) + A22/8 /8 δδ(f + f(f + fCC))GGPAMPAM(f(f) = (A/2)) = (A/2)2 2 / r/ rb b sincsinc22 ((ππ f / rf / rbb))

Carrier 20 kHz

rb = 1 kHzSVU Figure 3.6SVU Figure 3.6

•• The The bandwidthbandwidth of a BASK signal as a percentage of total of a BASK signal as a percentage of total power is power is double double that for the same bit rate rthat for the same bit rate rb b = 1/T= 1/Tbb binary binary rectangular PAM (SVU Table 2.1 p. 66 and SVU Table 3.1 rectangular PAM (SVU Table 2.1 p. 66 and SVU Table 3.1 p. 152).p. 152).

Bandwidth (Hz) Percentage of Total PowerBandwidth (Hz) Percentage of Total Power

2/T2/Tbb 90%90%3/T3/Tbb 93%93%4/T4/Tbb 95%95%6/T6/Tbb 96.5%96.5%8/T8/Tbb 97.5%97.5%10/T10/Tbb 98%98%

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• Binary Phase Shift KeyingBinary Phase Shift Keying

•• Pages 219Pages 219--225225

•• The analytical signal for binary phase shift keying The analytical signal for binary phase shift keying (BPSK) is:(BPSK) is:

ssBPSKBPSK(t(t) =) = ssbasebandbaseband sin (2sin (2ππ ffCC t + t + θθ) (S&M Eq. 5.11)) (S&M Eq. 5.11)ssbasebandbaseband(t(t) = + A b) = + A bii = 1 = 1 ssbasebandbaseband(t(t) = ) = –– A bA bii = 0 = 0

+ 180°0°

•• The binary phase shift keying (BPSK) signal can be The binary phase shift keying (BPSK) signal can be simulated in simulated in SystemVueSystemVue..

Baseband binary PAM signal 0 → 1 V, rb = 1 kb/sec

PM Modulator

BPSK signal

•• The Phase Modulator (PM) token is in the Function The Phase Modulator (PM) token is in the Function Library of Library of SystemVueSystemVue..

Modulation Gain = 0.5π/V

Initial phase = 3π/4

Frequency = 5 kHz (changed to 20 kHz later)

•• The spectrum of the BPSK signal is (S&M Eq. 5.13):The spectrum of the BPSK signal is (S&M Eq. 5.13):

SSBPSKBPSK(f(f) = ) = FF( ( ssPSKPSK(t(t) )) ) == FF(s(sbasebandbaseband(t(t) sin 2) sin 2ππ ffC C t)t)SSBPSKBPSK(f(f) = 1/2 j) = 1/2 j (Sbaseband(f – fc) + Sbaseband(f + fC) )

The analytical signal for the baseband binary PAM signal is:The analytical signal for the baseband binary PAM signal is:

ssbasebandbaseband(t(t) = ) = ssPAMPAM(t(t) ) (S&M Eq. 5.12)(S&M Eq. 5.12)SSbasebandbaseband(f(f) = ) = SSPAMPAM(f(f))

Note that there is no DC level in Note that there is no DC level in ssPAMPAM(t(t) and therefore by ) and therefore by substitution:substitution:

SSBPSKBPSK(f(f) = 1/ 2j ( ) = 1/ 2j ( SSPAMPAM(f(f –– ffCC) ) –– SSPAMPAM(f(f + f+ fCC) )) )

•• TheThe bibi--sided sided power spectral density PSD of the BPSK power spectral density PSD of the BPSK signal is (S&M Eq. 5.13)signal is (S&M Eq. 5.13)

GGBPSKBPSK(f) = 1/4 (f) = 1/4 GGPAMPAM(f(f –– ffCC) + 1/4 ) + 1/4 GGPAMPAM(f(f + f+ fCC))

For a rectangular polar PAM signal (For a rectangular polar PAM signal (±± A):A):

GGPAMPAM(f(f) = A) = A2 2 / r/ rb b sincsinc22 ((ππ f / rf / rbb) (S&M Eq. 5.8 modified)) (S&M Eq. 5.8 modified)

•• The The singlesingle--sidedsided power spectral density PSD of the BPSK power spectral density PSD of the BPSK signal is:signal is:

GGBPSKBPSK(f) = 1/2 (f) = 1/2 GGPAMPAM(f(f + f+ fCC))GGPAMPAM(f(f) = A) = A2 2 / r/ rb b sincsinc22 ((ππ f / rf / rbb))

No carrier

rb = 1 kHz SVU Figure 3.12SVU Figure 3.12

•• The The bandwidthbandwidth of a BPSK signal as a percentage of total of a BPSK signal as a percentage of total power is power is double double that for the same bit rate rthat for the same bit rate rb b = 1/T= 1/Tbb binary binary rectangular PAM (SVU Table 2.1 p. 66) and the same as rectangular PAM (SVU Table 2.1 p. 66) and the same as BASK (SVU Table 3.1 p. 152)BASK (SVU Table 3.1 p. 152)

2/T2/Tbb 90%90%3/T3/Tbb 93%93%4/T4/Tbb 95%95%6/T6/Tbb 96.5%96.5%8/T8/Tbb 97.5%97.5%10/T10/Tbb 98%98%

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• Binary Frequency Shift KeyingBinary Frequency Shift Keying

•• Pages 219Pages 219--225225

•• The analytical signal for binary frequency shift keying The analytical signal for binary frequency shift keying (BFSK) is:(BFSK) is:

ssBFSKBFSK(t(t) =) = A sin (2A sin (2ππ (f(fCC + + ∆∆f) t + f) t + θθ) if b) if bii = 1= 1ssBFSKBFSK(t(t) = A sin (2) = A sin (2ππ (f(fCC –– ∆∆f) t + f) t + θθ) if b) if bii = 0= 0

fc + ∆f fc – ∆f

•• The binary frequency shift keying (BFSK) signal can be The binary frequency shift keying (BFSK) signal can be simulated in simulated in SystemVueSystemVue..

Baseband binary PAM signal Baseband binary PAM signal ±± 1 V, r1 V, rbb = 1 kb/sec= 1 kb/sec

FM ModulatorFM ModulatorBFSK signalBFSK signal

•• The FM/VCO (frequency modulator/voltage controlled The FM/VCO (frequency modulator/voltage controlled oscillator) token is in the Function Library of oscillator) token is in the Function Library of SystemVueSystemVue..

Modulation Modulation gain = 2 kHz/Vgain = 2 kHz/V

•• The BFSK signal can be The BFSK signal can be decomposeddecomposed as (S&M Eq. 5.14):as (S&M Eq. 5.14):

ssBFSKBFSK(t(t) =) = ssbaseband1baseband1(t) sin (2(t) sin (2ππ (f(fCC + + ∆∆f) t + f) t + θθ) +) +ssbaseband2baseband2(t) sin (2(t) sin (2ππ (f(fCC –– ∆∆f) t + f) t + θθ))

fc – ∆ffc + ∆f

•• The BFSK signal is the The BFSK signal is the sum sum of two BASK signals:of two BASK signals:

ssBFSKBFSK(t(t) =) = ssbaseband1baseband1(t) sin (2(t) sin (2ππ (f(fCC + + ∆∆f) t + f) t + θθ) +) +ssbaseband2baseband2(t) sin (2(t) sin (2ππ (f(fCC –– ∆∆f) t + f) t + θθ))

From the From the linearity propertylinearity property, the resulting , the resulting singlesingle--sidedsided PSD PSD of the BFSK signal of the BFSK signal GGBFSKBFSK(f(f) is the ) is the sum sum of two of two GGBASKBASK(f(f) ) PSDs with f = fPSDs with f = fC C ±± ∆∆f:f:

GGBFSKBFSK(f(f) = (A/2)) = (A/2)22 / 2 r/ 2 rbb sincsinc22 ((ππ f / rf / rbb) + A) + A22/8 /8 δδ(f)(f)

fc + ∆f fc – ∆f

•• The The singlesingle--sidedsided power spectral density PSD of the BFSK power spectral density PSD of the BFSK signal is:signal is:

GGBFSKBFSK(f(f) = (A/2)) = (A/2)2 2 / 2r/ 2rbb sincsinc22 ((ππ (f(fCC + + ∆∆f)f) / r/ rbb) + A) + A22/8 /8 δδ(f(fCC + + ∆∆f)f)+ (A/2)+ (A/2)22 / 2r/ 2rbb sincsinc22 ((ππ (f(fCC –– ∆∆ff)/ r/ rbb) + A) + A22/8 /8 δδ(f(fCC –– ∆∆f)f)

Carriers

rb = 1 kHz SVU Figure 3.9SVU Figure 3.9∆∆ff = 2 kHz

•• Minimum frequency shift keying (MFSK) occurs when Minimum frequency shift keying (MFSK) occurs when ∆∆ff = 1/2Tb = rb/2 Hz.

GGBFSKBFSK(f(f) = (A/2)) = (A/2)2 2 / 2r/ 2rbb sincsinc22 ((ππ (f(fCC + + ∆∆f)f) / r/ rbb) + A) + A22/8 /8 δδ(f(fCC+ + ∆∆f) f) + (A/2)+ (A/2)22 / 2r/ 2rbb sincsinc22 ((ππ (f(fCC –– ∆∆ff)/ r/ rbb) + A) + A22/8 /8 δδ(f(fCC –– ∆∆f)f)

Carriers

rb = 1 kHz

SVU Figure 3.10SVU Figure 3.10∆∆ff = 500 Hz

•• This This carrier frequency separationcarrier frequency separation ∆∆ff = 1/2Tb = rb/2 Hz is the minimum possibleminimum possible because each carrier spectral impulse because each carrier spectral impulse is at the is at the null null of the PSD of the other decomposed BASK of the PSD of the other decomposed BASK signal.signal.

Carriers

rb = 1 kHz

SVU Figure 3.10SVU Figure 3.10∆∆ff = 500 Hz

•• The The bandwidthbandwidth of a BFSK signal as a percentage of total of a BFSK signal as a percentage of total power is power is greatergreater than that of either BASK or BPSK by 2than that of either BASK or BPSK by 2∆∆ffHz for the same bit rate rHz for the same bit rate rbb = 1/T= 1/Tb b (SVU Table 3.3 p. 160).(SVU Table 3.3 p. 160).

22∆∆f + f + 2/T2/Tbb 90%90%22∆∆f +f + 3/T3/Tbb 93%93%22∆∆f +f + 4/T4/Tbb 95%95%22∆∆f +f + 6/T6/Tbb 96.5%96.5%22∆∆f +f + 8/T8/Tbb 97.5%97.5%22∆∆f +f + 10/T10/Tbb 98%98%

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• Coherent Demodulation ofCoherent Demodulation ofBandpass SignalsBandpass Signals

•• Pages 225Pages 225--236236

•• The development of the optimum receiver for The development of the optimum receiver for bandpass bandpass signalssignals utilizes the concepts of the optimum utilizes the concepts of the optimum baseband baseband receiverreceiver::

Optimum FilterOptimum Filter −o bh (t) = k s(iT t)

Correlation ReceiverCorrelation Receiver

•• The optimum filter The optimum filter HHoo(f(f) and the correlation receiver ) and the correlation receiver are equivalent here also, with sare equivalent here also, with s11(t) = (t) = s(ts(t) for ) for symmetrical symmetrical signalssignals and and r(tr(t) = ) = γγ s(ts(t) + ) + n(tn(t) where ) where γγ is the communication is the communication channel attenuation and channel attenuation and n(tn(t) is AWGN. The energy per bit ) is AWGN. The energy per bit EEbb and the probability of bit error Pand the probability of bit error Pb b is (S&M p. 226):is (S&M p. 226):

∫ ∫b b

iT iT2 2

b(i-1)T (i-1)T

E = γ s(t) γ s(t) dt = γ s (t) dt

2 EP = QN

•• The matched filter or correlation receiver is a The matched filter or correlation receiver is a coherent coherent demodulationdemodulation process for bandpass signals because not process for bandpass signals because not only is bit time (Tonly is bit time (Tbb) as for baseband signals required but ) as for baseband signals required but carrier carrier synchronizationsynchronization is also needed. Carrier is also needed. Carrier synchronization requires an estimate of the synchronization requires an estimate of the transmitted transmitted frequencyfrequency (f(fCC) and the ) and the arrival phasearrival phase at the receiver (at the receiver (θθ):):

1 Cs (t) = sin(2π f t + θ)

•• BPSK signals are symmetrical with:BPSK signals are symmetrical with:ss1T1T(t) = (t) = –– ss2T2T(t) = A sin(2(t) = A sin(2ππ ffcctt) S&M ) S&M EqsEqs. 5.15. 5.15--5.195.19

[ ]−

b, BPSK C(i-1)T

iT 2 22 2b

b, BPSK C(i-1)T

E = ± A γ sin (2π f t) dt

γ A Tγ AE = 1 cos (4π f t) dt = 2 2

2 2b b

b, BPSKo o

2 E γ A T P = Q = QN N

•• For this analysis of EFor this analysis of Eb, PSKb, PSK for BPSK signals it is assumed for BPSK signals it is assumed that the transmitter produces an that the transmitter produces an integer numberinteger number of cycles of cycles within one bit period:within one bit period:

[ ]−

b, BPSK C(i-1)T

iT2 2 2 2b

b, BPSK C(i-1)T

γ AE = 1 cos (4π f t) dt2

γ A T γ AE = cos (4π f t) dt2 2

S&M Eq. 5.17S&M Eq. 5.17

and sand s11(t) = sin (2(t) = sin (2ππ ffcctt))

i b i 1(i-1)T

a (iT ) = γ s (t) s (t) dt

= 2 b 1 b

opta (iT )+ a (iT )τ = 0

S&M Eq. 4.67S&M Eq. 4.67

S&M Eq. 4.71S&M Eq. 4.71

=optτ 0

S&M Figure 4S&M Figure 4--1616

•• For this analysis of EFor this analysis of Ed, ASKd, ASK for BASK signals it is assumed for BASK signals it is assumed that the transmitter produces an that the transmitter produces an integer numberinteger number of cycles of cycles within one bit period:within one bit period:

[ ]−

d, BASK C(i-1)T

iT2 2 2 2b

d, BASK C(i-1)T

γ AE = 1 cos (4π f t) dt2

γ A T γ AE = cos (4π f t) dt2 2

S&M Eq. 5.27S&M Eq. 5.27

•• BASK OOK signals are BASK OOK signals are not not symmetrical with:symmetrical with:ss1T1T(t) = A sin(2(t) = A sin(2ππ ffcctt) ) ss2T2T(t) = 0(t) = 0 S&M S&M EqsEqs. 5.22, 5.23. 5.22, 5.23

and sand s11(t) = sin(2(t) = sin(2ππ ffcctt) s) s22(t) = 0(t) = 0

[ ]−∫

i b i 1 2(i-1)T

1 b 1 2 b(i-1)T

a (iT ) = γ s (t) s (t) s (t) dt

γ A Ta (iT ) = γ A s (t) dt = a (iT ) = 0 2

S&M Eq. 5.24S&M Eq. 5.24--5.265.26

2 b 1 b bopt

a (iT )+ a (iT ) γ A Tτ = = 2 4

•• BASK signals BASK signals in generalin general may may notnot be symmetrical with:be symmetrical with:ss1T1T(t) = A(t) = A11 sin(2sin(2ππ ffcctt) s) s2T2T(t) = A(t) = A22 sin(2sin(2ππ ffcctt))

and sand s11(t) (t) –– ss22(t) = sin (2(t) = sin (2ππ ffcctt) where the amplitude is arbitrary.) where the amplitude is arbitrary.

( ) = = 1 2 b2 b 1 b

γ A + A Ta (iT )+ a (iT )τ2 4

[ ]−∫

i b i 1 2(i-1)T

iT2 i b

i b i 1(i-1)T

a (iT ) = γ s (t) s (t) s (t) dt

γ A Ta (iT ) = γ A s (t) dt = 2

S&M Eq. 4.67S&M Eq. 4.67

S&M Eq. 4.71S&M Eq. 4.71

•• BFSK signals are BFSK signals are notnot symmetrical with:symmetrical with:ssTT(t(t) = A sin(2) = A sin(2ππ (f(fc c ±± ∆∆f)f) t) S&M Eq. 5.30t) S&M Eq. 5.30

if and

∆ − − ∆

∆ − ∆

d, BFSK C C(i-1)T

2 2d, BFSK b C 1 b C 2 b

E = A γ (sin (2π f + f) t sin (2π f f) t )) dt

E = γ A T f + f = n / T f f = n / T

2 2d, FSK b

b, BFSKo o

E γ A T P = Q Q2 N 2 N

S&M Eq. 5.31S&M Eq. 5.31S&M Eq. 5.32S&M Eq. 5.32

•• For this analysis of EFor this analysis of Ed, FSKd, FSK for BFSK signals it is assumed for BFSK signals it is assumed that the transmitter produces an that the transmitter produces an integer numberinteger number of cycles of cycles within one bit period:within one bit period:

− ∆

− − ∆

− + ∆ − ∆

iT2 22 2

d, BFSK b C(i-1)T

C(i-1)T

C C(i-1)T

γ AE = γ A T cos (4π (f + f) t) dt2

γ A cos (4π (f f) t) dt2

γ A sin (2π (f f) t) sin (2π (f f) t) dt

00S&M Eq. 5.31S&M Eq. 5.31

•• BFSK signals are BFSK signals are notnot symmetrical with:symmetrical with:ssTT(t(t) = A sin(2) = A sin(2ππ (f(fc c ±± ∆∆f)f) t) S&M Eq. 5.30t) S&M Eq. 5.30

and sand s11(t) (t) –– ss22(t) = sin(2(t) = sin(2ππ (fc + (fc + ∆∆f) t)f) t) – sin(2sin(2ππ (fc (fc –– ∆∆f) t)f) t)

S&M Eq. 5.31S&M Eq. 5.31

S&M Eq. 5.33S&M Eq. 5.33

[ ]−∫

i b i 1 2(i-1)T

iT2 i b

i b i i(i-1)T

a (iT ) = γ s (t) s (t) s (t) dt

γ A Ta (iT ) = γ A s (t) dt = 2

− 1 2 b2 b 1 bopt

γ A A Ta (iT )+ a (iT )τ = = = 02 4

•• A comparison of coherent PSK, FSK and ASK illustrates A comparison of coherent PSK, FSK and ASK illustrates the functional differences, but FSK and ASK use Ethe functional differences, but FSK and ASK use Edd not Enot Ebb::

2 2b b

b, BPSKo o

b, BPSKγ A TE =

2 2d, FSK b

b, BFSKo o

E γ A T P = Q = Q2 N 2 N

2 2d, BFSK bE = γ A T

2 2d, ASK b

b, BASKo o

E γ A T P = Q = Q2 N 4 N

d, BASKγ A TE =

•• The normalized EThe normalized Eb, FSK b, FSK = E= Eb, PSK b, PSK = = γγ2 2 AA2 2 TTb b / 2 / 2 (S&M Eq. (S&M Eq. 5.24) and E5.24) and Eb, ASKb, ASK = = γγ22 AA2 2 TTbb / 4/ 4 (S&M Eq. 5.36) so that:(S&M Eq. 5.36) so that:

Thus there are Thus there are no practical advantagesno practical advantages for either coherent for either coherent BFSK or BASK and BPSK is preferred (S&M p. 236).BFSK or BASK and BPSK is preferred (S&M p. 236).

2 2b, PSK b

b, BPSKo o

b, BPSKγ A TE =

2 2b, FSK b

b, BFSKo o

E γ A T P = Q = QN 2 N

b, BFSKγ A TE =

2 2b, ASK b

b, BASKo o

b, BASKγ A TE =

•• For the same PFor the same Pbb BPSK uses the least amount of energy, BPSK uses the least amount of energy, BFSK requires twice as much and BASK four times as much BFSK requires twice as much and BASK four times as much energy:energy:

2 2b, PSK b

b, BPSKo o

2 2b, FSK b

b, BFSKo o

2 2b, ASK b

b, BASKo o

Argument of Argument of Q should be Q should be as as largelarge as as possible to possible to minimize Pminimize Pbb

Chapter 3Chapter 3

Bandpass Modulation and Bandpass Modulation and DemodulationDemodulation•• Optimum Bandpass Receiver:Optimum Bandpass Receiver:The Correlation ReceiverThe Correlation Receiver

•• Pages 140Pages 140--146146

•• The matched filter orThe matched filter orcorrelation receivercorrelation receiverfor for bandpass bandpass symmetrical signalssymmetrical signalscan be simulated incan be simulated inSystemVue:SystemVue:

SVU Figure 3.1SVU Figure 3.1

•• The matched filter orThe matched filter orcorrelation receivercorrelation receiverfor for bandpass bandpass asymmetrical signalsasymmetrical signalscan also be simulatedcan also be simulatedinin SystemVue:SystemVue:

•• The The alternatealternate but but universaluniversalstructure which can be usedstructure which can be usedfor both asymmetric orfor both asymmetric orsymmetric binary bandpasssymmetric binary bandpasssignals can be simulatedsignals can be simulatedinin SystemVue:SystemVue:

SVUSVUFigure 3.2Figure 3.2

Chapter 3Chapter 3

Bandpass Modulation and Bandpass Modulation and DemodulationDemodulation•• Binary Amplitude Shift KeyingBinary Amplitude Shift Keying

•• Pages 146Pages 146--154154

•• Binary ASK coherent digital communication system with Binary ASK coherent digital communication system with BER analysis:BER analysis:

ThresholdThreshold

•• The BER and PThe BER and Pbb comparison (SVU Table 3.5, p. 167): comparison (SVU Table 3.5, p. 167):

Table 3.5Table 3.5 Observed BER and Theoretical PObserved BER and Theoretical Pb b as a as a Function of Function of EEd d / N/ Noo in a Binary ASK Digital in a Binary ASK Digital Communication System with Optimum ReceiverCommunication System with Optimum Receiver

EEdd/N/Noo dBdB NNoo VV22--secsec BERBER PPbb∞∞ 00 00 001212 7.89 7.89 ×× 1010--44 1.9 1.9 ×× 1010--33 2.5 2.5 ×× 1010--33

1010 1.25 1.25 ×× 1010--33 1.23 1.23 ×× 1010--22 1.25 1.25 ×× 1010--22

88 1.98 1.98 ×× 1010--33 3.61 3.61 ×× 1010--22 3.75 3.75 ×× 1010--22

66 3.14 3.14 ×× 1010--33 8.15 8.15 ×× 1010--22 7.93 7.93 ×× 1010--22

44 4.98 4.98 ×× 1010--33 1.325 1.325 ×× 1010--11 1.314 1.314 ×× 1010--11

22 7.85 7.85 ×× 1010--3 3 1.884 1.884 ×× 1010--11 1.872 1.872 ×× 1010--11

00 1.25 1.25 ×× 1010--22 2.387 2.387 ×× 1010--11 2.393 2.393 ×× 1010--11

Chapter 3Chapter 3

Bandpass Modulation and Bandpass Modulation and DemodulationDemodulation•• Binary Phase Shift KeyingBinary Phase Shift Keying

•• Pages 162Pages 162--163163

•• Binary PSK coherent digital communication system with Binary PSK coherent digital communication system with BER analysis:BER analysis:

SVU Figure 3.11SVU Figure 3.11ThresholdThreshold

Table 3.5Table 3.5 Observed BER and Theoretical PObserved BER and Theoretical Pb b as a as a Function of Function of EEb b / N/ Noo in a Binary PSK Digital in a Binary PSK Digital Communication System with Optimum ReceiverCommunication System with Optimum Receiver

EEbb/N/Noo dBdB NNoo VV22--secsec BERBER PPbb∞∞ 00 00 001010 1.25 1.25 ×× 1010--33 00 4.05 4.05 ×× 1010--66

88 1.98 1.98 ×× 1010--33 2 2 ×× 1010--44 2.06 2.06 ×× 1010--44

66 3.14 3.14 ×× 1010--33 2.9 2.9 ×× 1010--44 2.41 2.41 ×× 1010--33

44 4.98 4.98 ×× 1010--33 1.24 1.24 ×× 1010--22 1.25 1.25 ×× 1010--22

22 7.85 7.85 ×× 1010--3 3 3.77 3.77 ×× 1010--22 3.75 3.75 ×× 1010--22

00 1.25 1.25 ×× 1010--22 8.02 8.02 ×× 1010--22 7.93 7.93 ×× 1010--22

Chapter 3Chapter 3

Bandpass Modulation and Bandpass Modulation and DemodulationDemodulation•• Binary Frequency Shift KeyingBinary Frequency Shift Keying

•• Pages 162Pages 162--163163

•• Binary FSK coherent digital communication system with Binary FSK coherent digital communication system with BER analysis:BER analysis:

ThresholdfC+∆f fC-∆f

Table 3.4Table 3.4 Observed BER and Theoretical PObserved BER and Theoretical Pb b as a as a Function of Function of EEd d / N/ Noo in a Binary FSK (MFSK) Digital in a Binary FSK (MFSK) Digital Communication System with Optimum ReceiverCommunication System with Optimum Receiver

EEdd/N/Noo dBdB NNoo VV22--secsec BERBER PPbb∞∞ 00 00 001212 1.58 1.58 ×× 1010--33 1.7 1.7 ×× 1010--33 2.5 2.5 ×× 1010--33

1010 2.50 2.50 ×× 1010--33 1.19 1.19 ×× 1010--22 1.25 1.25 ×× 1010--22

88 3.96 3.96 ×× 1010--33 3.82 3.82 ×× 1010--22 3.75 3.75 ×× 1010--22

66 6.28 6.28 ×× 1010--33 8.19 8.19 ×× 1010--22 7.93 7.93 ×× 1010--22

44 9.95 9.95 ×× 1010--33 1.382 1.382 ×× 1010--11 1.314 1.314 ×× 1010--11

22 1.58 1.58 ×× 1010--2 2 1.854 1.854 ×× 1010--11 1.872 1.872 ×× 1010--11

00 2.50 2.50 ×× 1010--22 2.363 2.363 ×× 1010--11 2.393 2.393 ×× 1010--11

•• The BER and PThe BER and Pbb performance comparison for BASK, performance comparison for BASK, BPSK and BFSK (MFSK): BPSK and BFSK (MFSK):

EEdd/N/Noo dBdB NNoo VV22--secsec BERBER PPbb1010 1.25 1.25 ×× 1010--33 1.23 1.23 ×× 1010--22 1.25 1.25 ×× 1010--2 2 BASKBASK88 1.98 1.98 ×× 1010--33 3.61 3.61 ×× 1010--22 3.75 3.75 ×× 1010--22

EEbb/N/Noo dBdB NNoo VV22--secsec BERBER PPbb1010 1.25 1.25 ×× 1010--33 00 4.05 4.05 ×× 1010--6 6 BPSKBPSK88 1.98 1.98 ×× 1010--33 2 2 ×× 1010--44 2.06 2.06 ×× 1010--44

EEdd/N/Noo dBdB NNoo VV22--secsec BERBER PPbb1010 2.50 2.50 ×× 1010--33 1.19 1.19 ×× 1010--22 1.25 1.25 ×× 1010--2 2 BFSKBFSK88 3.96 3.96 ×× 1010--33 3.82 3.82 ×× 1010--22 3.75 3.75 ×× 1010--22

•• BER and PBER and Pbb comparison using Ecomparison using Ebb with Ewith Ebb, , ASK ASK = = γγ22 AA22 TTbb2 2 / 4/ 4

and thus reduced by 10 log (0.5) and thus reduced by 10 log (0.5) ≈≈ ––3 dB or3 dB or::

BASK performs better than BFSK but BPSK is the best. BASK performs better than BFSK but BPSK is the best.

EEbb/N/Noo dBdB NNoo VV22--secsec BERBER PPbb77 1.25 1.25 ×× 1010--33 1.23 1.23 ×× 1010--22 1.25 1.25 ×× 1010--2 2 BASKBASK55 1.98 1.98 ×× 1010--33 3.61 3.61 ×× 1010--22 3.75 3.75 ×× 1010--22

EEbb/N/Noo dBdB NNoo VV22--secsec BERBER PPbb1010 1.25 1.25 ×× 1010--33 00 4.05 4.05 ×× 1010--6 6 BPSKBPSK88 1.98 1.98 ×× 1010--33 2 2 ×× 1010--44 2.06 2.06 ×× 1010--44

EEbb/N/Noo dBdB NNoo VV22--secsec BERBER PPbb1010 2.50 2.50 ×× 1010--33 1.19 1.19 ×× 1010--22 1.25 1.25 ×× 1010--2 2 BFSKBFSK88 3.96 3.96 ×× 1010--33 3.82 3.82 ×× 1010--22 3.75 3.75 ×× 1010--22

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• Differential (Noncoherent) PhaseDifferential (Noncoherent) PhaseShift KeyingShift Keying

•• Pages 267Pages 267--271271

•• Differential (noncoherent) phase shift keying (DPSK) is Differential (noncoherent) phase shift keying (DPSK) is demodulated by using the received signal to derive the demodulated by using the received signal to derive the reference signal. The DPSK reference signal. The DPSK protocolprotocol is:is:

Binary 1Binary 1: Transmit the carrier signal with the : Transmit the carrier signal with the same phasesame phaseas used for the previous bit.as used for the previous bit.

Binary 0Binary 0: Transmit the carrier signal with its : Transmit the carrier signal with its phase shifted phase shifted by 180by 180°° relative to the previous bit. relative to the previous bit.

•• The oneThe one--bit delayed reference signal rbit delayed reference signal rii--11(t) is derived from (t) is derived from the received signal the received signal rrii(t(t) and if the carrier frequency f) and if the carrier frequency fCC is an is an integral multiple of the bit rate rintegral multiple of the bit rate rbb::

The output of the integrator for a binary 0 and binary 1 The output of the integrator for a binary 0 and binary 1 then is then is z(iTz(iTbb) = ) = ±± γγ22 AA22 TTbb / 2 (S&M / 2 (S&M EqsEqs. 5.91 and 5.93). 5.91 and 5.93)

−i 1 C b

r (t) = γ A sin (2π f (t T ) + θ)r (t) = γ A sin (2π f t + θ)

S&M S&M EqsEqs..5.88 and 5.895.88 and 5.89

•• DPSK signals have an equivalent bit interval TDPSK signals have an equivalent bit interval TDPSKDPSK = 2 T= 2 Tbb. . The probability of bit error for DPSK signal is different thanThe probability of bit error for DPSK signal is different thanthat for coherent demodulation of symmetric or asymmetric that for coherent demodulation of symmetric or asymmetric signals and is: signals and is:

− −

b, DPSKDPSKb, DPSK

b, DPSK

EE1 1P = exp = exp 2 2 N 2 N

γ A TE =2

S&M Eq. 5.102S&M Eq. 5.102

Chapter 3Chapter 3

Bandpass Modulation and Bandpass Modulation and DemodulationDemodulation•• Differential Phase Shift KeyingDifferential Phase Shift Keying

•• Pages 202Pages 202--207207

•• Binary DPSK Binary DPSK noncoherentnoncoherent digital communication system digital communication system with BER analysis:with BER analysis:

OneOne--bit delaybit delay

XORXOR BPFBPF

•• The The SystemVue SystemVue Logic Library provides the exclusiveLogic Library provides the exclusive--OR OR (XOR):(XOR):

•• The XOR logic generates the DPSK The XOR logic generates the DPSK source codingsource coding::

Table 3.13Table 3.13 Input Binary Data Input Binary Data bbii, Differentially Encoded , Differentially Encoded Binary Data Binary Data ddii, and Transmitted Phase , and Transmitted Phase φφii (Radians) for a (Radians) for a DPSK Signal.DPSK Signal.

bbii ddii--11 ddii φφii OneOne--bit startupbit startup11 0 0

11 11 11 00 XOR logicXOR logic00 11 00 ππ 0 0 10 0 100 00 11 0 0 0 1 00 1 011 11 11 00 1 0 01 0 000 11 00 ππ 1 1 11 1 100 00 11 0011 11 11 0011 11 11 00

•• The The SystemVue SystemVue Logic Library can be configured as true Logic Library can be configured as true binary binary (0, 1(0, 1)) or Mor M--ary (0, 1ary (0, 1……MM−−1)1) logic functions or logic functions or ananactual actual binary logicbinary logic family (0 and 1.8, 3.3 or 5 V):family (0 and 1.8, 3.3 or 5 V):

•• The The SystemVue SystemVue Operator Library provides the bandpass Operator Library provides the bandpass filter (BPF):filter (BPF):

•• The The SystemVue SystemVue Operator Library provides the bandpass Operator Library provides the bandpass filter (BPF) for the noncoherent correlation receiver :filter (BPF) for the noncoherent correlation receiver :

•• The Butterworth BPF is used for the The Butterworth BPF is used for the noncoherent receivernoncoherent receiver. . The The coherent receivercoherent receiver uses only the uses only the integratorintegrator as a as a virtual virtual BPFBPF::

Table 3.15Table 3.15 Observed BER and Theoretical PObserved BER and Theoretical Pb b as a as a Function of Function of EEb b / N/ Noo in a Binary DPSK Digital in a Binary DPSK Digital Communication System with Noncoherent Correlation Communication System with Noncoherent Correlation Receiver Receiver

EEbb/N/Noo dBdB NNoo VV22--secsec BERBER PPbb∞∞ 00 00 001212 7.89 7.89 ×× 1010--44 00 6.6 6.6 ×× 1010--88

1010 1.25 1.25 ×× 1010--33 2 2 ×× 1010--44 2.3 2.3 ×× 1010--55

88 1.98 1.98 ×× 1010--33 4.9 4.9 ×× 1010--33 1.8 1.8 ×× 1010--33

66 3.14 3.14 ×× 1010--33 1.71 1.71 ×× 1010--22 9.3 9.3 ×× 1010--33

44 4.98 4.98 ×× 1010--33 4.48 4.48 ×× 1010--22 4.06 4.06 ×× 1010--22

22 7.89 7.89 ×× 1010--3 3 9.96 9.96 ×× 1010--22 1.025 1.025 ×× 1010--11

00 1.25 1.25 ×× 1010--22 1.813 1.813 ×× 1010--11 1.839 1.839 ×× 1010--11

Statistical variation dueStatistical variation dueto small sample sizeto small sample size

•• BER and PBER and Pbb comparison between noncoherent, source comparison between noncoherent, source coded DPSK and coherent BPSK:coded DPSK and coherent BPSK:

BPSK performs better than DPSK but requires a coherent BPSK performs better than DPSK but requires a coherent reference signal. DPSK performs nearly as well as BPSK reference signal. DPSK performs nearly as well as BPSK at high SNR. at high SNR.

EEbb/N/Noo dBdB NNoo VV22--secsec BERBER PPbb1212 7.89 7.89 ×× 1010--44 00 6.6 6.6 ×× 1010--8 8 DPSKDPSK1010 1.25 1.25 ×× 1010--33 2 2 ×× 1010--44 2.3 2.3 ×× 1010--55

88 1.98 1.98 ×× 1010--33 4.9 4.9 ×× 1010--33 1.8 1.8 ×× 1010--33

EEbb/N/Noo dBdB NNoo VV22--secsec BERBER PPbb1212 7.89 7.89 ×× 1010--44 00 8.12 8.12 ×× 1010--8 8 BPSKBPSK1010 1.25 1.25 ×× 1010--33 00 4.05 4.05 ×× 1010--66

88 1.98 1.98 ×× 1010--33 2 2 ×× 1010--44 2.06 2.06 ×× 1010--44

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• MM--ary Bandpass Techniques:ary Bandpass Techniques:Quaternary Phase Shift KeyingQuaternary Phase Shift Keying

•• Pages 274Pages 274--286286

•• Quaternary phase shift keying (MQuaternary phase shift keying (M--ary, M = 4 or QPSK) ary, M = 4 or QPSK) source codes source codes dibits dibits bbii--11bbii as a as a symbolsymbol with one possible with one possible protocolprotocol as:as:

bbii--11bbii = = 11 A sin(211 A sin(2ππ ffCCtt + 45+ 45°°))bbii--11bbii = = 10 A sin(210 A sin(2ππ ffCCtt + 135+ 135°°))bbii--11bbii = = 00 A sin(200 A sin(2ππ ffCCtt + 225+ 225°°))bbii--11bbii = = 01 A sin(201 A sin(2ππ ffCCtt + 315+ 315°°))

The The Gray codeGray code is used asis used asfor Mfor M--ary PAM to improve ary PAM to improve the BER performance bythe BER performance bymitigating adjacent symbolmitigating adjacent symbolerror. The symbols are besterror. The symbols are bestdisplayed as a constellation plotdisplayed as a constellation plot S&M Figure 5S&M Figure 5--36 36 modifiedmodified

•• The QPSK signal can be simulated in The QPSK signal can be simulated in SystemVueSystemVue. The . The complex subsystems are shown as MetaSystems.complex subsystems are shown as MetaSystems.

II--Q component generator Q component generator MetaSystemMetaSystem

Gray source symbol Gray source symbol encoder MetaSystemencoder MetaSystem

SVU Figure 3.20 SVU Figure 3.20 (Transmitter only)(Transmitter only)

Quadrature Quadrature ModulatorModulator

•• The PN data source is Gray encoded as dibits and The PN data source is Gray encoded as dibits and converted to an M = 4 level symbol by the MetaSystem.converted to an M = 4 level symbol by the MetaSystem.

Gray source symbol Gray source symbol encoder MetaSystemencoder MetaSystem

•• The PN data source is Gray coded as dibits and converted The PN data source is Gray coded as dibits and converted to an M = 4 level symbol (0, 1, 2, and 3 V).to an M = 4 level symbol (0, 1, 2, and 3 V).

DownsamplerDownsampler

Gray coder Gray coder UpsamplerUpsampler

BitsBits--toto--symbol symbol converter 0, 1, 2 converter 0, 1, 2 or 3 Vor 3 V

•• The InThe In--phase (I) and Quadrature (Q) component phase (I) and Quadrature (Q) component generator provides the Igenerator provides the I--Q input to the Quad Modulator .Q input to the Quad Modulator .

II--Q component generator Q component generator MetaSystemMetaSystem

•• The M = 4 symbol input of 0, 1, 2 or 3 V is converted to The M = 4 symbol input of 0, 1, 2 or 3 V is converted to I and Q components of I and Q components of ±± 1 V using 1 V using threshold logicthreshold logic with with comparatorscomparatorsand and ANDAND gatesgateswith a summingwith a summingoutput nodeoutput nodeas an AND gateas an AND gate(I) or as an(I) or as anAdder token (Q)Adder token (Q)(SVU p. 187).(SVU p. 187).

I outputI output

Q outputQ output

•• The The ±± 1 V I1 V I--Q input signal to the Q input signal to the Quadrature ModulatorQuadrature Modulatorgenerates: A /generates: A /√√22 [ I sin (2[ I sin (2ππ ffCCtt + + θθ) + Q ) + Q coscos (2(2ππ ffCCtt + + θθ) ]) ]

•• For example, the IFor example, the I--Q signal I = 1, Q = 1 for bQ signal I = 1, Q = 1 for bii--11bbii = = 11 11 generates the signal sgenerates the signal s11(t):(t):

ss11(t) = A sin(2(t) = A sin(2ππ ffCCtt + 45+ 45°°) =) =A /A /√√2 [ I sin (22 [ I sin (2ππ ffCCtt + + θθ))

+ Q + Q coscos (2(2ππ ffCCtt + + θθ) ] =) ] =A /A /√√2 [ sin (22 [ sin (2ππ ffCCtt + + θθ))

+ + coscos (2(2ππ ffCCtt + + θθ) ]) ]

S&M Figure 5S&M Figure 5--36 36 modifiedmodified

sin Insin In--phasephase

Quadrature

ConstellationConstellationPlotPlot

Decision boundaries

•• The QPSK signal is derived from Gray coded dibits with The QPSK signal is derived from Gray coded dibits with 00 00 →→ 00 (0), 01 00 (0), 01 →→ 01 (1), 10 01 (1), 10 →→ 11 (3) and 11 11 (3) and 11 →→ 10 (2)10 (2). .

PN binary source, rPN binary source, rbb = 1 kb/sec= 1 kb/sec

M = 4 symbol, Gray coded dibitsM = 4 symbol, Gray coded dibits

QPSK signalQPSK signal

0 0 →→ 1 V1 V

0, 1, 2, 3 V0, 1, 2, 3 V

±± 5 V, f5 V, fCC = 2 kHz, r= 2 kHz, rSS = 500 Hz= 500 Hz

10 11 10 11 1111 0101

•• The QPSK signal is derived from Gray coded dibits with The QPSK signal is derived from Gray coded dibits with 00 00 →→ 00 (0), 01 00 (0), 01 →→ 01 (1), 10 01 (1), 10 →→ 11 (3) and 11 11 (3) and 11 →→ 10 (2)10 (2). .

PN binary source, rPN binary source, rbb = 1 kb/sec= 1 kb/sec

M = 4 symbol, Gray coded dibitsM = 4 symbol, Gray coded dibits

QPSK signalQPSK signal

0 0 →→ 1 V1 V

0, 1, 2, 3 V0, 1, 2, 3 V

±± 5 V, f5 V, fCC = 2 kHz, r= 2 kHz, rSS = 500 Hz= 500 Hz

10 11 10 11 1111 0101

•• The QPSK signal can be decomposed into I and Q BPSK The QPSK signal can be decomposed into I and Q BPSK signals which are signals which are orthogonalorthogonal to each other.to each other.

QPSK signal, fQPSK signal, fCC = 2 kHz, r= 2 kHz, rSS = 500 b/sec= 500 b/sec

Binary PSK signal, sine carrier (I)Binary PSK signal, sine carrier (I)

Binary PSK signal, cosine carrier (Q)Binary PSK signal, cosine carrier (Q)

±± 5 V5 V

±± 5 / 5 / √√2 = 3.536 V2 = 3.536 V

•• The The orthogonalityorthogonality of the I and Q components of the QPSK of the I and Q components of the QPSK signal can be exploited by the signal can be exploited by the universal coherent receiveruniversal coherent receiver. . The orthogonal I and Q components actually occupy the The orthogonal I and Q components actually occupy the same spectrumsame spectrum without interference. The coherent without interference. The coherent reference signals are:reference signals are:

Quadrature InQuadrature In--phasephasess11(t) = (t) = coscos (2(2ππ ffCCtt + + θθ) s) s22(t) = sin (2(t) = sin (2ππ ffCCtt + + θθ))

•• The orthogonality of the QPSK signals can be shown by The orthogonality of the QPSK signals can be shown by observing the output of the quadrature correlator to the I observing the output of the quadrature correlator to the I and Q signal.and Q signal.

1 S I C Q C C(i-1)T

1 S I C C(i-1)T

Q C(i-1)T

S1 S Q

γ Az (nT ) = d sin(2π f t) + d cos(2π f t) cos (2π f t) dt2

γ Az (nT ) = d sin(2π f t) cos (2π f t) dt + 2

γ A d cos (2π f t) dt2

γ A Tz (nT ) = d2 2

zz11(nT(nTSS))

S&M Eq. 5.109S&M Eq. 5.109

•• The probability of bit error PThe probability of bit error Pbb and the energy per bit Eand the energy per bit Ebb for a for a QPSK signal is the same as that as for a BPSK signal but QPSK signal is the same as that as for a BPSK signal but with a I and Q carrier amplitude of A / with a I and Q carrier amplitude of A / √√22 ..

zz11(nT(nTSS))

S&M Eq. 5.117S&M Eq. 5.117

2 2b, PSK b

b, BPSKo o

2 2b, QPSK S

b, QPSKo o

2 E γ A T P = Q = QN 2 N

b, BPSK

b, QPSK

γ A TE = 2

γ A TE = 4

•• Since TSince TSS = 2 T= 2 Tb b BPSK and QPSK have the same PBPSK and QPSK have the same Pbb but but QPSK can have twice the data rate rQPSK can have twice the data rate rbb = 2 r= 2 rS S within the same within the same bandwidth because of the orthogonal I and Q components.bandwidth because of the orthogonal I and Q components.

zz11(nT(nTSS))

2 2b b

b, BPSK b, QPSKo o

2 E γ A T P = P = Q = QN N

b, BPSK b, QPSKγ A TE = E =

Chapter 5Chapter 5

Bandpass Modulation and Bandpass Modulation and DemodulationDemodulation•• Multilevel (MMultilevel (M--ary) Phase Shiftary) Phase ShiftKeyingKeying

•• Pages 184Pages 184--194194

•• QPSK coherent digital communication system with BER QPSK coherent digital communication system with BER analysis:analysis:

Gray sourceGray sourcesymbol encoder symbol encoder MetaSystemMetaSystem

II--Q component generator MetaSystemQ component generator MetaSystem

II--Q correlator MetaSystemQ correlator MetaSystem

Gray sourceGray sourcesymbolsymboldecoder decoder MetaSystemMetaSystem

•• QPSK coherent receiver uses an IQPSK coherent receiver uses an I--Q correlator.Q correlator.

II--Q correlator MetaSystemQ correlator MetaSystem

•• The IThe I--Q correlationQ correlationreceiver is thereceiver is theuniversal universal structurestructurewith an integrationwith an integrationtime equal to thetime equal to thesymbol time symbol time TTSS..

MSB 0 or 2 VMSB 0 or 2 V

LSB 0 or 1 VLSB 0 or 1 V

0, 1, 20, 1, 2or 3 Vor 3 V

•• The output of the IThe output of the I--Q correlator is Gray decoded:Q correlator is Gray decoded:

Gray sourceGray sourcesymbolsymboldecoder decoder MetaSystemMetaSystem

•• The output of the IThe output of the I--Q correlation receiver is an M = 4 level Q correlation receiver is an M = 4 level symbol (0, 1, 2, and 3 V) and converted to a symbol (0, 1, 2, and 3 V) and converted to a dibitdibit and Gray and Gray decoded. decoded.

DownsamplerDownsampler SymbolSymbol--toto--bits Gray decoderbits Gray decoderconverterconverter

•• The The singlesingle--sidedsided power spectral density PSD of the QPSK power spectral density PSD of the QPSK signal uses signal uses rrss = r= rbb/2 and is:/2 and is:

GGQPSKQPSK(f(f) = 1/2 ) = 1/2 GGPAMPAM(f(f -- ffCC))GGPAMPAM(f(f) = A) = A2 2 / / rrss sincsinc22 ((ππ f / f / rrss))

No carrier

rs = 500 s/sec, rb = 1 kb/sec

•• The The singlesingle--sidedsided power spectral density PSD of BPSK has power spectral density PSD of BPSK has double the bandwidth than that for QPSK for the same bit double the bandwidth than that for QPSK for the same bit rate rrate rbb = 1/T= 1/Tb,b,

No carrier

rb = 1 kHz

•• The The bandwidthbandwidth of a QPSK signal as a percentage of total of a QPSK signal as a percentage of total power is power is half half that for the same bit rate rthat for the same bit rate rb b = 1/T= 1/Tbb BPSK signal BPSK signal since since rrss = r= rbb/2 or T/2 or Tss = 2T= 2Tb b (SVU Table 3.10). (SVU Table 3.10).

2/T2/Tss 1/T1/Tbb 90%90%3/T3/Tss 1.5/T1.5/Tbb 93%93%4/T4/Tss 2/T2/Tbb 95%95%6/T6/Tss 3/T3/Tbb 96.5%96.5%8/T8/Tss 4/T4/Tbb 97.5%97.5%10/T10/Tss 5/T5/Tbb 98%98%

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• MM--ary Bandpass Techniques:ary Bandpass Techniques:88--Phase Shift KeyingPhase Shift Keying

•• Pages 286Pages 286--292292

•• MM--ary phase shift keying (M = 8 or 8PSK) source codes ary phase shift keying (M = 8 or 8PSK) source codes tribitstribits bbii--22bbii--11bbii as a as a symbolsymbol with one possible with one possible protocolprotocol as:as:

bbii--22bbii--11bbii = = 000 A sin(2000 A sin(2ππ ffCCtt + 0+ 0°°))bbii--22bbii--11bbii = = 001 A sin(2001 A sin(2ππ ffCCtt + 45+ 45°°))bbii--22bbii--11bbii = = 011 A sin(2011 A sin(2ππ ffCCtt + 90+ 90°°))bbii--22bbii--11bbii = = 010 A sin(2010 A sin(2ππ ffCCtt + 135+ 135°°))bbii--22bbii--11bbii = = 110 A sin(2110 A sin(2ππ ffCCtt + 180+ 180°°))bbii--22bbii--11bbii = = 111 A sin(2111 A sin(2ππ ffCCtt + 225+ 225°°))bbii--22bbii--11bbii = = 101 A sin(2101 A sin(2ππ ffCCtt + 270+ 270°°))bbii--22bbii--11bbii = = 100 A sin(2100 A sin(2ππ ffCCtt + 315+ 315°°))

θθI, Q = 0, I, Q = 0, ±± 1/1/√√2, 2, ±± 1 1 s(ts(t) =) = A [ I sin (2A [ I sin (2ππ ffCCtt) + ) +

Q Q coscos (2(2ππ ffCCtt) ]) ] S&M Figure 5S&M Figure 5--4343

sin Insin In--phasephase

Quadrature

Constellation PlotConstellation Plot

•• The correlation receiverThe correlation receiverfor 8for 8--PSK uses fourPSK uses fourreference signals:reference signals:

ssrefref nn(t(t) = ) = sin (2sin (2ππ ffCCtt + +

n 45n 45°° + 22.5+ 22.5°°))φφ

n = 0, 1, 2, 3n = 0, 1, 2, 3φφ = 22.5= 22.5°°, 67.5, 67.5°°,,

112.5112.5°°, 157.5, 157.5°°

S&M Eq. 5.124S&M Eq. 5.124

•• The output from any one of the four correlators is:The output from any one of the four correlators is:

− −

∫ ∫

1 S C C(i-1)T

1 S C(i-1)T (i-1)T

z (nT ) = γ A sin (2π f t + θ) sin (2π f t + φ)dt

γ Az (nT ) = cos (θ φ) dt cos (4π f t + θ + φ) dt2

γ A Tz (nT ) = cos (θ φ)2

zz11((nnTTSS))

S&M Eq. 5.125S&M Eq. 5.125

•• The correlator output is > 0 if | The correlator output is > 0 if | θθ –– φφ | < 90| < 90°° and < 0 if notand < 0 if notbecause of the because of the coscos ((θθ –– φφ) term. For example, if s) term. For example, if s66(t) is (t) is received, the received, the ABCDABCD correlator sign output is: correlator sign output is: –– –– –– +. The +. The patterns of signs are unique and can be decoded to bpatterns of signs are unique and can be decoded to bii--22bbii--11bbii

(S&M Tab(S&M Table 5le 5--7 and 57 and 5--8)8)

zz11((nnTTSS))

A: sref 1(t)

B: sref 2(t)

C: sref 3(t)

D: sref 4(t)

•• The probability of The probability of symbol errorsymbol error PPSS for coherently for coherently demodulated Mdemodulated M--ary PSK is:ary PSK is:

≈ ≥

S coherent M-ary PSKo

2bS coherent M-ary PSK 2

A T πP 2Q sin M 4N M

E πP 2Q 2 log M sin M 4N M

S&M Eq. 5.126S&M Eq. 5.126

EEbb//NNoodBdB

•• The probability of The probability of symbol errorsymbol error PPSS for coherently for coherently demodulated Mdemodulated M--ary PSK is:ary PSK is:

EEbb//NNoodBdB

•• The probability of symbol error PThe probability of symbol error PSS must be related to must be related to probability of bit error Pprobability of bit error Pbb for consistency. If for consistency. If Gray codingGray coding is is used, assume that errors will only be due to used, assume that errors will only be due to adjacent adjacent symbolssymbols. Thus each symbol error produces only one bit in. Thus each symbol error produces only one bit inerror and logerror and log22 (M (M –– 1) correct bits or:1) correct bits or:

However for MHowever for M--ary PSK with M > 4 the assumption of errors ary PSK with M > 4 the assumption of errors being due to only adjacent symbols is invalid. For the being due to only adjacent symbols is invalid. For the worst case there are M worst case there are M –– 1 incorrect symbols and in M / 2 of 1 incorrect symbols and in M / 2 of these a bit will different from the correct bit so that:these a bit will different from the correct bit so that:

b errors due to adjacent symbols S2

1P = Plog M S&M Eq. 5.127S&M Eq. 5.127

≤ ≤−S b S

1 M P P Plog M 2 (M 1)

S&M Eq. 5.129S&M Eq. 5.129

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• MM--ary Bandpass Techniques:ary Bandpass Techniques:Quaternary Frequency Shift KeyingQuaternary Frequency Shift Keying

•• Pages 292Pages 292--298298

•• The analytical signal for quaternary (MThe analytical signal for quaternary (M--ary, M = 4) ary, M = 4) frequency shift keying (QFSK) is:frequency shift keying (QFSK) is:

ssQFSKQFSK(t(t) =) = A sin (2A sin (2ππ (f(fCC + 3+ 3∆∆f) t + f) t + θθ) if b) if bii--11bbii = 11= 11ssQFSKQFSK(t(t) = A sin (2) = A sin (2ππ (f(fCC + + ∆∆f) t + f) t + θθ) if b) if bii--11bbii = 10= 10ssQFSKQFSK(t(t) = A sin (2) = A sin (2ππ (f(fCC –– ∆∆f) t + f) t + θθ) if b) if bii--11bbii = 00 = 00 ssQFSKQFSK(t(t) = A sin (2) = A sin (2ππ (f(fCC –– 33∆∆f) t + f) t + θθ) if b) if bii--11bbi i = 01= 01

ffCC + 3+ 3∆∆ff

ffCC –– 33∆∆ff

ffCC + + ∆∆ff

ffCC –– ∆∆ff

•• Chose fChose fCC and and ∆∆f so that if there are a whole number of half f so that if there are a whole number of half cycles of a sinusoid within a symbol time Tcycles of a sinusoid within a symbol time TSS for M = 4 forfor M = 4 fororthogonality of the signals so that a correlation receiver orthogonality of the signals so that a correlation receiver can be utilized.can be utilized.

ssQFSKQFSK(t(t) =) = A sin (2A sin (2ππ (f(fCC + 3+ 3∆∆f) t + f) t + θθ) if b) if bii--11bbii = 11= 11ssQFSKQFSK(t(t) = A sin (2) = A sin (2ππ (f(fCC + + ∆∆f) t + f) t + θθ) if b) if bii--11bbii = 10= 10ssQFSKQFSK(t(t) = A sin (2) = A sin (2ππ (f(fCC –– ∆∆f) t + f) t + θθ) if b) if bii--11bbii = 00 = 00 ssQFSKQFSK(t(t) = A sin (2) = A sin (2ππ (f(fCC –– 33∆∆f) t + f) t + θθ) if b) if bii--11bbi i = 01= 01

ffCC + 3+ 3∆∆ff

ffCC –– 33∆∆ff

ffCC + + ∆∆ff

ffCC –– ∆∆ff

•• The correlation receiverThe correlation receiverfor QFSK uses fourfor QFSK uses fourreference signals:reference signals:

ssrefref nn(t(t) = ) = sin (2sin (2ππ (f(fC C + n + n ∆∆f)f) t)t)

n = n = ±±1, 1, ±±33

•• The probability of The probability of symbol errorsymbol error PPSS for coherently for coherently demodulated Mdemodulated M--ary FSK is:ary FSK is:

≤ − ≥

− ≥

S coherent M-ary FSKo

S coherent M-ary FSK

A TP (M 1) Q M 42 N

E(M 1) Q log M M 4N

S&M Eq. 5.132S&M Eq. 5.132

EEbb//NNoodBdB

•• The probability of The probability of symbol errorsymbol error PPSS for coherently for coherently demodulated Mdemodulated M--ary FSK is:ary FSK is:

EEbb//NNoodBdB

Chapter 5Chapter 5

Bandpass Modulation and Bandpass Modulation and DemodulationDemodulation•• Multilevel (MMultilevel (M--ary) Frequencyary) FrequencyShift KeyingShift Keying

•• Pages 177Pages 177--184184

•• 44--FSK coherent digital communication system with BER FSK coherent digital communication system with BER analysis:analysis:

SymbolSymbolencoderencoderMetaSystemMetaSystem

Frequency Frequency ModulatorModulator

•• The dibits are converted to a symbol and scaled. The data The dibits are converted to a symbol and scaled. The data is is not not Gray encoded. Symbol errors are equally likely Gray encoded. Symbol errors are equally likely among the M among the M –– 1 correlators and there is 1 correlators and there is no advantageno advantage to to Gray encoding.Gray encoding.

DownsamplerDownsampler BitsBits--toto--symbol symbol convconvertererter

Upsampler

44--FSK correlator MetaSystemFSK correlator MetaSystem

SymbolSymboldecoder decoder MetaSystemMetaSystem

44--FSK correlator MetaSystemFSK correlator MetaSystem

•• The 4The 4--FSK correlationFSK correlationreceiver has fourreceiver has fourcorrelators with ancorrelators with anintegration time equalintegration time equalto the to the symbol time symbol time TTSS..

SymbolSymboldecoder decoder MetaSystemMetaSystem

•• The symbols are converted to dibits. The original data is The symbols are converted to dibits. The original data is not not Gray encoded and is therefore Gray encoded and is therefore notnot Gray decodedGray decoded..

DownsamplerDownsampler SymbolSymbol--toto--bits bits converterconverter

•• The The singlesingle--sidedsided power spectral density PSD of the 4power spectral density PSD of the 4--FSK FSK with a minimum carrier frequency deviation in MFSK is with a minimum carrier frequency deviation in MFSK is ∆∆f = 1/Tf = 1/TS S = r= rbb/2. /2. The carriers should be spaced at The carriers should be spaced at multiples of n/Tmultiples of n/TS S = nr= nrbb/2 (S&M Eq. 5.131 is/2 (S&M Eq. 5.131 is incorrectincorrect).).

No carrier

rs = 500 s/sec, rb = 1 kb/sec

∆f = rb/2 = 500 Hz

•• The The bandwidthbandwidth of a Mof a M--ary FSK signal as a percentage of ary FSK signal as a percentage of total power (SVU Table 3.8 total power (SVU Table 3.8 typographical error correctedtypographical error corrected). ).

2( M 2( M –– 1) 1) ∆∆f + f + 4/T4/Tss 95%95%2 (M 2 (M –– 1) 1) ∆∆f + f + 6/T6/Tss 96.5%96.5%2 (M 2 (M –– 1) 1) ∆∆f + f + 8/T8/Tss 97.5%97.5%2 (M 2 (M –– 1) 1) ∆∆f + f + 10/T10/Tss 98%98%

For MFSK: For MFSK: ∆∆f = 1/Tf = 1/TSS = r= rbb/2/2

Chapter 5Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques•• MM--ary Bandpass Techniques:ary Bandpass Techniques:Quadrature Amplitude ModulationQuadrature Amplitude Modulation

•• Pages 298Pages 298--301301

•• The analytical signal for quadrature amplitude modulation The analytical signal for quadrature amplitude modulation (QAM) has I(QAM) has I--Q components:Q components:

ssQAMQAM(t(t) = I sin (2) = I sin (2ππ ffCCtt) + Q ) + Q coscos (2(2ππ ffCCtt))

A QAM signal has A QAM signal has both both amplitudeamplitude and and phasephasecomponents which can becomponents which can beshown in the shown in the constellationconstellationplotplot..

S&M Figure 5S&M Figure 5--5353I

16-ary QAM

•• An MAn M--ary PSK signal also has Iary PSK signal also has I--Q components but the Q components but the amplitude is amplitude is constantconstant and only the phase and only the phase variesvaries::

ssQAMQAM(t(t) = I sin (2) = I sin (2ππ ffCCtt) + Q ) + Q coscos (2(2ππ ffCCtt))

1616--ary ary PSKPSK

1616--ary ary QAMQAM

•• The The orthogonalityorthogonality of the I and Q components of the QAM of the I and Q components of the QAM signal can be exploited by the signal can be exploited by the universal coherent receiveruniversal coherent receiver. . The orthogonal I and Q components actually occupy the The orthogonal I and Q components actually occupy the same spectrumsame spectrum without interference. The coherent without interference. The coherent reference signals are:reference signals are:

Quadrature InQuadrature In--phasephasess11(t) = (t) = coscos (2(2ππ ffCCtt) s) s22(t) = sin (2(t) = sin (2ππ ffCCtt))

•• An An upperupper--boundbound for the probability of for the probability of symbol errorsymbol error PPSS for for coherently demodulated Mcoherently demodulated M--ary QAM is:ary QAM is:

≤ − s

S coherent M-ary QAMo

3 EP Q(M 1) N S&M Eq. 5.135S&M Eq. 5.135

QAM BER curveQAM BER curve M = 256M = 256

M = 4M = 4

•• An MAn M--ary QAM constellation plot shows the ary QAM constellation plot shows the stabilitystability of the of the signaling and the signaling and the transitiontransition from one signal to another:from one signal to another:

256256--ary QAMary QAM1616--ary QAMary QAM

Chapter Chapter 55

Bandpass Modulation and Bandpass Modulation and DemodulationDemodulation•• Quadrature AmplitudeQuadrature AmplitudeModulationModulation

•• Pages 194Pages 194--202202

QAMQAMMapperMapper

Quadrature ModulatorQuadrature Modulator

•• 44--FSK coherent digital communication transmitter utilizes FSK coherent digital communication transmitter utilizes the QAM Mapper and an external file.the QAM Mapper and an external file.

QAMQAMMapperMapper

II--Q correlatorQ correlatorMetaSystemMetaSystem

QAMQAMDeDe--mappermapper

II--Q correlatorQ correlatorMetaSystemMetaSystem

•• 44--FSK coherent digital communication receiver utilizes the FSK coherent digital communication receiver utilizes the QAM DeQAM De--Mapper and an external file:Mapper and an external file:

QAMQAMDemapperDemapper

•• The The singlesingle--sidedsided power spectral density PSD of the 16power spectral density PSD of the 16--ary ary QAM has a bandwidth of 1/M that of a PSK signal with the QAM has a bandwidth of 1/M that of a PSK signal with the same data rate rsame data rate rbb..

rs = 250 s/sec, rb = 1 kb/secSVU Figure 3.27SVU Figure 3.27

BPSK PSDBPSK PSDrrb b = 1 kb/sec= 1 kb/sec

1616--ary QAM PSDary QAM PSDrrbb = 1 kb/sec= 1 kb/secM = 4 rM = 4 rSS = 250 s/sec= 250 s/sec

1 kHz1 kHz

250 Hz250 Hz

•• The The bandwidthbandwidth of an Mof an M--ary QAM signal as a percentage of ary QAM signal as a percentage of total power is 1/Mtotal power is 1/M that for the same bit rate rthat for the same bit rate rb b = 1/T= 1/Tbb BPSK BPSK signal since signal since rrss = r= rbb/M or T/M or Tss = = MTMTbb (SVU Table 3.12). (SVU Table 3.12).

2/T2/Tss 2/MT2/MTbb 90%90%3/T3/Tss 3/MT3/MTbb 93%93%4/T4/Tss 4/MT4/MTbb 95%95%6/T6/Tss 6/MT6/MTbb 96.5%96.5%8/T8/Tss 8/MT8/MTbb 97.5%97.5%10/T10/Tss 10/MT10/MTbb 98%98%

•• 44--FSK coherent digital communication system received IFSK coherent digital communication system received I--Q Q components can be displayed on a constellation plot.components can be displayed on a constellation plot.

QAMQAMDeDe--mappermapper

•• The 16The 16--ary QAM Iary QAM I--Q component constellation plot withQ component constellation plot withEEbb/N/No o →→ ∞∞..

Signal points

Signal transitionsSignal transitions

Decision boundariesDecision boundaries

•• The 16The 16--ary QAM Iary QAM I--Q component constellation plot withQ component constellation plot withEEbb/N/No o →→ 12 dB, P12 dB, Pbb ≈≈ 0.00010.0001..

•• The 16The 16--ary QAM Iary QAM I--Q component constellation plot withQ component constellation plot withEEbb/N/No o →→ 10 dB, P10 dB, Pbb = 0.0024= 0.0024..

•• The 16The 16--ary QAM Iary QAM I--Q component constellation plot withQ component constellation plot withEEbb/No = 6 dB, P/No = 6 dB, Pbb = 0.0367.= 0.0367.

End of Chapter 5End of Chapter 5

Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques

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