9. Bandpass Modulation and Demodulation Techniques · Bandpass Modulation and Demodulation...

Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary

9. Bandpass Modulation and DemodulationTechniques

Y. Yoganandam, Runa Kumari, and S. R. Zinka

Department of Electrical & Electronics EngineeringBITS Pilani, Hyderbad Campus

October 21 – 30, 2015

9. Bandpass Modulation and Demodulation Techniques Communication Systems, Dept. of EEE, BITS Hyderabad


Outline

1 Introduction

2 Phase Shift Keying (PSK)

3 Differential PSK (DPSK)

4 Frequency Shift Keying (FSK)

5 ASK & APK

6 Summary



Outline

1 Introduction




5 ASK & APK

6 Summary



Symbols Symbols Symbols Symbols Symbols Symbols Symbols

Symbols Symbols Symbols Symbols Symbols Symbols SymbolsSymbols Symbols Symbols Symbols Symbols Symbols





Symbols Symbols Symbols Symbols Symbols









Symbols Symbols Symbols Symbols Symbols



Analog Modulation Techniques

Carrier Signal:Ac cos (ωct + θ)

Amplitude Modulation:AAM = f (m (t))

Frequency Modulation:ωFM = g (m (t))

Phase Modulation:θPM = h (m (t))































Digital Modulation Techniques

The main difference between analog and digital modulation techniques ischaracterized by the message signal. In analog case m (t) is an analog signal,

whereas in digital case m (t) is a digital signal.



Digital Modulation Techniques

The main difference between analog and digital modulation techniques ischaracterized by the message signal. In analog case m (t) is an analog signal,

whereas in digital case m (t) is a digital signal.



A Slight Modified Representation of Carrier Signal

A sinusoidal signal S (t) = A cos ωt can be represented in terms of it’s poweras

S (t) =√

2P cos ωt (1)

because P = A2/2.

The above equation can be further re-written as

S (t) =

√2Es

TScos ωt (2)

because P watts can be replaced by Es joules/TS seconds.��

��

As we have seen in the previous chapter, energy of a received signal is thekey parameter in determining the error performance of the detection process.

That is the reason why we are using the notation√

2Es/TS in this chapter.





S (t) =√

2P cos ωt (1)

because P = A2/2.


S (t) =

√2Es

TScos ωt (2)


��








S (t) =√

2P cos ωt (1)

because P = A2/2.


S (t) =

√2Es

TScos ωt (2)

because P watts can be replaced by Es joules/TS seconds.

��

��








S (t) =√

2P cos ωt (1)

because P = A2/2.


S (t) =

√2Es

TScos ωt (2)


��






So, Digital Modulation Techniques are ...

Carrier Signal: √2Es/TS cos (ωct + φ)

Phase Shift Keying:φPSK = φi (t)

Frequency Shift Keying:ωFSK = ωi (t)

Amplitude Shift Keying:EASK = Ei (t)

ASK + PSK:EAPK = Ei (t) & φAPK = φi (t)











































Digital Modulation Techniques – Examples

ASK




ASK




ASK

PSK




ASK

PSK

FSK




ASK

PSK

FSK

ASK/PSK (APK)



Bits vs Symbols



Bits vs Symbols



Bits vs Symbols

BitwiseASK



Bits vs Symbols

BitwiseASK



Bits vs Symbols

BitwiseASK

SymbolwiseASK



Why Digital Modulation?

Digital modulation is a process of translating the symbols into waveforms.

In baseband modulation, the waveforms are shaped pulses.

We can not send baseband signals over wireless. We need to translate thesignals to higher frequencies for ease of transmission.

Higher frequencies give us the benefit of multiplexing multiple channelsaround the same carrier.































Symbol Energy vs bit Energy

Es = nEb (3)



Symbol Energy vs bit Energy

Es = nEb (3)



Spectral Properties of ASK

Amplitude shift keyingor

On-off keying

0 1 0 1 1 0 1 0 1 1





On-off keying

0 1 0 1 1 0 1 0 1 1





On-off keying

0 1 0 1 1 0 1 0 1 1

Carrier





On-off keying

0 1 0 1 1 0 1 0 1 1

Carrier

On-offline-coded

data

1 V

0 V





On-off keying

0 1 0 1 1 0 1 0 1 1

Carrier

On-offline-coded

data

+1 V

0 V





On-off keying

0 1 0 1 1 0 1 0 1 1

Carrier

On-offline-coded

data

+1 V

0 V

2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω

On-offLine-codespectrum





On-off keying

0 1 0 1 1 0 1 0 1 1

Carrier

On-offline-coded

data

+1 V

0 V

2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω


ω

ASKspectrum

-ω0 +ω0



Spectral Properties of PSK

Phase shift keying

0 1 0 1 1 0 1 0 1 1




Phase shift keying

0 1 0 1 1 0 1 0 1 1




Phase shift keying

0 1 0 1 1 0 1 0 1 1

Carrier




Phase shift keying

0 1 0 1 1 0 1 0 1 1

Carrier

NRZ-Lline-coded

data

1 V

-1 V




Phase shift keying

0 1 0 1 1 0 1 0 1 1

Carrier

NRZ-Lline-coded

data

+1 V

-1 V




Phase shift keying

0 1 0 1 1 0 1 0 1 1

Carrier

NRZ-Lline-coded

data

+1 V

-1 V

2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω

NRZ-LLine-codespectrum




Phase shift keying

0 1 0 1 1 0 1 0 1 1

Carrier

NRZ-Lline-coded

data

+1 V

-1 V

2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω

NRZ-LLine-codespectrum

ω

PSKspectrum

-ω0 +ω0



Spectral Properties of FSK

Frequency shift keying

0 1 0 1 1 0 1 0 1 1





0 1 0 1 1 0 1 0 1 1





0 1 0 1 1 0 1 0 1 1

ASK 1





0 1 0 1 1 0 1 0 1 1

ASK 1

ASK 2

1 V

-1 V





0 1 0 1 1 0 1 0 1 1

ASK 1

ASK 2

+

1 V

-1 V




2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω



0 1 0 1 1 0 1 0 1 1

ASK 1

ASK 2

+

1 V

-1 V




2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω


ω

FSKspectrum

-ω2 +ω1


0 1 0 1 1 0 1 0 1 1

ASK 1

ASK 2

+

1 V

-1 V



Outline

1 Introduction




5 ASK & APK

6 Summary



Phase Shift Keying (MPSK)

If a digital signal can be completely represented by M different levels, thenthose M levels can be converted into M symbols (or alphabets). Each of theseM symbols can be associated with a unique phase φi, where i = 1, 2, · · · , M.Since 0 ≤ φ ≤ 2π, if we distribute these phases uniformly,

φi =2πiM

, i = 1, 2, · · · , M. (4)

So, the modulated PSK signal is given by

Si (t) =

√2Es

TSsin(

ω0t +2πiM

), i = 1, 2, · · · , M

=

√2Es

TS

[sin (ω0t) cos

(2πiM

)+ cos (ω0t) sin

(2πiM

)]=√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], (5)

where ψ1 (t) =√

2/TS sin ω0t and ψ2 (t) =√

2/TS cos ω0t.



Phase Shift Keying (MPSK)If a digital signal can be completely represented by M different levels, thenthose M levels can be converted into M symbols (or alphabets).

Each of theseM symbols can be associated with a unique phase φi, where i = 1, 2, · · · , M.Since 0 ≤ φ ≤ 2π, if we distribute these phases uniformly,

φi =2πiM

, i = 1, 2, · · · , M. (4)


Si (t) =

√2Es

TSsin(

ω0t +2πiM

), i = 1, 2, · · · , M

=

√2Es

TS

[sin (ω0t) cos

(2πiM

)+ cos (ω0t) sin

(2πiM

)]=√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], (5)

where ψ1 (t) =√


2/TS cos ω0t.



Phase Shift Keying (MPSK)If a digital signal can be completely represented by M different levels, thenthose M levels can be converted into M symbols (or alphabets). Each of theseM symbols can be associated with a unique phase φi, where i = 1, 2, · · · , M.

Since 0 ≤ φ ≤ 2π, if we distribute these phases uniformly,

φi =2πiM

, i = 1, 2, · · · , M. (4)


Si (t) =

√2Es

TSsin(

ω0t +2πiM

), i = 1, 2, · · · , M

=

√2Es

TS

[sin (ω0t) cos

(2πiM

)+ cos (ω0t) sin

(2πiM

)]=√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], (5)

where ψ1 (t) =√


2/TS cos ω0t.



Phase Shift Keying (MPSK)If a digital signal can be completely represented by M different levels, thenthose M levels can be converted into M symbols (or alphabets). Each of theseM symbols can be associated with a unique phase φi, where i = 1, 2, · · · , M.Since 0 ≤ φ ≤ 2π, if we distribute these phases uniformly,

φi =2πiM

, i = 1, 2, · · · , M. (4)


Si (t) =

√2Es

TSsin(

ω0t +2πiM

), i = 1, 2, · · · , M

=

√2Es

TS

[sin (ω0t) cos

(2πiM

)+ cos (ω0t) sin

(2πiM

)]=√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], (5)

where ψ1 (t) =√


2/TS cos ω0t.




φi =2πiM

, i = 1, 2, · · · , M. (4)


Si (t) =

√2Es

TSsin(

ω0t +2πiM

), i = 1, 2, · · · , M

=

√2Es

TS

[sin (ω0t) cos

(2πiM

)+ cos (ω0t) sin

(2πiM

)]=√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], (5)

where ψ1 (t) =√


2/TS cos ω0t.




φi =2πiM

, i = 1, 2, · · · , M. (4)


Si (t) =

√2Es

TSsin(

ω0t +2πiM

), i = 1, 2, · · · , M

=

√2Es

TS

[sin (ω0t) cos

(2πiM

)+ cos (ω0t) sin

(2πiM

)]

=√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], (5)

where ψ1 (t) =√


2/TS cos ω0t.




φi =2πiM

, i = 1, 2, · · · , M. (4)


Si (t) =

√2Es

TSsin(

ω0t +2πiM

), i = 1, 2, · · · , M

=

√2Es

TS

[sin (ω0t) cos

(2πiM

)+ cos (ω0t) sin

(2πiM

)]=√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], (5)

where ψ1 (t) =√


2/TS cos ω0t.



MPSK – Modulation Mechanism

Digitaldata(or)

stream




Digitaldata(or)

stream




Digitaldata(or)

stream

Pickone symbol

at a time




Digitaldata(or)

stream

Identifythe

symbol,i.e.,

i

Pickone symbol

at a time




Digitaldata(or)

stream

Identifythe

symbol,i.e.,

i

Pickone symbol

at a time




π/2

Digitaldata(or)

stream

Identifythe

symbol,i.e.,

i

Pickone symbol

at a time




Σ

π/2

Digitaldata(or)

stream

Identifythe

symbol,i.e.,

i

Pickone symbol

at a timeSi (t)



MPSK – Signal Space Representation of Si (t)

Si (t)

{ {

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]




Si (t)

{ {

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]




Si (t)

{ {Si (t) =

√Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]



MPSK – Constellation Diagram

S1(t)S2(t)

S4(t)S3(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]



Binary PSK (BPSK/2PSK)

S1(t) S2(t) M = 2

1 0 1 1 0 1 1 1 0 0

(i = 1) (i = 2)

S1(t) S2(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], M = 2 for BPSK

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t (6)

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t (7)




S1(t) S2(t) M = 2

1 0 1 1 0 1 1 1 0 0

(i = 1) (i = 2)

S1(t) S2(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], M = 2 for BPSK

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t (6)

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t (7)




S1(t) S2(t) M = 2

1 0 1 1 0 1 1 1 0 0

(i = 1) (i = 2)

S1(t) S2(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], M = 2 for BPSK

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t (6)

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t (7)




S1(t) S2(t) M = 2

1 0 1 1 0 1 1 1 0 0

(i = 1) (i = 2)

S1(t) S2(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], M = 2 for BPSK

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t (6)

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t (7)




S1(t) S2(t) M = 2

1 0 1 1 0 1 1 1 0 0

(i = 1) (i = 2)

S1(t) S2(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], M = 2 for BPSK

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t (6)

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t (7)



Quadri PSK (QPSK)

00(i = 1)

01(i = 2)

10(i = 3)

11(i = 4)

S1(t)

S2(t)

S3(t)

S4(t)M = 4

S1(t) S2(t) S4(t)S3(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], M = 4 for QPSK (8)



Quadri PSK (QPSK)

00(i = 1)

01(i = 2)

10(i = 3)

11(i = 4)

S1(t)

S2(t)

S3(t)

S4(t)M = 4

S1(t) S2(t) S4(t)S3(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], M = 4 for QPSK (8)



Quadri PSK (QPSK)

00(i = 1)

01(i = 2)

10(i = 3)

11(i = 4)

S1(t)

S2(t)

S3(t)

S4(t)M = 4

S1(t) S2(t) S4(t)S3(t)

Si (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

], M = 4 for QPSK (8)



8PSK

The constellations become denser as M increases.



8PSK




8PSK




MPSK – Coherent Demodulation

The demodulator is expected to give back the symbol.

Since the Symbol has been encoded into phase of the carrier, we need toextract the phase of the received carrier, during a particular symbol duration

and drop the carrier.

Since phase extraction is involved, the process has to be a coherent one.

Further, the received signal is noisy:

r (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]+ n (t) (9)









r (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]+ n (t) (9)









r (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]+ n (t) (9)









r (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]+ n (t) (9)









r (t) =√

Es

[cos

(2πiM

)ψ1 (t) + sin

(2πiM

)ψ2 (t)

]+ n (t) (9)




Apply minimum distance criterion with respect to the pre determined set.
















Apply minimum distance criterion with respect to the pre-determined set.




Apply minimum distance criterion with respect to the pre-determined set.




r(t)




r(t)




∫0

Ts

r(t)

∫0

Ts




∫0

Ts

r(t)

∫0

Ts




∫0

Ts

r(t)

∫0

Ts




∫0

Ts

r(t)

∫0

Ts




φ = tan−1 YX

(10)




φ = tan−1 YX

(10)



BPSK – Coherent Demodulation

∫0

Ts

r(t)

∫0

Ts




∫0

Ts

r(t)

∫0

Ts




∫0

Ts

r(t)

∫0

Ts




S1(t) S2(t)

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t




S1(t) S2(t)

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t




S1(t) S2(t)

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t




S1(t) S2(t)

S1 (t) = −√

Esψ1 (t) = −√

2Es/TS sin ω0t

S2 (t) = +√

Esψ1 (t) = +√

2Es/TS sin ω0t




∫0

Tsr(t) X γ0><




∫0

Tsr(t) X γ0><



BPSK – Probability of Symbol (Bit) Error

S1(t) S2(t)

The threshold γ0 is 0, and decision rule is to declare S1 or S2 based on thethreshold crossing. This is similar to NRZ binary baseband problem.

So, for AWGN with one sided spectral density of N0/2 Watts/Hz, probabilityof symbol error is given as

PBPSKS = Q

(√2Es

N0

). (11)




S1(t) S2(t)



PBPSKS = Q

(√2Es

N0

). (11)




S1(t) S2(t)

The threshold γ0 is 0, and decision rule is to declare S1 or S2 based on thethreshold crossing.

This is similar to NRZ binary baseband problem.


PBPSKS = Q

(√2Es

N0

). (11)




S1(t) S2(t)



PBPSKS = Q

(√2Es

N0

). (11)




S1(t) S2(t)



PBPSKS = Q

(√2Es

N0

). (11)



QPSK – Probability of Symbol Error

Probability of symbol error is given (proof will be provided later) as

PQPSKS = 2Q

(√Es

N0

)−[

Q

(√Es

N0

)]2

≈ 2Q

(√Es

N0

), for large SNR.





PQPSKS = 2Q

(√Es

N0

)−[

Q

(√Es

N0

)]2

≈ 2Q

(√Es

N0

), for large SNR.





PQPSKS = 2Q

(√Es

N0

)−[

Q

(√Es

N0

)]2

≈ 2Q

(√Es

N0

), for large SNR.





PQPSKS = 2Q

(√Es

N0

)−[

Q

(√Es

N0

)]2

≈ 2Q

(√Es

N0

), for large SNR.



MPSK – Probability of Symbol Error

For MPSK, probability of symbol error is given as

PMPSKS ≈ 2Q

(dmin√

2N0

)= 2Q

(√2Es

N0sin

π

M

), for large SNR.





PMPSKS ≈ 2Q

(dmin√

2N0

)= 2Q

(√2Es

N0sin

π

M

), for large SNR.





PMPSKS ≈ 2Q

(dmin√

2N0

)= 2Q

(√2Es

N0sin

π

M

), for large SNR.



Symbol Error vs Bit Error

00

11

10

01

Q

I

000

110100

010

Q

I

001

111

101

011

An error in phase might cause symbol error ... and one symbol error cancause atmost n bit errors.




00

11

10

01

Q

I

000

110100

010

Q

I

001

111

101

011





00

11

10

01

Q

I

000

110100

010

Q

I

001

111

101

011

An error in phase might cause symbol error ...

and one symbol error cancause atmost n bit errors.




00

11

10

01

Q

I

000

110100

010

Q

I

001

111

101

011




Symbol Mapping using Gray Code

01

11

10

00

Q

I

110

101000

011

Q

I

010

111

100

001

An error in phase might cause symbol error ... and Gray coding makes surethat symbols corresponding to adjacent phases change only one bit at a time.




01

11

10

00

Q

I

110

101000

011

Q

I

010

111

100

001





01

11

10

00

Q

I

110

101000

011

Q

I

010

111

100

001

An error in phase might cause symbol error ...

and Gray coding makes surethat symbols corresponding to adjacent phases change only one bit at a time.




01

11

10

00

Q

I

110

101000

011

Q

I

010

111

100

001




PS vs PB using Gray Code

01

11

10

00

Q

I

110

101000

011

Q

I

010

111

100

001

PB ≈PSk

=PS

log2 M(for PS � 1) (12)




01

11

10

00

Q

I

110

101000

011

Q

I

010

111

100

001

PB ≈PSk

=PS

log2 M(for PS � 1) (12)




01

11

10

00

Q

I

110

101000

011

Q

I

010

111

100

001

PB ≈PSk

=PS

log2 M(for PS � 1) (12)



Generation of Gray Code

01

n=1




01

n=1




0110

01

n=1




0110

01

0001

n=2n=1




0110

01

00011110

n=2n=1




0110

00011110

01

00011110

00011110

n=2n=1




0110

00011110

01

00011110

00011110

000001011010

n=2n=3

n=1




0110

00011110

01

00011110

00011110

000001011010

100101111110

n=2n=3

n=1



Generation of QPSK using BPSK

10 11

0100

Q

I

Sin + Cos

- Sin + Cos- Sin - Cos

Sin - Cos




10 11

0100

Q

I

Sin + Cos


Sin - Cos




10 11

0100

Q

I

Sin + Cos


Sin - Cos




+1 TS = 2Tb

10 11

0100

Q

I

Sin + Cos


Sin - Cos




+1 TS = 2Tb

10 11

0100

Q

I

Sin + Cos


Sin - Cos




+1

-1

TS = 2Tb

TS = 2Tb

10 11

0100

Q

I

Sin + Cos


Sin - Cos




+1

-1

TS = 2Tb

TS = 2Tb

10 11

0100

Q

I

Sin + Cos


Sin - Cos




+1

-1

TS = 2Tb

TS = 2Tb

10 11

0100

Q

I

Sin + Cos


Sin - Cos



Bit & Symbol Error Rates – BPSK vs QPSK

Although QPSK can be viewed as a quaternary modulation, it is easier to seeit as two independently modulated quadrature carriers. So, BPSK demod-ulation can be applied on both the carriers independantly. As a result, theprobability of bit-error for QPSK is the same as for BPSK:

PBPSK/QPSKB = Q

(√2EbN0

). (13)

Since we know the probability of bit-error, we can calculate probability of sym-bol error as

PQPSKS = 1−

(1− PBPSK

B

)2

= 2Q

(√2EbN0

)−[

Q

(√2EbN0

)]2

≈ 2Q

(√2EbN0

)= 2Q

(√2EbN0

). (14)




Although QPSK can be viewed as a quaternary modulation, it is easier to seeit as two independently modulated quadrature carriers.

So, BPSK demod-ulation can be applied on both the carriers independantly. As a result, theprobability of bit-error for QPSK is the same as for BPSK:

PBPSK/QPSKB = Q

(√2EbN0

). (13)


PQPSKS = 1−

(1− PBPSK

B

)2

= 2Q

(√2EbN0

)−[

Q

(√2EbN0

)]2

≈ 2Q

(√2EbN0

)= 2Q

(√2EbN0

). (14)




Although QPSK can be viewed as a quaternary modulation, it is easier to seeit as two independently modulated quadrature carriers. So, BPSK demod-ulation can be applied on both the carriers independantly.

As a result, theprobability of bit-error for QPSK is the same as for BPSK:

PBPSK/QPSKB = Q

(√2EbN0

). (13)


PQPSKS = 1−

(1− PBPSK

B

)2

= 2Q

(√2EbN0

)−[

Q

(√2EbN0

)]2

≈ 2Q

(√2EbN0

)= 2Q

(√2EbN0

). (14)





PBPSK/QPSKB = Q

(√2EbN0

). (13)


PQPSKS = 1−

(1− PBPSK

B

)2

= 2Q

(√2EbN0

)−[

Q

(√2EbN0

)]2

≈ 2Q

(√2EbN0

)= 2Q

(√2EbN0

). (14)





PBPSK/QPSKB = Q

(√2EbN0

). (13)


PQPSKS = 1−

(1− PBPSK

B

)2

= 2Q

(√2EbN0

)−[

Q

(√2EbN0

)]2

≈ 2Q

(√2EbN0

)= 2Q

(√2EbN0

). (14)





PBPSK/QPSKB = Q

(√2EbN0

). (13)


PQPSKS = 1−

(1− PBPSK

B

)2

= 2Q

(√2EbN0

)−[

Q

(√2EbN0

)]2

≈ 2Q

(√2EbN0

)= 2Q

(√2EbN0

). (14)





PBPSK/QPSKB = Q

(√2EbN0

). (13)


PQPSKS = 1−

(1− PBPSK

B

)2

= 2Q

(√2EbN0

)−[

Q

(√2EbN0

)]2

≈ 2Q

(√2EbN0

)= 2Q

(√2EbN0

). (14)





PBPSK/QPSKB = Q

(√2EbN0

). (13)


PQPSKS = 1−

(1− PBPSK

B

)2

= 2Q

(√2EbN0

)−[

Q

(√2EbN0

)]2

≈ 2Q

(√2EbN0

)= 2Q

(√2EbN0

). (14)



MPSK - PS and PB

Eb/N0 (dB)

Sym

bol e

rror

pro

babi

lity,

PS

Bit e

rror

pro

babi

lity,

PB

Eb/N0 (dB)

M = 2M = 4

M = 8

M = 16

M = 32

M = 64k = 1, 2

k = 3 k = 4 k = 5 k = 6



MPSK - PS and PB

Eb/N0 (dB)

Sym

bol e

rror

pro

babi

lity,

PS

Bit e

rror

pro

babi

lity,

PB

Eb/N0 (dB)

M = 2M = 4

M = 8

M = 16

M = 32

M = 64k = 1, 2

k = 3 k = 4 k = 5 k = 6



Bandwidth & Bandwidth Efficiency of MPSK

For MPSK, the baseband PSD is given by

Sbase−bandMPSK (f ) = 2Ebk

(sin πfkTb

πfkTb

)2. (15)

From the above equation it is clear that null-to-null bandwidth correspondingto M ary PSK is given by

Bnull−to−nullT =

2kTb

=2Rb

k. (16)

However, if we take 3 dB bandwidth as the criteria, then the correspondingbandwidth is given by

B3 dBT =

1kTb

=Rbk

. (17)

So, bandwidth efficiency is defined as

Bandwidth efficiency =Rb

B3 dBT

= k. (18)






(sin πfkTb

πfkTb

)2. (15)



2kTb

=2Rb

k. (16)


B3 dBT =

1kTb

=Rbk

. (17)



B3 dBT

= k. (18)






(sin πfkTb

πfkTb

)2. (15)



2kTb

=2Rb

k. (16)


B3 dBT =

1kTb

=Rbk

. (17)



B3 dBT

= k. (18)






(sin πfkTb

πfkTb

)2. (15)



2kTb

=2Rb

k. (16)


B3 dBT =

1kTb

=Rbk

. (17)



B3 dBT

= k. (18)






(sin πfkTb

πfkTb

)2. (15)



2kTb

=2Rb

k. (16)


B3 dBT =

1kTb

=Rbk

. (17)



B3 dBT

= k. (18)



Outline

1 Introduction




5 ASK & APK

6 Summary



Coherence Requirement for MPSK

In MPSK, since the information is in the phase of the transmitted waveform,coherence between Tx& Rx is a must.

For BPSK, if RF phase changes by angle θ, it can be shown that the probabilityof bit error changes to

P′B = Q

[cos θ

√2EbN0

]. (19)

In order to make PSK amenable to non coherent detection, Differential PSK(DPSK) is used.

In DPSK, phase of the current symbol is not transmitted, but the phase differ-ence between current and previous symbols is transmitted. This avoids thenecessity of a coherent carrier at the receiver.






P′B = Q

[cos θ

√2EbN0

]. (19)








P′B = Q

[cos θ

√2EbN0

]. (19)








P′B = Q

[cos θ

√2EbN0

]. (19)








P′B = Q

[cos θ

√2EbN0

]. (19)





Differential Phase Shift Keying (D8PSK)

Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

∆φk = φk − φk−1 π π/4 π 3π/4Recovered symbol 110 001 110 010




Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π





Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π





Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010

∆φk π π/4 π 3π/4φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π





Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π





Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π





Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π





Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π





Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

∆φk = φk − φk−1 π π/4 π 3π/4

Recovered symbol 110 001 110 010




Symbol ∆φ

000 0001 π/4011 2π/4010 3π/4110 4π/4111 5π/4101 6π/4100 7π/4

Modulation ref

Symbol 110 001 110 010∆φk π π/4 π 3π/4

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π

Demodulation ref

φk = φk−1 + ∆φk 0 π 5π/4 π/4 π




Binary Differential Phase Shift Keying (DBPSK)

Symbol ∆φ

1 00 π

Modulation ref

Symbol / Bit 1 0 1 1 0 0∆φk 0 π 0 0 π π

φk = φk−1 + ∆φk 0 0 π π π 0 π

Demodulation

φk = φk−1 + ∆φk 0 0 π π π 0 π

∆φk = φk − φk−1 0 π 0 0 π π

Recovered Symbol / Bit 1 0 1 1 0 0




Symbol ∆φ

1 00 π

Modulation ref

Symbol / Bit 1 0 1 1 0 0∆φk 0 π 0 0 π π

φk = φk−1 + ∆φk 0 0 π π π 0 π

Demodulation

φk = φk−1 + ∆φk 0 0 π π π 0 π

∆φk = φk − φk−1 0 π 0 0 π π





Symbol ∆φ

1 00 π

Modulation ref

Symbol / Bit 1 0 1 1 0 0∆φk 0 π 0 0 π π

φk = φk−1 + ∆φk 0 0 π π π 0 π

Demodulation

φk = φk−1 + ∆φk 0 0 π π π 0 π

∆φk = φk − φk−1 0 π 0 0 π π





Symbol ∆φ

1 00 π

Modulation ref

Symbol / Bit 1 0 1 1 0 0∆φk 0 π 0 0 π π

φk = φk−1 + ∆φk 0 0 π π π 0 π

Demodulation

φk = φk−1 + ∆φk 0 0 π π π 0 π

∆φk = φk − φk−1 0 π 0 0 π π





Modulation ref

Message ak 1 0 1 1 0 0 0 1 1Encoding dk = ak ⊕ dk−1 1 1 0 0 0 1 0 1 1 1

Signal phase φ 0 0 π π π 0 π 0 0 0Transmitted NRZ-L signal 1 1 -1 -1 -1 1 -1 1 1 1

Demodulation

Sk (t) 1 1 -1 -1 -1 1 -1 1 1 11

Eb

´ (k+1)TSkTS

Sk (t) Sk−1 (t) dt 1 -1 1 1 -1 -1 -1 1 1

Demodulator output ak 1 0 1 1 0 0 0 1 1




Modulation ref



Demodulation

Sk (t) 1 1 -1 -1 -1 1 -1 1 1 11

Eb

´ (k+1)TSkTS

Sk (t) Sk−1 (t) dt 1 -1 1 1 -1 -1 -1 1 1





Modulation ref



Demodulation

Sk (t) 1 1 -1 -1 -1 1 -1 1 1 11

Eb

´ (k+1)TSkTS

Sk (t) Sk−1 (t) dt 1 -1 1 1 -1 -1 -1 1 1




DBPSK Modulation & Demodulation

DelayT

LogicdeviceBinary data

{ak}

(0, 1)

dk-1

Binary DPSK Modulator

dk = ak ⊕ dk−1




DelayT


{ak}

(0, 1)

dk-1


dk = ak ⊕ dk−1




DelayT


{ak}

(0, 1)

dk-1


dk = ak ⊕ dk−1



DBPSKModulation & Demodulation

DelayT


{ak}

(0, 1)

dkdk-1

{dk}(0, 1)


dk = ak ⊕ dk−1




DelayT

Levelgenerator

Logicdevice Polar NRZ

d(t)Binary data

{ak}

(0, 1)

dkdk-1

{dk}(0, 1) (-1, 1)


dk = ak ⊕ dk−1




DelayT

Levelgenerator


d(t)Binary data

{ak}

(0, 1)

dkdk-1

{dk}(0, 1) (-1, 1)


dk = ak ⊕ dk−1




r(t)

DelayT

Binary DPSK Demodulator

DelayT

Levelgenerator


d(t)Binary data

{ak}

(0, 1)

dkdk-1

{dk}(0, 1) (-1, 1)


dk = ak ⊕ dk−1




∫0

Tr(t)

DelayT


DelayT

Levelgenerator


d(t)Binary data

{ak}

(0, 1)

dkdk-1

{dk}(0, 1) (-1, 1)


dk = ak ⊕ dk−1




si

>∫0

Tr(t)

Decisionstage

DelayT


DelayT

Levelgenerator


d(t)Binary data

{ak}

(0, 1)

dkdk-1

{dk}(0, 1) (-1, 1)


dk = ak ⊕ dk−1



Two Different Types of DBPSK Demodulators

si

>∫0

Tr(t)

Decisionstage

DelayT

Sub-Optimal Demodulator




si

>∫0

Tr(t)

Decisionstage

DelayT





si

>∫0

Tr(t)

Decisionstage

DelayT

si

>∫0

Tr(t)

Decisionstage

DelayT


Optimal DemodulatorFrom local

osicllator



DBPSK – Probability of Errors

0 5 10 1510 8

10 7

10 6

10 5

10 4

10 3

0.01

0.1

1

SuboptimumDBPSK

Optimum DBPSK

CoherentDEBPSK

Coherent BPSK

Pb

E b/ No (dB)



DBPSK – Probability of Errors

0 5 10 1510 8

10 7

10 6

10 5

10 4

10 3

0.01

0.1

1

SuboptimumDBPSK

Optimum DBPSK

CoherentDEBPSK

Coherent BPSK

Pb

E b/ No (dB)



Outline

1 Introduction




5 ASK & APK

6 Summary



A Simple Introduction to FSK (BFSK)

2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω


ω

FSKspectrum

-ω2 +ω1


0 1 0 1 1 0 1 0 1 1

ASK 1

ASK 2

+

1 V

-1 V



Generating BFSK on Emona Kit

IN 2

CON 1

Dual AnalogSwitch

CON 2

IN 1

NoiseGenerator

DC 5VFixed

Exor Module

X

VCO

Multipliermodule

BasebandLPF

TuneableLPF

Multipliermodule

To Ch. 1

To Ch. 2

100 kHzcos

~ 85 kHz

Ref

Baseband Data FSK Modulation Adding Noise FSK Demodulation / Detection



Coherent vs Non-Coherent FSK

In its most general form, the binary FSK scheme uses two signals with differ-ent frequencies to represent binary 1 and 0.

S1 (t) =

√2ESTS

cos (2πf1t + Φ1) , kTb ≤ t ≤ (k + 1)TS, for 1

S2 (t) =

√2ESTS

cos (2πf2t + Φ2) , kTb ≤ t ≤ (k + 1)TS, for 0

where Φ1 and Φ2 are initial phases at t = 0, and TS is the symbol period of thedigital data.

When Φ1 = Φ2, this form of FSK is called coherent FSK.

When Φ1 6= Φ2, this form of FSK is called noncoherent or discontinuous FSK.



Condition for Orthogonality – Noncoherent FSK

If signals S1 (t) and S2 (t) are orthogonalˆ TS

0S1 (t) S2 (t) dt = 0

⇒ˆ TS

0cos (2πf1t + Φ1) cos (2πf2t + Φ2) dt = 0

⇓

sin [2π (f1 + f2)TS]

2π (f1 + f2)cos (Φ1 + Φ2) +

sin [2π (f1 − f2)TS]

2π (f1 − f2)cos (Φ1 −Φ2)

+cos [2π (f1 + f2)TS]− 1

2π (f1 + f2)sin (Φ1 + Φ2)+

cos [2π (f1 − f2)TS]− 12π (f1 − f2)

sin (Φ1 −Φ2) = 0

Both sin x = 0 and cos x = 1 occur simultaneously when x = 2mπ. So,f1 + f2 = m/TS. Similarly it can be proved that f1 − f2 = n/TS.



Condition for Orthogonality – Coherent FSK

For coherent FSK, Φ1 = Φ2 = Φ. So, the equation given in the previous slidebecomes

⇓

sin [2π (f1 + f2)TS]

2π (f1 + f2)cos 2Φ +

sin [2π (f1 − f2)TS]

2π (f1 − f2)

+cos [2π (f1 + f2)TS]− 1

2π (f1 + f2)sin 2Φ = 0

Both sin x = 0 and cos x = 1 occur simultaneously when x = 2mπ. So,f1 + f2 = m/TS. Similarly it can be proved that f1 − f2 = n/2TS.



So, Conditions for Orthogonality are

Non-Coherent FSK:

f1 + f2 = m/TS

f1 − f2= n/TS

Coherent FSK:

f1 + f2 = m/TS

f1 − f2= n/2TS



BFSK vs 4FSK

BFSK



BFSK vs 4FSK

BFSK



BFSK vs 4FSK

BFSK

4FSK



MFSK Generation

Multiplexer

Oscillator 1

Oscillator 2

Oscillator M

Control lines

Binary input data S / PConverter

bnb1 b2



MFSK Generation

Multiplexer

Oscillator 1

Oscillator 2

Oscillator M

Control lines

Binary input data S / PConverter

bnb1 b2



Coherent MFSK Detection

∫0

Ts

r(t)

∫0

Ts

Logic circuitselects Si(t)

whosecomponentsaij best match

{zj(T)}



Coherent MFSK Detection – PS & PB

The symbol error probability for a coherently detected MFSK system has anupper bound, given as

PS ≤ (M− 1)Q

(√ESN0

). (20)

Can you visualize the above equation geometrically?

In MFSK, the ratio of the bit error probability to the symbol error probabilityis

PBPS

=M/2

M− 1. (21)





PS ≤ (M− 1)Q

(√ESN0

). (20)



PBPS

=M/2

M− 1. (21)





PS ≤ (M− 1)Q

(√ESN0

). (20)



PBPS

=M/2

M− 1. (21)





PS ≤ (M− 1)Q

(√ESN0

). (20)



PBPS

=M/2

M− 1. (21)



Non-Coherent MFSK Detection

z γ0>< si

>Σ z(T)

z1(T)∫0

T

r(t)

z2(T)∫0

T

z21

( )2.

( )2.z2

2

I channel

Q channelΣ

z3(T)∫0

T

z4(T)∫0

T

z23

( )2.

( )2.z2

4

I channel

Q channelΣ

+




z γ0>< si

>Σ z(T)

z1(T)∫0

T

r(t)

z2(T)∫0

T

z21

( )2.

( )2.z2

2

I channel

Q channelΣ

z3(T)∫0

T

z4(T)∫0

T

z23

( )2.

( )2.z2

4

I channel

Q channelΣ

+




r(t)

z1(T)

Filterf1

Envelopdetector

z3(T)

Envelopdetector

FilterfM

si

>

Decisionstage

z2(T)

Envelopdetector

Filterf2

Bandpass filters centeredat fi with bandwidth Wf = 1/T

In no noise case, assuming that the filters preserve the shape of the envelopeof the input signal waveforms, the output is of the type

z (T) =√

2ES/TS. (22)




r(t)

z1(T)

Filterf1

Envelopdetector

z3(T)

Envelopdetector

FilterfM

si

>

Decisionstage

z2(T)

Envelopdetector

Filterf2



z (T) =√

2ES/TS. (22)




r(t)

z1(T)

Filterf1

Envelopdetector

z3(T)

Envelopdetector

FilterfM

si

>

Decisionstage

z2(T)

Envelopdetector

Filterf2



z (T) =√

2ES/TS. (22)



Non-Coherent MFSK Detection – PS & PB

In the case of coherent detection, the pdf of the (signal + noise), at the inputof the decision stage, remained Gaussian.

In the case of non coherent detection, the pdf of the envelops of (signal +noise), at the input of the decision stage, do not remain Gaussian.

So, first, we need the knowledge of pdf of (signal + noise) at the decisionstage to evaluate the symbol / bit error probabilities.

The symbol error probability for a non coherently detected MFSK system isgiven as

PS =1M

exp(− ES

N0

) M

∑j=2

(−1)j(

Mj

)exp

(ES

jN0

), where

(Mj

)=

M!j! (M− j)!








PS =1M

exp(− ES

N0

) M

∑j=2

(−1)j(

Mj

)exp

(ES

jN0

), where

(Mj

)=

M!j! (M− j)!








PS =1M

exp(− ES

N0

) M

∑j=2

(−1)j(

Mj

)exp

(ES

jN0

), where

(Mj

)=

M!j! (M− j)!








PS =1M

exp(− ES

N0

) M

∑j=2

(−1)j(

Mj

)exp

(ES

jN0

), where

(Mj

)=

M!j! (M− j)!








PS =1M

exp(− ES

N0

) M

∑j=2

(−1)j(

Mj

)exp

(ES

jN0

), where

(Mj

)=

M!j! (M− j)!



Spectrum of MFSK

Frequency

Spec

trum

of M

FSK



Spectrum of MFSK

Frequency

Spec

trum

of M

FSK



Bandwidth & Bandwidth Efficiency of MFSK

Bandwidth

Non-Coherent MFSK: BT = M/TS

Coherent MFSK: BT = M/2TS

Bit rate Rb = 1/Tb = log2 M/TS = k/TS

Bandwidth Efficiency

(R/W)NCMFSK = Rb/BT = (1/M) log2 M

(R/W)CMFSK = Rb/BT = (2/M) log2 M

So, (R/W)NCMFSK is better than (R/W)NCMFSK.




Bandwidth











Bandwidth










Probability of Bit Errors for Various BinaryModulations

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

ba

bili

ty o

f B

it E

rro

r

BPSK(QPSK)

ASK/FSK

NC FSK

DPSK



Probability of Bit Errors for Various BinaryModulations

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

ba

bili

ty o

f B

it E

rro

r

BPSK(QPSK)

ASK/FSK

NC FSK

DPSK



Outline

1 Introduction




5 ASK & APK

6 Summary



Amplitude Shift Keying (MASK)

(Bipolar MASK)

(Unipolar MASK)

M-ary ASK waveform is given as

SMASKi (t) = Ai

√2Es

TSsin (ω0t + φ) , i = 1, 2, · · · , M (23)

where√

ES is the energy of the nominal carrier in a symbol duration TS.

For unipolar binary ASK , it is called on-off keying (OOK). Binary 1 and 0 arerepresented by

√ES and 0, respectively, in the signal space.




(Bipolar MASK)

(Unipolar MASK)


SMASKi (t) = Ai

√2Es

TSsin (ω0t + φ) , i = 1, 2, · · · , M (23)

where√







(Bipolar MASK)

(Unipolar MASK)


SMASKi (t) = Ai

√2Es

TSsin (ω0t + φ) , i = 1, 2, · · · , M (23)

where√







(Bipolar MASK)

(Unipolar MASK)


SMASKi (t) = Ai

√2Es

TSsin (ω0t + φ) , i = 1, 2, · · · , M (23)

where√






MASK – PS & PB

It can be shown that the symbol error probability for bipolar MASK is

PS =2 (M− 1)

MQ

(√6Eavg

(M2 − 1)N0

). (24)

Similarly, the symbol error probability for unipolar MASK is

PS =2 (M− 1)

MQ

(√3Eavg

(2M2 − 3M + 1)N0

). (25)

Eavg (average symbol energy) can be written in terms of average bit energy,Eb, as

Eavg = (log2 M)Eb = kEb. (26)

At high signal-to-noise ratios, the most likely errors are the adjacent symbols(when we use Gray code). So in this case

Pb ≈PS

log2 M=

PSk

. (27)



MASK – PS & PB


PS =2 (M− 1)

MQ

(√6Eavg

(M2 − 1)N0

). (24)


PS =2 (M− 1)

MQ

(√3Eavg

(2M2 − 3M + 1)N0

). (25)




Pb ≈PS

log2 M=

PSk

. (27)



MASK – PS & PB


PS =2 (M− 1)

MQ

(√6Eavg

(M2 − 1)N0

). (24)


PS =2 (M− 1)

MQ

(√3Eavg

(2M2 − 3M + 1)N0

). (25)




Pb ≈PS

log2 M=

PSk

. (27)



MASK – PS & PB


PS =2 (M− 1)

MQ

(√6Eavg

(M2 − 1)N0

). (24)


PS =2 (M− 1)

MQ

(√3Eavg

(2M2 − 3M + 1)N0

). (25)




Pb ≈PS

log2 M=

PSk

. (27)



MASK – PS & PB


PS =2 (M− 1)

MQ

(√6Eavg

(M2 − 1)N0

). (24)


PS =2 (M− 1)

MQ

(√3Eavg

(2M2 − 3M + 1)N0

). (25)




Pb ≈PS

log2 M=

PSk

. (27)



MASK – Bandwidth & Bandwidth Efficiency

Power spectral density of MASK will have the same shape as that of MPSK.So, transmission bandwidth is given as

BT =1

kTb=

Rbk

. (28)

So, bandwidth efficiency is for MASK is

Bandwidth efficiency =RbBT

= k = log2 M. (29)





BT =1

kTb=

Rbk

. (28)



= k = log2 M. (29)





BT =1

kTb=

Rbk

. (28)



= k = log2 M. (29)



Why ASK is Bad?

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

ba

bili

ty o

f B

it E

rro

r

BPSK(QPSK)

ASK/FSK

NC FSK

DPSK

From the above figure, it clear that MASK is usually not preferred due to itspoor power efficiency.

We also know that MPSK constellation diagram gets crowded as M increases.

So, we need to come up with a modulation technique which shares features ofboth MPSK and MASK.



Why ASK is Bad?

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

ba

bili

ty o

f B

it E

rro

rBPSK(QPSK)

ASK/FSK

NC FSK

DPSK






Why ASK is Bad?

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

ba

bili

ty o

f B

it E

rro

rBPSK(QPSK)

ASK/FSK

NC FSK

DPSK






Why ASK is Bad?

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

ba

bili

ty o

f B

it E

rro

rBPSK(QPSK)

ASK/FSK

NC FSK

DPSK






Why ASK is Bad?

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

ba

bili

ty o

f B

it E

rro

rBPSK(QPSK)

ASK/FSK

NC FSK

DPSK






ASK + PSK = APK

ASK

PSK

ASK/PSK (APK)



ASK + PSK = APK

ASK

PSK

ASK/PSK (APK)



MAPK (Quadrature Amplitude Modulation)

For MAPK,

SMAPKi (t) = Ai

√2Es

TSsin (ω0t + φi) , i = 1, 2, · · · , M. (30)

To get x and y coordinates corresponding to the signal in the constellationdiagram, the above equation can be re-written as

SMAPKi (t) = Ai

√Es cos φi︸︷︷︸

xi

ψ1 (t) + Ai√

Es sin φi︸︷︷︸yi

ψ2 (t) . (31)




For MAPK,

SMAPKi (t) = Ai

√2Es

TSsin (ω0t + φi) , i = 1, 2, · · · , M. (30)


SMAPKi (t) = Ai


xi

ψ1 (t) + Ai√


ψ2 (t) . (31)




For MAPK,

SMAPKi (t) = Ai

√2Es

TSsin (ω0t + φi) , i = 1, 2, · · · , M. (30)


SMAPKi (t) = Ai


xi

ψ1 (t) + Ai√


ψ2 (t) . (31)



16 QAM

Q

I

0000 0100 1100 1000

0001 0101 1101 1001

0011 0111 1111 1011

0010 0110 1110 1010

Q

I



16 QAM

Q

I

0000 0100 1100 1000

0001 0101 1101 1001

0011 0111 1111 1011

0010 0110 1110 1010

Q

I



8 QAM

I

Q

I

Q

I

Q

(1+√3,0)

(1,1)

I

Q



8 QAM

I

Q

I

Q

I

Q

(1+√3,0)

(1,1)

I

Q



QAM – Modulation & Demodulation

X

Y




X

Y




X

Y



MQAM – PS & PB

It can be shown that the symbol error probability for bipolar MQAM is giveninterms of symbol average energy Eavg as

PS ≤ 4Q

(√3kEb

(M− 1)N0

), (32)

where Eavg = (log2 M)Eb = kEb.


Pb ≈PS

log2 M=

PSk

. (33)



MQAM – PS & PB


PS ≤ 4Q

(√3kEb

(M− 1)N0

), (32)



Pb ≈PS

log2 M=

PSk

. (33)



MQAM – PS & PB


PS ≤ 4Q

(√3kEb

(M− 1)N0

), (32)



Pb ≈PS

log2 M=

PSk

. (33)



MQAM – Bandwidth & Bandwidth Efficiency

Power spectral density of MQAM will have the same shape as that of MPSK/ MASK. So, transmission bandwidth is given as

BT =1

kTb=

Rbk

. (34)

So, bandwidth efficiency is for MQAM is


= k = log2 M. (35)





BT =1

kTb=

Rbk

. (34)



= k = log2 M. (35)





BT =1

kTb=

Rbk

. (34)



= k = log2 M. (35)



Comparison of MPSK, MQAM, and MFSK



Comparison of MPSK, MQAM, and MFSK



Outline

1 Introduction




5 ASK & APK

6 Summary



MPSK – Summary

SMPSKi (t) =

√2Es

TSsin(

ω0t +2πiM

), i = 1, 2, · · · , M

PMPSKS ≈ 2Q

(dmin√

2N0

)= 2Q

(√2Es

N0sin

π

M

), for large SNR

PMPSKB ≈ PS

log2 M(for PS � 1 and Gray coding)

Bandwidth efficiencyMPSK =RbBT

= log2 M


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