Bandpass Digital TX

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

8. Bandpass Modulation and DemodulationTechniques

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

Department of Electrical & Electronics EngineeringBITS Pilani, Hyderbad Campus

October 21 – 30, 2015

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


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Outline

1 Introduction

2 Phase Shift Keying (PSK)

3 Differential PSK (DPSK)

4 Frequency Shift Keying (FSK)

5 ASK & APK

6 Summary



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Outline

1 Introduction




5 ASK & APK

6 Summary



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

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

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

Symbols Symbols Symbols Symbols Symbols SymbolsSymbols Symbols Symbols Symbols Symbols



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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))



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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 .


d h h f ( ) ff l ( ) h f ( )


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A Slight Modied Representation of Carrier Signal

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

S (t) = √ 2P cos ω t (1) because P = A2/2.

The above equation can be further re-written as

S (t) = 2EsT S cos ω t (2) because P watts can be replaced by E

s joules/ T

S 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/ T S in this chapter.


I t d ti Ph Shift K i (PSK) Diff ti l PSK (DPSK) F Shift K i (FSK) ASK & APK S


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So, Digital Modulation Techniques are ...

Carrier Signal:

2Es/ T S 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)




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Digital Modulation Techniques – Examples




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y g ( ) ( ) q y y g ( ) y

Bits vs Symbols




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Bits vs Symbols




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Bits vs Symbols




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Bits vs Symbols




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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 benet of multiplexing multiple channelsaround the same carrier.




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Symbol Energy vs bit Energy

Es = nEb (3)




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Spectral Properties of ASK




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Spectral Properties of PSK




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Spectral Properties of FSK




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Outline

1 Introduction




5 ASK & APK

6 Summary




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Phase Shift Keying (MPSK)If a digital signal can be completely represented by M different levels , then

those M levels can be converted into M symbols (or alphabets ). Each of these M 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 = 2π i M

, i = 1, 2,· · · , M. (4)So, the modulated PSK signal is given by

Si (t) = 2EsT S sin ω0t + 2π i M , i = 1, 2,· · · , M= 2EsT S sin (ω0t) cos 2π i M + cos (ω0t) sin 2π i M= √ Es cos 2π i M ψ1 (t) + sin

2π i M

ψ2 (t) , (5)

where ψ1 (t) = √ 2/ T S sin ω0t and ψ2 (t) = √ 2/ T S cos ω0t.8. Bandpass Modulation and Demodulation Techniques Communication Systems, Dept. of EEE, BITS Hyderabad



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MPSK – Modulation Mechanism




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MPSK – Signal Space Representation of Si (t)

Si (t) = √ Es cos 2π i M ψ1 (t) + sin2π i M

ψ2 (t)




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MPSK – Constellation Diagram


ψ2 (t)



( )


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Binary PSK (BPSK/2PSK)


ψ2 (t) , M = 2 for BPSK

S1 (t) = −√ Esψ1 (t) = − 2Es/ T S sin ω0t (6)S2 (t) = + √ Esψ1 (t) = + 2Es/ T S sin ω0t (7)



Q d i PSK (QPSK)


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Quadri PSK (QPSK)


ψ2 (t) , M = 4 for QPSK (8)



8PSK


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8PSK

The constellations become denser as M increases.



MPSK C h t D d l ti


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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 2π i M ψ1 (t) + sin2π i M

ψ2 (t) + n (t) (9)



MPSK C h t D d l ti


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



MPSK Coherent Demodulation


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





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φ̂ = tan −1 YX

(10)



BPSK – Coherent Demodulation


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BPSK Coherent Demodulation





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S1 (t) = −√ Esψ1 (t) = − 2Es/ T S sin ω0tS2 (t) = + √ Esψ1 (t) = + 2Es/ T S sin ω0t





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BPSK – Probability of Symbol (Bit) Error


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y y ( )

The threshold γ 0 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 N 0/2 Watts/Hz, probabilityof symbol error is given as

PBPSKS = Q 2EsN 0 . (11)8. Bandpass Modulation and Demodulation Techniques Communication Systems, Dept. of EEE, BITS Hyderabad


QPSK – Probability of Symbol Error


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y y

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

PQPSKS = 2Q EsN 0 − Q EsN 02

≈2Q EsN 0 , for large SNR.8. Bandpass Modulation and Demodulation Techniques Communication Systems, Dept. of EEE, BITS Hyderabad


MPSK – Probability of Symbol Error


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For MPSK, probability of symbol error is given as

PMPSKS ≈2Q dmin√ 2N 0 = 2Q 2EsN 0 sin π M , for large SNR.



Symbol Error vs Bit Error


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



Symbol Mapping using Gray Code


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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 .



PS vs PB using Gray Code


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PB ≈ PS

k = PS

log2 M (for PS 1) (12)



Generation of Gray Code


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Generation of QPSK using BPSK


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8. Bandpass Modulation and Demodulation Techniques Communication Systems, Dept. of EEE, BITS Hyderabad Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary

Bit & Symbol Error Rates – BPSK vs QPSK


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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 2EbN 0 . (13)Since we know the probability of bit-error, we can calculate probability of symbol error as

PQPSKS = 1− 1 −PBPSKB2

= 2Q 2EbN 0 − Q 2EbN 02

≈ 2Q 2EbN 0 = 2Q 2EbN 0 . (14)8. Bandpass Modulation and Demodulation Techniques Communication Systems, Dept. of EEE, BITS Hyderabad Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary

MPSK - PS and PB


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8 Bandpass Modulation and Demodulation Techniques Communication Systems Dept of EEE BITS Hyderabad Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary

Bandwidth & Bandwidth Efciency of MPSKF MPSK th b b d PSD i i b


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For MPSK, the baseband PSD is given by

S base− bandMPSK ( f ) = 2Ebk sin π fkT b

π fkT b

2. (15)

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

Bnull −to−nullT = 2kT b= 2Rb

k . (16)

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

B3 dBT = 1

kT b= Rb

k . (17)

So, bandwidth efciency is dened as

Bandwidth efciency = RbB3 dBT

= k . (18)


Outline


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1 Introduction




5 ASK & APK

6 Summary


Coherence Requirement for MPSK


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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

PB = Q cos θ 2EbN 0 . (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.


Differential Phase Shift Keying (D8PSK)


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Symbol ∆φ

000 0001 π /4

011 2π /4010 3π /4110 4π /4111 5π /4101 6π /4

100 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


Binary Differential Phase Shift Keying (DBPSK)


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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


Binary Differential Phase Shift Keying (DBPSK)


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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 0

Transmitted 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 1

1Eb ´ (k + 1)T

SkT S Sk (t) Sk −1 (t) dt 1 -1 1 1 -1 -1 -1 1 1Demodulator output âk 1 0 1 1 0 0 0 1 1


DBPSK Modulation & Demodulation


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dk = ak ⊕dk −1


DBPSKModulation & Demodulation


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dk = ak ⊕dk −1




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dk = ak ⊕dk −1




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dk = ak ⊕dk −1




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dk = ak ⊕dk −1

8 B d M d l ti d D d l ti T h iq C i ti S t D t f EEE BITS H d b d Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary



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dk = ak ⊕dk −1

8 B d M d l ti d D d l ti T h i C i ti S t D t f EEE BITS H d b d Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary



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dk = ak ⊕dk −1

8 B d M d l ti d D d l ti T h i C i ti S t D t f EEE BITS H d b d Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary

Two Different Types of DBPSK Demodulators


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8 B d M d l i d D d l i T h i C i i S D f EEE BITS H d b d Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary

Two Different Types of DBPSK Demodulators


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8 B d M d l i d D d l i T h i C i i S D f EEE BITS H d b d Introduction Phase Shift Keying (PSK) Differential PSK (DPSK) Frequency Shift Keying (FSK) ASK & APK Summary

DBPSK – Probability of Errors


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Outline


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1 Introduction




5 ASK & APK

6 Summary



A Simple Introduction to FSK (BFSK)


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Generating BFSK on Emona Kit


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Coherent vs Non-Coherent FSK


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

S1 (t) = 2EST S cos (2π f 1t + Φ 1) , kT b ≤t ≤ (k + 1) T S, for 1S2 (t) = 2EST S cos (2π f 2t + Φ 2) , kT b ≤t ≤ (k + 1) T S, for 0

where Φ 1 and Φ 2 are initial phases at t = 0, and T S is the symbol period of thedigital data.

When Φ 1 = Φ 2, this form of FSK is called coherent FSK .When Φ 1 = Φ 2, this form of FSK is called noncoherent or discontinuous FSK.



Condition for Orthogonality – Noncoherent FSK

If signals S1 (t) and S2 (t) are orthogonal


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If signals S1 (t) and S2 (t) are orthogonal

ˆ T S

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

⇒ˆ T S

0cos (2π f 1t + Φ 1) cos (2π f 2t + Φ 2) dt = 0

⇓sin [2π ( f 1 + f 2) T S]2π ( f 1 + f 2)

cos (Φ 1 + Φ 2) + sin [2π ( f 1 − f 2) T S]

2π ( f 1 − f 2) cos (Φ 1 −Φ 2)

+ cos [2π ( f 1 + f 2) T S]−1

2π ( f 1 + f 2) sin (Φ 1 + Φ 2) +

cos [2π ( f 1 − f 2) T S]−12π ( f 1 − f 2)

sin (Φ 1 −Φ 2) = 0

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



Condition for Orthogonality – Coherent FSK


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For coherent FSK, Φ 1 = Φ 2 = Φ . So, the equation given in the previous slide becomes

⇓

sin [2π ( f 1 + f 2) T S]2π ( f 1 + f 2) cos2

Φ + sin [2π ( f 1

− f 2) T S]

2π ( f 1 − f 2)+ cos [2π ( f 1

+ f 2) T S]−12π ( f 1 + f 2)

sin2Φ = 0

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



So, Conditions for Orthogonality are


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Non-Coherent FSK:

f 1 + f 2 = m/ T S f 1

− f 2= n/ T S

Coherent FSK:

f 1 + f 2 = m/ T S f 1 − f 2= n/2 T S



BFSK vs 4FSK


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BFSK vs 4FSK


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MFSK Generation


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Coherent MFSK Detection


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Coherent MFSK Detection – PS & PB


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

PS ≤( M −1) Q ESN 0 . (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)



Non-Coherent MFSK Detection


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Non-Coherent MFSK Detection


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In no noise case, assuming that the lters preserve the shape of the envelope

of the input signal waveforms, the output is of the type

z (T ) = 2ES/ T S. (22)8. Bandpass Modulation and Demodulation Techniques Communication Systems, Dept. of EEE, BITS Hyderabad


Non-Coherent MFSK Detection – PS & PB

In the case of coherent detection, the pdf of the (signal + noise) , at the input


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of 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, rst, 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 = 1 M

exp −ESN 0

M∑

j= 2(−1) j

M j exp

ES jN 0

, where M j = M!

j! ( M − j)!



Spectrum of MFSK


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Bandwidth & Bandwidth Efciency of MFSK

Bandwidth


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Non-Coherent MFSK: BT = M/ T S

Coherent MFSK: BT = M/2 T S

Bit rate Rb = 1/ T b = log2 M/ T S = k / T S

Bandwidth Efciency

(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 .



Probability of Bit Errors for Various BinaryModulations


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Outline

1 Introduction


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5 ASK & APK

6 Summary



Amplitude Shift Keying (MASK)


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M-ary ASK waveform is given as

SMASKi (t) = Ai 2EsT S sin (ω0t + φ) , i = 1, 2,· · · , M (23)where √ ES is the energy of the nominal carrier in a symbol duration T S.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.



MASK – PS & PBIt can be shown that the symbol error probability for bipolar MASK is


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PS = 2 ( M −1) M Q 6Eavg( M2 −1) N 0 . (24)Similarly, the symbol error probability for unipolar MASK is

PS = 2 ( M

−1)

M Q 3Eavg

(2 M2 −3 M + 1) N 0 . (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 ≈ PSlog2 M

= PSk

. (27)



MASK – Bandwidth & Bandwidth Efciency


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

BT = 1kT

b

= Rbk

. (28)

So, bandwidth efciency is for MASK is

Bandwidth efciency = RbBT

= k = log2 M. (29)



Why ASK is Bad?


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

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 of both MPSK and MASK .



ASK + PSK = APK


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MAPK (Quadrature Amplitude Modulation)


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For MAPK,

SMAPKi (t) = Ai 2EsT S sin (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)



16 QAM


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8 QAM


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QAM – Modulation & Demodulation


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QAM – Modulation & Demodulation


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MQAM – PS & PB

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


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interms of symbol average energy Eavg as

PS ≤4Q 3kEb( M −1) N 0 , (32)where Eavg = (log2 M) Eb = kEb.

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 = PS

k . (33)



MQAM – Bandwidth & Bandwidth Efciency


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

BT = 1kT b

= Rbk

. (34)

So, bandwidth efciency is for MQAM is

Bandwidth efciency = RbBT

= k = log2 M. (35)



Comparison of MPSK, MQAM, and MFSK


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Outline

1 Introduction


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5 ASK & APK

6 Summary



MPSK – Summary


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SMPSKi (t) = 2EsT S sin ω0t + 2π i M , i = 1, 2,· · · , MPMPSKS ≈2Q

dmin√ 2N 0 = 2Q 2EsN 0 sin π M , for large SNRPMPSKB ≈

PSlog2 M

(for PS 1 and Gray coding )

Bandwidth efciency MPSK = RbBT

= log2 M


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