Passband Data Transmission I References Phase-shift keying Chapter 4.1-4.3, S. Haykin, Communication Systems, Wiley.
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Introduction
In baseband pulse transmission, a data stream represented in the form of a discrete pulse-amplitude modulated (PAM) signal is transmitted over a low- pass channel.
In digital passband transmission, the incoming data stream is modulated onto a carrier with fixed frequency and then transmitted over a band-pass channel.
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1 Passband digital transmission allows more efficient use of the allocated RF bandwidth, and flexibility in accommodating different baseband signal formats.
Example Mobile Telephone Systems GSM: GMSK modulation is used (a variation of FSK) IS-54: π/4-DQPSK modulation is used (a variation of PSK)
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The modulation process making the transmission possible involves switching (keying) the amplitude, frequency, or phase of a sinusoidal carrier in accordance with the incoming data.
There are three basic signaling schemes: Amplitude-shift keying (ASK) Frequency-shift keying (FSK) Phase-shift keying (PSK)
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2 ASK
PSK
FSK
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Unlike ASK signals, both PSK and FSK signals have a constant envelope.
PSK and FSK are preferred to ASK signals for passband data transmission over nonlinear channel (amplitude nonlinearities) such as micorwave link and satellite channels.
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3 Classification of digital modulation techniques
Coherent and Noncoherent
Digital modulation techniques are classified into coherent and noncoherent techniques, depending on whether the receiver is equipped with a phase- recovery circuit or not.
The phase-recovery circuit ensures that the local oscillator in the receiver is synchronized to the incoming carrier wave (in both frequency and phase).
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Phase Recovery (Carrier Synchronization)
Two ways in which a local oscillator can be synchronized with an incoming carrier wave
transmit a pilot carrier
use a carrier-recovery circuit such as a phase-locked loop (PPL)
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4 M-ary signaling
In an M-ary signaling scheme, there are M possible signals during each signaling interval of duration T. Usually, n M = 2 and T = nTb where Tb is the bit duration.
In passband transmission, we have M-ary ASK, M-ary PSK, and M-ary FSK digital modulation schemes. We can also combine different methods: M-ary amplitude-phase keying (APK) M-ary quadrature-amplitude modulation (QAM)
In baseband transmission, we have M-ary PAM J.9
M-ary signaling schemes are preferred over binary signaling schemes for transmitting digital information over band-pass channels when the requirement is to conserve bandwidth at the expense of increased power.
The use of M-ary signaling enables a reduction in
transmission bandwidth by the factor n = log2 M over binary signaling.
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5 Coherent PSK The functional model of passband data transmission system is
Signal Signal si si (t) x(t) x mi transmission Modulator Channel Detector transmission encoder decoder mˆ
Carrier signal
• mi is a sequence of symbol emitted from a message source. • The channel is linear, with a bandwidth that is wide enough to transmit the modulated signal and the channel noise is Gaussian distributed with zero
mean and power spectral density No / 2. J.11
The following parameters are considered for a signaling scheme: Probability of error A major goal of passband data transmission systems is the optimum design of the receiver so as to minimize the average probability of symbol error in the presence of additive white Gaussian noise (AWGN)
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6 Power spectra Use to determine the signal bandwidth and co-channel interference in multiplexed systems. In practice, the signalings are linear operation, therefore, it is sufficient to evaluate the baseband power spectral density.
Bandwidth Efficiency R Bandwidth efficiency ρ = b bits/s/Hz B
where Rb is the data rate and B is the used channel bandwidth.
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In a coherent binary PSK system, the pair of signals s1 (t) and s2 (t) used to represent binary symbols 1 and 0, respectively, is defined by
2Eb s1 (t) = cos(2πfct) Tb
2Eb 2Eb s2 (t) = cos(2πfct +π ) = − cos(2πfct) Tb Tb
where 0 ≤ t ≤ Tb , and Eb is the transmitted signal energy per bit.
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7 For example,
T T b 2 2Eb b 2 2Eb Tb E = []s1 (t) dt = cos (2πfct)dt = ⋅ = Eb ∫0 ∫0 Tb Tb 2
To ensure that each transmitted bit contains an integral number of cycles of the carrier wave, the carrier frequency fc is chosen equal to n /Tb for some fixed integer n.
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The transmitted signal can be written as
s1 (t) = Ebφ(t) and
s2 (t) = − Ebφ(t)
2 where φ(t) = cos(2πfct) 0 ≤ t < Tb Tb
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8 Generation of coherent binary PSK signals To generate a binary PSK signal, we have to represent the input binary sequence in polar form with symbols 1 and 0 represented by constant amplitude levels of + Eb and
− Eb , respectively.
Signal si (t) 10101 si Product transmission Modulator encoder
2 φ(t) = cos(2πfct) Tb J.17
• This signal transmission encoder is performed by a polar nonreturn-to-zero (NRZ) encoder.
+ Eb input symbol is 1 si = − Eb input symbol is 0
• The carrier frequency fc = n /Tb where n is a fixed integer.
2Eb s1 (t) = cos(2πfct) if si = Eb Tb si (t) = 2Eb s2 (t) = − cos(2πfct) if si = − Eb Tb
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9 Detection of coherent binary PSK signals To detect the original binary sequence of 1s and 0s, we apply the noisy PSK signal to a correlator. The correlator output is compared with a threshold of zero volts.
Tb x1 Decision 1 if x1 > 0 x(t) X ∫0 device 0 if x1 < 0
φ(t) 0
Correlator
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Example: If the transmitted symbol is 1,
2Eb x(t) = cos(2πfct) Tb and the correlator output is
Tb x1 = x(t)φ(t)dt ∫0 T b 2Eb 2 = cos(2πfct)⋅ cos(2πfct)dt ∫0 Tb Tb T 2 b 2 = Eb ⋅ cos (2πfct)dt ∫0 Tb
= Eb
Similarly, If the transmitted symbol is 0, x1 = − Eb . J.20
10 Error probability of binary PSK We can represent a coherent binary system with a signal constellation consisting of two message points. • The coordinates of the message points are all the possible correlator output under a noiseless condition.
• The coordinates for BPSK are Eb and − Eb . Decision boundary
φ(t)
− Eb Eb J.21
There are two possible kinds of erroneous decision:
• Signal s2 (t) is transmitted, but the noise is such that
the received signal point inside region with x1 > 0
and so the receiver decides in favor of signal s1 (t).
• Signal s1 (t) is transmitted, but the noise is such that
the received signal point inside region with x1 < 0
and so the receiver decides in favor of signal s2 (t).
Tb x1 Decision 1 if x1 > 0 X ∫0 device < si (t) + w(t) 0 if x1 0
φ(t) 0 J.22
11 For the first case, the observable element x1 is related to the received signal x(t) by
Tb x1 = x(t)φ(t)dt ∫0
Tb = []si (t) + w(t) φ(t)dt ∫0
Tb = − Eb + w(t)φ(t)dt ∫0
x1 is a Gaussian process with mean x1 :
xi = E[xi ]
Tb = E[− Eb + w(t)φ(t)dt] ∫0
= − Eb J.23
and variance σ : 2 2 σ = E[(xi − xi ) ] 2 Tb = E w(t)φ(t)dt ∫0
TTbb = E w(t)w(u)φ(t)φ(u)dtdu ∫∫00
TTbb = E[w(t)w(u)]φ(t)φ(u)dtdu ∫∫00
TTbbN = o δ (t − u)φ(t)φ(u)dtdu ∫∫00 2 N Tb = o φ 2 (t)dt 2 ∫0 N = o 2 J.24
12 Therefore, the conditional probability density function of x1, given that symbol 0 was transmitted is 2 1 (x1 − x1 ) f (x1 | 0) = exp− 2 2πσ 2σ 2 1 (x1 + Eb ) = exp− πNo No
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and the probability of error is ∞ p10 = f (x1 | 0)dx1 ∫0 2 ∞ 1 (x1 + Eb ) = exp− dx1 ∫0 πNo No
1 Putting z = (x + Eb ), we have No ∞ 1 2 p10 = exp[]− z dz π ∫ Eb / No 0 1 E 2 ∞ = erfc b erfc(u) = exp(−z 2 )dz π ∫ 2 No u J.26
13 Similarly, the error of the second kind 1 E b p01 = p10 = erfc and hence 2 No 1 E b pe = erfc 2 No
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Quadriphase-shift keying (QPSK)
QPSK has twice the bandwith efficiency of BPSK, since 2 bits are transmitted in a single modulation symbol. The data input dk (t) is devided into an in- phase stream d I (t), and a quadrature stream dQ (t).
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14 1 0 0 1 dk (t)
t 1 0 dI (t)
t
0 1 dQ (t)
t
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The phase of the carrier takes on one of four equally spaced values, such as π/4, 3π/4, 5π/4, and 7π/4. 2E cos[2πfct + (2i −1)π / 4] 0 ≤ t ≤ T si (t) = T 0 elsewhere where i =1,2,3,4. E is the transmitted signal energy per symbol; T is the symbol duration;
fc = n /T ;
(Note : T = 2Tb )
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15 Each possible value of the phase corresponds to a unique dibit.
For example, 10 for i=1, 00 for i=2, 01 for i=3 and 11 for i=4. (only a single bit is change from one dibit to the next)
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The transmitted signal can be written as 2E s (t) = cos[2πf t + (2i −1)π / 4] i T c 2E = cos[2πf t]cos[(2i −1)π / 4] T c 2E − sin[2πf t]sin[(2i −1)π / 4] T c
= si1φ1(t) + si2φ2 (t) where 2 2 φ (t) = cos[2πf t]; φ (t) = sin[2πf t] 1 T c 2 T c
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16 Input dibit Phase of QPSK si1 si2 10 π/4 E / 2 − E / 2 00 3π/4 − E / 2 − E / 2 01 5π/4 − E / 2 E / 2 11 7π/4 E / 2 E / 2
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17 The constellation of QPSK is
φ2 (t)
(00) E / 2 (10) E / 2 φ1(t) − E / 2 (11) (01) − E / 2
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Generation of coherent QPSK signals
The incoming binary data sequence is first transformed into polar form by a nonreturn-to-zero level encoder. The binary wave is next divided by means of a demultiplexer into two separate binary sequences.
The result can be regarded as a pair of binary PSK signals, which may be detected independently due to the orthogonality of φ1 (t) and φ2 (t).
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18 2 φ1(t) = cos(2πfct) Tb
X s1i
10101 si Demulti- Polar NRZ plexer + s(t)
s2i X
2 φ1(t) = sin(2πfct) Tb J.37
Detection of coherent QPSK signals
T x1 Decision 1 if x1 > 0 x(t) X ∫0 device 0 if x1 < 0
φ (t) 1 In-phase channel 0
multiplexer
Quadrature channel
T x 2 Decision 1 if x > 0 X 2 ∫0 device 0 if x2 < 0
φ2 (t) 0 J.38
19 Error probability of QPSK The received signal is
x(t) = si (t) + w(t) and the observation elements are
Tb x1 = x(t)φ1(t)dt ∫0
Tb = ± Eb + w(t)φ1(t)dt ∫0
Tb x2 = x(t)φ2 (t)dt ∫0
Tb = ± Eb + w(t)φ2 (t)dt ∫0
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As a coherent QPSK is equivalent to two coherent binary PSK systems working in parallel and using two carriers that are in phase quadrature.
Hence, the average probability of bit error in each channel of the coherent QPSK system is 1 E / 2 1 E p = erfc = erfc 2 No 2 2No
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20 Error probability of QPSK
As the bit error in the in-phase and quadrature channels of the coherent QPSK system are statistically independent, the average probability of a correct decision resulting from the combined action of the two channels is 2 pc = (1− p) 2 1 E = 1− erfc 2 2No E 1 E 2 =1− erfc + erfc 2No 4 2No
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The average probability of symbol error for coherent QPSK is therefore
pe =1− pc E 1 2 E = erfc − erfc 2No 4 2No E ≈ >> erfc if E / 2No 1 2No
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21 In a QPSK system, since there are two bits per symbol, the transmitted signal energy per symbol is twice the signal energy per bit,
E = 2Eb
and then 1 0 0 1 dk (t) E b pe ≈ erfc 2No t 1 0 dI (t)
t
0 1 dQ (t)
t
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With Gray encoding, the bit error rate of QPSK is
1 Eb BER = erfc 2 2No
Therefore, a coherent QPSK system achieves the same average probability of bit error as a coherent binary PSK
system for the same bit rate and the same Eb / No but uses only half the channel bandwidth.
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22 M-ary PSK
During each signaling interval of duration T, one of the M possible signals 2E 2π si (t) = cos2πfct + (i −1) i =1,2,..., T M
is sent.
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M-ary PSK
The signal constellation of M-ary PSK consists of M message points which are equally spaced on a circle of radius E . For example, the constellation of octaphase-shift keying is
E π Pe ≈ erfc sin M ≥ 4 No M J.46
23 Power spectra of M-ary PSK signals The symbol function is 2E 0 ≤ t ≤ T g(t) = T 0 otherwise where T = Tb log2 M and Tb is the bit duration.
As the energy spectral density is the magnitude of the signal’s Fourier transform, the baseband power spectral density is sin 2 (πTf ) S( f ) = 2E (πTf )2 2 = 2Eb log2 M sinc (Tb f log2 M ) J.47
(Normalized to fTb ) J.48
24 Bandwidth efficiency The bandwidth required to pass M-ary signal (main lobe) is given by 2 B = sinc(2) = 0 T Q 2 = Tb log2 M 2R = b log2 M Therefore, the bandwidth efficiency is R ρ = b B log M = 2 2 J.49
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