EE4512 Analog and Digital Communications Chapter 5. Chapter 5 Digital Bandpass Modulation and Demodulation Techniques


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1 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques
2 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Binary Amplitude Shift Keying Pages
3 The analytical signal for binary amplitude shift keying (BASK) is: s BASK (t (t)) = s baseband (t) ) sin (2π f C t) (S&M Eq. 5.1) The signal s baseband (t) ) can be any two shapes over a bit time T b but it is usually a rectangular signal of amplitude 0 for a binary 0 and amplitude A for binary 1. Then BASK is also known as onoff off keying (OOK). MS Figure 3.5 T b
4 The binary amplitude shift keying (BASK) signal can be simulated in Simulink. s BASK (t) ) = s baseband (t) ) sin 2π2 f C t (S&M Eq. 5.1) Sinusoidal carrier f C = 20 khz, A c = 5 V Multiplier BASK signal baseband binary PAM signal 0,1 1 V, r b = 1 kb/sec
5 A BASK signal is a baseband binary PAM signal multiplied by a carrier (S&M Figure 53). 5 Unmodulated sinusoidal carrier Baseband binary PAM signal BASK signal
6 The unipolar binary PAM signal can be decomposed into a polar PAM signal and DC level (S&M Figure 54). 5 Unipolar binary PAM signal 0 1 V Polar binary PAM signal ± 0.5 V DC level 0.5 V
7 The spectrum of the BASK signal is (S&M Eq. 5.2): S BASK (f S BASK (f (f)) = F( s ASK (t) ) ) = F( s baseband (t) ) sin (2π f C t) ) (f)) = 1/2 j (S baseband (f f C ) + S baseband (f + f C ) ) The analytical signal for the baseband binary PAM signal is: s baseband (t S baseband (f (t)) = s PAM (t) ) + A/2 (S&M Eq. 5.3) (f)) = S PAM (f) ) + A/2 δ(f) (S&M Eq. 5.4) Therefore by substitution (S&M Eq. 5.5): S BASK (f (f)) = 1/ 2j ( S PAM (f f C ) + A/2 δ(f f C ) S PAM (f + f C ) A/2 δ(f + f C ) )
8 The bisided power spectral density PSD of the BASK signal is (S&M Eq. 5.7): G BASK (f (f)) = 1/4 G PAM (f f C ) + 1/4 G PAM (f + f C ) + A 2 /16 δ(f f C ) + A 2 /16 δ(f + f C ) For a rectangular polar PAM signal (±( A): G PAM (f (f)) = (A/2) 2 / r b sinc 2 (π f / r b ) (S&M Eq. 5.8) MS Figure 3.7
9 The singlesided sided power spectral density PSD of the BASK signal is: G PAM (f G BASK (f (f)) = (A/2) 2 / r b sinc 2 (π f / r b ) (f)) = 1/2 G PAM (f + f C ) + A 2 /8 δ(f + f C ) Carrier 20 khz MS Figure khz r b = 1 khz sinc 2
10 The bandwidth of a BASK signal as a percentage of total power is double that for the same bit rate r b = 1/T b binary rectangular PAM (MS Table 2.1 p. 22) (MS Table 3.1 p. 91). Bandwidth (Hz) Percentage of Total Power 2/T b 90% 3/T b 93% 4/T b 95% 6/T b 96.5% 8/T b 97.5% 10/T b 98%
11 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Binary Phase Shift Keying Pages
12 The analytical signal for binary phase shift keying (BPSK) is: s BPSK (t) = s baseband (t ) = s baseband sin (2π f C t + θ) ) (S&M Eq. 5.11) (t)) = + A b i = 1 s baseband (t) ) = A b i = 0 0 MS Figure T b
13 The BPSK signal initial phase θ = 0, 0, +A is a phase shift = 0 and A A is a phase shift = +180 s BPSK (t) = s baseband (t ) = s baseband sin (2π f C t) (S&M Eq. 5.11) (t)) = + A b i = 1 s baseband (t) ) = A b i = 0 0 MS Figure T b
14 The binary phase shift keying (BPSK) signal can be simulated in Simulink. PM modulator BPSK signal Fig312.mdl baseband binary PAM signal 0,1 V, r b = 1 kb/sec
15 The Phase Modulator block is in the Modulation, Communication Blockset but as an analog passband modulator not a digital baseband modulator.
16 The Phase Modulator block has the parameters of a carrier frequency f C in Hz, initial phase in radians and the phase deviation constant in radians per volt (rad( / V). f C = 20 khz initial phase φ o = π phase deviation k p = π / V
17 The Random Integer Generator outputs 0,1 V and with a initial phase = π and a phase deviation constant = π/v, the phase output φ of the BPSK signal is: b i = 0 φ = π + 0(π/V) = π b i = 1 φ = π + 1(π/V) = 2π 2 = 0
18 The spectrum of the BPSK signal is (S&M Eq. 5.13): S BPSK (f S BPSK (f (f)) = F( s PSK (t) ) ) = F(s baseband (t) ) sin 2π2 f C t) (f)) = 1/2 j (S baseband (f f c ) + S baseband (f + f C ) ) The analytical signal for the baseband binary PAM signal is: s baseband (t S baseband (f (t)) = s PAM (t) (S&M Eq. 5.12) (f)) = S PAM (f) Note that there is no DC level in s PAM (t) ) and therefore by substitution: S BPSK (f (f)) = 1/ 2j ( S PAM (f f C ) S PAM (f + f C ) )
19 The bisided power spectral density PSD of the BPSK signal is (S&M Eq. 5.13) G BPSK (f) = 1/4 G PAM (f f C ) + 1/4 G PAM (f + f C ) For a rectangular polar PAM signal (±( A): G PAM (f (f)) = A 2 / r b sinc 2 (π f / r b ) (S&M Eq. 5.8 modified) MS Figure 3.14
20 The singlesided sided power spectral density PSD of the BPSK signal is: G BPSK (f) = 1/2 G PAM (f + f C ) (f)) = A 2 / r b sinc 2 (π f / r b ) G PAM (f No carrier MS Figure 3.14 r b = 1 khz sinc 2
21 The bandwidth of a BPSK signal as a percentage of total power is double that for the same bit rate r b = 1/T b binary rectangular PAM (MS Table 2.1 p. 22) and the same as BASK (MS Table 3.1 p. 91) (MS Table e 3.5 p. 100) Bandwidth (Hz) Percentage of Total Power 2/T b 90% 3/T b 93% 4/T b 95% 6/T b 96.5% 8/T b 97.5% 10/T b 98%
22 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Binary Frequency Shift Keying Pages
23 The analytical signal for binary frequency shift keying (BFSK) is: s BFSK (t s BFSK (t f c f (t)) = A sin (2π (f C + f) ) t + θ) ) if b i = 1 (t)) = A sin (2π (f C f) ) t + θ) ) if b i = 0 f c + f f c + f f c f T b MS Figure 3.9 f c f
24 The BFSK signal initial phase θ = 0 0 s BFSK (t s BFSK (t (t)) = A sin (2π (f C + f) ) t) if b i = 1 (t)) = A sin (2π (f C f) ) t) if b i = 0 f c f f c + f f c + f f c f T b MS Figure 3.9 f c f
25 The binary frequency shift keying (BFSK) signal can be simulated in Simulink. BFSK signal FM Modulator Fig38.mdl baseband binary PAM signal 0,1 V, r b = 1 kb/sec
26 The Frequency Modulator block is in the Modulation, Communication Blockset but as an analog passband modulator not a digital baseband modulator.
27 The Frequency Modulator block has the parameters of a carrier frequency f C in Hz, initial phase in radians and the frequency deviation constant in Hertz per volt (Hz/V). f C = 20 khz initial phase = 0 frequency deviation = 2000
28 The Random Integer Generator outputs 0,1 V but is offset to ±1 and with a initial phase = 0 and a frequency deviation constant = 2000 Hz/V, the frequency shift f of the BFSK signal is: b i = 0 d i = 1 f = 0 1(2000 Hz/V) = 2000 Hz b i = 1 d i = +1 f = 0 + 1(2000 Hz/V) = Hz
29 The BFSK signal can be decomposed as (S&M Eq. 5.14): s BFSK (t (t)) = s baseband1 (t) sin (2π (f C + f) ) t + θ) ) + baseband2 (t) sin (2π (f C f) ) t + θ) s baseband f c + f f c f f c f f c + f f c + f
30 The BFSK signal is the sum of two BASK signals: s BFSK (t (t)) = s baseband1 (t) sin (2π (f C + f) ) t + θ) ) + baseband2 (t) sin (2π (f C f) ) t + θ) s baseband2 From the linearity property,, the resulting singlesided sided PSD of the BFSK signal G BFSK (f) ) is the sum of two G BASK (f) PSDs with f = f C ± f: G BFSK (f (f)) = (A/2) 2 / 2 r b sinc 2 (π f / r b ) + A 2 /8 δ(f) f c f f c + f
31 The singlesided sided power spectral density PSD of the BFSK signal is: G BFSK (f (f)) = (A/2) 2 / 2r2 b sinc 2 (π (f C + f) / r b ) + A 2 /8 δ(f C + f) + (A/2) 2 / 2r2 b sinc 2 (π (f C f)/ r b ) + A 2 /8 δ(f C f) carriers MS Figure 3.10 f = 2 khz r b = 1 khz sinc 2
32 Minimum frequency shift keying (MFSK) for BFSK occurs when f = 1/2T b = r b /2 Hz. G BFSK (f (f)) = (A/2) 2 / 2r2 b sinc 2 (π (f C + f) / r b ) + A 2 /8 δ(f C + f) + (A/2) 2 / 2r2 b sinc 2 (π (f C f)/ r b ) + A 2 /8 δ(f C f) carriers MS Figure 3.11 f = 500 Hz r b = 1 khz sinc 2
33 This BFSK carrier frequency separation 2 f = 1/T b = r b Hz is the minimum possible because each carrier spectral impulse is at the null of the PSD of the other decomposed BASK signal and thus is called minimum frequency shift keying (MFSK). carriers MS Figure f = 1000 Hz r b = 1 khz sinc 2
34 The bandwidth of a BFSK signal as a percentage of total power is greater than that of either BASK or BPSK by 2 f2 Hz for the same bit rate r b = 1/T b (MS Table 3.3 p. 95). For MFSK 2 f 2 = r b = 1/T b Hz. Bandwidth (Hz) Percentage of Total Power 2 f + 2/T b 90% 2 f + 3/T b 93% 2 f + 4/T b 95% 2 f + 6/T b 96.5% 2 f + 8/T b 97.5% 2 f + 10/T b 98%
35 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Coherent Demodulation of Bandpass Signals Pages
36 The development of the optimum receiver for bandpass signals utilizes the same concepts as that for the optimum baseband receiver: Optimum Filter h (t) = k s(it t) o b Correlation Receiver
37 The optimum filter H o (f) ) and the correlation receiver are equivalent here also, with s 1 (t) = s(t) ) for symmetrical signals and r(t) ) = γ s(t) ) + n(t) ) where γ is the communication channel attenuation and n(t) ) is AWGN. The energy per bit E b and the probability of bit error P b is (S&M p. 226): E b it b (i1)t b itb 2 2 = γ s(t) γ s(t) dt = γ s(t) dt P b 2 E b = Q No (i1)t b
38 The matched filter or correlation receiver is a coherent demodulation process for bandpass signals because not only is bit time (T( b ) as for baseband signals required but carrier synchronization is also needed. Carrier synchronization requires an estimate of the transmitted frequency (f C ) and the arrival phase at the receiver (θ):( s(t)=sin(2π 1 f C t + θ)
39 BPSK signals are symmetrical with: s 1T (t) = s 2T (t) = A sin(2π f c t) S&M Eqs E E it [ f ] = ± A γ sin (2π t) dt b, BPSK C (i1) T itb γ A γ A T = [ 1 cos (4π f t) ] dt = 2 2 b, BPSK C (i1) T P b, BPSK b b b E b γ A Tb = Q = Q No No 2 b
40 For this analysis of E b, PSK for BPSK signals it is assumed that the transmitter produces an integer number of cycles within one bit period T b : S&M Eq itb γ A E = b, BPSK [ 1 cos (4π fc t) ] dt 2 0 E (i1) T b b γ A Tb γ A = cos (4π f t) dt 2 2 b, BPSK C (i1) T it b
41 However, even for a noninteger number of cycles within one bit period T b if 1 / f C = T C << T b : S&M Eq it b (i1) T E b cos (4π f t) dt << T C b γ A Tb γ A = cos (4π f t) dt 2 2 b, BPSK C (i1) T b it b insignificant
42 BPSK signals are symmetrical with: s 1T (t) = s 2T (t) = A sin(2π f c t) S&M Eqs and s 1 (t) = sin (2π f c t) it b a(i T )= γ s(t) s(t) dt i b i 1 (i1) T b a(i T = 2 b)+a(i T ) τ 1 b opt = 0 2 S&M Eq S&M Eq. 4.71
43 BPSK signals are symmetrical with: s 1T (t) = s 2T (t) = A sin(2π f c t) S&M Eqs τ = 0 opt S&M Figure 4164
44 For this analysis of E d, ASK for BASK signals it is assumed that the transmitter produces an integer number of cycles within one bit period: S&M Eq itb γ A E = d, BASK [ 1 cos (4π fc t) ] dt 2 0 E (i1) T b b γ A Tb γ A = cos (4π f t) dt 2 2 d, BASK C (i1) T it b
45 BASK OOK signals are not symmetrical with: s 1T (t) = A sin(2π f c t) s 2T (t) = 0 S&M Eqs , 5.23 and s 1 (t) = sin(2π f c t) ) s 2 (t) = 0 it b [ ] a(i T )= γ s (t) s (t) s (t) dt i b i 1 2 (i1) T b itb 2 γ A Tb 1 Tb (i1) T a(i )= γ A s (t) dt = a (i T ) = 0 b S&M Eq b a(i T )+a(i T ) γ A T b 1 b b τ opt = =
46 BASK signals in general may not be symmetrical with: s 1T (t) = A 1 sin(2π f c t) s 2T (t) = A 2 sin(2π f c t) and s 1 (t) s 2 (t) = sin (2π f c t) where the amplitude is arbitrary. it b [ ] a(i T )= γ s (t) s (t) s (t) dt i b i 1 2 (i1) T b itb 2 γ Ai i Tb i 1 2 (i1) T a(i )= γ A s (t) dt = b τ opt T b S&M Eq S&M Eq ( ) a(i 2 T )+a(i γ A +A 1 T ) = = 2 4 b b 1 2 T b
47 BFSK signals are not symmetrical with: s T (t) ) = A sin(2π (f c ± f) t) S&M Eq it b b [ ] E = A γ (sin (2π f + f) t sin (2π f f) t )) dt d, BFSK C C (i1)t 2 2 E = γ A T if f + f = n / T and f f = n / T d, BFSK b C 1 b C 2 b P b, BFSK 2 2 E d, FSK γ A T = b = Q Q 2 No 2 No 2 S&M Eq S&M Eq. 5.32
48 For this analysis of E d, FSK for BFSK signals it is assumed that the transmitter produces an integer number of cycles within one bit period: S&M Eq it 0 b 2 2 γ A E d, BFSK = γ A Tb cos (4π ( fc + f) t) dt 2 (i1) Tb itb γ A cos (4π ( fc f) t) dt 2 itb γ A sin (2π ( f + f) t) sin (2π ( f f) t) dt (i1) T b (i1) T b C C
49 BFSK signals are not symmetrical with: s T (t) ) = A sin(2π (f c ± f) t) S&M Eq and s 1 (t) s 2 (t) = sin(2π (f c + f) ) t) sin(2π (f c f) ) t) it b [ ] a(i T )= γ s (t) s (t) s (t) dt i b i 1 2 (i1) T b itb 2 γ Ai i Tb i i 2 (i1) T a(i )= γ A s (t) dt = b T b S&M Eq S&M Eq ( A ) a(i 2 T γ A b)+a(i 1 T ) 1 2 T b τ opt = = b = if A = A 1 2
50 A comparison of coherent BPSK, BFSK and BASK illustrates the functional differences, but BFSK and BASK uses E d and not E b : P b, BPSK E b γ A Tb = Q = Q No No E b, BPSK = γ A T b P b, BFSK 2 2 E d, FSK γ A Tb = Q = Q 2 No 2 No E = γ A T 2 2 d, BFSK b P b, BASK 2 2 E d, ASK γ A Tb = Q = Q 2 No 4 No E d, BASK = γ A T b
51 The normalized E b, FSK = E b, PSK = γ 2 A 2 T b / 2 (S&M Eq. 5.24) and E b, ASK = γ 2 A 2 T b / 4 (S&M Eq. 5.36) so that: P b, BPSK E b, PSK γ A Tb = Q = Q No No E b, BPSK = 2 2 γ A T 2 b P b, BFSK 2 2 E b, FSK γ A Tb = Q = Q No 2 No E b, BFSK = 2 2 γ A T 2 b P b, BASK 2 2 E b, ASK γ A Tb = Q = Q No 4 No 2 2 γ A T 4 Thus there are no practical advantages for either coherent BFSK or BASK and BPSK is preferred (S&M p. 236). E b, BASK = b
52 For the same P b BPSK uses the least amount of energy, BFSK requires twice as much and BASK four times as much energy: E b, PSK γ A Tb Pb, BPSK = Q = Q No No Argument of 2 2 E b, FSK γ A Tb P = Q = Q Q should be b, BFSK No 2 No as large as possible to 2 2 E γ A b, ASK T minimize P b b Pb, BASK = Q = Q No 4 No
53 Chapter 3 Bandpass Modulation and Demodulation Optimum Bandpass Receiver: The Correlation Receiver Pages
54 The matched filter or correlation receiver for bandpass symmetrical signals can be simulated in Simulink: MS Figure 3.1
55 The matched filter or correlation receiver for bandpass asymmetrical signals can also be simulated in Simulink: MS Figure 3.2
56 The alternate but universal structure which can be used for both asymmetric or symmetric binary bandpass signals can be simulated in Simulink: MS Figure 3.3
57 Chapter 3 Bandpass Modulation and Demodulation Binary Amplitude Shift Keying Pages 8692
58 Binary ASK (OOK) coherent digital communication system with BER analysis: Threshold MS Figure 3.4
59 EE4512 Analog and Digital Communications Chapter 4 The BER and P b comparison (MS Table 3.2, p. 91): Table 3.2 Observed BER and Theoretical P b as a Function of E d / N o in a Binary ASK Digital Communication System with Optimum Receiver E d / N o db BER P b
60 Chapter 3 Bandpass Modulation and Demodulation Binary Phase Shift Keying Pages
61 Binary PSK coherent digital communication system with BER analysis: MS Figure 3.12
62 EE4512 Analog and Digital Communications Chapter 4 The BER and P b comparison (SVU Table 3.5, p. 167): Table 3.5 Observed BER and Theoretical P b as a Function of E b / N o in a Binary PSK Digital Communication System with Optimum Receiver E b / N o db BER P b
63 Chapter 3 Bandpass Modulation and Demodulation Binary Frequency Shift Keying Pages
64 Binary FSK coherent digital communication system with BER analysis: f C f f C + f MS Figure 3.9
65 The BER and P b comparison (MS Table 3.4, p. 98): Table 3.4 Observed BER and Theoretical P b as a Function of E d / N o in a Binary FSK (MFSK) Digital Communication System with Optimum Receiver E d /N o db BER P b
66 The BER and P b performance comparison for BASK, BPSK and BFSK (MFSK): E d / N o db BER P b BASK E b / N o db BER P b BPSK E d / N o db BER P b BFSK
67 BER and P b comparison using E b with E b, ASK = γ 2 A 2 T 2 b / 4 and thus reduced by 10 log (0.5) 33 db or: E b / N o db BER P b BASK E b / N o db BER P b BPSK E b / N o db BER P b BFSK BASK performs better than BFSK but BPSK is the best.
68 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Differential (Noncoherent) Phase Shift Keying Pages
69 Differential (noncoherent) phase shift keying (DPSK) is demodulated by using the received signal to derive the reference signal. The DPSK protocol is: Binary 1: Transmit the carrier signal with the same phase as used for the previous bit. Binary 0: Transmit the carrier signal with its phase shifted by 180 relative to the previous bit.
70 The onebit delayed reference signal r i1 (t) is derived from the received signal r i (t) ) and if the carrier frequency f C is an integral multiple of the bit rate r b : r (t) = γ A sin (2π f (t T ) + θ) i 1 r (t) = γ A sin (2π f t+ θ) i 1 C C The output of the integrator for a binary 0 and binary 1 then is z(it b ) = ± γ 2 A 2 T b / 2 (S&M Eqs and 5.93) b S&M Eqs and 5.89
71 DPSK signals have an equivalent bit interval T DPSK = 2 T b. The probability of bit error for DPSK signal is different than that for coherent demodulation of symmetric or asymmetric signals and is: P b, DPSK 1 E 1 E DPSK b, DPSK = exp = exp 2 2 No 2 No 2 2 γ A Tb Eb, DPSK = S&M Eq
72 Chapter 3 Bandpass Modulation and Demodulation Differential Phase Shift Keying Pages
73 Binary DPSK noncoherent digital communication system with BER analysis: onebit continuous BPF delay MS Figure 3.33
74 Binary DPSK noncoherent digital communication system differential binary encoder Simulink Subsystem: XOR one bit sample delay MS Figure 3.34
75 Simulink Logic and Bit Operations provides the Logical Operator block:
76 Simulink Logical Operator blocks can be selected to provide multiple input AND, OR, NAND, NOR, XOR, NXOR, and NOT functions:
77 The Logic and Bit Operations can be configured as scalar Boolean binary (0, 1) or Mary (0, 1 M 1) 1 1) vector logic functions. Here scalar Boolean binary data is used.
78 The XOR logic generates the DPSK source coding: Table 3.16 Input Binary Data b i, Differentially Encoded Binary Data d i, and Transmitted Phase φ i (Radians) for a DPSK Signal. b i d i1 d i φ i onebit startup XOR logic π π
79 The Signal Processing Blockset provides the Filtering, Analog Filter Design block:
80 The Signal Processing Blockset provides the analog bandpass filter (BPF) specified as a 9pole 9 Butterworth filter with cutoff frequencies of 19 khz and 21 khz centered around the carrier frequency f C = 20 khz. rad/s
81 The Butterworth BPF is used for the noncoherent receiver. The coherent receiver uses the integrator as a virtual BPF: MS Figure 3.33 DPSK MS Figure 3.12 PSK
82 The BER and P b comparison (MS Table 3.17, p. 134): Table 3.17 Observed BER and Theoretical P b as a Function of E b / N o in a Binary DPSK Digital Communication System with Noncoherent Correlation Receiver Statistical variation due to small sample E b / N o db BER P b size
83 BER and P b comparison between noncoherent, source coded DPSK and coherent BPSK: E b / N o db BER P b DPSK E b / N o db BER P b BPSK BPSK performs better than DPSK but requires a coherent reference signal. DPSK performs nearly as well as BPSK at high SNR.
84 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Mary Bandpass Techniques: Quaternary Phase Shift Keying Pages
85 Quaternary phase shift keying (Mary, M = 2 n = 4 or QPSK) source codes dibits b i1 b i as a symbol with one possible protocol as: b i1 b i = 11 A sin(2π f C t + 45 ) b i1 b i = 10 A sin(2π f C t ) b i1 b i = 00 A sin(2π f C t ) b i1 b i = 01 A sin(2π f C t ) The Gray code is used as for Mary M PAM to improve + the BER performance by mitigating adjacent symbol error. The symbols are best displayed as a constellation plot S&M Figure modified
86 Quaternary phase shift keying (QPSK) displayed as a constellation plot Note the signs on the sine reference axes. cos + Constellation Gemini constellation points sin S&M Figure modified
87 Chapter 3 Bandpass Modulation and Demodulation Multilevel (Mary) Phase Shift Keying Pages
88 The QPSK signal can be simulated in Simulink using Subsystems to simplify the design. QPSK IQ I Q correlation receiver 4level Gray coded bit to symbol 4level Gray coded symbol to bit MS Figure 3.22
89 The random binary data source is converted to an M = 4 level Gray encoded symbol by a Simulink Subsystem. MS Figure 3.22
90 The M = 4 level symbol (0, 1, 2 and 3) is Gray encoded by a Lookup Table Block from the Simulink Blockset. MS Figure 3.22
91 The M = 4 level symbol (0, 1, 2 and 3) is Gray encoded by a Lookup Table Block from the Simulink Blockset.
92 The M = 4 level symbol (0, 1, 2, and 3) is Gray encoded by a Lookup Table Block by mapping [0, 1, 2, 3] to [0, 1, 3, 2] Gray coding
93 The M = 4 level symbol is inputted to the Phase Modulator block with a carrier frequency f C = 20 khz, initial phase φ o = π/4 and a phase deviation factor k p = π/2 / V
94 The M = 4 level symbol (0, 1, 2, 3) and the carrier frequency f C = 20 khz, initial phase φ o = π/4 and a phase deviation factor k p = π/2 / V produces the phase shifts: d i = 0 φ = π/4 + 0(π/2) = π/4 d i = 1 φ = π/4 + 1(π/2) = 3π/43 d i = 2 φ = π/4 + 2(π/2) = 5π/45 d i = 3 φ = π/4 + 3(π/2) = 7π/47 MS Figure 3.22
95 The modulation phase shifts are the phase angle φ of the sinusoidal carrier A sin (2π f C t + φ) ) in QPSK. d i = 0 φ = π/4 + 0(π/2) = π/4 45 d i = 1 φ = π/4 + 1(π/2) = 3π/ d i = 2 φ = π/4 + 2(π/2) = 5π/ d i = 3 φ = π/4 + 3(π/2) = 7π/ MS Figure 3.22
96 The QPSK signal can be resolved into Inphase (I, cosine) and Quadrature (Q, sine) constellation components. For example, Plot if φ = π/4 = 45 : Quadrature s 1 (t) = A sin(2π f C t + 45 ) ) = cos A / 2/ 2 [ I sin (2π f C t) + Q cos (2π f C t) ] = A / 2/ 2 [ sin (2π f C t) + cos (2π f C t) ] + sin Inphase S&M Figure modified
97 The QPSK signal is derived from Gray coded dibits with (0), (1), (3) and (2) r b = 1 kb/sec M = 4 Delay ± 5 V, f C = 2 khz, r S = 500 Hz QPSK signal T S = 2 msec
98 The QPSK signal can be decomposed into I and Q BPSK signals which are orthogonal to each other. ± 5 V QPSK signal, f C = 2 khz, r S = 500 b/sec ± 5 / 2 2 = V Binary PSK signal, sine carrier (I) ± 5 / 2 2 = V Binary PSK signal, cosine carrier (Q) T S = 2 msec
99 The orthogonality of the I and Q components of the QPSK signal can be exploited by the universal coherent receiver. The orthogonal I and Q components actually occupy the same spectrum without interference. The coherent reference signals are: Quadrature Inphase s 1 (t) = cos (2π f C t + θ) ) s 2 (t) = sin (2π f C t + θ) S&M Figure 5405
100 The orthogonality of the QPSK signals can be shown by observing the output of the quadrature correlator to the I and Q signal. its γ A z(n 1 TS)= d I sin(2π fct) + dq cos(2π fct) cos (2π fct) dt 2 (i1) TS 0 its γ A z(n 1 TS)= d I sin(2π fct) cos (2π fct) dt + 2 (i1) TS S&M Eq its γ A 2 d Q cos (2π fc t) dt 2 γ A T z(n T )= d 2 2 S 1 S Q (i1) T S z 1 (nt S )
101 The probability of bit error P b and the energy per bit E b for a QPSK signal is the same as that as for a BPSK signal but with a I and Q carrier amplitude of A / 2. P P b, BPSK b, QPSK E b, PSK γ A Tb = Q = Q No No E b, QPSK γ A TS = Q = Q No 2 No note T S S&M Eq E E b, BPSK b, QPSK = = γ A T γ A T b S note T S z 1 (nt S )
102 Since T S = 2 T b BPSK and QPSK have the same P b but QPSK can have twice the data rate r b = 2 r S within the same bandwidth because of the orthogonal I and Q components. P E b γ A Tb = P = Q = Q No No b, BPSK b, QPSK E = E = b, BPSK b, QPSK γ A T b z 1 (nt S )
103 QPSK coherent digital communication system with BER analysis: 4Level Gray coded bit to symbol MS Figure Level Gray coded symbol to bit
104 QPSK coherent digital communication system uses a 4level Gray coded bit to symbol converter Simulink Subsystem. MS Figure 2.43
105 QPSK coherent receiver uses an IQ I Q correlator Simulink Subsystem MS Figure 3.24
106 The IQ I Q correlation receiver is the universal structure with an integration time equal to the symbol time T S. MS Figure 3.24 correlation receiver
107 The output of the IQ I Q correlation receiver is a dibit and converted to an M = 2 n = 4 level symbols (0, 1, 2, and 3). dibits Mary M scaling MS Figure 3.24 correlation receiver
108 The BER and P b comparison for 4PSK 4 (QPSK): Table 3.11 Observed BER and Theoretical UpperBound of P b as a Function of E b / N o in a Gray coded 4PSK 4 (QPSK) Digital Communication System with Optimum Receiver E d /N o db BER P b
109 The singlesided sided power spectral density PSD of the QPSK signal uses r s = r b /2 and is: G QPSK (f G PAM (f (f)) = 1/2 G PAM (f  f C ) (f)) = A 2 / r s sinc 2 (π f / r s ) No carrier r s = 500 s/sec, r b = 1 kb/sec Sinc 2 MS Figure 3.25
110 The singlesided sided power spectral density PSD of BPSK has double the bandwidth than that for QPSK for the same bit rate r b = 1/T b, No carrier r b = 1 khz Sinc 2 MS Figure 3.14
111 The bandwidth of a QPSK signal as a percentage of total power is half that for the same bit rate r b = 1/T b BPSK signal since r s = r b /2 or T s = 2T2 b (MS Table 3.9). Bandwidth (Hz) Percentage of Total Power 2/T s 1/T b 90% 3/T s 1.5/T b 93% 4/T s 2/T b 95% 6/T s 3/T b 96.5% 8/T s 4/T b 97.5% 10/T s 5/T b 98%
112 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Mary Bandpass Techniques: 8Phase Shift Keying Pages
113 Mary phase shift keying (M = 8 or 8PSK) source codes tribits b i2 b i1 b i as a symbol with one possible protocol as: b i2 b i1 b i = 000 A sin(2π f C t + 0 ) 0 Constellation Plot b i2 b i1 b i = 001 A sin(2π f C t + 45 ) b i2 b i1 b i = 011 A sin(2π f C t + 90 ) b i2 b i1 b i = 010 A sin(2π f C t ) b i2 b i1 b i = 110 A sin(2π f C t ) b i2 b i1 b i = 111 A sin(2π f C t ) b i2 b i1 b i = 101 A sin(2π f C t ) b i2 b i1 b i = 100 A sin(2π f C t ) θ I, Q = 0, ± 1/ 2, ± 1 s(t) ) = A [ I sin (2π f C t) + Q cos (2π f C t) ] S&M Figure Quadrature cos sin Inphase
114 The correlation receiver for 8PSK 8 uses four reference signals: (t)) = sin (2π f C t + n ) φ s ref n (t n = 0, 1, 2, 3 φ = 22.5,, 67.5, 112.5,, S&M Eq
115 The output from any one of the four correlators is: 1 it z(n T )= γ A sin (2π f t + θ) sin (2π f t + φ)dt (i1) TS 0 its its γ A z(n 1 TS)= cos (θ φ) dt cos (4π fct + θ + φ) dt 2 γ A TS z(n 1 TS )= cos (θ φ) 2 S S C C (i1) T S S&M Eq (i1) T S z 1( n T S)
116 The correlator output is > 0 if θ φ < 90 and < 0 if not because of the cos (θ( φ) ) term. For example, if s 6 (t) is received, the ABCD correlator sign output is: +. The patterns of signs are unique and can be decoded to b i2 b i1 b i (S&M Tables and 58) 5 D: s ref 4 (t) C: s ref 3 (t) A: s ref 1 (t) B: s ref 2 (t) z 1( n T S)
117 The probability of symbol error P S for coherently demodulated Mary M PSK is: 2 A T S 2 π PS coherent Mary PSK 2Q sin M 4 No M E 2 b π P 2Q 2 log S coherent Mary PSK 2 M sin M 4 N M o P s S&M Eq S&M Figure E b / N o db
118 The probability of symbol error P S for coherently demodulated Mary M PSK is: S&M Figure P s E b /N o db
119 The probability of symbol error P S must be related to probability of bit error P b for consistency. If Gray coding is used, assume that errors will only be due to adjacent symbols.. Thus each symbol error produces only one bit in error and log 2 (M 1) correct bits or: 1 Pb errors due to adjacent symbols = PS S&M log M Eq However for Mary M PSK with M > 4 the assumption of errors being due to only adjacent symbols is invalid. For the worst case there are M 1 incorrect symbols and in M / 2 of these a bit will different from the correct bit so that: 2 1 M PS P b PS S&M Eq log M 2 (M 1) 2
120 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Mary Bandpass Techniques: Quaternary Frequency Shift Keying Pages
121 The analytical signal for quaternary (Mary, M = 2 n = 4) frequency shift keying (QFSK or 4FSK) 4 is: FSK(t) = A sin (2π (f C + 3 f) 3 ) t + θ) ) if b i1 b i = 11 FSK(t) = A sin (2π (f C + f) ) t + θ) ) if b i1 b i = 10 FSK(t) = A sin (2π (f C f) ) t + θ) ) if b i1 b i = 00 FSK(t) = A sin (2π (f C 3 f) ) t + θ) ) if b i1 b i = 01 s 4FSK s 4FSK s 4FSK s 4FSK MS Figure 3.19 f C + 3 f3 f C f f C 3 f f C + f f C + 3 f
122 Chose f C and f so that if there are a whole number of half cycles of a sinusoid within a symbol time T S for M = 4 for orthogonality of the signals so that a correlation receiver can be utilized. FSK(t) = A sin (2π (f C + 3 f) 3 ) t + θ) ) if b i1 b i = 11 FSK(t) = A sin (2π (f C + f) ) t + θ) ) if b i1 b i = 10 FSK(t) = A sin (2π (f C f) ) t + θ) ) if b i1 b i = 00 FSK(t) = A sin (2π (f C 3 f) ) t + θ) ) if b i1 b i = 01 s 4FSK s 4FSK s 4FSK s 4FSK MS Figure 3.19
123 The correlation receiver for 4FSK 4 uses four reference signals: (t)) = sin (2π (f C + n f) t) s ref n (t n = ±1, ±3 S&M Figure 5495
124 The probability of symbol error P S for coherently demodulated Mary M FSK is: 2 A T s P S coherent Mary FSK (M 1)Q M 4 2 No P s P = S coherent Mary FSK E b (M 1) Q log 2 M M 4 N o S&M Figure E b /N o db S&M Eq
125 The probability of symbol error P S for coherently demodulated Mary M FSK is: S&M Figure P s E b /N o db
126 Chapter 3 Bandpass Modulation and Demodulation Multilevel (Mary) Frequency Shift Keying Pages
127 4FSK coherent digital communication system with BER analysis: MS Figure 3.18
128 The dibits are converted to a symbol and scaled. The data is not Gray encoded. For Mary M FSK symbol errors are equally likely among the M 1 correlators and there is no advantage to Gray encoding. MS Figure 3.18
129 4FSK coherent digital communication system with BER analysis: 4FSK correlation receiver MS Figure 3.18
130 4FSK coherent digital communication system with BER analysis: 4FSK correlation receiver MS Figure 3.20
131 The 4FSK 4 correlation receiver has four correlators with an integration time equal to the symbol time T S.
132 The symbols are converted to dibits. The original data is not Gray encoded and is therefore not Gray decoded. MS Figure 3.18
133 The BER and P b comparison for 4FSK: 4 Table 3.10 Observed BER and Theoretical Upper Bound of P b as a Function of E b / N o in 4level 4 FSK Digital Communication System with Optimum Receiver E d /N o db BER P b
134 The singlesided sided power spectral density PSD with a minimum carrier frequency deviation (MFSK) for Mary M FSK is f = 1/2T S = r S /2. For MFSK the carriers should be spaced at multiples of 2 f 2 f = 1/T S = r S (S&M Eq is incorrect). Here f = 2 r S = 1 khz M = 4 f = 1 khz r s = 500 s/sec, r b = 1 kb/sec Sinc 2 MS Figure 3.21
135 The bandwidth of a Mary M FSK signal as a percentage of total power (MS Table 3.9). Bandwidth (Hz) Percentage of Total Power 2( M 1) f + 4/T s 95% 2 (M 1) f + 6/T s 96.5% 2 (M 1) f + 8/T s 97.5% 2 (M 1) f + 10/T s 98% For MFSK: f = 1/2T S = r S /2 M = 2 n and r S = r b /n
136 Chapter 5 Digital Bandpass Modulation and Demodulation Techniques Mary Bandpass Techniques: Quadrature Amplitude Modulation Pages
137 The analytical signal for quadrature amplitude modulation (QAM) has IQ I Q components: s QAM (t (t)) = I sin (2π f C t) ) + Q cos (2π f C t) A QAM signal has both amplitude and phase components which can be shown in the constellation plot. 16ary QAM Q S&M Figure I
138 An Mary M PSK signal also has IQ I Q components but the amplitude is constant and only the phase varies: s QAM (t (t)) = I sin (2π f C t) ) + Q cos (2π f C t) 16ary PSK S&M Figure ary QAM Q Q I I
139 The orthogonality of the I and Q components of the QAM signal can be exploited by the universal coherent receiver. The orthogonal I and Q components actually occupy the same spectrum without interference. The coherent reference signals are: Quadrature Inphase s 1 (t) = cos (2π f C t) ) s 2 (t) = sin (2π f C t) S&M Figure
140 An upperbound for the probability of symbol error P S for coherently demodulated Mary M QAM is: P S coherent Mary QAM 4 3 E s Q (M 1) No S&M Eq QAM BER curve M = 256 M = 4
141 An Mary M QAM constellation plot shows the stability of the signaling and the transition from one signal to another: 256ary QAM 16ary QAM
142 Chapter 3 Bandpass Modulation and Demodulation Quadrature Amplitude Modulation Pages
143 QAM coherent digital communication system with BER analysis: 4 bit to I,Q symbol 16QAM correlation QAM receiver MS Figure level symbol to bit
144 QAM coherent digital communication system with BER analysis: 4 bit to IQ I Q symbol Simulink S subsystem. MS Figure 3.27
145 QAM coherent digital communication system with BER analysis: Table 3.12 I and Q output amplitudes Input I Q Input I Q Input I Q I LUT ± 1 to ± 3 Q LUT symbol 0 to 15 MS Figure 3.27
146 QAM coherent digital communication system with BER analysis: QAM modulator Simulink S subsystem. MS Figure 3.27
147 QAM coherent digital communication system with BER analysis: 16QAM correlation receiver Simulink S subsystem. MS Figure 3.30
148 QAM coherent digital communication system with BER analysis: Table 3.14 I, Q Symbol LUT 16level output amplitudes I Q Output I Q Output
149 QAM coherent digital communication system with BER analysis: 16 level symbol to 4 bit Simulink S subsystem. MS Figure 3.31
150 QAM coherent digital communication system with BER analysis: 16 level symbol to 4 bit Simulink S subsystem. MS Figure 3.31
151 The singlesided sided power spectral density PSD of the 16ary QAM has a bandwidth of 1/M that of a PSK signal with the same data rate r b. r s = 250 s/sec, r b = 1 kb/sec Sinc 2 no discrete component at f C = 20 khz MS Figure 3.32
152 1 khz BPSK PSD r b = 1 kb/sec 16ary QAM PSD 250 Hz r b = 1 kb/sec M = 4 r S = 250 s/sec
153 The bandwidth of an Mary M QAM signal as a percentage of total power is 1/n that for the same bit rate r b = 1/T b BPSK signal since r s = r b /n or T s = ntn b where M = 2 n (MS Table 3.14). Bandwidth (Hz) Percentage of Total Power 2/T s 2/nT b 90% 3/T s 3/nT b 93% 4/T s 4/nT b 95% 6/T s 6/nT b 96.5% 8/T s 8/nT b 97.5% 10/T s 10/nT b 98%
154 16QAM coherent digital communication system received IQ Q components can be displayed on as a signal trajectory or constellation plot. realimaginary (a + b j) conversion to complex polar (M exp(jθ)) conversion Figure 3.42
155 The RealImaginary to Complex conversion block is in the Math Operations, Simulink Blockset Figure 3.42
156 The constellation plot (scatter plot) and signal trajectory are Comm Sinks blocks from the Communications Blockset Figure 3.42
157 The 16ary QAM IQ I Q component constellation plot with E b /N o (MS Figures 3.43, 3.45). signal transitions signal points decision boundaries
158 The 16ary QAM IQ I Q component constellation plot with E b /N o = 12 db, P b 104 (MS Figures 3.44, 3.46 (top))
159 The 16ary QAM IQ I Q component constellation plot with E b /N o = 6 db, P b = 3.67 x 104 (MS Figures 3.44, 3.46 (bot( bot))
160 End of Chapter 5 Digital Bandpass Modulation and Demodulation Techniques
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