S. BASK. (f) = 1/2 j. ) = 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.
EE4512 Analog and Digital Communications
Chapter 5 Digital Bandpass Modulation and Demodulation Techniques
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5 Digital Bandpass Modulation and Demodulation Techniques • Binary Amplitude Shift Keying • Pages 212-219
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The analytical signal for binary amplitude shift keying (BASK) is: sBASK(t) = sbaseband(t) sin (2π fC t) (S&M Eq. 5.1) The signal sbaseband(t) can be any two shapes over a bit time Tb 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 on-off keying (OOK). MS Figure 3.5 Tb
0
1
1
0
1
EE4512 Analog and Digital Communications
Chapter 5
• The binary amplitude shift keying (BASK) signal can be simulated in Simulink. sBASK(t) = sbaseband(t) sin 2π fC t (S&M Eq. 5.1) Sinusoidal carrier fC = 20 kHz, Ac = 5 V
Multiplier
BASK signal
baseband binary PAM signal 0,1 V, rb = 1 kb/sec
EE4512 Analog and Digital Communications
Chapter 5
• A BASK signal is a baseband binary PAM signal multiplied by a carrier (S&M Figure 5-3). Unmodulated sinusoidal carrier
Baseband binary PAM signal
BASK signal
EE4512 Analog and Digital Communications
Chapter 5
• The unipolar binary PAM signal can be decomposed into a polar PAM signal and DC level (S&M Figure 5-4). Unipolar binary PAM signal
0→1V
Polar binary PAM signal
± 0.5 V
DC level
0.5 V
EE4512 Analog and Digital Communications
Chapter 5
• The spectrum of the BASK signal is (S&M Eq. 5.2): SBASK(f) = F( sASK(t) ) = F( sbaseband(t) sin (2π fC t) ) SBASK(f) = 1/2 j (Sbaseband(f – fC) + Sbaseband(f + fC) ) The analytical signal for the baseband binary PAM signal is: sbaseband(t) = sPAM(t) + A/2 Sbaseband(f) = SPAM(f) + A/2 δ(f)
(S&M Eq. 5.3) (S&M Eq. 5.4)
Therefore by substitution (S&M Eq. 5.5): SBASK(f) = 1/ 2j ( SPAM(f – fC) + A/2 δ(f – fC) – SPAM(f + fC) – A/2 δ(f + fC) )
EE4512 Analog and Digital Communications
Chapter 5
• The bi-sided power spectral density PSD of the BASK signal is (S&M Eq. 5.7): GBASK(f) = 1/4 GPAM(f – fC) + 1/4 GPAM(f + fC) + A2/16 δ(f – fC) + A2/16 δ(f + fC) For a rectangular polar PAM signal (± A): GPAM(f) = (A/2)2 / rb sinc2 (π f / rb)
(S&M Eq. 5.8) MS Figure 3.7
EE4512 Analog and Digital Communications
Chapter 5
• The single-sided power spectral density PSD of the BASK signal is: GPAM(f) = (A/2)2 / rb sinc2 (π f / rb) GBASK(f) = 1/2 GPAM(f + fC) + A2/8 δ(f + fC) Carrier 20 kHz
MS Figure 3.7 1 kHz sinc2
rb = 1 kHz
EE4512 Analog and Digital Communications
Chapter 5
• The bandwidth of a BASK signal as a percentage of total power is double that for the same bit rate rb = 1/Tb binary rectangular PAM (MS Table 2.1 p. 22) (MS Table 3.1 p. 91). Bandwidth (Hz) Percentage of Total Power 2/Tb 3/Tb 4/Tb 6/Tb 8/Tb 10/Tb
90% 93% 95% 96.5% 97.5% 98%
EE4512 Analog and Digital Communications
Chapter 5 Digital Bandpass Modulation and Demodulation Techniques • Binary Phase Shift Keying • Pages 219-225
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The analytical signal for binary phase shift keying (BPSK) is: sBPSK(t) = sbaseband sin (2π fC t + θ) (S&M Eq. 5.11) sbaseband(t) = + A bi = 1 sbaseband(t) = – A bi = 0 0° MS Figure 3.13 Tb +180° 0° +180° 0°
0
0
1
1
0
EE4512 Analog and Digital Communications
Chapter 5
• The BPSK signal initial phase θ = 0°, +A is a phase shift = 0° and –A is a phase shift = +180° sBPSK(t) = sbaseband sin (2π fC t) (S&M Eq. 5.11) sbaseband(t) = + A bi = 1 sbaseband(t) = – A bi = 0 0° MS Figure 3.13 Tb +180° 0° +180° 0°
0
0
1
1
0
EE4512 Analog and Digital Communications
Chapter 5
• 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, rb = 1 kb/sec
EE4512 Analog and Digital Communications
• The Phase Modulator block is in the Modulation, Communication Blockset but as an analog passband modulator not a digital baseband modulator.
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The Phase Modulator block has the parameters of a carrier frequency fC in Hz, initial phase in radians and the phase deviation constant in radians per volt (rad / V). fC = 20 kHz initial phase φo = π phase deviation kp = π / V
EE4512 Analog and Digital Communications
Chapter 5
• 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: bi = 0 φ = π + 0(π/V) = π bi = 1 φ = π + 1(π/V) = 2π = 0
EE4512 Analog and Digital Communications
Chapter 5
• The spectrum of the BPSK signal is (S&M Eq. 5.13): SBPSK(f) = F( sPSK(t) ) = F(sbaseband(t) sin 2π fC t) SBPSK(f) = 1/2 j (Sbaseband(f – fc) + Sbaseband(f + fC) ) The analytical signal for the baseband binary PAM signal is: sbaseband(t) = sPAM(t) Sbaseband(f) = SPAM(f)
(S&M Eq. 5.12)
Note that there is no DC level in sPAM(t) and therefore by substitution: SBPSK(f) = 1/ 2j ( SPAM(f – fC) – SPAM(f + fC) )
EE4512 Analog and Digital Communications
Chapter 5
• The bi-sided power spectral density PSD of the BPSK signal is (S&M Eq. 5.13) GBPSK(f) = 1/4 GPAM(f – fC) + 1/4 GPAM(f + fC) For a rectangular polar PAM signal (± A): GPAM(f) = A2 / rb sinc2 (π f / rb) (S&M Eq. 5.8 modified)
MS Figure 3.14
EE4512 Analog and Digital Communications
Chapter 5
• The single-sided power spectral density PSD of the BPSK signal is: GBPSK(f) = 1/2 GPAM(f + fC) GPAM(f) = A2 / rb sinc2 (π f / rb) MS Figure 3.14
No carrier rb = 1 kHz
sinc2
EE4512 Analog and Digital Communications
Chapter 5
• The bandwidth of a BPSK signal as a percentage of total power is double that for the same bit rate rb = 1/Tb binary rectangular PAM (MS Table 2.1 p. 22) and the same as BASK (MS Table 3.1 p. 91) (MS Table 3.5 p. 100) Bandwidth (Hz) Percentage of Total Power 2/Tb 3/Tb 4/Tb 6/Tb 8/Tb 10/Tb
90% 93% 95% 96.5% 97.5% 98%
EE4512 Analog and Digital Communications
Chapter 5 Digital Bandpass Modulation and Demodulation Techniques • Binary Frequency Shift Keying • Pages 219-225
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The analytical signal for binary frequency shift keying (BFSK) is: sBFSK(t) = sBFSK(t) =
A sin (2π (fC + ∆f) t + θ) A sin (2π (fC – ∆f) t + θ)
if bi = 1 if bi = 0
fc – ∆f
fc + ∆f
fc + ∆f
fc – ∆f Tb
0
1
1
0
MS Figure 3.9 fc – ∆f
0
EE4512 Analog and Digital Communications
Chapter 5
• The BFSK signal initial phase θ = 0° sBFSK(t) = A sin (2π (fC + ∆f) t) if bi = 1 sBFSK(t) = A sin (2π (fC – ∆f) t) if bi = 0
fc – ∆f
fc + ∆f
fc + ∆f
fc – ∆f Tb
0
1
1
0
MS Figure 3.9 fc – ∆f
0
EE4512 Analog and Digital Communications
Chapter 5
• 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, rb = 1 kb/sec
EE4512 Analog and Digital Communications
• The Frequency Modulator block is in the Modulation, Communication Blockset but as an analog passband modulator not a digital baseband modulator.
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The Frequency Modulator block has the parameters of a carrier frequency fC in Hz, initial phase in radians and the frequency deviation constant in Hertz per volt (Hz/V). fC = 20 kHz initial phase = 0 frequency deviation = 2000
EE4512 Analog and Digital Communications
Chapter 5
• 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: bi = 0 di = –1 ∆f = 0 – 1(2000 Hz/V) = –2000 Hz bi = 1 di = +1 ∆f = 0 + 1(2000 Hz/V) = +2000 Hz
EE4512 Analog and Digital Communications
Chapter 5
• The BFSK signal can be decomposed as (S&M Eq. 5.14): sBFSK(t) = sbaseband1(t) sin (2π (fC + ∆f) t + θ) + sbaseband2(t) sin (2π (fC – ∆f) t + θ)
1
0
0
1
1
fc + ∆f
fc – ∆f
fc – ∆f
fc + ∆f
fc + ∆f
EE4512 Analog and Digital Communications
Chapter 5
• The BFSK signal is the sum of two BASK signals: sBFSK(t) = sbaseband1(t) sin (2π (fC + ∆f) t + θ) + sbaseband2(t) sin (2π (fC – ∆f) t + θ) From the linearity property, the resulting single-sided PSD of the BFSK signal GBFSK(f) is the sum of two GBASK(f) PSDs with f = fC ± ∆f: GBFSK(f) = (A/2)2 / 2 rb sinc2 (π f / rb) + A2/8 δ(f) fc – ∆f
fc + ∆f
EE4512 Analog and Digital Communications
Chapter 5
• The single-sided power spectral density PSD of the BFSK signal is: GBFSK(f) = (A/2)2 / 2rb + (A/2)2 / 2rb
sinc2 (π (fC + ∆f) / rb) + A2/8 δ(fC + ∆f) sinc2 (π (fC – ∆f)/ rb) + A2/8 δ(fC – ∆f) carriers
∆f = 2 kHz
MS Figure 3.10 rb = 1 kHz sinc2
EE4512 Analog and Digital Communications
Chapter 5
• Minimum frequency shift keying (MFSK) for BFSK occurs when ∆f = 1/2Tb = rb/2 Hz. GBFSK(f) = (A/2)2 / 2rb + (A/2)2 / 2rb
sinc2 (π (fC + ∆f) / rb) + A2/8 δ(fC+ ∆f) sinc2 (π (fC – ∆f)/ rb) + A2/8 δ(fC – ∆f) carriers
∆f = 500 Hz
MS Figure 3.11 rb = 1 kHz sinc2
EE4512 Analog and Digital Communications
Chapter 5
• This BFSK carrier frequency separation 2∆f = 1/Tb = rb 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
2∆f = 1000 Hz
MS Figure 3.11 rb = 1 kHz sinc2
EE4512 Analog and Digital Communications
Chapter 5
• The bandwidth of a BFSK signal as a percentage of total power is greater than that of either BASK or BPSK by 2∆f Hz for the same bit rate rb = 1/Tb (MS Table 3.3 p. 95). For MFSK 2∆f = rb = 1/Tb Hz. Bandwidth (Hz) Percentage of Total Power 2∆f + 2/Tb 2∆f + 3/Tb 2∆f + 4/Tb 2∆f + 6/Tb 2∆f + 8/Tb 2∆f + 10/Tb
90% 93% 95% 96.5% 97.5% 98%
EE4512 Analog and Digital Communications
Chapter 5 Digital Bandpass Modulation and Demodulation Techniques • Coherent Demodulation of Bandpass Signals • Pages 225-236
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The development of the optimum receiver for bandpass signals utilizes the same concepts as that for the optimum baseband receiver: Optimum Filter ho (t) = k s(iTb − t)
Correlation Receiver
EE4512 Analog and Digital Communications
Chapter 5
• The optimum filter Ho(f) and the correlation receiver are equivalent here also, with s1(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 Eb and the probability of bit error Pb is (S&M p. 226): iTb
Eb =
∫
(i-1)Tb
iTb
γ s(t) γ s(t) dt = γ 2 Pb = Q
2 Eb No
∫
(i-1)Tb
s2 (t) dt
EE4512 Analog and Digital Communications
Chapter 5
• The matched filter or correlation receiver is a coherent demodulation process for bandpass signals because not only is bit time (Tb) as for baseband signals required but carrier synchronization is also needed. Carrier synchronization requires an estimate of the transmitted frequency (fC) and the arrival phase at the receiver (θ):
s1(t) = sin(2π fC t + θ)
EE4512 Analog and Digital Communications
Chapter 5
• BPSK signals are symmetrical with: s1T(t) = – s2T(t) = A sin(2π fc t) S&M Eqs. 5.15-5.19 iTb
∫ [ ± A γ sin (2π f t)]
Eb, BPSK =
2
C
dt
(i-1)Tb 2
Eb, BPSK
γ A = 2
Pb, BPSK
2
iTb
∫
(i-1)Tb
= Q
γ 2 A 2Tb [1− cos (4π fC t)] dt = 2 2 Eb No
= Q
γ 2 A 2Tb No
EE4512 Analog and Digital Communications
Chapter 5
• For this analysis of Eb, PSK for BPSK signals it is assumed that the transmitter produces an integer number of cycles within one bit period Tb: S&M Eq. 5.17 2
Eb, BPSK
γ A = 2 2
Eb, BPSK
2
iTb
∫ [1− cos (4π f t)] dt C
(i-1)Tb 2
2
γ A Tb γ A = − 2 2
2
0
iTb
∫
(i-1)Tb
cos (4π fC t) dt
EE4512 Analog and Digital Communications
Chapter 5
• However, even for a non-integer number of cycles within one bit period Tb if 1 / fC = TC 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: 1 M PS ≤ Pb ≤ PS log2 M 2 (M − 1)
S&M Eq. 5.129
EE4512 Analog and Digital Communications
Chapter 5 Digital Bandpass Modulation and Demodulation Techniques • M-ary Bandpass Techniques: Quaternary Frequency Shift Keying • Pages 292-298
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The analytical signal for quaternary (M-ary, M = 2n = 4) frequency shift keying (QFSK or 4-FSK) is: s4-FSK(t) = s4-FSK(t) = s4-FSK(t) = s4-FSK(t) =
A sin (2π (fC + 3∆f) t + θ) A sin (2π (fC + ∆f) t + θ) A sin (2π (fC – ∆f) t + θ) A sin (2π (fC – 3∆f) t + θ)
if bi-1bi = 11 if bi-1bi = 10 if bi-1bi = 00 if bi-1bi = 01
MS Figure 3.19
fC + 3∆f 11
fC – ∆f 00
fC – 3∆f 01
fC + ∆f 10
fC + 3∆f 11
EE4512 Analog and Digital Communications
Chapter 5
• Chose fC and ∆f so that if there are a whole number of half cycles of a sinusoid within a symbol time TS for M = 4 for orthogonality of the signals so that a correlation receiver can be utilized. s4-FSK(t) = s4-FSK(t) = s4-FSK(t) = s4-FSK(t) =
A sin (2π (fC + 3∆f) t + θ) A sin (2π (fC + ∆f) t + θ) A sin (2π (fC – ∆f) t + θ) A sin (2π (fC – 3∆f) t + θ)
if bi-1bi = 11 if bi-1bi = 10 if bi-1bi = 00 if bi-1bi = 01
MS Figure 3.19
EE4512 Analog and Digital Communications
• The correlation receiver for 4-FSK uses four reference signals: sref n(t) = sin (2π (fC + n ∆f) t) n = ±1, ±3
S&M Figure 5-49
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The probability of symbol error PS for coherently demodulated M-ary FSK is: A 2T s PS coherent M-ary FSK ≤ (M − 1) Q 2 No Ps PS coherent M-ary FSK = (M − 1) Q
M ≥ 4
Eb log2 M M ≥ 4 No
S&M Eq. 5.132 S&M Figure 5-51 Eb/No dB
EE4512 Analog and Digital Communications
Chapter 5
• The probability of symbol error PS for coherently demodulated M-ary FSK is: S&M Figure 5-51 Ps
Eb/NodB
EE4512 Analog and Digital Communications
Chapter 3 Bandpass Modulation and Demodulation • Multilevel (M-ary) Frequency Shift Keying • Pages 110-116
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• 4-FSK coherent digital communication system with BER analysis:
MS Figure 3.18
EE4512 Analog and Digital Communications
Chapter 5
• The dibits are converted to a symbol and scaled. The data is not Gray encoded. For M-ary FSK symbol errors are equally likely among the M – 1 correlators and there is no advantage to Gray encoding.
MS Figure 3.18
EE4512 Analog and Digital Communications
Chapter 5
• 4-FSK coherent digital communication system with BER analysis: 4-FSK correlation receiver
MS Figure 3.18
EE4512 Analog and Digital Communications
Chapter 5
• 4-FSK coherent digital communication system with BER analysis: 4-FSK correlation receiver
MS Figure 3.20
EE4512 Analog and Digital Communications
• The 4-FSK correlation receiver has four correlators with an integration time equal to the symbol time TS.
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The symbols are converted to dibits. The original data is not Gray encoded and is therefore not Gray decoded.
MS Figure 3.18
EE4512 Analog and Digital Communications
Chapter 5
• The BER and Pb comparison for 4-FSK: Table 3.10 Observed BER and Theoretical Upper Bound of Pb as a Function of Eb / No in 4-level FSK Digital Communication System with Optimum Receiver Ed/No dB ∞ 12 10 8 6 4 2 0
BER 0 0 0 1 × 10-4 5.1 × 10-3 2.26 × 10-2 5.97 × 10-2 1.209 × 10-1
Pb 0 ≈10-8 ≈10-6 ≈10-4 4.8 × 10-3 2.52 × 10-2 7.54 × 10-2 1.586 × 10-1
EE4512 Analog and Digital Communications
Chapter 5
• The single-sided power spectral density PSD with a minimum carrier frequency deviation (MFSK) for M-ary FSK is ∆f = 1/2TS = rS/2. For MFSK the carriers should be spaced at multiples of 2∆f = 1/TS = rS (S&M Eq. 5.131 is incorrect). Here ∆f = 2 rS = 1 kHz M=4 ∆f = 1 kHz
rs = 500 s/sec, rb = 1 kb/sec Sinc2
MS Figure 3.21
EE4512 Analog and Digital Communications
Chapter 5
• The bandwidth of a M-ary FSK signal as a percentage of total power (MS Table 3.9). Bandwidth (Hz) Percentage of Total Power 2( M – 1) ∆f + 4/Ts 2 (M – 1) ∆f + 6/Ts 2 (M – 1) ∆f + 8/Ts 2 (M – 1) ∆f + 10/Ts
95% 96.5% 97.5% 98%
For MFSK: ∆f = 1/2TS = rS/2 M = 2n and rS = rb/n
EE4512 Analog and Digital Communications
Chapter 5 Digital Bandpass Modulation and Demodulation Techniques • M-ary Bandpass Techniques: Quadrature Amplitude Modulation • Pages 298-301
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The analytical signal for quadrature amplitude modulation (QAM) has I-Q components: sQAM(t) = I sin (2π fCt) + Q cos (2π fCt) A QAM signal has both amplitude and phase components which can be shown in the constellation plot.
16-ary QAM Q
S&M Figure 5-53 I
EE4512 Analog and Digital Communications
Chapter 5
• An M-ary PSK signal also has I-Q components but the amplitude is constant and only the phase varies: sQAM(t) = I sin (2π fCt) + Q cos (2π fCt) S&M Figure 5-54
16-ary PSK
16-ary QAM
Q
I
Q
I
EE4512 Analog and Digital Communications
Chapter 5
• 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 In-phase s2(t) = sin (2π fCt) s1(t) = cos (2π fCt) S&M Figure 5-55
EE4512 Analog and Digital Communications
Chapter 5
• An upper-bound for the probability of symbol error PS for coherently demodulated M-ary QAM is: PS coherent M-ary QAM ≤ 4Q
3 Es (M − 1) No
S&M Eq. 5.135
M = 256
QAM BER curve M=4
EE4512 Analog and Digital Communications
Chapter 5
• An M-ary QAM constellation plot shows the stability of the signaling and the transition from one signal to another:
256-ary QAM
16-ary QAM
EE4512 Analog and Digital Communications
Chapter 3 Bandpass Modulation and Demodulation • Quadrature Amplitude Modulation • Pages 123-130
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• QAM coherent digital communication system with BER analysis: 4 bit to I,Q symbol 16-QAM correlation receiver QAM
16-level symbol to bit MS Figure 3.26
EE4512 Analog and Digital Communications
Chapter 5
• QAM coherent digital communication system with BER analysis: 4 bit to I-Q symbol Simulink subsystem.
MS Figure 3.27
EE4512 Analog and Digital Communications
Chapter 5
• QAM coherent digital communication system with BER analysis: Table 3.12 I and Q output amplitudes Input 0 1 2 3 4
I –1 –3 –1 –3 1
Q 1 1 –3 3 1
Input I 5 1 6 3 7 3 8 –1 9 –1
Q 3 1 3 –1 –3
Input I 10 –3 11 –3 12 1 13 3 14 1 15 3
Q –1 –3 –1 –1 –3 –2
I LUT ± 1 to ± 3 Q LUT symbol 0 to 15
MS Figure 3.27
EE4512 Analog and Digital Communications
Chapter 5
• QAM coherent digital communication system with BER analysis: QAM modulator Simulink subsystem. MS Figure 3.27
EE4512 Analog and Digital Communications
Chapter 5
• QAM coherent digital communication system with BER analysis: 16-QAM correlation receiver Simulink subsystem. MS Figure 3.30
EE4512 Analog and Digital Communications
•
Chapter 5
QAM coherent digital communication system with BER analysis: Table 3.14 I, Q Symbol LUT 16-level output amplitudes I 1 1 1 1 2 2 2 2
Q 1 2 3 4 1 2 3 4
Output 11 9 14 15 10 8 12 13
I 3 3 3 3 4 4 4 4
Q 1 2 3 4 1 2 3 4
Output 1 0 4 6 3 2 5 7
EE4512 Analog and Digital Communications
Chapter 5
• QAM coherent digital communication system with BER analysis: 16 level symbol to 4 bit Simulink subsystem. MS Figure 3.31
EE4512 Analog and Digital Communications
Chapter 5
• QAM coherent digital communication system with BER analysis: 16 level symbol to 4 bit Simulink subsystem.
MS Figure 3.31
EE4512 Analog and Digital Communications
Chapter 5
• The single-sided power spectral density PSD of the 16-ary QAM has a bandwidth of 1/M that of a PSK signal with the same data rate rb. rs = 250 s/sec, rb = 1 kb/sec Sinc2
no discrete component at fC = 20 kHz
MS Figure 3.32
EE4512 Analog and Digital Communications
1 kHz
Chapter 5
BPSK PSD rb = 1 kb/sec
16-ary QAM PSD 250 Hz
rb = 1 kb/sec M = 4 rS = 250 s/sec
EE4512 Analog and Digital Communications
Chapter 5
• The bandwidth of an M-ary QAM signal as a percentage of total power is 1/n that for the same bit rate rb = 1/Tb BPSK signal since rs = rb/n or Ts = nTb where M = 2n (MS Table 3.14). Bandwidth (Hz) Percentage of Total Power 2/Ts 3/Ts 4/Ts 6/Ts 8/Ts 10/Ts
2/nTb 3/nTb 4/nTb 6/nTb 8/nTb 10/nTb
90% 93% 95% 96.5% 97.5% 98%
EE4512 Analog and Digital Communications
Chapter 5
• 16-QAM coherent digital communication system received I-Q components can be displayed on as a signal trajectory or constellation plot . real-imaginary (a + b j) conversion to complex polar (M exp(jθ)) conversion
Figure 3.42
EE4512 Analog and Digital Communications
• The Real-Imaginary to Complex conversion block is in the Math Operations, Simulink Blockset
Figure 3.42
Chapter 5
EE4512 Analog and Digital Communications
• The constellation plot (scatter plot) and signal trajectory are Comm Sinks blocks from the Communications Blockset
Figure 3.42
Chapter 5
EE4512 Analog and Digital Communications
Chapter 5
• The 16-ary QAM I-Q component constellation plot with Eb/No → ∞ (MS Figures 3.43, 3.45). signal transitions
signal points
decision boundaries
EE4512 Analog and Digital Communications
Chapter 5
• The 16-ary QAM I-Q component constellation plot with Eb/No = 12 dB, Pb ≈ 10-4 (MS Figures 3.44, 3.46 (top))
EE4512 Analog and Digital Communications
Chapter 5
• The 16-ary QAM I-Q component constellation plot with Eb/No = 6 dB, Pb = 3.67 x 10-4 (MS Figures 3.44, 3.46 (bot))
EE4512 Analog and Digital Communications
End of Chapter 5 Digital Bandpass Modulation and Demodulation Techniques
Chapter 5
