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TEMPLE EE 4512 - Digital Bandpass Modulation and Demodulation Techniques

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EE4512 Analog and Digital Communications Chapter 5Chapter 5Chapter 5Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniquesEE4512 Analog and Digital Communications Chapter 5Chapter 5Chapter 5Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques••Binary Amplitude Shift KeyingBinary Amplitude Shift Keying••Pages 212Pages 212--219219EE4512 Analog and Digital Communications Chapter 5••The analytical signal for binary amplitude shift keying The analytical signal for binary amplitude shift keying (BASK) is:(BASK) is:ssBASKBASK(t(t) =) =ssbasebandbaseband(t(t) sin (2) sin (2ππffCCt)t)(S&M Eq. 5.1)(S&M Eq. 5.1)The signal The signal ssbasebandbaseband(t(t) can be ) can be any twoany twoshapes over a bit time shapes over a bit time TTbbbutbutit is usually a rectangular signal of amplitude 0 for a it is usually a rectangular signal of amplitude 0 for a binary 0 and amplitude A for binary 1. Then BASK is also binary 0 and amplitude A for binary 1. Then BASK is also known as known as onon--off keyingoff keying(OOK).(OOK).TTbb0 1 1 0 0 1 1 0 1 1 MS Figure 3.5MS Figure 3.5EE4512 Analog and Digital Communications Chapter 5••The binary amplitude shift keying (BASK) signal can be The binary amplitude shift keying (BASK) signal can be simulated in simulated in SimulinkSimulink..ssBASKBASK(t(t) = ) = ssbasebandbaseband(t(t) sin 2) sin 2ππffCCt (S&M Eq. 5.1)t (S&M Eq. 5.1)baseband binary PAM signal 0,baseband binary PAM signal 0,1 V, 1 V, rrbb= 1 kb/sec= 1 kb/secSinusoidal carrier Sinusoidal carrier ffCC= 20 kHz, A= 20 kHz, Acc= 5 V= 5 VMultiplierMultiplierBASK signalBASK signalEE4512 Analog and Digital Communications Chapter 5••A BASK signal is a baseband binary PAM signal multiplied A BASK signal is a baseband binary PAM signal multiplied by a carrier (S&M Figure 5by a carrier (S&M Figure 5--3). 3). Unmodulated sinusoidal carrierUnmodulated sinusoidal carrierBaseband binary PAM signalBaseband binary PAM signalBASK signalBASK signalEE4512 Analog and Digital Communications Chapter 5••The unipolar binary PAM signal can be decomposed into a The unipolar binary PAM signal can be decomposed into a polar PAM signal and DC level (S&M Figure 5polar PAM signal and DC level (S&M Figure 5--4). 4). Unipolar binary PAM signalUnipolar binary PAM signalPolar binary PAM signalPolar binary PAM signalDC levelDC level0 0 →→1 V1 V±±0.5 V0.5 V0.5 V0.5 VEE4512 Analog and Digital Communications Chapter 5••The spectrum of the BASK signal is (S&M Eq. 5.2):The spectrum of the BASK signal is (S&M Eq. 5.2):SSBASKBASK(f(f) = ) = FF( ( ssASKASK(t(t) )) )==FF( ( ssbasebandbaseband(t(t) sin (2) sin (2ππffCCt) )t) )SSBASKBASK(f(f) = 1/2 j) = 1/2 j (Sbaseband(f – fC) + Sbaseband(f + fC) )The analytical signal for the baseband binary PAM signal is:The analytical signal for the baseband binary PAM signal is:ssbasebandbaseband(t(t) = ) = ssPAMPAM(t(t) + A/2 ) + A/2 (S&M Eq. 5.3)(S&M Eq. 5.3)SSbasebandbaseband(f(f) = ) = SSPAMPAM(f(f) + A/2 ) + A/2 δδ(f) (f) (S&M Eq. 5.4)(S&M Eq. 5.4)Therefore by substitution (S&M Eq. 5.5):Therefore by substitution (S&M Eq. 5.5):SSBASKBASK(f(f) = 1/ 2j ( ) = 1/ 2j ( SSPAMPAM(f(f––ffCC) + A/2 ) + A/2 δδ(f (f ––ffCC) ) ––SSPAMPAM(f(f+ + ffCC) ) ––A/2 A/2 δδ(f + (f + ffCC) )) )EE4512 Analog and Digital Communications Chapter 5••TheThebibi--sided sided power spectral density PSD of the BASK power spectral density PSD of the BASK signal is (S&M Eq. 5.7):signal is (S&M Eq. 5.7):GGBASKBASK(f(f) = 1/4 ) = 1/4 GGPAMPAM(f(f––ffCC) + 1/4 ) + 1/4 GGPAMPAM(f(f+ + ffCC))+ A+ A22/16 /16 δδ(f (f ––ffCC) + A) + A22/16 /16 δδ(f + (f + ffCC))For a rectangular polar PAM signal (For a rectangular polar PAM signal (±±A):A):GGPAMPAM(f(f) = (A/2)) = (A/2)2 2 / r/ rb b sincsinc22((ππf / f / rrbb))(S&M Eq. 5.8)(S&M Eq. 5.8)MS Figure 3.7MS Figure 3.7EE4512 Analog and Digital Communications Chapter 5••The The singlesingle--sidedsidedpower spectral density PSD of the BASK power spectral density PSD of the BASK signal is:signal is:GGPAMPAM(f(f) = (A/2)) = (A/2)22/ r/ rb b sincsinc22((ππf / rf / rbb))GGBASKBASK(f(f) = 1/2 ) = 1/2 GGPAMPAM(f(f+ + ffCC) + A) + A22/8 /8 δδ(f + (f + ffCC))Carrier 20 kHzCarrier 20 kHzsincsinc22rrbb= 1 kHz= 1 kHzMS Figure 3.7MS Figure 3.71 kHz1 kHzEE4512 Analog and Digital Communications Chapter 5••The The bandwidthbandwidthof a BASK signal as a percentage of total of a BASK signal as a percentage of total power is power is double double that for the same bit rate that for the same bit rate rrbb= 1/= 1/TTbbbinary binary rectangular PAM (MS Table 2.1 p. 22)rectangular PAM (MS Table 2.1 p. 22)(MS Table 3.1 p. 91).(MS Table 3.1 p. 91).Bandwidth (Hz) Percentage of Total PowerBandwidth (Hz) Percentage of Total Power2/2/TTbb90%90%3/3/TTbb93%93%4/4/TTbb95%95%6/6/TTbb96.5%96.5%8/8/TTbb97.5%97.5%10/10/TTbb98%98%EE4512 Analog and Digital Communications Chapter 5Chapter 5Chapter 5Digital Bandpass ModulationDigital Bandpass Modulationand Demodulationand DemodulationTechniquesTechniques••Binary Phase Shift KeyingBinary Phase Shift Keying••Pages 219Pages 219--225225EE4512 Analog and Digital Communications Chapter 5••The analytical signal for binary phase shift keying The analytical signal for binary phase shift keying (BPSK) is:(BPSK) is:ssBPSKBPSK(t(t) =) =ssbasebandbasebandsin (2sin (2ππffCCt + t + θθ) (S&M Eq. 5.11)) (S&M Eq. 5.11)ssbasebandbaseband(t(t) = + A b) = + A bii= 1 = 1 ssbasebandbaseband(t(t) = ) = ––A bA bii= 0 = 0 00°°+180+180°°+180+180°°0 0 1 1 0 0 1 1 00MS Figure 3.13MS Figure 3.13TTbb00°°00°°EE4512 Analog and Digital Communications Chapter 5••The BPSK signal initial phase The BPSK signal initial phase θθ= 0= 0°°, +A is a phase shift = , +A is a phase shift = 00°°and and ––A is a phase shift = +180A is a phase shift = +180°°ssBPSKBPSK(t(t) =) =ssbasebandbasebandsin (2sin (2ππffCCt) (S&M Eq. 5.11)t) (S&M Eq. 5.11)ssbasebandbaseband(t(t) = + A b) = + A bii= 1 = 1


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