Chapter 10. Quadrature Multiplexing and Frequency Division Multiplexing

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1 Chapter 10. Quadrature Multiplexing and Frequency Division Multiplexing 1

2 Goals Design a modulation and demodulation system for quadrature multiplexing (QM) amplitude modulation (AM). Design a frequency division multiplexing (FDM) system. Analyze the effects of phase and frequency errors in QM systems. Create stereo sound effects using intentional frequency error in QM methods.

3 Quadrature multiplexing (QM) Multiplexing : a method which combines multiple information signals into one signal over a shared media Examples: Quadrature multiplexing (QM), Frequency division multiplexing (FDM), Time division multiplexing (TDM), Code division multiplexing (CDM) QM Modulated signal (tt)=ff 1 tt cos ωω cc tt + ff tt sin ωω cc tt ff 1 tt, ff tt : two individual information signals ωω cc = carrier frequency. Transmitter(modulator) and Receiver(demodulator) 3

4 Quadrature multiplexing (QM) [CH10.1A3] Demodulator output ee 1 (tt) Neglecting noise and path loss(=signal power decrease by distance) ee 1 (tt)=lpf gg 1 (tt) =LPF tt cos ωω cc tt =LPF ff 1 tt cos ωω cc tt + ff tt sin ωω cc tt cos ωω cc tt =LPF ff 1 tt cos ωω cc tt + ff tt sin ωω cc tt cos ωω cc tt =LPF ff 1 tt =LPF ff 1 tt 1+cos ωω cc tt +LPF ff 1 tt cos ωω cc tt + ff tt sin(ωω cctt) +LPF ff tt sin(ωω cc tt) In frequency domain, EE 1 (ωω)=lpf 1 FF 1 ωω +LPF 1 4 FF 1 ωω ωω cc FF 1 ωω + ωω cc +LPF 1 4jj FF ωω ωω cc 1 4jj FF ωω + ωω cc = 1 FF 1 ωω ee 1 (tt)= ff 1 tt 4

5 Quadrature multiplexing (QM) [CH10.1A3] Demodulator output ee (tt) Neglecting noise and path loss(=signal power decrease by distance) ee (tt) =LPF gg (tt) =LPF tt sin ωω cc tt =LPF ff 1 tt sin(ωω cctt) =LPF ff tt =LPF ff 1 tt cos ωω cc tt sin ωω cc tt + ff tt sin ωω cc tt + ff tt LPF ff tt cos ωω cc tt 1 cos ωω cc tt +LPF ff 1 tt sin(ωω cc tt) In frequency domain, EE (ωω)=lpf 1 FF ωω -LPF 1 FF 4 ωω ωω cc + 1 FF 4 ωω + ωω cc +LPF 1 4jj FF 1 ωω ωω cc = 1 FF ωω ee (tt)= ff tt 1 4jj FF 1 ωω + ωω cc 5

6 Effect of phase error [CH10.3B] What if there exists a phase error between the carrier and the local carrier? Say, local carrier =cos ωω cc tt + θθ not cos ωω cc tt. Then, gg 1 (tt)= tt cos ωω cc tt + θθ = ff 1 tt cos ωω cc tt + ff tt sin ωω cc tt cos ωω cc tt + θθ =ff 1 tt cos ωω cc tt cos ωω cc tt + θθ + ff tt sin ωω cc tt cos ωω cc tt + θθ =ff 1 tt cos ωω cctt+θθ +cos θθ + ff tt sin ωω cctt+θθ +sin θθ = 1 ff 1 tt cos ωω cc tt + θθ + 1 ff 1 tt cos θθ + 1 ff tt sin ωω cc tt + θθ 1 ff tt sin θθ Spectra of two terms 1 ff 1 tt cos ωω cc tt + θθ and 1 ff tt sin ωω cc tt + θθ are centered at ωω = ±ωω cc, i.e., they are pass-band signals. Spectra of two terms 1 ff 1 tt cos θθ and - 1 ff tt sin θθ are centered at ωω = 0, i.e., they are baseband signals. ee 1 (tt)=lpf gg 1 (tt) =LPF ff 1 tt cos ωω cc tt+θθ =ff 1 tt cos θθ ff tt +LPF ff 1 tt cos θθ sin θθ +LPF ff tt sin ωω cc tt+θθ +LPF ff tt sin θθ 6

7 Effect of phase error (Continued) By using similar derivations ee (tt)=ff 1 tt +ff tt Phase error θθ results in cross interference between ff 1 tt and ff tt. Special case: Case 1: If phase error θθ is small, i.e., θθ 0, then sin θθ 0,cos θθ 1 and thus, Case : If phase error θθ = ππ 4 ee 1 (tt)= ff 1 tt cos ππ 4 sin θθ ee 1 (tt) 1 ff 1 tt, ee (tt) 1 ff tt ff tt sin ππ 4 = 1 ff 1 tt ff tt, ee (tt)= ff 1 tt sin ππ 4 + ff tt cccccc ππ 4 = 1 ff 1 tt + ff tt. If ff 1 tt and ff tt are audio signals, they are added in ee 1 (tt) (or ee (tt)). we can hear both signals only by ee 1 (tt) (or ee (tt)). [CH10.3B,3B4] Especially if ff 1 tt and ff tt are a pair of stereo sound, we can hear both of them in mono signal ee 1 (tt) (or ee (tt)). Case 3: If phase error θθ = ππ,? [CH10.3B1] 4 Case 4: If phase error θθ = ππ cos θθ ππ ee 1 (tt)= ff 1 tt cos ππ ff tt sin ππ = 1 ff tt, ee (tt)= ff 1 tt sin ππ + ff tt cccccc ππ = 1 ff 1 tt, phase error results in interchange(switch) of the demodulated signals. 7

8 Effect of frequency error Now we consider the case when local carrier =cos ωω cc + ωω tt not cos ωω cc tt. We can rewrite local carrier = cos ωω cc tt + ωωtt =cos ωω cc tt + θθ where θθ= ωωtt. This implies that the frequency error results in linearly increasing phase error. So, we only need to change θθ into ωωtt in all the derivations for the phase error case. cos ωωtt sin ωωtt ee 1 tt = ff 1 tt ff tt, sin ωωtt cos ωωtt ee tt = ff 1 tt + ff tt If the frequency offset is small ωω, cos ωωtt and sin ωωtt are slowly changing, i.e., instantaneously constant. 8

9 Effect of frequency error [CH10.3C] Surrounding sound effect by frequency offset Power (volume) of ff 1 tt in ee 1 tt = Power (volume) of ff tt in ee 1 tt = cos ωωtt sin ωωtt Sound volumes of ff 1 tt and ff tt repeatedly increase and decrease in ee 1 (tt) (or ee (tt)). More specifically, if sound volume ff 1 tt increases, then ff tt decreases and vice versa., Power (volume) of ff 1 tt in ee tt = Power (volume) of ff tt in ee tt = sin ωωtt cos ωωtt, If the sound volume of ff 1 tt increases in ee 1 tt, then the sound volume of ff 1 tt decreases in ee tt and vice versa. Similarly, if the sound volume ff 1 tt increases in ee tt, then the sound volume ff 1 tt decreases in ee tt and vice versa. If we hear ee 1 tt and ee tt by left and right ear, respectively, we feel like that ff 1 tt and ff 1 tt are turning around. The frequency error ωω determines the turning speed. 9

10 Frequency division multiplexing (FDM) Modulated signal (tt)=ff 1 tt cos ωω 1 tt + ff tt cos ωω tt +ff 3 tt cos ωω 3 tt +. ff 1 tt, ff tt, ff 3 tt, : individual information signals ωω 1, ωω, ωω 3 = their carrier frequencies. Carrier frequency requirement : the carrier difference between two closest spectra should be larger than the sum of BW of those two spectra Φ( ω) ω = ω ω 1 W 1 + W ω ω1 ω ω 1 W1 W Why? To avoid spectrum overlapping. For example, if ωω 1 <ωω <ωω 3 and BWs of ff 1 tt, ff tt, ff 3 tt, are WW 1, WW, WW 3, respectively. Then, ωω ωω 1 WW 1 + WW ωω 3 ωω WW + WW 3 10

11 Frequency division multiplexing (FDM) Receiver(demodulator) Assuming the received signal = tt, i.e, no path loss, no noise. To demodulate ff 1 tt : ee 1 (tt) =LPF (tt) cos ωω 1 tt To demodulate ff tt : ee (tt) =LPF (tt) cos ωω tt To demodulate ff 3 tt : ee 3 (tt) =LPF (tt) cos ωω 3 tt and etc. What if the multiplexed signals have different phases: (tt)=ff 1 tt cos ωω 1 tt + θθ 1 + ff tt cos ωω tt + θθ +ff 3 tt cos ωω 3 tt + θθ 3 +. No problem because each spectrum component of FDM are disjoint. We only need to synchronize the phase in the receiver side, i.e., To demodulate ff 1 tt : ee 1 (tt) =LPF (tt) cos ωω 1 tt + θθ 1 To demodulate ff tt : ee (tt) =LPF (tt) cos ωω tt + θθ and etc. 11

12 Frequency division multiplexing (FDM) Demodulation example: g ( t 1 ) LPF BW=W 1 1 f 1 ( t) φ(t) cos( ωt + θ ) 1 1 g ( t) LPF BW=W 1 ( ) f t cos( ω t + θ ) g () t = φ()cos( t ωt+ θ ) ( )cos ( ω θ ) ( )cos( ω θ )cos( ω θ ) = f t t+ + f t t+ t cos(ωt+ θ ) 1 1 f ( t) = f ( t) cos (( ω + ω ) t+ θ + θ 1 1 ) f () t + cos(( ω ω 1 ) t + θ θ 1 ) f () t f ()cos( t ωt+ θ ) f () t f () t cos(( 1 ) t 1 ) = + + ω + ω + θ + θ + cos(( ω ω 1 ) t + θ θ 1 ) [CH10.B5] We have four terms whose center frequencies=0, ±ωω 1, ±(ωω 1 +ωω 1 ), ±(ωω 1 -ωω 1 ) Finally, LPF[gg 1 (tt)]= 1 ff(tt). 1

13 Frequency division multiplexing (FDM) Comparison between QM and FDM Bandwidth efficiency : [CH10.1C3] DSB-SC: By frequency spectrum shifting to passband, BW of DSB-SC signal is twice of BW of the information signal. For example, BW of ff 1 tt =WW 1 then, BW of ff 1 tt cos ωω cc tt =WW 1 QM : Identical carrier frequency for ff 1 tt and ff tt [CH10.1C3] Spectrum of ff 1 tt cos ωω cc tt in passband. and ff tt sin ωω cc tt overlap Share the same band. [CH10.1A] Total BW of QM is equal to single DSB-SC signal. [CH10.1C4] Bandwidth Efficient. FDM: Disjoint spectrum in passband Requires separate BW for each signals Bandwidth inefficient Phase error: QM : Cross interference between ff 1 tt and ff tt in addition to SNR loss. FDM: No cross interference among the multiplexed signals. SNR loss only. 13

Chapter 9. Amplitude Modulation

Chapter 9. Amplitude Modulation 1 Goals Modulate and demodulate the double sideband-suppressed carrier (DSB-SC) amplitude modulation (AM) signals. Investigate the effects of frequency and phase errors