Carrier Transmission. The transmitted signal is y(t) = k a kh(t kt ). What is the bandwidth? More generally, what is its Fourier transform?

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1 The transmitted signal is y(t) = k a kh(t kt ). What is the bandwidth? More generally, what is its Fourier transform?

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3 The baseband signal is y(t) = k a kh(t kt ). The power spectral density of the transmission is H(f) H(f) f, [khz] The carrier modulated signal is y m (t) = y(t) cos(2πtf c ) But bandwidth gets twice as large!

4 The baseband signal is y(t) = k a kh(t kt ). The power spectral density of the transmission is H(f) H(f) 2 H m (f) f, [khz] The carrier modulated signal is y m (t) = y(t) cos(2πtf c ) But bandwidth gets twice as large!

5 The baseband signal is y(t) = k a kh(t kt ). The power spectral density of the transmission is H(f) BW H(f) 2 H m (f) 2 BW f, [khz] The carrier modulated signal is y m (t) = y(t) cos(2πtf c ) But bandwidth gets twice as large!

6 Where did the energy go? Basic Fourier relations: cos(2πf c t)h(t) 1 2 H(f f c) H(f + f c) sin(2πf c t)h(t) i 2 H(f f c) i 2 H(f + f c) The 1/2 factor corresponds to a 1/4 of the energy. Since there are two terms, 1/2 of the energy is preserved. What about the increased bandwidth? Important

7 Where did the energy go? Basic Fourier relations: cos(2πf c t)h(t) 1 2 H(f f c) H(f + f c) sin(2πf c t)h(t) i 2 H(f f c) i 2 H(f + f c) The 1/2 factor corresponds to a 1/4 of the energy. Since there are two terms, 1/2 of the energy is preserved. What about the increased bandwidth? Important

8 Where did the energy go? Basic Fourier relations: cos(2πf c t)h(t) 1 2 H(f f c) H(f + f c) sin(2πf c t)h(t) i 2 H(f f c) i 2 H(f + f c) Important The 1/2 factor corresponds to a 1/4 of the energy. Since there are two terms, 1/2 of the energy is preserved. What about the increased bandwidth?

9 Assume two independent baseband transmissions Two transmissions f, [khz]

10 Assume two independent baseband transmissions After modulation with cos(2πtf c ) and sin(2πtf c ) we get Modulated by cos(2π tf c ) Modulated by sin(2π tf c ) f, [khz]

11 Assume two independent baseband transmissions After demodulation with cos(2πtf c ) we get There is a red one here Cut away these with low pass filter f, [khz] The red spectras cancel out, thus, we can detect the blue independently from the red Similar for demodulation with sin(2πtf c )

12 The block diagram of the transmitter is cos(2 tfc) yi(t) yq(t) y(t) -sin(2 tfc) y(t) = y I (t) cos(2πf c t) y Q (t) sin(2πf c t)

13 The block diagram of the receiver is cos(2 tfc) r(t) LPF LPF yi(t)+ni(t) yq(t)+nq(t) -sin(2 tfc) The in-phase and the quadrature components can be independently detected! The LPF (low pass filters) can be taken as a matched filter to h(t)

14 The signals at both rails are baseband signals, and conventional processing follows: matched filter sampling every T s second decision unit Sample: kts yi(t)+ni(t) MF ri[k]=ai[k]+ni[k] Sample: kts yq(t)+nq(t) MF rq[k]=aq[k]+nq[k]

15 What is a complex-valued symbol 1 + i? In QPSK, we transmit complex valued symbols. In one symbol interval, we have y(t) = h(t) }{{} y I (t) cos(2πf c t) h(t) sin(2πf }{{} c t) y Q (t)

16 What is a complex-valued symbol 1 + i? In QPSK, we transmit complex valued symbols. In one symbol interval, we have y(t) = h(t) }{{} y I (t) Real part goes here cos(2πf c t) h(t) sin(2πf }{{} c t) y Q (t) and imaginary here

17 We can alternatively express the signal y(t) as y(t) = y I (t) cos(2πf c t) y Q (t) sin(2πf c t) = e(t) cos(2πf c t + θ(t)) where e(t) is the envelope and θ(t) is the phase For QPSK, e(t) = 2h(t) and θ(t) {π/4, 3π/4, 5π/4, 7π/4} We can further manipulate y(t) into y(t) = Re{(y I (t) + iy Q (t))e 2πf ct } = Re{ỹ(t)e i2πf ct } where ỹ(t) = y I (t) + iy Q (t)

18 Example Assume that we have two bits to transmit, say +1 and -1. We can either do this as y(t) = h(t) cos(2πf c t) ( h(t)) sin(2πf c t) or as or as y(t) = 2h(t) cos(2πf c t + 7π/4) y(t) = Re{(1 i)h(t)e 2πf ct }

19 Example Assume that we have two bits to transmit, say +1 and -1. We can either do this as y(t) = h(t) cos(2πf c t) ( h(t)) sin(2πf c t) or as or as y(t) = 2h(t) cos(2πf c t + 7π/4) y(t) = Re{(1 i)h(t)e 2πf ct }

20 Example Assume that we have two bits to transmit, say +1 and -1. We can either do this as y(t) = h(t) cos(2πf c t) ( h(t)) sin(2πf c t) or as or as y(t) = 2h(t) cos(2πf c t + 7π/4) y(t) = Re{(1 i)h(t)e 2πf ct }

21 Example Assume that we have two bits to transmit, say +1 and -1. We can either do this as y(t) = h(t) cos(2πf c t) ( h(t)) sin(2πf c t) or as or as y(t) = 2h(t) cos(2πf c t + 7π/4) y(t) = Re{(1 i)h(t)e 2πf ct }

22 In the last representation, we can change the receiver processing into Sample: kts yi(t)+ni(t) MF ri[k]=ai[k]+ni[k] Sample: kts r[k]+n[k] yq(t)+nq(t) MF rq[k]=aq[k]+nq[k] i

23 System model y (t) r (t ) We want to represent the outputs as functions of the inputs Note that the receiver and transmitter are not synchronous Lund University / Faculty / Department / Unit / Document / Date

24 Models of input and channel I Q The transmitted signal y(t) equals y(t) =y I (t)cos(ω c t) y Q (t)sin(ω c t). Similarily, the channel impulse response can be expressed as h(t) =h I (t)cos(ω c t) h Q (t)sin(ω c t). Lund University / Faculty / Department / Unit / Document / Date

25 Channel output in the Fourier domain To evaluate r(t) = y(t) h(t), we consider the signals in the Fourier domain: R(f) = Y (f)h(f) = 1 4 [Y I(f + f c )+Y I (f f c )+jy Q (f + f c ) jy Q (f f c )] [H I (f + f c )+H I (f f c )+jh Q (f + f c ) jh Q (f f c )] Now observe that a product of the type Y I/Q (f ± f c )H I/Q (f f c )=0 Lund University / Faculty / Department / Unit / Document / Date

26 Channel output in the Fourier domain R(f) = 1 4 [Y I(f + f c )H I (f + f c )+jy I (f + f c )H Q (f + f c )+Y I (f f c )H I (f f c ) jy I (f f c )H Q (f f c )+jy Q (f + f c )H I (f + f c ) Y Q (f + f c )H Q (f + f c ) jy Q (f f c )H I (f + f c ) Y Q (f f c )H Q (f f c )] Lund University / Faculty / Department / Unit / Document / Date

27 Channel output in the Fourier domain R(f) = 1 4 [Y I(f + f c )H I (f + f c )+jy I (f + f c )H Q (f + f c )+Y I (f f c )H I (f f c ) jy I (f f c )H Q (f f c )+jy Q (f + f c )H I (f + f c ) Y Q (f + f c )H Q (f + f c ) jy Q (f f c )H I (f + f c ) Y Q (f f c )H Q (f f c )] By identifying terms, we get that r(t) = r I (t)cos(ω c t) r Q (t)sin(ω c t), with r I (t) = 1 2 [y I(t) h I (t) y Q (t) h Q (t)] and r Q (t) = 1 2 [y I(t) h Q (t)+y Q (t) h I (t)]. Lund University / Faculty / Department / Unit / Document / Date

28 Channel output in the Fourier domain R(f) = 1 4 [Y I(f + f c )H I (f + f c )+jy I (f + f c )H Q (f + f c )+Y I (f f c )H I (f f c ) jy I (f f c )H Q (f f c )+jy Q (f + f c )H I (f + f c ) Y Q (f + f c )H Q (f + f c ) jy Q (f f c )H I (f + f c ) Y Q (f f c )H Q (f f c )] By identifying terms, we get that r(t) = r I (t)cos(ω c t) r Q (t)sin(ω c t), with r I (t) = 1 2 [y I(t) h I (t) y Q (t) h Q (t)] and r Q (t) = 1 2 [y I(t) h Q (t)+y Q (t) h I (t)]. Lund University / Faculty / Department / Unit / Document / Date

29 Channel output in the Fourier domain R(f) = 1 4 [Y I(f + f c )H I (f + f c )+jy I (f + f c )H Q (f + f c )+Y I (f f c )H I (f f c ) jy I (f f c )H Q (f f c )+jy Q (f + f c )H I (f + f c ) Y Q (f + f c )H Q (f + f c ) jy Q (f f c )H I (f + f c ) Y Q (f f c )H Q (f f c )] By identifying terms, we get that r(t) = r I (t)cos(ω c t) r Q (t)sin(ω c t), with r I (t) = 1 2 [y I(t) h I (t) y Q (t) h Q (t)] and r Q (t) = 1 2 [y I(t) h Q (t)+y Q (t) h I (t)]. Lund University / Faculty / Department / Unit / Document / Date

30 Channel output in the Fourier domain R(f) = 1 4 [Y I(f + f c )H I (f + f c )+jy I (f + f c )H Q (f + f c )+Y I (f f c )H I (f f c ) jy I (f f c )H Q (f f c )+jy Q (f + f c )H I (f + f c ) Y Q (f + f c )H Q (f + f c ) jy Q (f f c )H I (f + f c ) Y Q (f f c )H Q (f f c )] By identifying terms, we get that r(t) = r I (t)cos(ω c t) r Q (t)sin(ω c t), with r I (t) = 1 2 [y I(t) h I (t) y Q (t) h Q (t)] and r Q (t) = 1 2 [y I(t) h Q (t)+y Q (t) h I (t)]. Lund University / Faculty / Department / Unit / Document / Date

31 Interlude Interlude y (t) r (t ) r(t) = r I (t)cos(ω c t) r Q (t)sin(ω c t), r I (t) = 1 2 [y I(t) h I (t) y Q (t) h Q (t)] r Q (t) = 1 2 [y I(t) h Q (t)+y Q (t) h I (t)]. Lund University / Faculty / Department / Unit / Document / Date

32 Interlude Interlude y (t) r (t ) Now, lets multiply by cos(wt + φ) Lund University / Faculty / Department / Unit / Document / Date

33 Basic trig properties cos(x)cos(y) = 1 2 [cos(x + y)+cos(x y)] sin(x)cos(y) = 1 2 [sin(x + y)+sin(x y)] sin(x)sin(y) = 1 2 [cos(x + y) cos(x y)] Signal at upper rail equals Lund University / Faculty / Department / Unit / Document / Date

34 Basic trig properties cos(x)cos(y) = 1 2 [cos(x + y)+cos(x y)] sin(x)cos(y) = 1 2 [sin(x + y)+sin(x y)] sin(x)sin(y) = 1 2 [cos(x + y) cos(x y)] Signal at upper rail equals Remove by low pass filtering Lund University / Faculty / Department / Unit / Document / Date

35 Channel output in the Fourier domain By using the trigonometric identities we get that cos(x)cos(y) = 1 2 [cos(x + y)+cos(x y)] sin(x)cos(y) = 1 2 [sin(x + y)+sin(x y)] sin(x)sin(y) = 1 2 [cos(x + y) cos(x y)] r I (t) = 1 2 [ r I(t)cos(φ)+ r Q (t)sin(φ)] = 1 4 [(y I(t) h I (t) y Q (t) h Q (t)) cos(φ)+(y I (t) h Q (t)+y Q (t) h I (t))) sin(φ)] and r Q (t) = 1 4 [ (y I(t) h I (t) y Q (t) h Q (t)) sin(φ)+(y I (t) h Q (t)+y Q (t) h I (t)) cos(φ)] Lund University / Faculty / Department / Unit / Document / Date

36 Final result Now construct the two complex valued signals y c (t) y I (t)+jy Q (t) and r c (t) r I (t)+jr Q (t). By identyifing some terms we can conclude that r c (t) =y c (t) h c (t), with h c (t) h I (t)cos(φ)+h Q (t)sin(φ)+j(h Q (t)cos(φ) h I (t)sin(φ)). Lund University / Faculty / Department / Unit / Document / Date

37 Final result y (t) r (t ) This can be modeled in the complex baseband by Lund University / Faculty / Department / Unit / Document / Date

38 Final result h c (t) h I (t)cos(φ)+h Q (t)sin(φ)+j(h Q (t)cos(φ) h I (t)sin(φ)). Channel This can be modeled in the complex baseband by this Lund University / Faculty / Department / Unit / Document / Date

39 No channel case Leakage between the Inphase and the qudrature components! Lund University / Faculty / Department / Unit / Document / Date

40 No channel case Leakage between the Inphase and the qudrature components! In complex baseband, this shows up as a rotation! Lund University / Faculty / Department / Unit / Document / Date

41 Effect of ϕ What is the effect of ϕ? h c (t) h I (t)cos(φ)+h Q (t)sin(φ)+j(h Q (t)cos(φ) h I (t)sin(φ)). Energy of the impulse response = Energy is independent of ϕ!! Doesn t matter if tx and rx are not synchronous Lund University / Faculty / Department / Unit / Document / Date

42 Summary We can always work in the complex baseband domain with the input/ output relation And we do not care about ϕ (it must be estimated though) Lund University / Faculty / Department / Unit / Document / Date