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1B Paper 6: Communications Pulse Amplitude Modulation (Recap) 1B Paper 6: Communications Handout 5: Digital Passband Modulation Ramji Venkataramanan Signal Processing and Communications Lab Department of Engineering [email protected] Lent Term 2016 1 / 22 Pulse Amplitude Modulation (recap) Recall that the PAM signal carrying the information symbols X1, X2,... is x (t) = X p(t kT ) b k − ∑k x (t) is a baseband signal • b (Recall: its bandwidth is the same as that of the pulse p(t)) If the channel is a baseband channel, e.g., ethernet cable, we • can directly transmit xb(t) But most channels are passband — we are only allowed to transmit our signal over a fixed frequency band centred around a carrier frequency fc . E.g., a wireless channel may have fc = 2 GHz and channel bandwidth = 10 MHz How do you “up-convert” the PAM signal to passband ? 2 / 22 Baseband to Passband The natural thing to do is to modulate the amplitude of a high-frequency carrier with xb(t): x(t) = xb(t) cos(2πfc t) = Xk p(t kT ) cos(2πfc t) [ − ] ∑k x(t) y(t) n(t) For lack of a standard name, we’ll call this passband • modulation scheme up-converted PAM. Note the similarity with analogue modulation (DSB-SC): here • the information signal is a “digital” waveform, which is determined by the constellation symbols and a pulse. 3 / 22 Up-converted PAM with a rectangular pulse 1 for t (0, T ] p(t) = √T ∈ { 0 otherwise is also called “Amplitude Shift Keying” (ASK). In this case, the transmitted passband waveform is x(t) = X p(t kT ) cos(2πf t) k − c ∑k 1 = Xk cos(2πfc t) for t [kT , (k + 1)T ) √T ∈ where X is chosen from a PAM constellation, say A, A . k {− } Note that a rectangular pulse (and consequently, ASK) is not bandwidth efficient. (You may encounter the terminology ASK in past Tripos questions. You will not be examined on it, but recognise that it is just up-converted PAM with a rectangular pulse shape.) 4 / 22 Demodulation of up-converted PAM at the Rx First, down-convert via product modulator + low-pass filter v(t) Low-pass y(t) y (t) × filter [−W; W ] b cos(2πfct) y (t) = x (t) + n (t) = X p(t kT ) + n (t) b b b k − b ∑k where nb(t) is baseband noise. Next demodulate baseband waveform yb(t). We already know how to do this: pass yb(t) through matched filter, and then sample at times mT m Z { } ∈ t = mT Filter r(t) yb(t) r(mT ) h(t) = p(−t) 5 / 22 Detection The sampled output of the matched filter at time mT is • Ym = Xm + Nm where N = n (τ)p(τ mT ) dτ (see prev. handout for details) m b − N is Gaussian∫ with zero mean and variance σ2 • m (σ2 can be empirically estimated) Maximum-likelihood detection: Choose Xˆ to be the • m constellation symbol closest to Ym 6 / 22 Spectrum of up-converted PAM The transmitted waveform x(t) = xb(t) cos(2πfc t) has spectrum 1 X (f ) = [X (f f ) + X (f + f )] 2 b − c b c Note that x (t) = X p(t kT ) b k − ∑k is a real signal since both the pulse p(t) and the symbols X are { k } real-valued X ( f ) = X (f ) ⇒ b − b∗ The spectrum X (f ) for f 0 determines X (f ) for f < 0 • ≥ jXb(f)j f −W 0 W Lower Sideband Upper Sideband jX(f)j −fc − W −fc −fc + W fc − W fc fc + W Sending both sidebands is redundant, since all the information • is contained in one 7 / 22 A bandwidth-efficient alternative to up-converted PAM One way to save bandwidth is to send only one sideband, just • like SSB-SC (see Amplitude Modulation handout) But since we are transmitting digital information, there is a • better way: make the information symbols complex-valued The baseband waveform is again x (t) = X p(t kT ) b k − ∑k but now the constellation from which the symbols Xk are drawn is complex-valued (i.e., Xk are now two-dimensional) xb(t) is now complex, but the passband signal we transmit has to be real. It is generated as j2πfc t x(t) = Re xb(t) e = Re([x (t))cos(2π]f t) Im(x (t))sin(2πf t) b c − b c This is called Quadrature Amplitude Modulation (QAM) 8 / 22 Quadrature Amplitude Modulation The upconverted QAM waveform that we transmit is x(t) = p(t kT )[Re(X ) cos(2πf t) Im(X ) sin(2πf t)] − k c − k c ∑k = p(t kT ) X cos(2πf t + ϕ ) − | k| c k ∑k where X and ϕ denote the magnitude and phase of the | k | k complex symbol Xk Thus, one can understand QAM in two ways: 1. QAM has two carriers, the cosine carries Re(Xk ) and the sin carries the Im(Xk ). 2. In QAM, the information symbol modulates both the amplitude and phase of the carrier; in up-converted PAM the information symbol is real and only modulates the amplitude Pulse shape p(t) is the same as that for PAM • The main difference between QAM and PAM is the • constellation. In PAM, Im(Xk ) = 0. 9 / 22 Some typical QAM Constellations Im(Xk) Im(Xk) Re(Xk) Re(Xk) −A A −A A BPSK QPSK −A A 8-PSK 16-QAM In “Phase Shift Keying”, the magnitude of Xk is constant, and the information is in the phase of the symbol. In an constellation with M symbols, each symbol corresponds to log M bits 2 10 / 22 Average Energy per Symbol Im(Xk) Re(Xk) −A A −A A −A A BPSK QPSK 8-PSK 2 For all the PSK constellations, average symbol energy Es = A Im(Xk) d Average energy per symbol for 16-QAM Re(X ) k 40d2 2 Es = 16 = 2.5d 16-QAM Average energy per bit E = E /log M b s 2 11 / 22 The transmitted waveform is x(t) = p(t kT ) X r cos(2πf t) X i sin(2πf t) − k c − k c k ∑ [ ] r i where Xk := Re(Xk ) and Xk := Im(Xk ) x(t) y(t) n(t) The energy per symbol is important because the power of • x(t) Es ∝ T At the Rx, we have to down-convert y(t) to a baseband • waveform via product modulators and low-pass filter For QAM, we need two product modulators, one for the • cosine and the other for sine 12 / 22 At the receiver, we have y(t) = p(t kT ) X r cos(2πf t) X i sin(2πf t) + n(t) − k c − k c k ∑ [ ] Low-pass y(t) yr(t) × filter [−W; W ] cos(2πfct) Low-pass y(t) yi(t) × filter [−W; W ] − sin(2πfct) After down-converting, we get y r (t) = X r p(t kT ) + nr (t) k − ∑k y i (t) = X i p(t kT ) + ni (t) k − ∑k where nr (t) and ni (t) are filtered (baseband) versions of n(t). Next perform matched-filter demodulation of y r (t) and y i (t) 13 / 22 Demodulation t = mT Filter r(t) yr(t) Y r h(t) = p(−t) m t = mT Filter r(t) yi(t) Y i h(t) = p(−t) m The sampled outputs of the matched filters for m = 0, 1,... • are: r r r Ym = Xm + Nm i i i Ym = Xm + Nm It can be shown that Nr and Ni are each independent • m m Gaussians distributed as (0, σ2) for each m N For each m, we now have to detect the complex-valued • r i r i constellation symbol Xm = (Xm, Xm) from Ym = (Ym, Ym) 14 / 22 Detection The discrete-time channel is Y = X + N where X , Y , N are now two-dimensional vectors, i.e., complex numbers with X = (X r , X i ) and Y = (Y r , Y i ) The optimal detection rule (assuming all constellation symbols are equally likely) is the maximum-likelihood detection rule Xˆ = arg max f (Y x) x | ∈C “Choose the symbol from which y is most likely to have occurred” The conditional distribution of Y = (Y r , Y i ) given x = (xr , xi ) is 1 (Y r xr )2/2σ2 1 (Y i xi )2/2σ2 f (Y x) = e− − e− − | √2πσ2 √2πσ2 1 [(Y r xr )2+(Y i xi )2]/2σ2 = 2πσ2 e− − − 15 / 22 The optimal detector is therefore Xˆ = arg min (Y r xr )2 + (Y i xi )2 = arg min Y x 2 x − − x | − | ∈C ∈C Choose the constellation symbol x closest to observed output Y (Same detection principle as PAM, but the symbols are complex in QAM) Example 1: The decision regions for QPSK Im(Y ) X^ = p1 p4 p1 A Re(Y ) p3 p2 16 / 22 Xˆ = arg min Y x 2 x | − | ∈C Example 2: The decision regions for 8-PSK Im(Y ) p7 p6 p8 X^ = p1 A Re(Y ) p5 p1 p4 p2 p3 Can similarly sketch the decision regions for other • constellations We can also calculate the probability of detection error • P = P(Xˆ = X ) for various constellations e ̸ 17 / 22 Im(Xk) Im(Xk) d d Re(Xk) Re(Xk) −A A 16-QAM 8-PSK We won’t do an exact calculation of QAM probability of error Pe , but can be shown that: P is a function depending on d , where d is the • e Q− σ separation between adjacent constellation points and σ2 is the variance of the Gaussian noise P decays exponentially with d2/σ2 • e Suppose we increase the number of constellation points, e.g., 16 QAM 64 QAM 256 QAM, while keeping E constant → → s (Es = average energy per symbol) : Transmission rate increases: 4 bits/symbol 6 bits/sym.
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