Digital Band-pass Modulation PROF. MICHAEL TSAI 2011/11/10

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1 Digital Band-pass Modulation PROF. MICHAEL TSAI 211/11/1

2 Band-pass Signal Representation a t g t General form: 2πf c t + φ t g t = a t cos 2πf c t + φ t Envelope Phase Envelope is always non-negative, or we can switch the phase by 18 degree This is called the canonical representation of a bandpass signal

3 Band-pass Signal Representation g t = a t cos 2πf c t + φ t can be re-arranged into g t = g I t cos 2πf c t g Q t sin 2πf c t g I t = a t cos φ t and g Q t = a t sin φ t g I t and g Q t are called inphase and quadrature components of the signal g(t), respectively Then a t = g I 2 t + g Q 2 t and φ t = tan 1 g Q t g I t

4 Band-pass Signal Representation We can also represent g(t) as g t = Re g t exp j2πf c t g t = g I t + jg Q t a t φ t g t g t is called the complex envelope of the band-pass signal. This is to remove the annoying exp j2πf c t in the analysis.

5 5 Sinusoidal Functions Fourier Transform Complex exponential function F exp j2πf c t = δ(f f c ). Sinusoidal functions: cos 2πf c t = 1 2 exp j2πf ct + exp j2πf c t F cos 2πf c t = 1 2 δ f f c + δ f + f c sin 2πf c t = 1 2 exp j2πf ct exp j2πf c t F sin 2πf c t = 1 2 δ f f c δ f + f c G(f) G(f) G(f) f c f f c f c f f c f c f

6 g t = g I t cos 2πf c t + g Q t sin 2πf c t Band-pass Signal Transmitter g I (t) Maps each bit into g I t and g Q t cos (2πf c t) Message Source Signal Encoder 9 degree shift Σ + + Band-pass Signal g(t) sin (2πf c t) g Q (t)

7 Assumption The channel is linear: flat-fading channel. B c > B s Negligible distortion to g(t) The received signal s(t) is perturbed by AWGN noise w(t) ~N, N 2 N 2 is the PSD of the noise and also its variance (since it s white)

8 AWGN Channel Channel path loss or attenuation Add additive Band-pass Signal g(t) Channel A c + + Σ Received Signal plus Noise s t + w(t) White Gaussian Noise w t x t = s t + w t = A c g t + w(t)

9 Received Signal plus Noise Band-pass Signal Receiver Filters out outof-band signals and noises Band-pass Filter x t = s t + n(t) g Q (t) Mixer 9 degree shift Low-pass Filter cos (2πf c t) sin (2πf c t) Low-pass Filter 1 2 A cg I t + n I t Signal Detector Message Sink 1 2 A cg Q t + n Q t

10 Band-pass Filter Band-pass Filter The band-pass filter at the frontend filters out out-of-band signals and noises 1. Signal s(t) is within the band not affected 2. White noise w(t) becomes narrowband noise n(t) Much smaller since now we only include noises within the band Still white over the bandwidth of the signal 3. Other signal (out-of-band) is filtered out Other signals s(t) Noise f c B 2 f c f c + B 2 f

11 Up-conversion (TX) cos (2πf c t) In time domain A t exp jθ t A c cos 2πf c t In frequency domain Convolution f f c f c f f c f c f

12 Low-pass Filter Down-conversion (RX) cos (2πf c t) In time domain s t A c cos (2πf c t + φ) A c cos 2πf c t In frequency domain Convolution Low-pass Filter f c f c f f c f c f 2f c 2f c f

13 Signal Detector Signal Detector The signal detector: Observes complex representation of the received signal, g I t + n I t + j[g Q t + n Q t ], For a duration of T seconds (symbol/bit period) And the make its best estimate of the corresponding transmitted signal g I t + jg Q t g I t + jg Q t bit stream

14 Time synchronization To simplify, we assume we have time synchronization between the TX and the RX Where does each symbol start and end? t Symbol boundary needs to be same for TX and RX In practice, a timing recovery circuit is required

15 Coherent & non-coherent Sometimes, the receiver is phase-locked to the transmitter That means, the in TX and in RX generate cos(2πf c t) with no phase difference. RX looks at the received signal to lock onto TX s carrier When that happens, we say The receiver is a coherent receiver, carrying out coherent detection Otherwise, we say The receiver is a non-coherent receiver, carrying out non-coherent detection

16 Basic forms of digital modulation Amplitude Shift Keying Frequency Shift Keying Phase Shift Keying Keying == Switching

17 (Binary) Amplitude Shift Keying (BASK) Fixed Amplitude/fixed frequency for a duration of to represent 1 No transmission to represent Or, more formally, s 1 t = A c cos(2πf c t) s t = for a duration of

18 (Binary) Phase Shift Keying (BPSK) Same amplitude, same frequency Send the original carrier to represent 1 Send an inverted carrier (phase difference 18 degrees) to represent Or, more formally, s 1 t = A c cos(2πf c t) s t = A c cos 2πf c t + π = A c cos(2πf t)

19 (Binary) Phase Shift Keying (BPSK)

20 (Binary) Frequency Shift Keying (BFSK) Same amplitude Send a carrier at f 1 to represent 1 Send a carrier at f to represent Or, more formally, s 1 t = A c cos(2πf 1 t) s t = A c cos(2πf t) for a duration of

21 (Binary) Frequency Shift Keying (BFSK) Usually we have f 1 = f c + Δf, f = f c Δf s 1 t = A c cos[2π f c + Δf t] s t = A c cos[2π f c Δf t] Then, s 1 t = Re A c exp j2π f c + Δf t = g t exp j2πf c t s t = Re A c exp j2π f c Δf t = g t exp j2πf c t So, For 1, g t = g I t + jg Q t = A c exp [ j2πδft] For,g t = g I t + jg Q t = A c exp [+j2πδft] Q I

22 Coherent Detection of FSK and PSK signals Since f c is large compared to 1 (symbol rate, or bit rate), we can say that the same signal energy E b is transmitted in a bit interval : E b = s 2 t dt = s 2 1 t dt = A c 2 2

23 Two-path correlation receiver (general case) s 1 (t) Correlator: see how similar x t and s 1 (t) are x(t): received signal x(t) dt dt + Σ l Choose 1 if l > Detection device Otherwise, choose s (t) Correlator: see how similar x t and s (t) are

24 Coherent Detection w(t): AWGN, N, N 2 H : x t = s t + w(t) H 1 : x t = s 1 t + w(t) Receiver output: Decision level: l = x t s 1 t s t dt If l is larger than 1, than x(t) is more similar to s 1 (t) If l is smaller than 1, than x(t) is more similar to s (t)

25 Coherent Detection l = x t s 1 t s t dt H 1 : l = s 1 t s 1 t s t dt w t s 1 t s t dt Since the noise w(t) is zero-mean, L: the random variable whose value is l E L H 1 = s 1 t s 1 t s t dt = E b (1 ρ) ρ: the correlation coefficient of the signals s (t) and s 1 t ρ = s t s 1 t dt s 2 t dt s 2 1 t dt 1 2 = 1 s E t s 1 t dt b ρ 1

26 Coherent Detection Similarly, E L H = E b 1 ρ L s variance is the same for H 1 and H. Since s 1 (t) and s (t) is deterministic given the transmitted bit, we have Var L = E L E L 2 = E w t w u s 1 t s t s 1 u s u dt du = E w t w u s 1 t s t s 1 u s u dt du = δ(t u) s 1 t s t s 1 u s u dt du

27 = δ(t u) s 1 t s t s 1 u s u dt du = N 2 s 1 t s t 2 dt = N E b (1 ρ) Therefore, we know that L conditioned on H is a Gaussian distributed random variable: N E b 1 ρ, N E b 1 ρ

28 Q Function Q function is defined over the CDF of Gaussian distribution N(, 1) Q x = 1 2π x exp u2 2 du = 1 Φ(x) CDF of N(, 1) f(u) N(,1) s PDF x u Integration (Area under the curve)

29 Bit Error Rate L H 1 ~N E b 1 ρ, N E b 1 ρ ς 2 ~N E b 1 ρ E b 1 ρ Integration (Area under the curve) How to express this area with Q function? Divide u by N E b 1 ρ Shift left by E b 1 ρ N, N E b 1 ρ N(, 1) E b 1 ρ E b 1 ρ N E b 1 ρ

30 Bit Error Rate N(, 1) N(, 1) E b 1 ρ N E b 1 ρ E b 1 ρ N E b 1 ρ P e = Q E b 1 ρ For BPSK, ρ = 1 For BFSK, ρ = N P e = Q 2E b N P e = Q E b N

31 Signal Space - BPSK sin (2πf c t) Quadrature Bit A c Energy Noise s 1 t = A c cos (2πf c t) s t = A c cos (2πf c t) Bit 1 Inphase A c cos (2πf c t)

32 Signal Space - QPSK sin (2πf c t) Quadrature Bit 1 Bit 11 A c 2 s 11 t = A c cos 2πf c t + π 4 s 1 t = A c cos 2πf c t + 3π 4 A c A c 2 φ A c 2 Inphase cos (2πf c t) s t = A c cos 2πf c t + 5π 4 s 1 t = A c cos 2πf c t + 7π 4 Bit A c 2 Bit 1

33 M-ary Modulation Quadrature 4-PSK 8-PSK Quadrature Inphase A c φ Inphase Increasing M would increase the data rate (given the same signal bandwidth)

34 M-ary Modulation 16-QAM Quadrature Inphase

35 M-PAM BER versus SNR SNR = E b ς n 2 BER P e M Q 3 SNR M 2 1

36 M-QAM BER versus SNR BER P e M Q 3 SNR M 1 BER P e M Q 3 SNR 31 M 32 1

37 M-PSK BER versus SNR BER P e 2 Q 2 SNR sin π M

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