Outline. EECS 3213 Fall Sebastian Magierowski York University. Review Passband Modulation. Constellations ASK, FSK, PSK.

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1 EECS 3213 Fall 2014 L12: Modulation Sebastian Magierowski York University 1 Outline Review Passband Modulation ASK, FSK, PSK Constellations 2 1

2 Underlying Idea Attempting to send a sequence of digits through a continuous channel channel Not easy modulerize the design 3 Continuous-Time Communication Map symbols into analog waveforms that match characteristics of channel spectral shape location (carrier) channel coder symbols tx filter channel rx filter waveforms channel decoder 4 2

3 Simple Baseband Modulator (Pulse Shaper) Basic implementation coder tx filter channel rx filter decoder coder sinc FIR channel sampler decoder Time and spectral characteristics (Nyquist criterion satisfied) 5 [Razavi12] Spectral Efficiency What is the bit- relative to the spectrum used: bit- v = bandwidth 2f s log 2 (M) bits v = [ s Hz ] Simple 2-level baseband modulation 2 bits/s Hz or 2 symbols/s Hz f s 6 3

4 Passband Modulation What if the channel is not baseband? 0 f c f s f c f c + f s Replace DC 1/0 representation time with AC representation time Have 3 ways to modulate a sinusoid: A c cos(ω c t + φ) 7 Modulator in a Communication System line coder tx filter modulator channel demodulator rx filter sampler line decoder 8 4

5 Amplitude Modulation A c cos(ω c t + φ) Map bits into amplitude of sinusoid: 1 send sinusoid; 0 no sinusoid Demodulator looks for signal vs. no signal Information Amplitude Shift Keying +1-1 t 9 Frequency Modulation A c cos(ω c t + φ) Map bits into frequency: 1 send frequency f c + δ ; 0 send frequency f c - δ Demodulator looks for power around f c + δ or f c - δ Information Frequency Shift Keying +1-1 t 10 5

6 (Binary) Phase Modulation Information Binary Phase Shift Keying (BPSK) -1 Map bits into phase of sinusoid: 1 send A cos(2πft), i.e. phase is 0 0 send A cos(2πft+π), i.e. phase is π t Equivalent to multiplying cos(2πft) by +A or -A 1 send A cos(2πft), i.e. multiply by 1 0 send A cos(2πft+π) = - A cos(2πft), i.e. multiply by -1 We will focus on phase modulation 11 AM Modulator Simplest and most dominant case, multiply symbol by a carrier Modulate cos(2πf c t) by multiplying by A k for T seconds: A k x Y i (t) = A k cos(2πf c t) cos(2πf c t) Transmitted signal during kth interval A k Y i (t) coder modulator channel demodulator decoder 12 6

7 AM Demodulator Demodulate (recover A k ) by multiplying by 2cos(2πf c t) for T seconds and lowpass filtering (smoothing): Y i (t) = A k cos(2πf c t) Received signal during kth interval x 2cos(2πf c t) Lowpass Filter (Smoother) X i (t) 2A k cos 2 (2πf c t) = A k {1 + cos(2π2f c t)} coder modulator channel Y i (t) demodulator X i (t) decoder 13 Example of Modulation Information Baseband Signal +A -A Modulated Signal x(t) +A -A A cos(2πft) -A cos(2πft) 14 7

8 Example of Demodulation A {1 + cos(4πft)} -A {1 + cos(4πft)} After multiplication at receiver x(t)2cos(2πf c t) +A -A Baseband signal discernable after smoothing +A -A Recovered Information Signaling Rate and Transmission Bandwidth Fact from modulation theory: If Baseband signal x(t) with bandwidth f s /2 Hz then f s /2 f Modulated signal x(t)cos(2πf c t) has bandwidth f s Hz f c f s /2 f c f c + f s /2 f If bandpass channel has bandwidth f s Hz, It s baseband version has bandwidth of f s /2 Hz, so modulation system supports f s /2 x 2 = f s symbols/second That is, f s symbols/second per f s Hz = 1 symbols/s Hz Recall baseband transmission system supports 2 symbols/s Hz!!! 16 8

9 Quadrature Amplitude Modulation (QAM) QAM uses two-dimensional signaling A k modulates in-phase cos(2πf c t) B k modulates quadrature phase cos(2πf c t + π/4) = sin(2πf c t) Transmit sum of in-phase & quadrature phase components A k x Y i (t) = A k cos(2πf c t) cos(2πf c t) + Y(t) B k x Y q (t) = B k sin(2πf c t) Transmitted Signal sin(2πf c t) Y i (t) and Y q (t) both occupy the bandpass channel QAM sends 2 symbols/s Hz 17 QAM Signal Mapping Serial-to-parallel converter distributes input bits Two bits sent simultaneously 18 [Razavi12] 9

10 QAM Demodulation Y(t) x Lowpass filter (smoother) A k 2cos(2πf c t) x 2sin(2πf c t) 2A k cos 2 (2πf c t)+2b k cos(2πf c t)sin(2πf c t) = A k {1 + cos(4πf c t)}+b k {0 + sin(4πf c t)} Lowpass filter (smoother) B k smoothed to zero 2B k sin 2 (2πf c t)+2a k cos(2πf c t)sin(2πf c t) = B k {1 - cos(4πf c t)}+a k {0 + sin(4πf c t)} smoothed to zero 19 Signal Constellations Convenient to write modulated signals in quadrature form y(t) =y i (t) cos(2 f c t)+y q (t)sin(2 f c t) in-phase and quadrature components And to plot it as a signal constellation in the complex place Plot samples of in-phase terms along real axis Plot samples of quadrature terms along imaginary axis 20 10

11 BPSK Constellation For example y BPSK (t) = a n cos(ω c t), a n = ±1 only the in-phase term is present two possible amplitudes 21 [Razavi12] 4-QAM Constellation Both terms present In-phase and quadrature y QAM (t) = a n cos(ω c t) + b n sin(ω c t), a n,b n = ±1 b n (-a,b) (a, b) a n (-a,-b) (a,-b) 22 11

12 Quadrature Amplitude Modulation Quadrature Amplitude Modulation Larger QAM Constellations Quadrature With 4-QAM QPSK 2 bits /symbol Symbol Rate = 1/2 bit QPSK 2 bits /symbol Symbol Rate = 1/2 bit 64 QAM 6 bits /symbol Symbol Rate = 1/6 bit Amplitude Modulation bits represented per symbol: N = 2 Constellation points: QPSK M = 2N = 4 16 QAM 16 QAM 4 bits /symbol Quadrature Amplitude Modulation Many other possibilities (rectangular constellation) Quadrature Amplitude Modulation N = 3,4,5,... 4 bits/symbol /symbol 2 bits Symbol Rate = 1/2 bit Symbol Rate = 1/4 bit 64 QAM 6 bits /symbol Symbol Rate = 1/6 bit Symbol Rate = 1/4 bit 64 QAM 6 bits /symbol Symbol Rate = 1/6 bit QPSK N preferred (easier coder) Even 2 bits /symbol Symbol Rate = 1/2 bit QPSK 2 bits /symbol Symbol Rate = 1/2 bit M=4 L=2 16 QAM 4 bits /symbol M = 16 L=4 QAM Symbol 16 Rate = 1/4 bit 4 bits /symbol Symbol Rate = 1/4 bit 16 QAM 32 QAM 4 bits /symbol 5 bits /symbol Symbol Rate = 1/5 bit Symbol Rate = 1/4 bit 32 QAM 5 bits /symbol Symbol Rate = 1/5 bit M = 32 L=6 M = 64 L=8 32 QAM 5 bits /symbol Symbol Rate = 1/5 bit 64 QAM 6 bits /symbol Symbol Rate = 1/6 bit 64 QAM 6 bits /symbol Symbol Rate = 1/6 bit 128 QAM 7 bits /symbol Symbol Rate = 1/7 bit M = 128 L = QAM 7 bits /symbol Symbol Rate = 1/7 bit Increasing N requires more power EECS 3213, F14 Page 9 32 QAM 5 bits /symbol Symbol Rate = 1/5 bit 32 QAM Page 9 5 bits /symbol Symbol Rate = 1/5 bit Page 9 Page 9 Page 9 L12: Modulation 128 QAM 7 bits /symbol Symbol Rate = 1/7 bit QAM 7 bits /symbol Spectrum Analyzers Seminar X-Series Spectrum Analyzers X-Series Seminar 2010 Symbol Rate = 1/7 bit 128 QAM 7 bits /symbol Symbol Rate = 1/7 bit X-Series Spectrum Analyzers Seminar 2010 Phase-Shift Keying (PSK) X-Series Spectrum Analyzers Seminar 2010 A common variant X-Series Spectrum Analyzers Seminar 2010 Conceptually, vary phase of signal based on symbol (already saw this) ancos(ωct + kπ/n), k = integer bn 4-PSK QPSK QAM 4-QAM bn an an 8-PSK π/4-qpsk Constant amplitude EECS 3213, F14 L12: Modulation 24 12

13 π/4-qpsk Modulator Just 2 QAM modulators phase shifted relative to each other top modulator has both cos/sin always active bottom modulator has either cos/sin active 25 [Razavi12] 13

Lecture #11 Overview. Vector representation of signal waveforms. Two-dimensional signal waveforms. 1 ENGN3226: Digital Communications L#

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