EE445S Real-Time Digital Signal Processing Lab Spring 2014 Quadrature Amplitude Modulation (QAM) Transmitter Prof. Brian L. Evans Dept. of Electrical and Computer Engineering The University of Texas at Austin Lecture 15 Introduction • Digital Pulse Amplitude Modulation (PAM) Modulates digital information onto amplitude of pulse May be later upconverted (e.g. to radio frequency) • Digital Quadrature Amplitude Modulation (QAM) Two-dimensional extension of digital PAM Baseband signal requires sinusoidal amplitude modulation May be later upconverted (e.g. to radio frequency) • Digital QAM modulates digital information onto pulses that are modulated onto Amplitudes of a sine and a cosine, or equivalently Amplitude and phase of single sinusoid 15 - 2 Review Amplitude Modulation by Cosine Y1 (w ) • y1(t) = x1(t) cos(wc t) 1 1 X 1 (w + wc ) + X 1 (w - wc ) 2 2 Assume x1(t) is an ideal lowpass signal with bandwidth w1 Assume w1 << wc Y1(w) is real-valued if X1(w) is real-valued X1(w) Y1(w) ½X1(w - wc) ½X1(w + wc) 1 ½ -w1 0 w1 Baseband signal w -wc - w1 -wc -wc + w1 0 wc - w1 Upconverted signal wc wc + w1 • Demodulation: modulation then lowpass filtering 15 - 3 w Review Amplitude Modulation by Sine j j Y2 (w ) X 2 (w + wc ) - X 2 (w - wc ) 2 2 • y2(t) = x2(t) sin(wc t) Assume x2(t) is an ideal lowpass signal with bandwidth w2 Assume w2 << wc Y2(w) is imaginary-valued if X2(w) is real-valued X2(w) Y2(w) -j ½X2(w - wc) j ½X2(w + wc) 1 j -w2 0 w2 Baseband signal w ½ wc – w2 -wc – w2 -wc wc wc + w2 -wc + w2 -j ½ Upconverted signal • Demodulation: modulation then lowpass filtering 15 - 4 w Baseband Digital QAM Transmitter • Continuous-time filtering and upconversion Impulse modulator i[n] Index Bits 1 Serial/ parallel converter gT(t) Pulse shapers (FIR filters) Map to 2-D constellation J q[n] Q Impulse modulator s(t) Delay Local Oscillator + 90o gT(t) d -d d -d 4-level QAM Constellation I Delay matches delay through 90o phase shifter Delay required but often omitted in diagrams 15 - 5 Phase Shift by 90 Degrees • 90o phase shift performed by Hilbert transformer cosine => sine sine => – cosine 1 1 cos( 2 f 0 t ) ( f + f 0 ) + ( f - f 0 ) 2 2 j j sin( 2 f 0 t ) ( f + f 0 ) - ( f - f 0 ) 2 2 • Frequency response H ( f ) - j sgn( f ) Magnitude Response Phase Response H ( f ) | H( f )| 90o f All-pass except at origin f -90o 15 - 6 Hilbert Transformer • Continuous-time ideal Hilbert transformer • Discrete-time ideal Hilbert transformer H (w ) - j sgn( w ) H ( f ) - j sgn( f ) 1/( t) if t 0 h(t) = h[n] = 0 if t = 0 h(t) 2 sin 2 (n / 2) n if n0 0 if n=0 h[n] t n Even-indexed samples are zero 15 - 7 Discrete-Time Hilbert Transformer • Approximate by odd-length linear phase FIR filter Truncate response to 2 L + 1 samples: L samples left of origin, L samples right of origin, and origin Shift truncated impulse response by L samples to right to make it causal L is odd because every other sample of impulse response is 0 • Linear phase FIR filter of length N has same phase response as an ideal delay of length (N-1)/2 (N-1)/2 is an integer when N is odd (here N = 2 L + 1) • Matched delay block on slide 15-5 would be an ideal delay of L samples 15 - 8 Baseband Digital QAM Transmitter i[n] Impulse modulator Index Bits 1 Serial/ parallel converter J Pulse shapers (FIR filters) i[n] Index Bits 1 Serial/ parallel converter J L samples/symbol (upsampling factor) Impulse modulator L s(t) Delay Local Oscillator Map to 2-D constellation q[n] 100% discrete time gT(t) + 90o gT(t) gT[m] s[m] Pulse shapers (FIR filters) Map to 2-D constellation q[n] L gT[m] cos(w0 m) sin(w0 m) + s(t) D/A 15 - 9 Performance Analysis of PAM • If we sample matched filter output at correct time instances, nTsym, without any ISI, received signal x(nTsym ) s (nTsym ) + v(nTsym ) v(nT) ~ N(0; 2/Tsym) where transmitted signal is s (nTsym ) an (2i - 1)d for i = -M/2+1, …, M/2 3d v(t) output of matched filter Gr(w) for input of channel additive white Gaussian noise N(0; 2) Gr(w) passes frequencies from -wsym/2 to wsym/2 , where wsym = 2 fsym = 2 / Tsym d -d -3 d 4-level PAM • Matched filter has impulse response gr(t) Constellation 15 - 10 Performance Analysis of PAM d PI (e) P( v(nTsym ) d ) 2 Q Tsym d PO- (e) P(v(nTsym ) d ) Q Tsym d PO+ (e) P(v(nTsym ) -d ) P(v(nTsym ) d ) Q Tsym • Decision error for inner points • Decision error for outer points • Symbol error probability M -2 1 1 2( M - 1) d P ( e) PI (e) + PO+ (e) + PO- (e) Q Tsym M M M M 8-level PAM Constellation O- I I I I I I -7d -5d -3d -d d 3d 5d O+ 7d 15 - 11 Performance Analysis of QAM • If we sample matched filter outputs at correct time instances, nTsym, without any ISI, received signal x(nTsym ) s (nTsym ) + v(nTsym ) Q d • Transmitted signal s(nTsym ) an + j bn (2i - 1)d + j (2k - 1)d where i,k { -1, 0, 1, 2 } for 16-QAM -d d I -d 4-level QAM • Noise v(nTsym ) vI (nTsym ) + j vQ (nTsym ) Constellation For error probability analysis, assume noise terms independent and each term is Gaussian random variable ~ N(0; 2/Tsym) In reality, noise terms have common source of additive noise in channel 15 - 12 Performance Analysis of 16-QAM Q • Type 1 correct detection 3 P1 (c) P( vI (nTsym ) d & vQ (nTsym ) d ) ( ) ( P vI (nTsym ) d P vQ (nTsym ) d ( ( )) ( d T) ) d 1 - 2Q Tsym d 1 1 )) 3 3 2 2 ( 2Q ( 2 2 1 - P vI (nTsym ) d 1 - P vQ (nTsym ) d 2Q ( 2 I 2 1 1 2 2 3 16-QAM T) 2 1 - interior decision region 2 - edge region but not corner 3 - corner 15 -region 13 Performance Analysis of 16-QAM • Type 2 correct detection P2 (c) P(vI (nTsym ) d & vQ (nTsym ) d ) P(vI (nTsym ) d ) P( vQ (nTsym ) d ) d d 1 - Q Tsym 1 - 2Q Tsym • Type 3 correct detection Q 3 2 2 1 1 2 2 2 3 3 I 2 1 1 2 2 3 16-QAM P3 (c) P(vI (nTsym ) d & vQ (nTsym ) -d ) P(vI (nTsym ) d ) P(vQ (nTsym ) -d ) d 1 - Q Tsym 2 1 - interior decision region 2 - edge region but not corner 3 - corner 15 -region 14 Performance Analysis of 16-QAM • Probability of correct detection 2 4 4 d d P(c) 1 - 2Q Tsym + 1 - Q Tsym 16 16 + 2 8 d d 1 - Q Tsym 1 - 2Q Tsym 16 d 9 d 1 - 3Q Tsym + Q 2 Tsym 4 • Symbol error probability (lower bound) d 9 2 d P(e) 1 - P(c) 3Q Tsym - Q Tsym 4 • What about other QAM constellations? 15 - 15 Average Power Analysis 3d • Assume each symbol is equally likely • Assume energy in pulse shape is 1 • 4-PAM constellation Amplitudes are in set { -3d, -d, d, 3d } Total power 9 d2 + d2 + d2 + 9 d2 = 20 d2 Average power per symbol 5 d2 d -d -3 d 4-level PAM Constellation • 4-QAM constellation points Points are in set { -d – jd, -d + jd, d + jd, d – jd } Total power 2d2 + 2d2 + 2d2 + 2d2 = 8d2 Average power per symbol 2d2 Q d -d d -d 4-level QAM 15 - 16 Constellation I