Communication Systems Engineering Episode 1 Part 6 potx

Lectures 6&7 Modulation Eytan Modiano AA Dept. Eytan Modiano Slide 1 Modulation • Representing digital signals as analog waveforms • Baseband signals – Signals whose frequency components are concentrated around zero • Passband signals – Signals whose frequency components are centered at some frequency fc away from zero • Baseband signals can be converted to passband signals through modulation – Multiplication by a sinusoid with frequency fc Eytan Modiano Slide 2 Baseband signals • The simplest signaling scheme is pulse amplitude modulation (PAM) – With binary PAM a pulse of amplitude A is used to represent a “1” and a pulse with amplitude -A to represent a “0” • The simplest pulse is a rectangular pulse, but in practice other type of pulses are used – For our discussion we will generally assume a rectangular pulse • If we let g(t) be the basic pulse shape, than with PAM we transmit g(t) to represent a “1” and -g(t) to represent a “0” g(t) 1 => S(t) = g(t) A 0 => S(t) = -g(t) Tb Eytan Modiano Slide 3 M-ary PAM • Use M signal levels, A 1 …A M – Each level can be used to represent Log 2 (M) bits • E.g., M = 4 => A 1 = -3, A 2 = -1, A 3 = 1, A 4 = 3 – S i (t) = A i g(t) • Mapping of bits to signals Si b1b2 S 1 00 S 2 01 S 3 11 S 4 10 Eytan Modiano Slide 4 gt tT Signal Energy T T E m = ∫ 0 ((t)) 2 dt = ( A m ) 2 ∫ 0 () 2 dt = ( A m ) 2 E g S m g t • The signal energy depends on the amplitude • E g is the energy of the signal pulse g(t) • For rectangular pulse with energy E g => T E g = ∫ 0 A 2 dt = TA 2 => A = E g 2 / g(t)  E g / 2 0 ≤≤ () =   0 otherwise E g /2 Eytan Modiano T Slide 5 Symmetric PAM • Signal amplitudes are equally distant and symmetric about zero -7 -5 -3 -1 0 1 3 5 7 A m = (2m-1-M), m=1…M E M E ave = g ∑ (2 m − 1 − M ) 2 = E g ( M 2 − 1)/ 3 M m =1 Eytan Modiano Slide 6 Gray Coding • Mechanism for assigning bits to symbols so that the number of bit errors is minimized – Most likely symbol errors are between adjacent levels – Want to MAP bits to symbols so that the number of bits that differ between adjacent levels is mimimized • Gray coding achieves 1 bit difference between adjacent levels • Example M= 8 (can be generalized) A 1 000 A 2 001 A 3 011 A 4 010 A 5 110 A 6 111 A 7 101 Eytan Modiano A 8 100 Slide 7 Bandpass signals • To transmit a baseband signal S(t) through a bandpass channel at some center frequency f c , we multiply S(t) by a sinusoid with that frequency S m (t) U m (t)= S m (t)Cos(2πf c t) = A m g(t) Cos(2πf c t) Cos(2πf c t) F[Cos(2πf c t)] = (δ(f-f c )+δ(f+f c ))/2 Eytan Modiano -f c f c Slide 8 Passband signals, cont. F[A m g(t)] = depends on g() A m -w w F[A m g(t) Cos(2πf c t)] A m /2 A m /2 -f c f c Eytan Modiano Slide 9 Recall: Multiplication in time = convolution in frequency () Energy content of modulated signals ∞ ∞ E m = ∫ −∞ U 2 t dt = ∫ −∞ A m g 2 () m () 2 t Cos 2 ( 2 π f c t ) dt Cos 2 α = 1 + cos( 2 α ) 2 A 2 ∞ A 2 ∞ 2 () + 2 m ∫ −∞ gt Cos 2 ( 4 π f c t ) dtE m = 2 m ∫ −∞ gt 2 () A 2 m E = E + ≈ 0 m 2 g • The cosine part is fast varying and integrates to 0 • Modulated signal has 1/2 the energy as the baseband signal Eytan Modiano Slide 10 [...]... energy m =1 After modulation Eu = Es = Eg ( M 2 − 1) / 6 2 (M 2 − 1) Eb = average energy per bit = Eg 6Log2 ( M ) • Average energy per bit increases as M increases BW efficiency energy efficiency Eytan Modiano Slide 14 M Two-dimensional signals • Si = (Si1, Si2) • Set of signal points is called a constellation S4 =( -1, 1) S2 =( -1, -1) S1= (1, 1) S3 = (1, -1) • 2-D constellations are commonly used • Large... some K 3 Signal levels on each axis are 1 the same as for PAM E.g., 4 − QAM ⇒ 16 − QAM ⇒ Eytan Modiano Slide 16 x y Am , Am x y Am , Am -3 ∈ {+ / − 1} ∈ {+ / − 1, + / − 3} -1 1 -1 -3 3 Bandwidth occupancy of QAM • When using a rectangular pulse, the Fourier transform is a Sinc |G(f)| AT g(t) A T • -2/T -1/ T First null BW is still 2/T – – – – Eytan Modiano Slide 17 K = Log2(M) bits per symbol Rb = Log2(M)/T... Eytan Modiano Slide 11 Bandwidth occupancy g(t) G(f) = F[g(t)] G( f ) = ∫ ∞ g(t)e − j 2πft dt = −∞ ∫ T A Ae − j 2πft dt 0 T G( f ) = ( AT )Sinc(πfT )e − jπfT |G(f)| AT -2/T • Eytan Modiano Slide 12 -1/ T 1/ T First “null” bandwidth = 2 (1/ T) = 2/T 2/T Bandwidth efficiency • Rs = symbol rate = 1/ T – • BW = 2/T = 2Rs – • M=2 => bandwidth efficiency = 1/ 2 M=4 => bandwidth efficiency = 1 M=8 => bandwidth efficiency... Log2(M)/2 => “Same as for PAM” 1/ T 2/T Energy efficiency x y Esm = [( Am )2 + ( Am )2 ]Eg x E[( Am )2 ] = Es = y E[( Am )2 ] = K2 1 M 1 , = 3 3 2(M − 1) Eg 3 Transmitted energy = Es (M − 1) = Eg 2 3 Eb (QAM) = Energy / bit = • K= M (M − 1) Eg 3Log2 ( M ) Compare to PAM: Eb increases with M, but not nearly as fast as PAM Eytan Modiano Slide 18 (M 2 − 1) Eb (PAM) = Eg 6Log2 ( M ) Bandpass QAM • Modulate... symbols as far apart as possible • Common constellations – QAM: Quadrature Amplitude Modulation PAM in two dimensions Eytan Modiano Slide 15 – PSK: Phase Shift Keying Special constellation where all symbols have equal power Symmetric M-QAM { } x y x y Sm = ( Am , Am ), Am , Am ∈ + / − 1, + / − 3, , + / − ( M − 1) M is the total number of signal points (symbols) M signal levels on each axis 16 -QAM Constellation... Modiano Slide 13 Bandwidth efficiency = Rb/BW = log2(M)/T * (T/2) = log2(M)/2 BPS/Hz Example: – – – • • Log2(M) bits per symbol => Rb = bit rate = log2(M)/T bits per second Need to increase symbol energy level in order to overcome errors Tradeoff between BW efficiency and energy efficiency Energy utilization Eave = Eg M M ∑ (2m − 1 − M)2 =Eg (M 2 − 1) / 3, Eg = basic pulse energy m =1 After modulation... when multiplied by cos Eytan Modiano Slide 21 Demodulation, cont U (t )2Cos(2πfc t) = 2 A x g(t )Cos 2 (2πfc t) + 2 A y g(t ) cos(2πfc t)sin(2πfc t ) Cos 2 (α ) = 1 + cos(2α ) 2 => U (t )2Cos(2πfc t) = S x (t ) + S x (t ) cos(4πfc t) ≈ S x (t ) = A x g(t) Similarly, t U(t)2Sin(2πfc t) = 2 A x g(t)Cos(2πfc t)Sin(2πfc t ) + 2 A y g(t)sin 2 (2πfc t ) Sin 2 (α ) = 1 − cos(2α ) 2 => U (t )2Sin(2πfc t) = S... or sometimes as AQ, and AI for quadrature or in-phase components • The transmitted signal, corresponding to the mth symbol is: x y U m (t ) = Am g(t )Cos(2πfc t) + Am g(t )Sin(2πfc t), m = 1 M Eytan Modiano Slide 19 Modulator g(t) Binary data Map k bits Into one of 2k symbols Am=(Ax, Ay) Cos(2πfct) Um(t) Sin(2πfct) g(t) Eytan Modiano Slide 20 Demodulation: Recovering the baseband signals U(t) LPF Sxr(t)... levels – I.e., they lie on a circle or radius Es Es • Symbols can be equally spaced to minimize likelihood of errors • E.g., Binary PSK Es s1 • Eytan Modiano Slide 23 4-PSK (above) same as 4-QAM Es s2 M-PSK Aix = Cos(2πi / M), Aiy = Sin(2πi / M), m = 0, , M − 1 x y U m (t ) = g(t ) Am Cos(2πfc t) − g(t ) Am Sin(2πfc t) Notice : Cos(α )Cos( β ) = Sin(α )Sin( β ) = Cos(α − β ) + Cos(α + β ) 2 Cos(α −... )Sin( β ) = Cos(α − β ) + Cos(α + β ) 2 Cos(α − β ) − Cos(α + β ) 2 Hence, U m (t) = g(t)Cos(2πfc t + 2πm / M) φ m = 2πm / M = phases shift of m th symbol U m (t ) = g(t )Cos(2πfc t + φ m ), m = 0 M − 1 Eytan Modiano Slide 24 M-PSK Summary • Constellation of M Phase shifted symbols – – • All have equal energy levels K = Log2(M) bits per symbol Modulation: g(t) Binary data Map k bits Into one of 2k symbols . coding achieves 1 bit difference between adjacent levels • Example M= 8 (can be generalized) A 1 000 A 2 0 01 A 3 011 A 4 010 A 5 11 0 A 6 11 1 A 7 10 1 Eytan Modiano A 8 10 0 Slide. PAM Eg x y ,4 − QAM ⇒ A m , A m ∈ { + / − 1 } x y 16 − QAM ⇒ A m , A m ∈ { + / − 1, + / − 3 } Eytan Modiano Slide 16 -3 -1 1 3 -3 -1 1 3 Bandwidth occupancy of QAM • When using. = (S i1 , S i2 ) • Set of signal points is called a constellation S 1 = (1, 1) S 2 =( -1, -1) S 3 = (1, -1) S 4 =( -1, 1) • 2-D constellations are commonly used • Large constellations can