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 f c 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 d t = ( A m ) 2 ∫ 0 () 2 d t = ( 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 d t = 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(2 π f c t) = A m g(t) Cos(2 π f c t) Cos(2 π f c t) F[Cos(2 π f 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(2 π f 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 d t = ∫ ?∞ 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 ) dt E 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 1 0 Demodulation ? How do we recover the baseband signal? U m (t)= S m (t)Cos(2 π f c t) = A m g(t) Cos(2 π f c t) U(t) LPF S r (t) 2Cos(2 π f c t) U(t)2Cos(2 π f c t)= 2S(t)Cos 2 (2 π f c t) = S(t) + S(t)Cos(4 π f c t) The high frequency component is rejected by the LPF and we are left with S(t). Eytan Modiano Slide 1 1 jf Bandwidth occupancy G(f) = F[g(t)] g(t) A ∞ T Gf 2 π () = ∫ ?∞ g ( t ) e ? jf t d t = ∫ 0 A e ? j 2 π f t d t Gf () = ( AT ) Sinc ( π fT ) e ? π T T |G(f)| AT -2/T -1/T 1/T 2 / T ? First “null ” bandwidth = 2(1/T) = 2/T Eytan Modiano Slide 1 2 Bandwidth efficiency ? Rs = symbol rate = 1/T – Log 2 (M) bits per symbol => R b = bit rate = log 2 (M)/T bits per second ? BW = 2/T = 2Rs – Bandwidth efficiency = R b /BW = log 2 (M)/T * (T/2) = log 2 (M)/2 BPS/Hz ? Example: – M=2 => bandwidth efficiency = 1/2 – M=4 => bandwidth efficiency = 1 – M=8 => bandwidth efficiency = 3/2 ? Increased BW efficiency with increasing M ? However, as M increase we are more prone to errors as symbols are closer together (for a given energy level) – Need to increase symbol energy level in order to overcome errors – Tradeoff between BW efficiency and energy efficiency Eytan Modiano Slide 1 3 Energy utilization E M E ave = g ∑ ( 2 m ? 1 ? M ) 2 = E g ( M 2 ? 1 ) / 3 , E g = basic pulse energy M m = 1 2 g ( ) / After modulation E u = E s = EM 2 ? 16 E b = average energy per bit = ( M 2 ? 1 ) E g 6 Log 2 () M ? Average energy per bit increases as M increases BW efficiency energy efficiency Eytan Modiano Slide 1 4 M Two-dimensional signals ? S i = (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 be used to transmit many bits per symbol – More bandwidth efficient – More error prone ? The “shape ” of the constellation can be used to minimize error probability by keeping symbols as far apart as possible ? Common constellations – QAM: Quadrature Amplitude Modulation PAM in two dimensions Eytan Modiano – PSK: Phase Shift Keying Slide 1 5 Special constellation where all symbols have equal power Symmetric M-QAM x y x y / S m = ( A m , A m ), A m , A m ∈ { + / ? 1 , + / ? 3 , ..., +? ( M ? 1 ) } M is the total number of signal points (symbols) signal levels on each axis M 16-QAM Constellation is symmetric ? M= K 2 , for some K Signal levels on each axis are the same as for PAM Eg x y .. , 4 ? QAM ? A m , A m ∈ { + / ? 1 } x y 16 ? QAM ? A m , A m ∈ { + / ? 1 , + / ? 3 } Eytan Modiano Slide 1 6 -3 -1 1 3 -3 -1 1 3 Bandwidth occupancy of QAM ? When using a rectangular pulse, the Fourier transform is a Sinc g(t) |G(f)| A AT T -2/T -1/T 1/T 2 / T ? First null BW is still 2/T – K = Log 2 (M) bits per symbol – Rb = Log 2 (M)/T – Bandwidth Efficiency = Rb/BW = Log 2 (M)/2 – => “Same as for PAM ” Eytan Modiano Slide 1 7 Energy efficiency x y E sm = [( A m ) 2 + ( A m ) 2 ] E g x EA m ) 2 ] = EA m ) 2 ] = K 2 ? 1 = M ? 1 , K = [( [( y 3 3 M E = 2 ( M ? 1 ) E s 3 g Transmitted energy = E s = ( M ? 1 ) E g 2 3 E b ( QAM ) = Energy / bit = ( M ? 1 ) E g 3 Log 2 () M ? Compare to PAM: Eb increases with M, but not nearly as fast as PAM E b ( PAM ) = ( M 2 ? 1 ) E g M Eytan Modiano 6 Log 2 () Slide 1 8 Bandpass QAM ? Modulate the two dimensional signal by multiplication by orthogonal carriers (sinusoids): Sin & C o s – This is accomplished by multiplying the A x component by Cos and the A y component by sin – Typically, people do not refer to these components as x,y but rather A c or A s for cos and sin or sometimes as A Q , and A I for quadrature or in-phase components ? The transmitted signal, corresponding to the m th symbol is: x Ut () y () m () = A m g t Cos ( 2 π f c t ) + A m g t Sin ( 2 π f c t ) , m = 1 . . . M Eytan Modiano Slide 1 9 Eytan Modiano Slide 2 0 Modulator Binary data Map k bits Into one of 2 k symbols A m =(A x , A y ) g(t) g(t) Cos(2 π f c t) Sin(2 π f c t) U m (t) Demodulation: Recovering the baseband signals U(t) LPF S x r (t) 2Cos(2 π f c t) U(t) LPF S y r (t) 2Sin(2 π f c t) ? Over a symbol duration, Sin(2 π f c t) and Cos(2 π f c t) are orthogonal – As long as the symbol duration is an integer number of cycles of the carrier wave (fc = n/T) for some n ? When multiplied by a sin, the cos component of U(t) disappears and Eytan Modiano similarly the sin component disappears when multiplied by cos Slide 2 1 () t () Demodulation, cont. () 2 Cos ( 2 π f c t ) = 2 A x g t Cos 2 ( 2 π f c t ) + 2 A y g t Ut () () cos( 2 π f c t ) sin ( 2 π f c t ) Cos 2 α = 1 + cos( 2 α ) 2 x x x x => Ut t () cos( 4 π f c t ) ≈ St () 2 Cos ( 2 π f c t ) = S () + St ( ) = A g ( t ) Similarly , U ( t ) 2 Sin ( 2 π f c t ) = 2 A x g ( t ) Cos ( 2 π f c t ) Sin ( 2 π f c t ) + 2 A y g ( t ) sin 2 ( 2 π f c t ) Sin 2 α = 1 ? cos( 2 α ) 2 => Ut t y () cos( 4 π f c t ) ≈ St () 2 Sin ( 2 π f c t ) = S y () ? St y ( ) = A y g ( t ) Eytan Modiano Slide 2 2 Phase Shift Keying (PSK) ? Two Dimensional signals where all symbols have equal energy levels – ? I.e., they lie on a circle or radius Symbols can be equally spaced to minimize likelihood of errors E s E s ? E.g., Binary PSK s 1 s 2 E s E s ? 4-PSK (above) same as 4-QAM Eytan Modiano Slide 2 3 () ) M-PSK A i x = Cos ( 2 π i / M ), A i y = Sin ( 2 π i / M ), m = 0 , ..., M ? 1 x Ut () () y m () = g t A m Cos ( 2 π f c t ) ? g t A m Sin ( 2 π f c t ) ? + Notice : Cos ( α ) Cos ( β ) = Cos ( αβ ) + Cos ( αβ ) 2 ? + Sin α Sin ( β ) = Cos ( αβ ) ? Cos ( αβ ) 2 Hence , U m ( t ) = g ( t ) Cos ( 2 π f c t + 2 π m / M ) / φ m = 2 π mM = phases shift of m th symbol Ut () m () = g t Cos ( 2 π f c t + φ m ) , m = 0 . . . M ? 1 Eytan Modiano Slide 2 4 M-PSK Summary ? Constellation of M Phase shifted symbols – All have equal energy levels – K = Log 2 (M) bits per symbol ? Modulation: Binary data U m (t) Map k bits Into one of 2 k symbols A m =(A x , A y ) g(t) g(t) Cos ( 2 π f c t) -Sin(2 π f c t) ? Notice that for PSK we subtract the sin component from the cos component – For convenience of notation only. If we added, the phase shift would have been negative but the end result is the same ? Demodulation is the same as for QAM Eytan Modiano Slide 2 5