4601 Communication Systems Week 8 Binary Modulated Signal Sets Non-Binary Signal Sets 0 c 2011, Georgia Institute of Technology (lect8 1)
Binary PSK (BPSK) With BPSK information is transmitted in the carrier phase. Two sinusoids are used having a relative phase different of π radians. That is s 1 (t) = s 2 (t) = 2 T cos(2πf ct) 2 T cos(2πf ct+π) for 0 t T. Note that s 2 (t) = = 2 T cos(2πf ct+π) 2 T {cos(2πf ct)cosπ sin(2πf c t)sinπ} = 2 T cos(2πf ct) = s 1 (t) 0 c 2011, Georgia Institute of Technology (lect8 2)
Binary PSK (BPSK) Assuming that f c T 1, the energy in s 1 (t) and s 2 (t) is T 0 s2 i(t)dt = The vector representation of binary PSK signals requires only a single basis function because s 1 (t) = s 2 (t), i.e., the two signals are linearly dependent. We have that f 1 (t) = s 1(t) = 2 T cos(2πf ct) Then s 1 (t) = f 1 (t) s 2 (t) = f 1 (t) 0 c 2011, Georgia Institute of Technology (lect8 3)
Binary PSK (BPSK) s 1 s 2-0 f (t) 1 The minimum distance decision rule is where r N(s 1,N 0 /2). The error probability is choose d 12 = 2 s 1 if r < 0 s 2 if r > 0 P e = P e s1 sentp(s 1 sent)+p e s2 sentp(s 2 sent) = P e s1 sent = P(r < 0) = Q 2 N o 0 c 2011, Georgia Institute of Technology (lect8 4)
Binary FSK (BFSK) With binary FSK signals, information is transmitted in the carrier frequency. Two sinusoids are used that have different carrier frequencies. s 1 (t) = s 2 (t) = 2 T cos(2πf ct) 2 T cos(2π(f c + f )t) for 0 t T. The frequency difference is f. Note that s 1 (t) and s) 2 (t) both have energy. 0 c 2011, Georgia Institute of Technology (lect8 5)
Binary FSK (BFSK) The vector representation of binary FSK signals requires two basis functions because s 1 (t) = s 2 (t), i.e., the two signals are linearly independent. We have that Then s 21 = T f 1 (t) = s 1(t) = 0 s 2(t)f 1 (t)dt = 2 T T 2 T cos(2πf ct) 0 cos(2π(f c + f )t)cos(2πf c t)dt 0 c 2011, Georgia Institute of Technology (lect8 6)
Binary FSK (BFSK) s 21 = = = T T T T 0 {cos(2π f)t+cos(2π(2f c + f )t)}dt T 0 cos(2π f)t dt sin(2π f t) 2π f = sin(2π ft) 2π f T = sinc(2 f T) T 0 Hence, f 2 (t) = 2 T cos(2π(f c+ f )t) sinc(2 f T) (1 sinc 2 2 f T) 2 T cos(2πf ct) 0 c 2011, Georgia Institute of Technology (lect8 7)
Binary Orthogonal FSK Suppose that f = 1/(2T). Then it follows that s 21 = sinc(2 f T) = 0 In this case, s 1 (t) and s 2 (t) are orthogonal. Hence, f 2 (t) = 2 T cos(2π(f c + f )t) 0 c 2011, Georgia Institute of Technology (lect8 8)
Binary Orthogonal FSK f (t) 2 s 2 0 0 s 1 f (t) 1 d 2 12 = 2 0 c 2011, Georgia Institute of Technology (lect8 9)
Binary Orthogonal FSK The minimum distance decision rule is choose s 1 s 2 if r is below dashed line if r is above dashed line Using the circular symmetric property of the noise vector n, the error probability is P e = P e s1 sent = Q /2 = Q No /2 N o Note that coherent BFSK requires a factor of 2 (3 db) increase in /N o to achieve the same error probability as BPSK. 0 c 2011, Georgia Institute of Technology (lect8 10)
M-ary PAM With M-ary Pulse Amplitude Modulation, information is transmitted in the carrier amplitude, such that the amplitude takes on one of M possible values. During any baud interval, the transmitted waveform is s m (t) = 2 0 T a mcos(2πf c t) where a m {±1,±3,±5,±(M 1)} and 0 is the energy of the signal with the lowest amplitude, i.e., when a m = ±1. Usually, M = 2 k for some k, i.e., M = 2,4,8,16, etc. During each baud interval of length T, k = log 2 M bits are transmitted. The baud rate R = 1/T and the bit rate is R b = kr. 0 c 2011, Georgia Institute of Technology (lect8 11)
M-ary PAM M-ary PAM signals can be expressed in terms of signal vectors. Since all the M signals are linearly dependent, there is only one basis function. Then f 1 (t) = 2 T cos(2πf ct), 0 t T s m (t) = a m 0 f 1 (t) Hence, the signal-space diagram for M-ary PAM is shown below. 000 001 011 010 110 111 101 100-7 -5-3 -1 +1 +3 +5 +7 X 0 0 c 2011, Georgia Institute of Technology (lect8 12)
M-ary QAM Quadrature Amplitude Modulation (QAM) signals can be thought of a independent PAM on the inphase (cosine) and quadrature (sine) carrier components. During any baud interval the transmitted waveform is where s m (t) = 2 0 T ( ) a c m cos(2πf ct) a s m sin(2πf ct) {±1,±3,±5,±(M 1)} and 2 0 is the energy of the signal with the lowest amplitude, i.e., when a c m,a s m = ±1. a {c,s} m 0 c 2011, Georgia Institute of Technology (lect8 13)
M-ary QAM QAM signals can be expressed in terms of signal vectors. Since the functions cos2πf c t and sin2πf c t, with f c T 1, are orthogonal over the interval (0,T), we have two basis functions Then f 1 (t) = 2 T cos2πf ct f 2 (t) = 2 T sin2πf ct s m (t) = a c m 0 f 1 (t)+a s m 0 f 2 (t), m = 1,...,M, 0 t T Hence ) s m (t) s m = 0 (a c m,a s m 0 c 2011, Georgia Institute of Technology (lect8 14)
M-ary QAM For the case when M = 2 k, k even, the resulting signal space diagram has a square constellation. In this case the QAM signal can be thought of as 2 PAM signals in quarature. For M = 2 k, k odd, the constellation takes on a cross form. For example, 16-QAM constellation is 0000 0001 0011 0010 0100 0101 0111 0110 1100 1101 1111 1110 1000 1001 1011 1010 0 c 2011, Georgia Institute of Technology (lect8 15)
M-ary PSK Phase shift keyed (PSK) signals transmit information in the carrier phase. During any baud interval, the transmitted waveform is s m (t) = 2 T cos(2πf ct+θ k ),0 t T where We can rewrite this in the form s m (t) = 2 T ( θ k 2π(m 1) M, m = 1,...,M ) cosθ m cos2πf c t sinθ m sin2πf c t,m = 1,...,M Using the same basis functions as QAM, we have s m (t) s m = 0 ( cosθ m,sinθ m ) 0 c 2011, Georgia Institute of Technology (lect8 16)
8-PSK Constellation f (t) 1 011 010 001 110 000 f (t) 2 111 100 101 0 c 2011, Georgia Institute of Technology (lect8 17)
M-ary FSK For Frequency shift keyed (FSK) signals, the transmitted signal during any given baud interval is s m (t) = Acos(2πf c t+2πf m t) where f m = (m 1) f, m = 1,...,M Wehaveseenbeforethatthechoice f = 1 2T giveswaveformsthatareorthogonal. f (t) 2 M = 3 f (t) 1 f (t) 3 0 c 2011, Georgia Institute of Technology (lect8 18)