EE6604 Personal & Mobile Communications Week 12a Power Spectrum of Digitally Modulated Signals 1
POWER SPECTRUM OF BANDPASS SIGNALS A bandpass modulated signal can be written in the form s(t) = R { s(t)e j(2πf ct) } = 1 j(2πf { s(t)e c t) + s (t)e j(2πf ct) } 2 The autocorrelation of a bandpass modulated signal is φ ss (τ) = E[s(t)s(t+τ)] = 1 4 E [( s(t)e j2πfct + s (t)e j2πf ct ) ( s(t+τ)e j(2πf ct+2πf c τ) + s (t+τ)e j(2πf ct+2πf c τ) )] = 1 4 E [ s(t) s(t+τ)e j(4πf ct+2πf c τ) + s(t) s (t+τ)e j2πf cτ + s (t) s(t+τ)e j2πfcτ + s (t) s (t+τ)e j(4πf ct+2πf c τ) ] = 1 [ E[ s(t) s(t+τ)]e j(4πf c t+2πf c τ) +E[ s(t) s (t+τ)]e j2πf cτ 4 +E[ s (t) s(t+τ)]e j2πfcτ +E[ s (t) s (t+τ)]e j(4πf ct+2πf c τ) ]. 2
If s(t) is a wide-sense stationary random process, then the exponential terms that involve t must vanish, i.e., E[ s(t) s(t+τ)] = 0 and E[ s (t) s (t+τ)] = 0. Substituting s(t) = s I (t)+j s Q (t) into the above expectations gives the result φ si ß I (τ) = E[ s I (t) s I (t+τ)] = E[ s Q (t) s Q (t+τ)] = φ sq s Q (τ) φ si s Q (τ) = E[ s I (t) s Q (t+τ)] = E[ s Q (t) s I (t+τ)] = φ sq s I (τ) Using these results, the autocorrelation is φ ss (τ) = 1 2 φ s s(τ)e j2πf cτ + 1 2 φ s s(τ)e j2πf cτ where φ s s (τ) = 1 2 E[ s (t) s(t+τ)] The power density spectrum is the Fourier transform of φ ss (τ): S ss (f) = 1 2 [S s s(f f c )+S s s( f f c )] S s s (f) is the power density spectrum of the complex envelope s(t), which is always realvalued but not necessarily even about f = 0. S ss (f) = 1 2 [S s s(f f c )+S s s ( f f c )] 3
POWER SPECTRAL DENSITY OF A COMPLEX ENVELOPE In general, the complex lowpass signal is of the form s(t) = A k b(t kt,x k ) The autocorrelation of s(t) is φ s s (t,t+τ) = 1 2 E[ s (t) s(t+τ)] = A2 2 i k E[b (t it,x i )b(t+τ kt,x k )]. Observe that s(t) is a cyclostationary random process, meaning that the autocorrelation function φ s s (t,t+τ) is periodic in t with period T. To see this property, first note that φ s s (t+t,t+t +τ) = A2 2 = A2 2 i k i E[b (t+t it,x i )b(t+t +τ kt,x k )] k E[b (t i T,x i +1)b(t+τ k T,x k +1)]. 4
Under the assumption that the information sequence is a stationary random process it follows that φ s s (t+t,t+t +τ) = A2 E[b (t i T,x i )b(t+τ k T,x k )] 2 i k = φ s s (t,t+τ). (1) where data blocks x i +1 and x k +1 are replaced by x i and x k, respectively. Therefore s(t) is cyclostationary. 5
Since s(t) is cyclostationary, the autocorrelation φ s s (τ) can be obtained by taking the time average of φ s s (t,t+τ), given by φ s s (τ) = < φ s s (t,t+τ) > = A2 1 T 2 T 0 E[b (t it,x i )b(t+τ kt,x k )]dt = A2 2T = A2 2T = A2 2T = A2 2T i k it+t i k it it+t i m it it+t i m m it E[b (z,x i )b(z +τ (k i)t,x k )]dz E[b (z,x i )b(z +τ mt,x m+i )]dz E[b (z,x 0 )b(z +τ mt,x m )]dz E[b (z,x 0 )b(z +τ mt,x m )]dz, where denotes time averaging and the second last equality used the stationary property of the data sequence {x k }. 6
The psd of s(t) is obtained by taking the Fourier transform of φ s s (τ), S s s (f) = E = E = E = A2 2T A 2 2T m A 2 2T m A 2 2T b (z,x 0 )b(z +τ mt,x m )dze j2πfτ dτ b (z,x 0 )e j2πfz dz b(z +τ mt,x m)e j2πf(z+τ mt) dτe j2πfmt] m b (z,x 0 )e j2πfz dz m E[B (f,x 0 )B(f,x m )]e j2πfmt, where B(f,x m ) is the Fourier transform of b(t,x m ). b(τ,x m )e j2πfτ dτ e j2πfmt Finally, where S s s (f) = A2 T m S b,m(f)e j2πfmt S b,m (f) = 1 2 E[B (f,x 0 )B(f,x m )] 7
Suppose that x m and x 0 are uncorrelated for m K. Then where S b,m (f) = S b,k (f), m K S b,k (f) = 1 2 E[B (f,x 0 )]E[B(f,x m )] m K = 1 2 E[B (f,x 0 )]E[B(f,x 0 )] m K = 1 2 E[B(f,x 0)] 2, m K. It follows that where S s s (f) = S c s s(f)+s d s s(f) A2 S c s s (f) = T S d s s(f) = A T m <K (S b,m (f) S b,k (f))e j2πfmt 2S b,k (f) n ( δ f n T Note that the spectrum consists of discrete and continuous parts. The discrete portion has spectral lines spaced at 1/T Hz apart. ) 8
ZERO MEAN SIGNALS If s(t) has zero mean, i.e., E[b(t,x 0 )] = 0, then E[B(f,x 0 )] = 0. Under this condition S b,k (f) = 1 2 E[B(f,x 0)] 2 = 0 Hence, S s s (f) has no discrete component and S s s (f) = A 2 T m <K S b,m (f)e j2πfmt 9
UNCORRELATED SOURCE SYMBOLS With uncorrelated source symbols the information symbols x m,k constituting data blocks x m = (x m,1,x m,2,...,x m,n ) are mutually uncorrelated. Under this condition x m and x 0 are obviously uncorrelated for m 1. Hence, S b,m (f) = S b,1 (f), for m 1, where S b,0 (f) = 1 2 E [ B(f,x 0 ) 2] S b,1 (f) = 1 2 E[B(f,x 0)] 2 Hence If s(t) has zero mean as well, then A2 S d s s (f) = T 2S b,1(f) n δ(f n/t) A2 S c s s (f) = T (S b,0(f) S b,1 (f)) S s s (f) = A2 T S b,0(f) 10
LINEAR FULL RESPONSE MODULATION Here it is assumed that b(t,x k ) = x k h a (t) B(f,x k ) = x k H a (f), where the x k may be correlated. Using the above leads to S b,m (f) = 1 2 E[B (f,x 0 )B(f,x m )] = 1 2 E[x 0Ha(f))x m H a (f))] = 1 2 E [ x 0x m H a (f) 2] = 1 2 E[x 0x m ] H a (f) 2 = φ xx (m) H a (f) 2 where φ xx (m) = 1 2 E[x kx k+m ] 11
The psd of the complex envelope is where S s s (f) = A2 T m S b,m(f)e j2πfmt = A2 T H a(f) 2 m φ xx(m)e j2πfmt = A2 T H a(f) 2 S xx (f) S xx (f) = m φ xx(m)e j2πfmt With uncorrelated source symbols where σ 2 x = 1 2 E[ x k 2 ], µ x = E[x k ]. If µ x = 0, then S b,1 (f) = 0 and S b,0 (f) = σ 2 x H a (f) 2 S b,m (f) = 1 2 µ x 2 H a (f) 2, m 1. S s s (f) = A2 T σ2 x H a(f) 2 12
POWER SPECTRAL DENSITY OF ASK 0-20 τ = 4T τ = 6T τ = 8T no truncation S vv (f) (db) -40-60 -80-100 0.0 0.5 1.0 1.5 2.0 Frequency, ft Psd of ASK with a truncated square root raised cosine pulse with various truncation lengths; β = 0.5. 13
OFDM Power Spectrum The data symbols x n,k,k = 0,...,N 1 that modulate the N sub-carriers are assumed to have zero mean, variance σ 2 x = 1 2 E[ x n,k 2 ], and they are mutually uncorrelated. In this case, the psd of the OFDM waveform is where and B(f,x 0 ) = N 1 k=0 S s s (f) = A2 T g S b,0 (f), S b,0 (f) = 1 2 E [ B(f,x 0 ) 2], x 0,k Tsinc(fT k)+ N 1 k=0 Using the above along with T = NT s yields the result S s s (f) = σ 2 x A2 T + α2 g 1+α g 1 1+α g N 1 k=0 N 1 k=0 sinc 2 (NfT s k) sinc 2 (α g (NfT s k)) + 2α g cos(2πnft s ) N 1 1+α g k=0 x 0,k α g Tsinc(α g (ft k))e j2πft. sinc(nft s k)sinc(α g (NfT s k)) Note that the Nyquist frequency in this case is 1/2T g s = (1+α g )/2T s.. 14
0 S ss (f) (db) 20 40 60 2 1 0 1 2 ft s Psd of OFDM with N = 16,α g = 0. 15
20 10 0 S ss (f) (db) 10 20 30 40 50 60 2 1 0 1 2 ft s Psd of OFDM with N = 16,α g = 0.25. 16
20 10 0 S ss (f) (db) 10 20 30 40 50 60 2 1 0 1 2 ft s Psd of OFDM with N = 1024,α g = 0. 17
20 10 0 S ss (f) (db) 10 20 30 40 50 60 2 1 0 1 2 ft s Psd of OFDM with N = 1024,α g = 0.25. 18
OFDM Power Spectrum -IFFT Implementation The output of the IDFT baseband modulator is{x g } = {X g n,m}, where m is the block index and Xn,m g = X n,(m) N = A N 1 x n,k e j2πkm N k=0, m = 0, 1,..., N +G 1 Thepowerspectrumofthesequence{X g }canbecalculatedbyfirstdeterminingthediscretetime autocorrelation function of the time-domain sequence {X g } and then taking a discretetime Fourier transform of the discrete-time autocorrelation function. The psd of the OFDM complex envelope with ideal DACs can be obtained by applying the resulting power spectrum to an ideal low-pass filter with a cutoff frequency of 1/(2T g s ) Hz. 19
Discrete-time Autocorrelation Function The time-domain sequence {X g } is a periodic wide-sense stationary sequence having the discrete-time autocorrelation function φ X g X g(m,l) = 1 2 E[Xg n,m (Xg n,m+l) ] = A 2 N 1 N 1 1 k=0 i=0 2 E[x n,kx n,i ]ej2π for m = 0,...,N +G 1. N (km im il), Thedatasymbols,x n,k,areassumedtobemutuallyuncorrelatedwithzeromeanandvariance σ 2 x = 1 2 E[ x n,k 2 ]. Using the fact, that X g n,m = X n,(m)n, it follows that m = 0,...,G 1,l = 0,N φ X g X g(m,l) = Aσx 2 m = G,...,N 1,l = 0. m = N,...,N +G 1,l = 0, N 0 otherwise Averaging over all time indices m in a block gives the time-average discrete-time autocorrelation function Aσ φ X g X g(l) = x 2 l = 0 G N+G Aσ2 x l = N,N. 0 otherwise 20
Power Spectrum Taking the discrete-time Fourier transform of the discrete-time autocorrelation function gives S X g X g(f) = m φ X g X g(l)e j2πfmntg s = Aσ 2 x = Aσ 2 x 1+ G g N +G e j2πfnt s + G N +G ej2πfnt 1+ 2G N +G cos(2πfntg s) Finally, assume that the sequence {X g } = {X g n,m} is passed through an ideal DACs.. The ideal DAC is a low-pass filter with cutoff frequency 1/(2T g s ). The OFDM complex envelope has the psd S s s (f) = Aσ 2 x 1+ 2G N +G cos(2πfntg s ) rect(ft g s ). g s 21
20 0 S ss (f) (db) 20 40 60 2 1.5 1 0.5 0 0.5 1 1.5 2 ft s Psd of IDFT-based OFDM with N = 16,G = 0. Note in this case that T g s = T s. 22
20 10 0 10 S ss (f) (db) 20 30 40 50 60 2 1.5 1 0.5 0 0.5 1 1.5 2 ft s g Psd of IDFT-based OFDM with N = 16,G = 4. 23
20 0 S ss (f) (db) 20 40 60 2 1.5 1 0.5 0 0.5 1 1.5 2 ft s g Psd of IDFT-based OFDM with N = 1024,G = 256. 24
20 0 h=0.25 h=0.5 h=0.75 S ss (f) (db) 20 40 60 80 4 3 2 1 0 1 2 3 4 ft Power spectral density of binary CPFSK for various modulation indices. 25
40 20 h=0.8 h=0.9 h=0.99 S ss (f) (db) 0 20 40 60 4 3 2 1 0 1 2 3 4 ft Psd of binary CPFSK as the modulation index h 1. 26
0-20 S vv (f) (db) -40-60 -80-100 MSK BT = 0.2 BT = 0.25 BT = 0.3-120 0.0 0.5 1.0 1.5 2.0 2.5 Frequency, ft Power spectral density of GMSK with various normalized filter bandwidths BT. 27