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b) It is apparent from S(f) that the terms |S k (f)| 2 decay proportionally as 1 (f−f k ) 2 . also note that |S k (f)| 2 = T b E b 2 sin πT b (f − f k ) π(f −f k ) 2 + sin πT b (f + f k ) π(f + f k ) 2 because the product sin πT b (f − f k ) π(f −f k ) × sin πT b (f + f k ) π(f + f k ) ≈ 0 due to the relation that the carrier frequency f c 1 T b . Problem 8.7 1) The autocorrelation function of the information symbols {a n } is R a (k)=E[a ∗ n a + n + k]= 1 4 ×|a n | 2 δ(k)=δ(k) Thus, the power spectral density of v(t)is S V (f)= 1 T S a (f)|G(f)| 2 = 1 T |G(f)| 2 where G(f)=F[g(t)]. If g(t)=AΠ( t− T 2 T ), we obtain |G(f)| 2 = A 2 T 2 sinc 2 (fT) and therefore, S V (f)=A 2 T sinc 2 (fT) In the next figure we plot S V (f) for T = A =1. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -5 -4 -3 -2 -1 0 1 2 3 45 frequency f Sv(f) 2) If g(t)=A sin( πt 2 )Π( t−T/2 T ), then G(f)=A 1 2j δ(f − 1 4 ) − 1 2j δ(f + 1 4 ) Tsinc(fT)e −j2πf T 2 = AT 2 [δ(f − 1 4 ) − δ(f + 1 4 )] sinc(fT)e −j(2πf T 2 + π 2 ) = AT 2 e −jπ[(f− 1 4 )T + 1 2 ] sinc((f − 1 4 )T ) −sinc((f − 1 4 )T )e −j πT 2 Thus, |G(f)| 2 = A 2 T 2 4 sinc 2 ((f + 1 4 )T ) + sinc 2 ((f − 1 4 )T ) −2sinc((f + 1 4 )T )sinc((f − 1 4 )T ) cos πT 2 218 and the power spectral of the transmitted signal is S V (f)= A 2 T 4 sinc 2 ((f + 1 4 )T ) + sinc 2 ((f − 1 4 )T ) −2sinc((f + 1 4 )T )sinc((f − 1 4 )T ) cos πT 2 In the next figure we plot S V (f) for two special values of the time interval T . The amplitude of the signal A was set to 1 for both cases. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 -5 -4 -3 -2 -1 0 1 2 3 45 frequency f Sv(f) T=1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 -5 -4 -3 -2 -1 0 1 2 3 45 frequency f Sv(f) T=2 3) The first spectral null of the power spectrum density in part 1) is at position W null = 1 T The 3-dB bandwidth is specified by solving the equation: S V (W 3dB )= 1 2 S V (0) Since sinc 2 (0) = 1, we obtain sinc 2 (W 3dB T )= 1 2 =⇒ sin(πW 3dB T )= 1 √ 2 πW 3dB T Solving the latter equation numerically we find that W 3dB = 1.3916 πT = 0.443 T To find the first spectral null and the 3-dB bandwidth for the signal with power spectral density in part 2) we assume that T = 1. In this case S V (f)= A 2 4 sinc 2 ((f + 1 4 )) + sinc 2 ((f − 1 4 )) and as it is observed there is no value of f that makes S V (f) equal to zero. Thus, W null = ∞.To find the 3-dB bandwidth note that S V (0) = A 2 4 2sinc( 1 4 )= A 2 4 1.6212 Solving numerically the equation S V (W 3dB )= 1 2 A 2 4 1.6212 we find that W 3dB =0.5412. As it is observed the 3-dB bandwidth is more robust as a measure for the bandwidth of the signal. 219 Problem 8.8 The transition probability matrix P is P = 1 2 0101 0011 1100 1010 Hence, P 2 = 1 4 1021 2110 0112 1201 and P 4 = 1 16 2446 4264 4624 6442 and therefore, P 4 γ = 1 16 2446 4264 4624 6442 100−1 01−10 0 −110 −1001 = 1 16 −4004 0 −440 04−40 400−4 = − 1 4 γ Thus, P 4 γ = − 1 4 γ and by pre-multiplying both sides by P k , we obtain P k+4 γ = − 1 4 P k γ Problem 8.9 a) Taking the inverse Fourier transform of H(f), we obtain h(t)=F −1 [H(f)] = δ(t)+ α 2 δ(t − t 0 )+ α 2 δ(t + t 0 ) Hence, y(t)=s(t) h(t)=s(t)+ α 2 s(t − t 0 )+ α 2 s(t + t 0 ) b) If the signal s(t) is used to modulate the sequence {a n }, then the transmitted signal is u(t)= ∞ n=−∞ a n s(t − nT ) The received signal is the convolution of u(t) with h(t). Hence, y(t)=u(t) h(t)= ∞ n=−∞ a n s(t − nT ) δ(t)+ α 2 δ(t − t 0 )+ α 2 δ(t + t 0 ) = ∞ n=−∞ a n s(t − nT )+ α 2 ∞ n=−∞ a n s(t − t 0 − nT )+ α 2 ∞ n=−∞ a n s(t + t 0 − nT ) 220 Thus, the output of the matched filter s(−t) at the time instant t 1 is w(t 1 )= ∞ n=−∞ a n ∞ −∞ s(τ − nT )s(τ −t 1 )dτ + α 2 ∞ n=−∞ a n ∞ −∞ s(τ − t 0 − nT )s(τ − t 1 )dτ + α 2 ∞ n=−∞ a n ∞ −∞ s(τ + t 0 − nT )s(τ − t 1 )dτ If we denote the signal s(t) s(t)byx(t), then the output of the matched filter at t 1 = kT is w(kT)= ∞ n=−∞ a n x(kT − nT ) + α 2 ∞ n=−∞ a n x(kT − t 0 − nT )+ α 2 ∞ n=−∞ a n x(kT + t 0 − nT ) c) With t 0 = T and k = n in the previous equation, we obtain w k = a k x 0 + n=k a n x k−n + α 2 a k x −1 + α 2 n=k a n x k−n−1 + α 2 a k x 1 + α 2 n=k a n x k−n+1 = a k x 0 + α 2 x −1 + α 2 x 1 + n=k a n x k−n + α 2 x k−n−1 + α 2 x k−n+1 The terms under the summation is the ISI introduced by the channel. Problem 8.10 a) Each segment of the wire-line can be considered as a bandpass filter with bandwidth W = 1200 Hz. Thus, the highest bit rate that can be transmitted without ISI by means of binary PAM is R =2W = 2400 bps b) The probability of error for binary PAM transmission is P 2 = Q 2E b N 0 Hence, using mathematical tables for the function Q[·], we find that P 2 =10 −7 is obtained for 2E b N 0 =5.2=⇒ E b N 0 =13.52=11.30 dB c) The received power P R is related to the desired SNR per bit through the relation P R N 0 = R E b N 0 Hence, with N 0 =4.1 × 10 −21 we obtain P R =4.1 × 10 −21 × 1200 × 13.52=6.6518 × 10 −17 = −161.77 dBW 221 Since the power loss of each segment is L s =50Km ×1 dB/Km = 50 dB the transmitted power at each repeater should be P T = P R + L s = −161.77+50=−111.77 dBW Problem 8.11 The pulse x(t) having the raised cosine spectrum is x(t) = sinc(t/T ) cos(παt/T ) 1 − 4α 2 t 2 /T 2 The function sinc(t/T ) is 1 when t = 0 and 0 when t = nT. On the other hand g(t)= cos(παt/T ) 1 − 4α 2 t 2 /T 2 = 1 t =0 bounded t =0 The function g(t) needs to be checked only for those values of t such that 4α 2 t 2 /T 2 =1orαt = T 2 . However, lim αt→ T 2 cos(παt/T ) 1 − 4α 2 t 2 /T 2 = lim x→1 cos( π 2 x) 1 − x and by using L’Hospital’s rule lim x→1 cos( π 2 x) 1 − x = lim x→1 π 2 sin( π 2 x)= π 2 < ∞ Hence, x(nT )= 1 n =0 0 n =0 meaning that the pulse x(t) satisfies the Nyquist criterion. Problem 8.12 Substituting the expression of X rc (f) in the desired integral, we obtain ∞ −∞ X rc (f)df = − 1−α 2T − 1+α 2T T 2 1 + cos πT α (−f − 1 − α 2T ) df + 1−α 2T − 1−α 2T Tdf + 1+α 2T 1−α 2T T 2 1 + cos πT α (f − 1 − α 2T ) df = − 1−α 2T − 1+α 2T T 2 df + T 1 − α T + 1+α 2T 1−α 2T T 2 df + − 1−α 2T − 1+α 2T cos πT α (f + 1 − α 2T )df + 1+α 2T 1−α 2T cos πT α (f − 1 − α 2T )df =1+ 0 − α T cos πT α xdx + α T 0 cos πT α xdx =1+ α T − α T cos πT α xdx =1+0=1 222 Problem 8.13 Let X(f) be such that Re[X(f)] = T Π(fT)+U(f) |f| < 1 T 0 otherwise Im[X(f)] = V (f) |f| < 1 T 0 otherwise with U(f ) even with respect to 0 and odd with respect to f = 1 2T Since x(t) is real, V (f)isodd with respect to 0 and by assumption it is even with respect to f = 1 2T . Then, x(t)=F −1 [X(f)] = 1 2T − 1 T X(f)e j2πft df + 1 2T − 1 2T X(f)e j2πft df + 1 T 1 2T X(f)e j2πft df = 1 2T − 1 2T Te j2πft df + 1 T − 1 T [U(f)+jV (f)]e j2πft df = sinc(t/T )+ 1 T − 1 T [U(f)+jV (f)]e j2πft df Consider first the integral 1 T − 1 T U(f)e j2πft df . Clearly, 1 T − 1 T U(f)e j2πft df = 0 − 1 T U(f)e j2πft df + 1 T 0 U(f)e j2πft df and by using the change of variables f = f + 1 2T and f = f − 1 2T for the two integrals on the right hand side respectively, we obtain 1 T − 1 T U(f)e j2πft df = e −j π T t 1 2T − 1 2T U(f − 1 2T )e j2πf t df + e j π T t 1 2T − 1 2T U(f + 1 2T )e j2πf t df a = e j π T t − e −j π T t 1 2T − 1 2T U(f + 1 2T )e j2πf t df =2j sin( π T t) 1 2T − 1 2T U(f + 1 2T )e j2πf t df where for step (a) we used the odd symmetry of U(f ) with respect to f = 1 2T , that is U(f − 1 2T )=−U(f + 1 2T ) For the integral 1 T − 1 T V (f)e j2πft df we have 1 T − 1 T V (f)e j2πft df = 0 − 1 T V (f)e j2πft df + 1 T 0 V (f)e j2πft df = e −j π T t 1 2T − 1 2T V (f − 1 2T )e j2πf t df + e j π T t 1 2T − 1 2T V (f + 1 2T )e j2πf t df 223 However, V (f ) is odd with respect to 0 and since V (f + 1 2T ) and V (f − 1 2T ) are even, the translated spectra satisfy 1 2T − 1 2T V (f − 1 2T )e j2πf t df = − 1 2T − 1 2T V (f + 1 2T )e j2πf t df Hence, x(t) = sinc(t/T )+2j sin( π T t) 1 2T − 1 2T U(f + 1 2T )e j2πf t df −2 sin( π T t) 1 2T − 1 2T U(f + 1 2T )e j2πf t df and therefore, x(nT )= 1 n =0 0 n =0 Thus, the signal x(t) satisfies the Nyquist criterion. Problem 8.14 The bandwidth of the channel is W = 3000 − 300 = 2700 Hz Since the minimum transmission bandwidth required for bandpass signaling is R, where R is the rate of transmission, we conclude that the maximum value of the symbol rate for the given channel is R max = 2700. If an M-ary PAM modulation is used for transmission, then in order to achieve a bit-rate of 9600 bps, with maximum rate of R max , the minimum size of the constellation is M =2 k = 16. In this case, the symbol rate is R = 9600 k = 2400 symbols/sec and the symbol interval T = 1 R = 1 2400 sec. The roll-off factor α of the raised cosine pulse used for transmission is is determined by noting that 1200(1 + α) = 1350, and hence, α =0.125. Therefore, the squared root raised cosine pulse can have a roll-off of α =0.125. Problem 8.15 Since the bandwidth of the ideal lowpass channel is W = 2400 Hz, the rate of transmission is R =2× 2400 = 4800 symbols/sec The number of bits per symbol is k = 14400 4800 =3 Hence, the number of transmitted symbols is 2 3 = 8. If a duobinary pulse is used for transmission, then the number of possible transmitted symbols is 2M − 1 = 15. These symbols have the form b n =0, ±2d, ±4d, ,±12d where 2d is the minimum distance between the points of the 8-PAM constellation. The probability mass function of the received symbols is P (b =2md)= 8 −|m| 64 ,m=0, ±1, ,±7 224 An upper bound of the probability of error is given by (see (8.4.33)) P M < 2 1 − 1 M 2 Q π 4 2 6 M 2 − 1 kE b,av N 0 With P M =10 −6 and M = 8 we obtain kE b,av N 0 =1.3193 × 10 3 =⇒E b,av =0.088 Problem 8.16 a) The spectrum of the baseband signal is S V (f)= 1 T S a (f)|X rc (f)| 2 = 1 T |X rc (f)| 2 where T = 1 2400 and X rc (f)= T 0 ≤|f |≤ 1 4T T 2 (1 + cos(2πT(|f|− 1 4T )) 1 4T ≤|f|≤ 3 4T 0 otherwise If the carrier signal has the form c(t)=A cos(2πf c t), then the spectrum of the DSB-SC modulated signal, S U (f), is S U (f)= A 2 [S V (f − f c )+S V (f + f c )] AsketchofS U (f) is shown in the next figure. 2 2 AT -fc-3/4T -fc+3/4T fcfc-3/4T fc+3/4T-fc b) Assuming bandpass coherent demodulation using a matched filter, the received signal r(t)is first passed through a linear filter with impulse response g R (t)=Ax rc (T −t) cos(2πf c (T −t)) The output of the matched filter is sampled at t = T and the samples are passed to the detector. The detector is a simple threshold device that decides if a binary 1 or 0 was transmitted depending on the sign of the input samples. The following figure shows a block diagram of the optimum bandpass coherent demodulator. ✓ ✂ . . . . . . . . . . . . . ❘ ❅ ❅ ✲✲✲ device) (Threshold Detector t = T r(t) g R (t) matched filter Bandpass 225 Problem 8.17 a) If the power spectral density of the additive noise is S n (f), then the PSD of the noise at the output of the prewhitening filter is S ν (f)=S n (f)|H p (f)| 2 In order for S ν (f) to be flat (white noise), H p (f) should be such that H p (f)= 1 S n (f) 2) Let h p (t) be the impulse response of the prewhitening filter H p (f). That is, h p (t)=F −1 [H p (f)]. Then, the input to the matched filter is the signal ˜s(t)=s(t) h p (t). The frequency response of the filter matched to ˜s(t)is ˜ S m (f)= ˜ S ∗ (f)e −j2πft 0 == S ∗ (f)H ∗ p (f)e −j2πft 0 where t 0 is some nominal time-delay at which we sample the filter output. 3) The frequency response of the overall system, prewhitening filter followed by the matched filter, is G(f)= ˜ S m (f)H p (f)=S ∗ (f)|H p (f)| 2 e −j2πft 0 = S ∗ (f) S n (f) e −j2πft 0 4) The variance of the noise at the output of the generalized matched filter is σ 2 = ∞ −∞ S n (f)|G(f)| 2 df = ∞ −∞ |S(f)| 2 S n (f) df At the sampling instant t = t 0 = T , the signal component at the output of the matched filter is y(T )= ∞ −∞ Y (f)e j2πfT df = ∞ −∞ s(τ)g(T − τ)dτ = ∞ −∞ S(f) S ∗ (f) S n (f) df = ∞ −∞ |S(f)| 2 S n (f) df Hence, the output SNR is SNR = y 2 (T ) σ 2 = ∞ −∞ |S(f)| 2 S n (f) df Problem 8.18 The bandwidth of the bandpass channel is W = 3300 − 300 = 3000 Hz In order to transmit 9600 bps with a symbol rate R = 2400 symbols per second, the number of information bits per symbol should be k = 9600 2400 =4 Hence, a 2 4 = 16 QAM signal constellation is needed. The carrier frequency f c is set to 1800 Hz, which is the mid-frequency of the frequency band that the bandpass channel occupies. If a pulse 226 with raised cosine spectrum and roll-off factor α is used for spectral shaping, then for the bandpass signal with bandwidth W R = 1200(1 + α) = 1500 and α =0.25 A sketch of the spectrum of the transmitted signal pulse is shown in the next figure. -1800 -300 -3300 300 3300 1800 1/2T f 600 3000 Problem 8.19 The channel bandwidth is W = 4000 Hz. (a) Binary PSK with a pulse shape that has α = 1 2 . Hence 1 2T (1 + α) = 2000 and 1 T = 2667, the bit rate is 2667 bps. (b) Four-phase PSK with a pulse shape that has α = 1 2 . From (a) the symbol rate is 1 T = 2667 and the bit rate is 5334 bps. (c) M = 8 QAM with a pulse shape that has α = 1 2 . From (a), the symbol rate is 1 T = 2667 and hence the bit rate 3 T = 8001 bps. (d) Binary FSK with noncoherent detection. Assuming that the frequency separation between the two frequencies is ∆f = 1 T , where 1 T is the bit rate, the two frequencies are f c + 1 2T and f c − 1 2T . Since W = 4000 Hz, we may select 1 2T = 1000, or, equivalently, 1 T = 2000. Hence, the bit rate is 2000 bps, and the two FSK signals are orthogonal. (e) Four FSK with noncoherent detection. In this case we need four frequencies with separation of 1 T between adjacent frequencies. We select f 1 = f c − 1.5 T , f 2 = f c − 1 2T , f 3 = f c + 1 2T , and f 4 = f c + 1.5 T , where 1 2T = 500 Hz. Hence, the symbol rate is 1 T = 1000 symbols per second and since each symbol carries two bits of information, the bit rate is 2000 bps. (f) M = 8 FSK with noncoherent detection. In this case we require eight frequencies with frequency separation of 1 T = 500 Hz for orthogonality. Since each symbol carries 3 bits of information, the bit rate is 1500 bps. Problem 8.20 1) The bandwidth of the bandpass channel is W = 3000 − 600 = 2400 Hz Since each symbol of the QPSK constellation conveys 2 bits of information, the symbol rate of transmission is R = 2400 2 = 1200 symbols/sec Thus, for spectral shaping we can use a signal pulse with a raised cosine spectrum and roll-off factor α = 1, that is X rc (f)= T 2 [1 + cos(πT|f|)] = 1 2400 cos 2 π|f| 2400 227 [...]... Received seq bn : Decoded seq dn : 0 -1 0 -1 1 1 1 2 1 0 0 -1 0 0 229 0 1 1 0 0 1 1 1 2 1 0 1 1 0 0 1 0 -1 -2 1 1 0 -1 -2 1 0 0 -1 0 0 0 0 -1 0 0 1 1 1 2 1 0 0 -1 0 0 Problem 8.25 Let X(z) denote the Z-transform of the sequence xn , that is xn z −n X(z) = n Then the precoding operation can be described as P (z) = D(z) X(z) mod − M where D(z) and P (z) are the Z-transforms of the data and precoded sequences... per bit and the error probability for the case of no ISI As it observed from the figure, the relative difference in SNR of the error probability of 10−6 is 2 dB -2 -2 .5 -3 log(P(e) -3 .5 -4 -4 .5 -5 -5 .5 -6 -6 .5 -7 6 7 8 9 10 11 SNR/bit, dB 232 12 13 14 Problem 8.30 The power spectral density of the noise at the output of the matched filter is Sν (f ) = Sn (f )|GR (f )|2 = N0 N0 1 πf |X(f )| = cos( ) 2... sequence 10010110010 modulates a duobinary transmitting filter Data seq dn : Precoded seq pn : Transmitted seq an : Received seq bn : Decoded seq dn : 0 -1 1 1 1 0 1 0 1 1 2 0 0 1 1 2 0 1 0 -1 0 1 0 0 -1 -2 0 1 1 1 0 1 1 0 -1 0 1 0 0 -1 -2 0 0 0 -1 -2 0 1 1 1 0 1 0 1 1 2 0 Problem 8.24 The following table shows the precoded sequence, the transmitted amplitude levels, the received signal levels and the... state The upper branch is associated with the transmission of −1, whereas the lower branch is associated with the transmission of 1 a (am−1 , am ) m+1 -1 -1 -1 u u 1 -1 1 u -1 u 1 -1 1 u 1 -1 u -1 u u 11 1 231 Problem 8.29 a) The output of the matched filter at the time instant mT is ym = k 1 am xk−m + νm = am + am−1 + νm 4 The autocorrelation... symbols/sec If a 8-QAM constellation is employed, then the required symbol rate is R = 9600/3 = 3200 If a signal pulse with raised cosine spectrum is used for shaping, the maximum allowable roll-off factor is determined by 1600(1 + α) = 2000 which yields α = 0.25 Since α is less than 50%, we consider a larger constellation With a 16-QAM constellation we obtain 9600 R= = 2400 4 and 120 0(1 + α) = 2000... minimum free distance between the paths is dfree = 3, whereas the minimum free distance between paths for the duobinary signal is 2 Modified Duobinary (pm−2 , pm−1 ) u 0 /-1 u E Eu Eu 00 d Duobinary pm−1 0 /-1 0 u Eu Eu 1/1 U U d 1/1 ¡ 10 0 /-1 11 ! ¡ ¡ d d u ¡ u u ¡ ¡ ¡ d d e ¡ d e¡ d ¡ d d ¡ e ¡ du u edu ¡ e e e Eu u
u 01 1/1 w u u u 1 E ! ¡ u u u Problem...If the desired spectral characteristic is split evenly between the transmitting filter GT (f ) and the receiving filter GR (f ), then 1 π|f | cos , 120 0 2400 GT (f ) = GR (f ) = |f | < 1 = 120 0 T A block diagram of the transmitter is shown in the next figure an E QPSK l E to Channel E× GT (f ) T cos(2πfc t) 2) If the bit rate is 4800 bps, then the symbol rate is R= 4800... = log2 λmax = log2 233 √ 3 = 7925 Problem 8.32 The state transition matrix of the (0,1) runlength-limited code is 1 1 1 0 D= The eigenvalues of D are the roots of det(D − λI) = −λ(1 − λ) − 1 = λ2 − λ − 1 The roots of the characteristic equation are λ1,2 √ 1± 5 = 2 Thus, the capacity of the (0,1) runlength-limited code is √ 1± 5 C(0, 1) = log2 ( ) = 0.6942 2 The capacity of a (1, ∞) code is found from... observed, the two codes have exactly the same capacity This result is to be expected since the (0,1) runlength-limited code and the (1, ∞) code produce the same set of code sequences of length n, N (n), with a renaming of the bits from 0 to 1 and vise versa For example, the (0,1) runlength-limited code with a renaming of the bits, can be described as the code with no minimum number of 1’s between 0’s... the characteristic equation are √ 1+ 5 λ1,2 = ± 2 1 2 λ3,4 √ 1− 5 =± 2 1 2 Thus, the capacity of the code is C = log2 λmax = log2 λ1 = log2 √ 1+ 5 2 1 2 = 0.3471 Problem 8.38 The state diagram of Fig P- 8-3 8 describes a runlength constrained code, that forbids any sequence containing a run of more than three adjacent symbols of the same kind The state transition matrix is 0 0 0 1 0 0 1 0 0 1 0 0 . relative difference in SNR of the error probability of 10 −6 is 2 dB. -7 -6 .5 -6 -5 .5 -5 -4 .5 -4 -3 .5 -3 -2 .5 -2 6 7 8 9 10 11 12 13 14 SNR/bit, dB log(P(e) 232 Problem 8.30 The power spectral density. 0010111000010 Transmitted seq. a n : -1 -1 1-1 11 1-1 - 1-1 -1 1-1 Received seq. b n : 2 002 0-2 -2 0 02 0 Decoded seq. d n : 10010110010 229 Problem 8.25 Let X(z) denote the Z-transform of the sequence x n ,. cases. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 -5 -4 -3 -2 -1 0 1 2 3 45 frequency f Sv(f) T=1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 -5 -4 -3 -2 -1 0 1 2 3 45 frequency f Sv(f) T=2 3) The first spectral
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