CHAPTER 9. x(t)e j2πft dt y(t)e j2πft dt = K exp( j2πft 0 )X(f) T 2 1 sin. πt β f. sin. 2 rect 2T. 1, f a 0, o.w

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1 CHAPER 9 Problem 9. : We want y(t) =Kx(t t 0 ). hen : herefore : X(f) = x(t)e jπft dt Y (f) = y(t)e jπft dt = K exp( jπft 0 )X(f) { A(f)e jθ(f) = Ke jπft 0 Note that nπ, n odd, results in a sign inversion of the signal. A(f) =K, for all f θ(f) =πft 0 ± nπ, n =0,,,... } Problem 9. : (a) Since cos(a + π/) = sin(a), we can write : {, 0 f β X(f) = [ ( )] sin π β f, β +β f hen, taking the first two derivatives with respect to f : { X π (f) = cos ( ) π β β f, β and : X (f) = { 3 π sin π β β f +β 0, otherwise ( ) f, β 0, otherwise herefore the second derivative can be expressed as : X (f) = π [ X(f) ( ) β β rect f where : rect(af) = {, f a 0, o.w f +β } } } ( )] +β rect f Since the Fourier transform of dx/dt is jπfx(f), we exploit the duality between (f,t), take the inverse Fourier transform of X (f) andobtain: 4π t x(t) = π [ x(t) β πt sin β ] πt +β sin πt πt 95 }

2 Solving for x(t) we obtain: x(t) = 4β t / [ πt/ [ ( = 4β t / πt/ sin πt ( sin β πt +sin+β πt)] )] πβt cos (b) When β =,X(f) is non-zero in f /, and : X(f) = ( + cos πf) he Hilbert transform is : { } j ( + cos πf), 0 f / ˆX(f) = j ( + cos πf), / f 0 hen : ˆx(t) = ˆX(f)exp(jπft)dt = 0 ˆX(f)exp(jπft)dt / + / 0 ˆX(f)exp(jπf t)dt Direct substitution for ˆX(f) yields the result : [ sin πt/ 4t / ] ˆx(t) = πt Note that ˆx(t) is an odd function of t. 4t / (c) No, since ˆx(0) = 0 and ˆx(n ) 0,forn 0. Also n= ˆX(f + n/ ) constant for f /. (d) he single-sideband signal is : x(t)cosπf c t ± ˆx(t)sinπf c t = Re [ (x(t) ± jˆx(t)) e jπfct] heenvelopeisa(t) = x (t)+ˆx (t). For β= : a(t) = ( 8t πt/ 4t / / )sin (πt/)+6t 4 / 4 Problem 9.3 : (a) k h(t k) =u(t) is a periodicsignal with period. Hence, u(t) can be expanded in the Fourier series : u(t) = u n e jπnt/ n= 96

3 where : u n = = / / u(t)exp( jπnt/ )dt / / k= h(t k)exp( jπnt/ )dt = / k= / h(t k)exp( jπnt/ )dt = h(t)exp( jπnt/ )dt = H ( ) n hen : u(t) = n= H ( ) n e jπnt/ U(f) = u(t)g(t), it follows that X(f) =U(t) G(f). Hence : X(f) = n= H n= H ( n ( ) ( n G f n ) ) ( ) δ f n. Since x(t) = (b) (i) (ii) (iii) Let Hence : But k= h(k) =u(0) = k= h(t k) =u(t) = v(t) =h(t) k= V (f) = δ(t k) = n= n= k= H h(k)e jπfk k= ( ) n H ( n ) e jπnt/ h(k)δ(t k) V (f) = H(f) Fourier transform of k= δ(t k) = H(f) n= δ(f n )= n= H(f n ) (c) he criterion for no intersymbol interference is {h(k) = 0, k 0 and h(0) = }. If the above condition holds, then from (iii) above we have : n= H(f n )= k= h(k)e jπfk = Conversely, if n= H(f n )=, f k= h(k)e jπfk =, f. his is possible only if the left-hand side has no dependence on f, which means h(k) =0, for k 0. hen k= h(k)e jπfk = h(0) =. 97

4 Problem 9.4 : x(t) =e πa t X(f) = a e πf /a Hence : X(0) = a,x(w )= a e πw /a We have : X(W ) X(0) =0.0 e πw /a But due to the condition for the reduced ISI : =0.0 W = a π ln(0.0) x( )=e πa =0.0 = πa ln(0.0) Hence W = ln(0.0) =.466 or : π W =.466 For the raised cosine spectral characteristic (with roll-off factor ) W = /. Hence, the Gaussian shaped pulse requires more bandwidth than the pulse having the raised cosine spectrum. Problem 9.5 : he impulse response of a square-root raised cosine filter is given by x S (t) = +β +β X rc (f)e jπft df where X rc (f) is given by (9.-6). Splitting the integral in three parts we obtain x S (t) = + + β +β β β +β β he second term () gives immediately π / +cos( β ( f β ) ) e jπft df () e jπft df () / +cos( π β (f β ) ) e jπft df (3) () = πt sin(π( β)t/ ) 98

5 he third term can be solved with the transformation λ = f β.hen β ) πλ (3) = / +cos( +β jπt(λ+ e ) dλ 0 β Using the relationship + cos A =cos A +cosa = cos A = cosa, wecan rewrite the above expression as (3) = β 0 ( ) πλ cos e β +β jπt(λ+ Since cos A = eja +e ja, the above integral simplifies to the sum of two simple exponential argument intregrals. Similarly to (3), the first term () can be solved with the transformation λ = f + β (notice that cos( π β ( f )) = cos( π β (f + ))). hen again, the integral simplifies to the sum of β β two simple exponential argument integrals. Proceeding with adding (),(),(3) we arrive at the desired result. ) dλ Problem 9.6 : (a)(b) In order to calculate the frequency response based on the impulse response, we need the values of the impulse response at t =0, ±/, which are not given directly by the expression of Problem 9.5. Using L Hospital s rule it is straightforward to show that: x(0) = + π, x(±/) = ( + π) π hen, the frequency response of the filters with N = 0, 5, 0 compared to the frequency response of the ideal square-root raised cosine filter are depicted in the following figure. 99

6 0 0 0 Frequency response of truncated SQR Raised Cosine filters Ideal N=0 N=5 N=0 0 db t/ As we see, there is no significant difference in the passband area of the filters, but the realizable, truncated filters do have spectral sidelobes outside their ( + β)/ nominal bandwidth. Still, depending on how much residual ISI an application can tolerate, even the N = 0 filter appears an acceptable approximation of the ideal (non-realizable) square-root raised cosine filter. Problem 9.7 : (a),(b) Given a mathematical package like MALAB, the implementation in software of the digital modulator of Fig P9.7 is relatively straightforward. One comment is that the interpolating filters should have a nominal passband of [ π/3,π/3], since the interpolation factor applied to the samples at the output of the shaping filter is 3. We chose our interpolation filters (designed with the MALAB fir function) to have a cutoff frequency (-3 db frequency) of π/5. his corresponds to the highest frequency with significant signal content, since with the spectrum of the baseband signal should be (approximately, due to truncation effects) limited to (+0.5)/, so samled at 6 it should be limited to a discrete frequency of ( π (+0.5)/ )/6 0. π. he plot with the power spectrum of the digital signal sequence is given in the following figure. We have also plotted the power spectrum of the baseband in-phase (I component) of the signal. 00

7 0 0 Spectrum of baseband (In phase part) and modulated bandpass signal Baseband (I) Modulated Frequency (Hz) We notice the rather significant sidelobe that is due to the non-completely eliminated image of the spectrum that was generated by the interpolating process. We could mitigate it by choosing an interpolation filter with lower cut-off frequency, but then, we would lose a larger portion of the useful signal as well. he best solution would be to use a longer interpolation filter. (c) By repeating the experiment for a total of 6 runs we get the following figure 0 Spectrum modulated bandpass signal over 6 runs Frequency (Hz) We notice the smoother shape of the PSD, and we can verify that indeed the spectrum is centered around 800 Hz. 0

8 Problem 9.8 : (a) he alternative expression for s(t) can be rewritten as s(t)? = R { n I nq(t n )} = R { n I n e jπfcn g(t n )[cos πf c (t n )+jsin(πf c (t n )] } = R{ n I n g(t n )[cos πf c n + j sin πf c n ][cos πf c (t n )+jsin(πf c (t n )]} = R{ n I n g(t n )[cos πf c n cos πf c (t n ) sin πf c n sin πf c (t n ) +j sin πf c n cos πf c (t n )+jcos πf c n sin πf c (t n )]} = R{ n I n g(t n )[cos πf c t + j sin πf c t]} = R { n I n g(t n )e πfct} = s(t) So, indeed the alternative expression for s(t) is a valid one. (b) e jpfn q(t) q(t) e -jpfn Inr I nr Ini I ni - ^ q(t) ^ q(t) o Detector Modulator (with phase rotator) Demodulator (with phase derotator) Problem 9.9 : (a) From the impulse response of the pulse having a square-root raised cosine characteristic, which is given in problem 9.5, we can see immediately that x SQ (t) =x SQ ( t), i.e. the pulse g(t) is an even function. We know that the product of an even function times and even function has even symmetry, while the product of even times odd has odd symmetry. Hence q(t) iseven, while ˆq(t) is odd. Hence, the product q(t)ˆq(t) has odd symmetry. We know that the (symettric around 0) integral of an odd function is zero, or q(t)ˆq(t)dt = (+β)/ (+β)/ 0 q(t)ˆq(t)dt =0

9 (b) We notice that when f c = k/, wherek is an integer, then the rotator/derotaror of a carrierless QAM system (described in Problem 9.8) gives a trivial rotation of an integer number of full circles (πkn), and the carrierless QAM/PSK is equivalent to CAP. Problem 9.0 : (a) (i) x 0 =,x =,x =, otherwise x n =0. hen : and : x(t) = sin(πwt) πwt + sin(πw(t /W )) πw(t /W ) sin(πw(t /W )) πw(t /W ) [ X(f) = W +e jπf/w e ] jπf/w, f W X(f) = W [ 6+cos πf W he plot of X(f) is given in the following figure : ] πf / 4cos W, f W W X(f) fw (ii) x =, x 0 =,x =, otherwise x n =0. hen : x(t) = sin(πwt) πwt sin(πw(t +/W )) πw(t +/W ) sin(πw(t /W )) πw(t /W ) and : X(f) = W [ e jπf/w e +jπf/w] = W [ cos πf W ] = W [ cos πf W ], f W he plot of X(f) is given in the following figure : 03

10 .5.5 W X(f) fw (b) Basedontheresultsobtainedinpart(a):.5.5 (i): x(t) tw.5 (ii): x(t) tw 04

11 (c) he possible received levels at the receiver are given by : (i) B n =I n + I n I n where I m = ±. Hence : (ii) where I m = ±. Hence : P (B n =0)=/4 P (B n = ) = /4 P (B n =)=/4 P (B n = 4) = /8 P (B n =4)=/8 B n =I n I n I n+ P (B n =0)=/4 P (B n = ) = /4 P (B n =)=/4 P (B n = 4) = /8 P (B n =4)=/8 Problem 9. : he bandwidth of the bandpass channel is W = 4 KHz. Hence, the rate of transmission should be less or equal to 4000 symbols/sec. If a 8-QAM constellation is employed, then the required symbol rate is R = 9600/3 = 300. If a signal pulse with raised cosine spectrum is used for shaping, the maximum allowable roll-off factor is determined by : 600( + β) = 000 which yields β =0.5. Since β is less than 50%, we consider a larger constellation. With a 6-QAM constellation we obtain : R = 9600 = and : 00( + β) = 000 or β = /3, which satisfies the required conditions. he probability of error for an M-QAM constellation is given by : P M = ( P M ) where : ( P M = ) Q M 05 [ 3Eav (M )N 0 ]

12 With P M =0 6 we obtain P M =5 0 7 and therefore using the last equation and the table of values for the Q( ) function, we find that the average transmitted energy is : E av = Note that if the desired spectral characteristic X rc (f) is split evenly between the transmitting and receiving filter, then the energy of the transmitting pulse is : g (t)dt = G (f) df = X rc (f)df = Hence, the energy E av = P av depends only on the amplitude of the transmitted points and the symbol interval.since =, the average transmitted power is : 400 P av = E av = = If the points of the 6-QAM constellation are evenly spaced with minimum distance between them equal to d, then there are four points with coordinates (± d, ± d ), four points with coordinates (± 3d, ± 3d), and eight points with coordinates (± 3d, ± d), or (± d, ± 3d ). hus, the average transmitted power is : P av = 6 6 i=(a mc + A ms) = 3 [ 4 d +4 9d ] 0d +8 = d Since P av = ,weobtain d = 4 P av 5 = Problem 9. : he channel (bandpass) bandwidth is W = 4000 Hz. Hence, the lowpass equivalent bandwidth will extend from - to KHz. (a) Binary PAM with a pulse shape that has β =. Hence : ( + β) = 000 so = 667, and since k = bit/symbols is transmitted, the bit rate is 667 bps. (b) Four-phase PSK with a pulse shape that has β =. From (a) the symbol rate is = 667 and the bit rate is 5334 bps. (c) M = 8 QAM with a pulse shape that has β =. From (a), the symbol rate is = 667 and hence the bit rate 3 = 800 bps. 06

13 (d) Binary FSK with noncoherent detection. Assuming that the frequency separation between the two frequencies is f =,where is the bit rate, the two frequencies are f c + and f c.sincew = 4000 Hz, we may select = 000, or, equivalently, = 000. Hence, the bit rate is 000 bps, and the two FSK signals are orthogonal. (e) Four FSK with noncoherent detection. In this case we need four frequencies with separation of between adjacent frequencies. We select f = f c.5, f = f c, f 3 = f c +,and f 4 = f c +.5,where = 500 Hz. Hence, the symbol rate is = 000 symbols per second and since each symbol carries two bits of information, the bit rate is 000 bps. (f) M = 8 FSK with noncoherent detection. In this case we require eight frequencies with frequency separation of = 500 Hz for orthogonality. Since each symbol carries 3 bits of information, the bit rate is 500 bps. Problem 9.3 : (a) he bandwidth of the bandpass channel is : W = = 400 Hz Since each symbol of the QPSK constellation conveys bits of information, the symbol rate of transmission is : R = = 400 = 00 symbols/sec hus, for spectral shaping we can use a signal pulse with a raised cosine spectrum and roll-off factor β =, since the spectral requirements will be = 00Hz. Hence : ( + β) = X rc (f) = [ + cos(π f )] = 00 cos ( π f 400 If the desired spectral characteristic is split evenly between the transmitting filter G (f) and the receiving filter G R (f), then ( ) π f G (f) =G R (f) = 00 cos, f < 400 = 00 A block diagram of the transmitter is shown in the next figure. ) a n QPSK G (f) to Channel cos(πf c t) (b) If the bit rate is 4800 bps, then the symbol rate is R = 4800 = 400 symbols/sec 07

14 In order to satisfy the Nyquist criterion, the the signal pulse used for spectral shaping, should have roll-off factor β = 0 with corresponding spectrum : X(f) =, f < 00 hus, the frequency response of the transmitting filter is G (f) =, f < 00. Problem 9.4 : he bandwidth of the bandpass channel is : W = = 3000 Hz In order to transmit 9600 bps with a symbor rate R = = 400 symbols per second, the number of information bits per symbol should be : k = =4 Hence, a 4 = 6 QAM signal constellation is needed. he carrier frequency f c is set to 800 Hz, which is the mid-frequency of the frequency band that the bandpass channel occupies. If a pulse with raised cosine spectrum and roll-off factor β is used for spectral shaping, then for the bandpass signal with bandwidth W : ( + β) =W = 500 β =0.5 A sketch of the spectrum of the transmitted signal pulse is shown in the next figure. / f Problem 9.5 : he SNR at the detector is : E b = P b = P b( + β) N 0 N 0 N 0 W =30dB 08

15 Since it is desired to expand the bandwidth by a factor of 0 while maintaining the same SNR, 3 the received power P b should increase by the same factor. hus the additional power needed is 0 P a =0log 0 =5.88 db 3 Hence, the required transmitted power is : P S = =.88 dbw Problem 9.6 : he pulse x(t) having the raised cosine spectrum given by (9--6/7) is : x(t) =sinc(t/ ) cos(πβt/) 4β t / he function sinc(t/ ) iswhent = 0and0whent = n. herefore, the Nyquist criterion will be satisfied as long as the function g(t) is: g(t) = cos(πβt/) { 4β t / = t =0 bounded t 0 he function g(t) needs to be checked only for those values of t such that 4β t / =or βt =. However : cos(πβt/) 4β t / = lim cos( π x) x x lim βt and by using L Hospital s rule : cos( π lim x) x x = lim π x sin(π x)=π < Hence : { n =0 x(n )= 0 n 0 meaning that the pulse x(t) satisfies the Nyquist criterion. Problem 9.7 : Substituting the expression of X rc (f) given by (8..) in the desired integral, we obtain : β [ X rc (f)df = +cos π β ( f β ] β ) df + df +β 09 β

16 = +β + β β + +β = + = + β +β 0 β β β [ +cos π β (f β ) ( ) β +β df + + β ] df df cos π β (f + β +β )df + cos π β xdx + β cos π β 0 β cos π β xdx xdx =+0= cos π β (f β )df Problem 9.8 : Let X(f) be such that Re[X(f)] = { Π(f)+U(f) f < 0 otherwise Im[X(f)] = { V (f) f < 0 otherwise with U(f) even with respect to 0 and odd with respect to f = Since x(t) isreal,v (f) isodd with respect to 0 and by assumption it is even with respect to f =. hen, x(t) = F [X(f)] = = X(f)e jπft df + e jπft df + = sinc (t/ )+ Consider first the integral X(f)e jπft df + [U(f)+jV (f)]e jπft df [U(f)+jV (f)]e jπft df U(f)e jπft df. Clearly, X(f)e jπft df U(f)e jπft df = 0 U(f)e jπft df + 0 U(f)e jπft df and by using the change of variables f = f + and f = f right hand side respectively, we obtain for the two integrals on the U(f)e jπft df 0

17 = e j π t a = ( e j π t e j π t) = j sin( π t) U(f )ejπf t df + e j π t U(f + )ejπf t df U(f + )ejπf t df U(f + )ejπf t df where for step (a) we used the odd symmetry of U(f ) with respect to f =,thatis U(f )= U(f + ) For the integral V (f)e jπft df we have = V (f)e jπft df 0 = e j π t V (f)e jπft df + 0 V (f)e jπft df V (f )ejπf t df + e j π t V (f + )ejπf t df However, V (f) is odd with respect to 0 and since V (f + ) are even, the translated spectra satisfy Hence, V (f )ejπf t df = x(t) = sinc (t/ )+j sin( π t) sin( π t) U(f + )ejπf t df and therefore, { n =0 x(n )= 0 n 0 hus, the signal x(t) satisfies the Nyquist criterion. V (f + )ejπf t df U(f + )ejπf t df Problem 9.9 : he bandwidth of the channel is : W = = 700 Hz

18 Since the minimum transmission bandwidth required for bandpass signaling is R, wherer is the rate of transmission, we conclude that the maximum value of the symbol rate for the given channel is R max = 700. 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 = k = 6. In this case, the symbol rate is : R = 9600 = 400 symbols/sec k and the symbol interval = = sec. he roll-off factor β of the raised cosine pulse used R 400 for transmission is is determined by noting that 00( + β) = 350, and hence, β =0.5. herefore, the squared root raised cosine pulse can have a roll-off of β =0.5. Problem 9.0 : Since the one-sided bandwidth of the ideal lowpass channel is W = 400 Hz, the rate of transmission is : R = 400 = 4800 symbols/sec (remember that PAM can be transmitted single-sideband; hence, if the lowpass channel has bandwidth from -W to W, the passband channel will have bandwidth equal to W ; on the other hand, a PSK or QAM system will have passband bandwidth equal to W ). he number of bits per symbol is k = =3 Hence, the number of transmitted symbols is 3 = 8. If a duobinary pulse is used for transmission, then the number of possible transmitted symbols is M = 5. hese symbols have the form b n =0, ±d, ±4d,...,±d where d is the minimum distance between the points of the 8-PAM constellation. he probability mass function of the received symbols is P (b =md) = 8 m, m =0, ±,...,±7 64 An upper bound of the probability of error is given by (see (9-3-8)) ( P M < ) M With P M =0 6 and M =8weobtain Q (π 4 ) 6 M ke b,av N 0 ke b,av N 0 = = E b,av =0.088

19 Problem 9. : (a) he spectrum of the baseband signal is (see (4-4-)) where = 400 and Φ V (f) = Φ ii(f) X rc (f) = X rc(f) 0 f 4 X rc (f) = ( + cos(π( f 4 )) f otherwise If the carrier signal has the form c(t) =Acos(πf c t), then the spectrum of the DSB-SC modulated signal, Φ U (f), is Φ U (f) = A [Φ V (f f c )+Φ V (f + f c )] AsketchofΦ U (f) is shown in the next figure. A -fc-3/4 -fc -fc+3/4 fc-3/4 fc fc+3/4 (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)cos(πf c ( t)) he output of the matched filter is sampled at t = and the samples are passed to the detector. he detector is a simple threshold device that decides if a binary or 0 was transmitted depending on the sign of the input samples. he following figure shows a block diagram of the optimum bandpass coherent demodulator. r(t) Bandpass matched filter g R (t) t =. Detector (hreshold device) 3

20 Problem 9. : (a) he power spectral density of X(t) is given by (see (4-4-)) he Fourier transform of g(t) is Φ x (f) = Φ a(f) G (f) G (f) =F[g(t)] = A sin πf πf e jπf Hence, and therefore, G (f) =(A ) sinc (f) Φ x (f) =A Φ a (f)sinc (f)=a sinc (f) (b) If g (t) is used instead of g(t) andthesymbolintervalis,then Φ x (f) = Φ a(f) G (f) = (A ) sinc (f )=4A sinc (f ) (c) If we precode the input sequence as b n = a n + αa n 3,then φ b (m) = and therefore, the power spectral density Φ b (f) is +α m =0 α m = ±3 0 otherwise Φ b (f) =+α +α cos(πf3 ) o obtain a null at f =, the parameter α should be such that 3 +α +α cos(πf3 ) f= 3 =0= α = (c) he answer to this question is no. his is because Φ b (f) is an analyticfunction and unless it is identical to zero it can have at most a countable number of zeros. his property of the analyticfunctions is also referred as the theorem of isolated zeros. 4

21 Problem 9.3 : he roll-off factor β is related to the bandwidth by the expression +β =W,orequivalently R( + β) =W. he following table shows the symbol rate for the various values of the excess bandwidth and for W = 500 Hz. β R he above results were obtained with the assumption that double-sideband PAM is employed, so the available lowpass bandwidth will be from W = 3000 to W Hz. If single-sideband transmission is used, then the spectral efficiency is doubled, and the above symbol rates R are doubled. Problem 9.4 : he following table shows the precoded sequence, the transmitted amplitude levels, the received signal levels and the decoded sequence, when the data sequence modulates a duobinary transmitting filter. Data seq. D n : Precoded seq. P n : ransmitted seq. I n : Received seq. B n : Decoded seq. D n : Problem 9.5 : he following table shows the precoded sequence, the transmitted amplitude levels, the received signal levels and the decoded sequence, when the data sequence modulates a modified duobinary transmitting filter. Data seq. D n : Precoded seq. P n : ransmitted seq. I n : Received seq. B n : Decoded seq. D n :

22 Problem 9.6 : Let X(z) denotethez-transform of the sequence x n,thatis X(z) = n x n z n hen 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 dequences respectively. For example, if M =andx(z) =+z (duobinary signaling), then P (z) = D(z) +z = P (z) =D(z) z P (z) which in the time domain is written as p n = d n p n and the subtraction is mod-. However, the inverse filter exists only if x X(z) 0, the first coefficient of X(z) is relatively prime with M. If this is not the case, then the precoded symbols p n cannot be determined uniquely from the data sequence d n. In the example given in the book, where x 0 = we note that whatever the value of d n (0 or ), the value of (d n mod ) will be zero, hence this precoding scheme cannot work. Problem 9.7 : (a) he frequency response of the RC filter is C(f) = jπrcf R + jπrcf he amplitude and the phase spectrum of the filter are : C(f) = heenvelopedelayis = +jπrcf ( ), θc (f) =arctan( πrcf) +4π (RC) f τ c (f) = dθ c (f) = πrc π df π +4π (RC) f = RC +4π (RC) f 6

23 Aplotofτ(f) withrc =0 6 is shown in the next figure : c(f) x Frequency (f) (b) he following figure is a plot of the amplitude characteristics of the RC filter, C(f). he values of the vertical axis indicate that C(f) can be considered constant for frequencies up to 000 Hz. Since the same is true for the envelope delay, we conclude that a lowpass signal of bandwidth f = KHz will not be distorted if it passes the RC filter. C(f) Frequency (f) Problem 9.8 : Let G (f) andg R (f) be the frequency response of the transmitting and receiving filter. hen, the condition for zero ISI implies, 0 f 4 G (f)c(f)g R (f) =X rc (f) = [ + cos(π( f )], f , f > 3 4 Since the additive noise is white, the optimum tansmitting and receiving filter characteristics 7

24 are given by (see (9--8)) G (f) = X rc(f) C(f), G R (f) = X rc(f) C(f) hus, G (f) = G R (f) = [ +0.3cosπf [ (+cos(π( f ) (+0.3cosπf) ], 0 f 4 ], f , otherwise Problem 9.9 : A 4-PAM modulation can accomodate k = bits per transmitted symbol. hus, the symbol interval duration is : = k 9600 = 4800 sec Since, the channel s bandwidth is W = 400 =, in order to achieve the maximum rate of transmission, R max =, the spectrum of the signal pulse should be : ( ) f X(f) = Π W hen, the magnitude frequency response of the optimum transmitting and receiving filter is (see (9--8)) G (f) = G R (f) = + ( f 400 ) 4 Π ( ) f = W [ + ( ) ] f 4, 400 f < 400 0, otherwise Problem 9.30 : We already know that σ v = Φ nn (f) G R (f) df W P av = d G (f) df W X rc (f) G (f) =, f W G R (f) C(f) 8

25 From these σ v d = P av W W Φ nn (f) G R (f) df W W X rc (f) df (4) G R (f) C(f) he optimum G R (f) can be found by applying the Cauchy-Schwartz inequality [ ] U (f) df U (f) df U (f) U (f) df where U (f), U (f) are defined as U (f) = Φ nn (f) G R (f) X rc (f) U (f) = G R (f) C(f) he minimum value of () is obtained when U (f) is proportional to U (f), i.e. U (f) = K U (f) or, equivalently, when G R (f) = X rc (f) / K, f W [N 0 /] /4 C(f) / where K is an arbitrary constant. By setting it appropriately, G R (f) = X rc(f) / C(f) /, f W he corresponding modulation filter has a magnitude characteristic of G (f) = X rc (f) G R (f) C(f) = X rc(f) / C(f) /, f W Problem 9.3 : In the case where the channel distortion is fully precompensated at the transmitter, the loss of SNR is given by W X rc (f) 0 log L, with L = W C(f) whereas in the case of the equally split filters, the loss of SNR is given by 0 log[l ], with L = W W X rc (f) C(f) Assuming that / = W, so that we have a raised cosine characteristic with β =0,wehave 9

26 hen L = W [ 0 W/ = X rc (f) = W 0 W [ [+cos πf W ] C(f) W +cos π f W ] [+cos πf W ] + W W/ W = 5π 6 π Hence, the loss for the first type of filters is 0 log L =.89 db. In a similar way, L = W [ 0 W/ = 0 = 3π π W [+cos πf W ] C(f) W [+cos πf W ] + W W/ W ] [+cos πf W ] /4 ] [+cos πf W ] / Hence, the loss for the second type of filters is 0 log[l ] =.45 db. As expected, the second type of filters which split the channel characteristics between the transmitter and the receiver exhibit a smaller SNR loss. Problem 9.3 : he state transition matrix of the (0,) runlength-limited code is : ( ) D = 0 he eigenvalues of D are the roots of he roots of the characteristic equation are : det(d λi) = λ( λ) =λ λ λ, = ± 5 hus, the capacity of the (0,) runlength-limited code is : C(0, ) = log ( ± 5 )=0.694 he capacity of a (, ) code is found from able 9-4- to be As it is observed, the two codes have exactly the same capacity. his result is to be expected since the (0,) runlengthlimited code and the (, ) code produce the same set of code sequences of length n, N(n), 0

27 with a renaming of the bits from 0 to and vise versa. For example, the (0,) runlength-limited code with a renaming of the bits, can be described as the code with no minimum number of s between 0 s in a sequence, and at most one between two 0 s. In terms of 0 s, this is simply the code with no restrictions on the number of adjacent 0 s and no consequtive s, that is the (, ) code. Problem 9.33 : Let S 0 represent the state that the running polarity is zero, and S the state that there exists some polarity (dc component). he following figure depicts the transition state diagram of the AMI code : 0/0 /s(t) S 0 S 0/0 / s(t) he state transition matrix is : D = ( he eigenvalues of the matrix D can be found from det(d λi) =0= ( λ) =0orλ( λ) =0 he largest real eigenvalue is λ max =,sothat: ) C =log λ max = Problem 9.34 : Let {b k } be a binary sequence, taking the values, 0 depending on the existence of polarization at the transmitted sequence up to the time instant k. For the AMI code, b k is expressed as b k = a k b k = a k a k a k... where denotes modulo two addition. hus, the AMI code can be described as the RDS code, with RDS (=b k ) denoting the binary digital sum modulo of the input bits.

28 Problem 9.35 : Defining the efficiency as : we obtain : efficiency = k n log 3 Code Efficiency B B B B Problem 9.36 : (a) he characteristic polynomial of D is : det(d λi) =det λ λ = λ λ he eigenvalues of D are the roots of the characteristic polynomial, that is λ, = ± 5 hus, the largest eigenvalue of D is λ max = + 5 and therefore : C =log + 5 =0.694 (b) he characteristic polynomial is det(d λi) = ( λ) with roots λ, =. Hence, C =log = 0. he state diagram of this code is depicted in the next figure. 0 S 0 S (c) As it is observed the second code has zero capacity. his result is to be expected since with thesecondcodewecanhaveatmostn + different sequences of length n, sothat C = lim n n log N(n) = lim n n log (n +)=0

29 he n + possible sequences are 0 }...0 {{} }... {{} k n k (n sequences) and the sequence..., which occurs if we start from state S. Problem 9.37 : (a) he two symbols, dot and dash, can be represented as 0 and 0 respectively, where denotes line closure and 0 an open line. Hence, the constraints of the code are : A 0 is always followed by. Only sequences having one or three repetitions of, are allowed. he next figure depicts the state diagram of the code, where the state S 0 denotes the reception of a dot or a dash, and state S i denotes the reception of i adjacent s. S S S S 0 (b) he state transition matrix is : D = (c) he characteristic equation of the matrix D is : he roots of the characteristic equation are : det(d λi) =0= λ 4 λ =0 λ, = ± ( + 5 ) λ 3,4 = ± ( 5 ) 3

30 hus, the capacity of the code is : ( ) + 5 C =log λ max =log λ =log =0.347 Problem 9.38 : he state diagram of Fig. P9-3 describes a runlength constrained code, that forbids any sequence containing a run of more than three adjacent symbols of the same kind. he state transition matrix is : D = he corresponding trellis is shown in the next figure : History Problem 9.39 : he state transition matrix of the (,7) runlength-limited code is the 8 8 matrix : D =