Problem Set Solution 18: Discrete-Time Processing of Continuous-Time Signals

18 Discrete-Time Processing of Continuous-Time Signals Solutions to Recommended Problems S18.1 (a) Since x,(t) = xc(t)p(t), then X,(w) is just a replication of Xe(w) centered at mul­ tiples of the sampling frequency, namely 8 kHz or 27r8 X 10' rad/s. The sam­ pling period is T = 1/8000. X,(W) 8000 T -21r X 8000 21r X 4000 27n X 8000 Figure S18.1-1 (b) X(Q) is just a rescaling of the frequency axis, where 21r8 X 103 becomes X(Q) is shown in Figure S18.1-2. 2 1r. (c) Y(Q) is the product G(Q)X(Q). Therefore, Y(Q) appears as in Figure S18.1-3. Y(2) 8000 6000 1 -2ff -- T 4 u 48 2f Figure S18.1-3 S18-1 Signals and Systems S18-2 (d) Y,(w) is a frequency-scaled version of Y(w) but only in the range 0 = --w to 7, as shown in Figure S18.1-4. Also note the gain of T. Ye(W) _ - 8000 r 4 S -8000 Figure S18.1-4 S18.2 (a) The maximum nonzero frequency component of H(w) is 5 00w. Therefore, this frequency can correspond to, at most, the maximum digital frequency before folding, i.e., Q = x. From the relation wT = Q,we get T. (b) Since w S18.2-1. = = - 500w =2 ms 500w maps to Q = 7r, the discrete-time filter G(Q) is as shown in Figure G _' .- .04 Figure S18.2-1 (c) The complete system is given by Figure S18.2-2. Note the need for an anti-alias­ ing filter. Discrete-Time Processing of Continuous-Time Signals / Solutions S18-3 S18.3 (a) Recall that Xc(w) is as given by Figure S18.3-1. Xc(w) 7r IT -10000r 100007T Figure S18.3-1 X(Q) is given by eq. (S18.3-1) and Figure S18.3-2. X(Q) 1 = - Tn 00 27rn) E = 20000 Xc[20000(9 - 27rn)] (S18.3-1) = -oo Ye(w) is given by eq. (S18.3-2) and Figure S18.3-3. TX(wT), IwI < T cI(W) elsewhere Thus x(t) = y(t) in this case. (S18.3-2) Signals and Systems S18-4 (b) Xc(w) is as given in Figure S18.3-4. Xc(w) co -54000r 540007r Figure S18.3-4 We now use eq. (S18.3-1), shown in Figure S18.3-5. 2000OXc(20000(2 - 2n)) n= 0 200007r 54T 20 n= -1 n =+ 1 iT 27r Figure S18.3-5 Thus, in the range ± r, X(Q) = 20000 E",,U X,[20000(9 Figure S18.3-6. X(2 ) 200007 -­ 20 F41T i4gr Figure S18.3-6 20 IS - 2irn)] is given as in Discrete-Time Processing of Continuous-Time Signals / Solutions S18-5 Using eq. (S18.3-2), we find Ye(w) as in Figure S18.3-7. YJ(w) -14000n 14000T Figure S18.3-7 Note aliasing since 27000 Hz is above half the sampling rate of 20000 Hz. (c) Xe(w) is as given in Figure S18.3-8. tT74 Xc(w) -34000ir 340007r Figure S18.3-8 Again we use eq. (S18.3-1), shown in Figure S18.3-9. C Signals and Systems S18-6 Thus X(Q) is given as in Figure S18.3-10. Finally, from eq. (S18.3-2) we have Y,(w) shown in Figure S18.3-11. Yc(w) 600r -6000i Figure S18.3-11 S18.4 It is required that we sample at a rate such that the discrete-time frequency 7r/ 2 will correspond to c. The relation between 9, and c is Q, = ocTo. Thus, we require = TO W"C As wc increases, demanding a wider filter, To decreases, and consequently the sam­ pling frequency must be increased. There are two ways to calculate Wa. First, since we are sampling at a rate of or (ir/2)/(Ac = 4(c ' we need an anti-aliasing filter that will remove power at frequencies higher than half the sampling rate; therefore wa = 2wc. Alternatively, we note that the "folding frequency," or the frequency at which aliasing begins, is 0 = 7r. Since 9 = r/2 cor­ responds to wc, then r must correspond to 2w,. Discrete-Time Processing of Continuous-Time Signals / Solutions S18-7 S18.5 2 (a) We sketch X(Q) by stretching the frequency axis so that 7 corresponds to the sampling frequency with a gain of 1/TO. We then repeat the spectrum, as shown in Figure S18.5-1. X(O2) 1 7T 7T Figure S18.5-1 After filtering, Y(Q) is given as in Figure S18.5-2. Y(E2) TO 3~ 3 Figure S18.5-2 (b) We see that Y(Q) looks like X(w) filtered and then sampled. The discrete-time frequency is -/3. Again, 2-x corresponds to 27r/To, so 7/3 corresponds to ir/3T. Thus, if x(t) is filtered by G(w) as given in Figure S18.5-3, then y[n] = z[n]. G(c) 3 To 3To Figure S18.5-3 Signals and Systems S18-8 Solutions to Optional Problems S18.6 (a) Since we are allowing all frequencies less than 1007r through the anti-aliasing filter, we need to sample at least twice 100r, or 2 00r. Thus, 2 007r = 2-x/T or To = 10 ms. To find K, recall that impulse sampling introduces a gain of 1/TO. To account for this, K must equal To, or K = 0.01. (b) (i) Since X(w) is bandlimited to 100-r, the anti-aliasing filter has no effect. The Fourier transform of x,(t), the modulated pulse train, is given in Figure S18.6-1. Xp(2) 1 - 00 To -4007 -1007T 1007 4007 Figure S18.6-1 Since To = 0.005, the sampling frequency is 4007r. After conversion to a discrete-time signal, X(Q) appears as in Figure S18.6-2. X(W) 200 -2w7r 2w 2 2 Figure S18.6-2 After filtering, Y(Q) is given by Figure S18.6-3. Y(n) 200 3 3~ Figure S18.6-3 Discrete-Time Processing of Continuous-Time Signals / Solutions S18-9 (ii) There are three effects to note in D/C conversion: (1) a gain of To, (2) a frequency scaling by a factor of T,, and (3) the removal of repeated spec­ tra. Thus, Y(w) is as shown in Figure S18.6-4. Y(W) IT IT 3To 3To Figure S18.6-4 S18.7 After the initial shock, you should realize that this problem is not as difficult as it seems. If instead of h[n] we had been given the frequency response H(Q), then He(w) would be just a scaled version of H(Q) bandlimited to ir/T. Let us find, then, H(Q). Using properties of the Fourier transform, we have 1 Y(Q) =- ei"YG) + X(Q) 2 1 Y(Q) X(Q) H(0) 1 - ie-j" Thus, IH(Q)I = 1 \ - cos 0 ( -< H() = -tan' sin g 1 - cos Therefore, the magnitude and phase of He(c) are as shown in Figure S18.7. 1 He(w)| = \/ -coswT' elsewhere 0, -tan-' 0, T' ( isinwT I,icoswT 1 - T elsewhere Signals and Systems S18-10 IHc(w) I r iT T T 4Hc(w) /00000* I /.A V 7V 7T T Figure S18.7 S18.8 The system under study is shown in Figure S18.8. covrsot x [n] -x en p(t) annipleConversion impulse -- T[ ) x:c(t)W yc~t W to T[n +ahst)uXnca T ypW ot) sont e -qr y [n] T ( 8(t-nT) n=­ Figure S18.8 From our previous study, we know that Xe(o) in the range ±ir/Tlooks just like X(Q) in the range ±ir. Similarly, Y,(w) between -- r/T and +r/T looks like Y(Q) in the range -ir to r. Although there is a factor of T, we can disregard it in analyzing this system because it is accounted for in the H(w) filter. The transformation of xc(t) to yc(t) will correspond to filtering x[n], yielding y[n]. In fact, the equivalent system will have a system function H(Q) given by H(Q) = He , |u| < r, where He(w) is the Fourier transform of h(t). Thus, we need to find Hc(w). The rela­ tion between yc(t) and xc(t) is governed by the following differential equation: d td + 4'(dyc dyt +y4t + ~ 3yc(tt)=t) ) = xc(t ) Discrete-Time Processing of Continuous-Time Signals / Solutions S18-11 Using the properties of the Fourier transform, we have (j) 2 Ye(w) + 4(j)Ye(w) + 3Y0(w) = Xco), 1 He(w) =1 2 (jco) + 4jw + 3 Therefore, H(Q) 2 + 4j + Jul < K + 4j- + 3 S18.9 (a) It is instructive to sketch a typical y,(t), which we have done in Figure S18.9-1. Let us suppose that T is changed by being reduced. Then the envelope of y,(t) seems to correspond to a higher-frequency cosine. At time kt, 2wk y,(t) = cos - N b(t - kT) = cos 2w(kT) 2wt 6(t - kT), NT NT 6(t - kT) = cos NT - where we use the sampling property of the impulse function. Thus, y,(t) = Y cos (t - kT) = cos wot ( ot - kT), k=-w0 k= -w0 where wo = 21/NT. If the minimum wo is wi, and since T = 21/N(o, Tmax= 2T A) I Similarly, T.in = NW2 Signals and Systems S18-12 (b) Recall that sampling with an impulse train repeats the spectrum with a period of 2r/T and a gain factor of 1/T. Since 5([cos(2t/NT)J is as given by Figure S18.9-2, Y,(w) is then given by Figure S18.9-3. 7T 7T 2,n NT 27r NT Figure S18.9-2 27r T T - 7rr 2n T T T NT Nr i 27/N - 1 N ~-WO 2| WO= r 2rN - 1\ T N 21r T Figure S18.9-3 (c) The minimum value of N is 2, corresponding to the impulses at o and (27r/T ­ wo) being superimposed at w/T. The lowpass filter cutoff frequency must be such that the (superimposed) impulses at 7r/T are in the passband and those at 37r/T are outside the passband. Consequently, ir T 3­ T (d) Comparing Y(w) and Y,(w) in Figures S18.9-2 and S18.9-3 respectively, we see that for N > 2 the cosine output will have an amplitude of 1/T = w/21r. If N = 2, then the output amplitude will be 2/T = w/7r. S18.10 (a) By sampling sc(t), we get s[n] = setnT) = x(nT) + ax(nT - TO) = x(nT) + ax[(n - 1)T] since T = To. Let x[n] = x(nT). Then s[n] = x[n] + ax[n - 1] Therefore x[n] = -ax[n - 11 + s[n] Discrete-Time Processing of Continuous-Time Signals / Solutions S18-13 This is a first-order difference equation, so given s[n], we can find x[n]. Since x(t) is appropriately bandlimited, we can then set y[n] = -ay[n - 1] + s[n] which will make yc(t) = A -x(t) T (b) From part (a) we see that T = A will make y(t) = x(t). (c) Since we do not want to alias, we still need T < 7r/wM. Now s(t) = x(t) + ax(t - TO) Taking the continuousFourier transform, we see that S(w) = X(w) + ae -iwTox(w) Thus, the continuous-time inverse system has frequency response 1 1 + ae ~-T0 We want to implement this in discrete time. Therefore, using the relation, we obtain (Q) H(Q) = H, A =i1 1 - + aaef1 jQ(T 0 /T)' ileM Again, the filter should be A = T. < M s < bM