Problem Set Solution 10: Discrete-Time Fourier Series

10 Discrete-Time Fourier Series Solutions to Recommended Problems S10.1 The output of a discrete-time linear, time-invariant system is given by y[n] = j h[kjx[n - k], k= ­ where h[n] is the impulse response and x[n] is the input. By substitution, we have the following. (a) y[n] = )k k=0 2 e jwn 1 3 i- k (b) y[n] = k=0 1 (c) y[n] = k)4 (1ke j,(n- ()k[ 0 - ej k=0 fjir/4) 2 - e 8 ) i(n -)/4) [e e (/8) e(rn/4) 1e +±2e 3 (w/4 ) 1 k Cos ( k)/41 2 where we have used Euler's relation e 2 ] k=0 S -j[(/8)+(rn/4)] -j(r/4) + 2e n + -) 2Cos (4n 4 ''r8e - '''"- +21-e -j(/ 8)e -j(rn/4) k=0 Iej[(1/ 8 )+(in/4)] 1 ] 2 Cr 4 -o - ­ 1 (,r/4) O s "4 S1O.2 (a) 1 1[n] = 1 + sin( 10f) To find the period of t 1[n], we set 2 1 [n] = 1 1[n + N] and determine N. Thus 27r 1 + sin [1 0 ( 1 + sin 21rn = 1 + sin n + N N) Since 2s sin (0n + 2r) =sin 27n 10n S10-1 Signals and Systems S10-2 (i the period of t1n] is 10. Similarly, setting t 2[n] ns 207 1+ sin (2 n +-2 12 = t 20ir 2 [n + N], we have = 1+ sin -1+ 7+ 20-) sin 20-x n + - + -N 2 12 (12 2(n +N) +-2 Hence, for -irN to be an integer multiple of 21r, N must be 6. u rn (b) 21[n] = 1 + sin (21 Using Euler's relation, we have x 1 [n] 1 ± + = 2j e27non _ J(2r/10)n e 2j (S10.2-1) Note that the Fourier synthesis equation is given by ake jk(2x/N)n 21l[n] =. T k= (N) where N = 10. Hence, by inspection of eq. (S10.2-1), we see that ao -1 = all = alk = 2j 1 and 2j 0, 2 s k - 8, -8 : k - -2 Similarly, t2[n] 1 e = 1 + e-j(20Y/12> -j(-/2> 2j Therefore, N = 12. a 2 0 = 1, a 2 -1 e -fir/2) 2j a2 ± 2, (c) The sequence alk 1 a 21 2' . . . , a2±= 1 - - 23 ear/2> 1 = -, 2 and 0 is periodic with period 10 and a 2k is periodic with period 12. S10.3 The Fourier series coefficients can be expressed as the samples of the envelope ak 1 N N sin[(2Ni + 1)9/2] sin (9/2) sin (39/2) sin (0/2) 0=2irk/N 0=2wk/N where N, = 1 (see Example 5.3 on page 302 of the text) (a) For N = 6, ak =- sin 3(2k)] 1 sin [26/) 6 sin 1 2r 26 1 sin (2) i (k sin -- ) Discrete-Time Fourier Series / Solutions S10-3 (b) For N = 12, ak .i [3 (27rk{ 1 sin 2 / = 1 ___________ - 12 [112 12 (c) For N = 60, (27rk 2n 60)] [1 (2ck .i [ k 1 60 1 s 60 20) sin 60 sin sin. -grk S10.4 (a) The discrete-time Fourier transform of the given sequence is X(Q) = x[n]e -'"" ( n= -o = leit + 1 + ie ­ = 1 + cos Q X(Q) is sketched in Figure S10.4. X(E2) IT IT Figure S10.4 (b) The first sequence can be thought of as 9,[n] = x[n] * T, 6[n - 3k]I Lk =0-Go Hence 21 YI(Q) = 0" X(Q) 2,rk - 3 , Therefore, the Fourier series of y 1 [n] is given by 2 k\ 1( 1 (2 k , ak=--Y1--3 kJ=-1+cos for all k Signals and Systems S1O-4 The second sequence is given by [n - 5k] y 2[n] = x[n]* Similarly, the Fourier series of this sequence is given by ak = 5- 1 1 + cos for all k , (5 This result can also be obtained by using the fact that the Fourier series coeffi­ cients are proportional to equally spaced samples of the discrete-time Fourier transform of one period (see Section 5.4.1 of the text, page 314). S10.5 (a) The given relation 3 x[n] ake k(2r'')" = k=o results in the following set of equations ao + ai + ao + aie^,fir/2> + a 2e'" + ao + alei" + a2e 2 ' ao + aie(3 / 2) + a2e 3' + a2 + a3 = x[O] = 1, aej(3 / 2) = x[1] = 0, + a3e 3 v = x[2] = 2, asej( 9/ 2) = x[3] = - 1 The preceding set of linear equations can be reduced to the form ao + a1 + a 2 + a3 = 1, ao + jai - a 2 -ja 3 ao - a1 + a 2 - a 3 ao - jai - a 2 + ja 3 = 0, = 2, -1 = Solving the resulting equations, we get ao = 1 , a- __1+j 2 4 , a 2 = +1, a= - 1-j 4 4 (S10.5-1) By the discrete-time Fourier series analysis equation, we obtain ak = 1 + 2e -ik which is the same as eq. (S10.5-1) for 0 - e -j(3rk/2)j, k - 3. S10.6 (a) ak = ak+10 for all k is true since t[n] is periodic with period 10. (b) ak = a_, for all k is false since I[n] is not even. (c) ake ik(21/) is real. This statement is true because it would correspond to the Four­ ier series of t[n + 2], which is a purely real and even sequence. (d) ao = 0 is true since the sum of the values of I[n] over one period is zero. Discrete-Time Fourier Series / Solutions S10-5 Solutions to Optional Problems S10.7 The Fourier series coefficients of x[n], which is periodic with period N, are given by ak = - j(wNf T x[n]e -jk(2w/N)n NI n=O ForN = 8, E x[n]e-k(/ ak = 4 (S10.7-1) )n 8n=0 (a) We are given that ak = cos (-)+ sin 4 3k, (S10.7-2) (4 ak = .1ej(wk/4) + -e -j(ik/4) + 1 e-t3rk/4) 2j 2 2 _ e -(3rk/4) - 2j Hence, by comparing eqs. (S10.7-1) and (S10.7-2) we can immediately write x[n] = 46[n - 1] + 46[n - 7] - 4jb[n - 3 ] + 4jb[n - 51, n !5 7 0 7 7 (b) x[n] k k(2r/8) jk(/4)n k=O k=O ei(kr/3) 1e-j(kr/3)] jk(/4)n II k=O k=O 7 wn/ 4)+(7r/3) 1 [e( 2j 1 4 - 1 1 [(7n/4)--(7r/3) 2 1 3 )+(-/ )] [eJ(rn/ j[(rn/4)-(/3) 4 {( -n/ )-(7-/3)j 7 S[(7n/4)+(7r/3)] 2j 1 ee[(rn/4 )+(v/ 7 7 Sx[n] 3 k=O 3 4 1 + ej(/ 4 )n + ej( -/ =1++ (-) j[(n/4)-(r/3)j akjk(w/4)n k jk(2/8) k=O 1 ) )n 7 4 + ejrn + ej(51/ 4 )n + ej( r/ )n + 2 cos +2cos (n n 0 (d) Using an analysis similar to that in part (c), we find x[n] = 2 + 2 cos4 n+ + co S10.8 The impulse response of the LTI system is h[n] = (i)'" Cos n, n 7 Signals and Systems S10-6 The discrete-time Fourier transform of h[n] is (- H(Q) = ( e -j"" + ~" ~) -- 1 n=-o n=O 1 e~" 1+ 3 1 1 -i -1 5 - 4 cos 0 (a) (i) 1j n 3r= -e3r!4)n x[n] = sin 2j 4 1 - 2j e -j(3,/4)n The period of x[n] is sin ( i 3 - (n + N) n) = sin Thus 3) sin 3r n + -rN) (4 We set 37rN/4 = 21rm to get N = 8 (m = 3). Hence, the period is 8. si 4r 7 T akejk(2w/8)n x[n] k=O Therefore, 1 a3 = =- a5 All other coefficients a, are zero. By the convolution property, the Fourier series representation of y[n] is given by bk, where bk = T aJH(C s0=(2k)/8 Thus 1 b 3 2j 5 - 4 cos(3r/4) =b* All other bk are zero in the range 0 (ii) I[n] = 6[n - k - 7. 4k] k= -o The Fourier series of t[n] is 3 t4n~e -jk<2/4> ak = = 1, f r all k n=O And the Fourier series of D[n] is bk akH(Q)I 1 4 -=5 = - 3 4 cos[(2r/2)k] 3 20 f or all k Discrete-Time Fourier Series / Solutions S10-7 (iii) The Fourier series of f[n] is (k), 1 + 2 cos ak = 0 s k - 5 and the Fourier series of 9[n] is bk = akH(Q) 3 + 2 cos Irk 1 (3)5 - 4 cos[(7r/3)k] (iv) x[n] = j" + (-1)" The period of k[n] is 4. x[n] can be rewritten as 2 x[n] = [ej(r/ )n + (e'")" 3 k T jk(2/4)n k=O Hence, ai = 1, a3 = 0 ao = 0, a2 = 1, Therefore, bo = b = 0 and bi b2 = 3 5 - 4 cos(7r/2) 3 3 5 - 4 cos r E 9 (b) h[n] is sketched in Figure S10.8. h [n ] -2 -1 012 Figure S10.8 H(Q) = h[n]e-"a" E -_,;2 - e'o + 1 + e-~" + e-j2Q n = -00 H(Q) = 1 - 2j sin 0 - 2j sin 20 It follows from part (a): 1 b3 = --j H(Q) =31r/4 1 2j 3-7r sin - 4 -sin All other coefficients bk are zero, in the range 0 -- = b* 2 k 5 7. Signals and Systems S10-8 (ii) H) bk = rk-j 1 j. 42 (iii) bk = - (iv) b0 = 0, sin- - 2 1 + 2 cos ( b 1 = H(Q) sin 2 k k =- 1 4 H(Q) - j .7rk -sin­ 2 2 f=rk/3 =1 - 2j, b 2 = H(Q) =1, 0 b= S10.9 x[n] ak (a) x[n (b) jk(2w/N)n0 k 50 -E2= ake-j(/Nn -no] x[n] - x[n - (c) x[n] - x [n ­ N ak[l - 5 11 . 2 (d) x[n] + x n + -], e -j(2rk/N) -jr ak(l - ei-*), 0, k even, 2ak, k odd period N 2 ( 2 (NI x[nI + x Nn=O = ) a* = L =*[- 2 a2k e -jk( 2 /N)n Nn=O N1 N-i1 jk( 2 /N)n 1-N+1 =- ~Lx[n]e -jk(2/N)n ak Therefore, ak = atk. S10.10 (a) N even i'i[n] = I[n] + 9[n], i[n + NM]= .[n + NM] + P[n + NM] = t[n] + 9[n] = f>[n] Hence, ?ID[n] is periodic with period NM. n + - e -jk(4/N)n Discrete-Time Fourier Series / Solutions S10-9 (b) Z NMn=0 NM-I Ck = 1 NM-1 n E jk(2W/NM)n [nje ZNMn4N 1 + -jk(2/NM)n N n=0 >3 n=0 M-1 N-1 NM '1 N jk(2T/NM)n NM=1 NM-1 = [N[n] + P[n]]e = e -jk(2./NM)(n+LN) [n] E 1=0 + n 1 akIM + jk(2/NM)n :P4nje-j(/N 1 M-1 N-1 NMnO 1=0 N ik(2/NM)(n+IM) for k a multiple of M and N, 1bk/N, 1 akIM, for k a multiple of M, 1 1 bk/N, for k a multiple of N, 0, otherwise S10.11 (a) t[n] = sin [(4 To find the period, we set g[n] = T[n + N]. Thus, sin [(n - 1)] sin [(n - 1)+ N + N - 1) i(n Let (7rN)/4 = 21i, when i is an integer. Then N = 8 and 1[n] 1 e~ir(n- 1 )/ 41 1 - 2j 1 e e -il(n -- 1)/41 2j j(w/4) -j(,rn/4) j(/4) j(rn/4) 2j 2j Therefore, e -j(i/4) 2j 2j All other coefficients ak are zero, in the range 0 : k - 7. The magnitude and phase of ak are plotted in Figure S10.11-1. lak I 0 1 2 3 4 5 6 4ak 0 - 31r1 7 3 2 I 3 4 5 6 7 1 Figure S10.11-1 k Signals and Systems S10-10 (b) The period N = 21 and the Fourier series coefficients are 1 a 7 = a 14 = a 3 = a, 2 1 8 = ­ 2j The rest of the coefficients ak are zero. The magnitude and phase of ak are given in Figure S10.11-2. |ak 0 1 2 3 4 k 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 4ak 2 3 0 1 2 4 5 6 7 8 9 10 11 12 13 -k 14 15 16 17 18 19 20 2 Figure S10.11-2 (c) The period N = 8. a 3 = a* = e -j(-/3) The rest of the coefficients ak are zero. The magnitude and phase of ak are given in Figure S10.11-3. 1 akk 1 0 1234567 4ak 3 3 0 1 2 0 4 5 k 6 Fr 3 Figure S10.11-3 7