HW11solution

# HW11solution - ¢ «e law 201 mm a EE 301 10.21 The...

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Unformatted text preview: ¢ «e; law 201 mm a; ,_ EE 301 10.21. The pole-zero plots are all shown in Figure \$10.21. (a) For :c[n] = 6{n + 5], X(z) = 25, All 2. The Fourier transform exists because the ROC includes the unit circle. (b) For :r[n] = 6[n — 5], Xe) = 2-5, All 2 except 0. The Fourier transform exists because the ROC includes the unit circle. (0) For min] = (~1)"uln]a 00 Z \$[n]z_" n=~—oo 00 Z(_l)nz~n 71:0 -_— 1/(1+z"), lz|>1:: ﬁ‘ ) The Fourier transform does not exist because the ROC does not include the unit circle. X(z) ll ‘ RM L ’(e) For = (~1/3)"u[—-n — 2], X(Z) = Z\$[n]z_” H M8 A I H \ 33 I NT ’2‘?" 3...» m 4- :3 i! .7.— l ,\ oo z “" n = Z(_1/3)—n~—2zn+2 '1‘le a: Ru 3 3:) ":0 D10 = 9z2/(1 + 32:), Izl < 1/3 32—2 32/(1 + (1/3)z_1), I2] <1/3 3" L2, _f_¥/3 jl‘glAYB‘ The Fourier transform does not exist because the ROC does not include the unit circle (g) Consider m1[n] = 2"u[—-n]. X1(Z) = Z (1:1[n]z_" n.0—oo = Z (2)nz-n I Mr i<=~n H M8 E gr N Iii N. A Ni V7, karo i‘ 1/(1 — (1/2)Z)» [2! < 2 _2z—1/(1—22—1)3 Consider 3:2[71] = (1/4)"u[n — 1]. X2(Z) H i(1/4)k+lz—k—l k=o (z‘1/4)[1/(1 - (1/4)Z‘1)], [2 > 1/4 The z-transform of the overall sequence \$[n] = 2:1[n] + x2[n] is 22—1 (1 —22‘1) X (z) C8) Figure \$10.21 + . 2‘1/4‘ 1— (1/4)z-1’ (‘53 (1/4) < < 2. 5°) __ 10.24. (a) We may write X(z) as 1 , V 1— 22—1 r «a r i: W‘ (1—%z“><1-2Z‘1) 3’" 'ﬁwi 2 gig/'52» Therefore, 1 l g W B _ 3% X05) -* 1_ %z~1 If is absolutely summable, then the ROC of X has to include the unit circle. Therefore, the R00 is |z| > 1/2. It follow that \$[n] = u[n], (b) Carrying out long division on X(z), we get 1 ‘ 1 X(z) = l — z"1+ 52—2 — 21—24 Using the analysis equation (10.3), we get a:[n] = 5m] — can—1 u[n -1]. (c) We may write X(z) as 324'1 3z~1 X z = ~——-—-———— = —————————————————-. ( ) l—ﬁz‘1 —£1—;z"‘2 (1 —— %z‘l)(1 + 12—1) The partial fraction expansion of X is 4 4 Xz=.—______...__._.‘ () 1—%2‘1 1+§z—1 Since \$[n] is absolutely summable, the ROC must be |z| > 1/2 in order to include the unit circle. It follows that i”) zz:[n] = 4 u[n] — 4 u[n]. - g” ('gilill-zil‘l) I“ W "l" 'erW 3(2)»: iii “if e m: WM 6-Wllutw‘1 «5 \$12; vi, The Fourier transform exists because the ROC includes the unit circle. ' .' 10.33. (a) Taking the z—transform of both sides of the given difference equation and simplifying, We get '2‘ H(z)vY(z)~_____1_____ w ,3" _ i X“) levee—2‘ -¥;Z“‘“%%"Tf The poles of H(z) are at (1/4) ij(\/§/4). Since h[n] is causal, the ROC has to be M > 1(1/4) +j(\/§/4)| = (1/2)~ .3 (b) We have '3 1 1 X(Z) -—— —1_—%Z:, [2| > Therefore, 1 (1 -i13z"1)(1—-21-z”1+ﬁz“2)' The ROC of Y(z) will be the intersection of the R005 of X(z) and H(z). This implies that the ROC of Y(z) is > 1/2. The partial fraction expansion of Y(z) is h< E II 5 3 l 1 2‘1/2 Y z = — + m. ( ) 1— %z”1 1 — %z“1 + %z"2 Using Table 10.2 we get y[n] = (97114711 + 5—3.) (gym u[n]. K3 FHt‘Eescm a may cal gmﬁslw exgmwéam": will”? Tl? ciaéiclmlc if???“ ,_,Q,_7,)Sggg2c>se ,i‘i‘ , if m,ul‘fl_u_‘),,(y_ lei ,, £4 ,_,&hcl _ rei’iSiclef ‘ P lin 10.37. (a) The block-diagram may be redrawn as shown in part (a) of the figure below. This may be treated as a cascade of the two systems shown within the dotted lines in Figure \$10.37. These two systems may be interchanged as shown in part (b) of the ﬁgure Figure 310.37 without changing the system function of the overall system. From the ﬁgure below, it is clear that Figure \$10.37 (b) Taking the z—transform of the above difference equation and simplifying, we get H(z) = Yiz) = ILL} '" gz—l X(z) 1+ Eli-2’1 — 32—2 ” (1+§z‘1)(1——%z—1)' H(z) has poles at z = 1/3 and z = —2/3. Since the system is causal, the ROC has to be > 2/3. The ROC includes the unit circle and hence the system is stable. ...
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HW11solution - ¢ «e law 201 mm a EE 301 10.21 The...

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