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A Few Schaums Probs on DTFT Stuff

# A Few Schaums Probs on DTFT Stuff - 326 FOURIER ANALYSIS OF...

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Unformatted text preview: 326 FOURIER ANALYSIS OF DISCRETE—TIME SIGNALS AND SYSTEMS [CHAR 6 N 1 and 1,1,[11]: Z 5[/c]+——>7T5(Q)+1_6_jn IQIS’IT k=~0° FREQUENCY RESPONSE A causal discrete—time LTI system is described by y[n] Zy[n 'l] +;y[n 2] —x[n.] (6.143) where x[n.] and y[n] are the input and output of the system, respectively (Prob. 4.32). (a) Determine the frequency response H(Q) of the system. (b) Find the impulse response h[n] of the system. (a) Taking the Fourier transform of Eq. (6.143), we obtain I A _. . _. M WM rm — ge Jar/(o) + ge Mir/(o) =X(n) I “Mutual 01‘ (1—%erm + pump/(o) =X(o) l‘ a 1 Thus, *4“ .unu Ym) 1 1 Hal) X(Q) 1—3e‘j0+1e”jm (1—1e“m)(1~\$e‘jn) 4 s 2 4 m“ I (b) Using partial-fraction expansions, we have ' “‘ll'lllrllll - 1 2 1 ‘ H .0 — . . . _ imm'mlhhmm ll ( ) (1 — %e‘1“)<1 * ie_’”) 1 — if“) 1 — i6 "1“ “Mill! . . . . Taklng the Inverse Fourier transform of H (.0), we obtain Wainwrle h[ 1 [ (1),, (1H [ ] l n = 2 5 — z u n ‘ minim r . . . . “will” Wthh 1s the same result obtained in Prob. 4.»32(b). \l- . / ll Consider a discrete—time LTI system described by ‘ y[n] —]§y[n—1]=x[n] +%x[n—1] (6.144) (a) Determine the frequency response H (0.) of the system. (b) Find the impulse response h[n] of the system. (c) Determine its response y[n] to the input 7T x[n] =cos—2—n (a) Taking the Fourier transform of Eq. (6.144), we obtain rm) — %e"mY(ﬂ) =X(Q) + §e~f“X(n) 327 CHAP. 6] FOURIER ANALYSIS OF DISCRETE—TIME SIGNALS AND SYSTEMS 31 Thus, Y(Q) 1+§e"“ H o =—»——=————.— ( ) X(Q) 1~%e”’n (b) Hm) 1 +1 [in d 1—%eﬁm 2 1—%e‘j0 Taking the inverse Fourier transform of H (0), we obtain _ 1 ll [1—1 I .— 1 n: h[n]—(E)u[n]+5(§) u[n—l]— (%));~1 n21 (c) From Eq. (6.137) 1T X(o) =7r[5(o— + 5(o+ 3)] lﬂlsrr lmi Then Yo XQHQ [so Tr +6().+1T]1+%eﬂj0 mm ()_ ( ) ( )‘Wi 2i i 2)1—§e-J‘n 1 + %e”j7T/Z 17 1 + %e_j7'/2 17- ‘ W 1—53—17” gin 2i+w 1—%e‘j”/2 5(Q+E) u 1 —jl 71' 1 +jl 7r Tr ,21 5(0 ) in f 5(o+*) 1 +1; 2 1 —]§ 2 I 7T . _ 7T . _ = 11' 5(Q — e—szn 1(1/2) + 11. 5(0 + eﬂtan l(1/2) M Taking the inverse Fourier transform of Y(Q) and using Eq. (6.135), we get 1 [ ]_ leKvT/Z)” e—j2taif‘(l/2) + 18*1'07/2) ethan“(l/2) “i y ” r 2 2 1111 ’TI' 1 = cos ~11 — 2ta11"1— llliil 2 2 «Ill1 6.33. Consider a discrete—time LTI system with impulse response sin(7rn/4 W] = #4 7m 1m Find the output y[n] if the input x[n] is a periodic sequence with fundamental period N0 = 5 as shown in Fig. 6—17. From Eq. (6.134) we have * 1 1m Sw/4 H<Q>ﬂ<0 17/4<IQIS'1T Since 00 = 21T/N0 = Z'IT/S and the ﬁlter passes only frequencies in the range IQI s 77/ 4, only the dc term is passed through. From Fig. 6—17 and Eq. (6.11) 1 4 3 C0=§ inniz‘g 11:0 CHAP. 6] FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS 335 Consider a three-point moving—average discrete—time ﬁlter described by the difference equation y[n] = +x[n — 1] +x[n—2]} (6.167) (a) Find and sketch the impulse response h[n] of the ﬁlter. (b) Find the frequency response H (0.) of the ﬁlter. (6) Sketch the magnitude response IH(Q)I and the phase reSponse 0(0) of the ﬁlter. (a) By the deﬁnition of h[n] [Eq. (2.30)] we have m}=§{a[n]+5[n—1]+3[n—2]} (6.168) or h[n] = % 03mg? 0 otherwrse which is sketched in Fig. 6—26(a). Note that h[n] satisﬁes the condition (6.163) with N = 3. ' (b) Taking the Fourier transform of Eq. (6.168), we have H(Q) = §{1 +e‘j0+ e_2jﬂ} lH(Q)| h[n] m— (a) (b) Fig. 6-26 lfti [Ht ill ll" m: [M 336 FOURIER ANALYSIS OF DISCRETE—TIME SIGNALS AND SYSTEMS [CHAR 6 By Eq. (1,90), with a = 6”“, we get 1 1* e—jBO. 1 e—j30/2(ej3I2/2 _ eaten/2) 1—K“) 3 1_e—jﬂ 3 e—jn/2(ejQ/2 _ e—jo/z) 1 , sin(30./2) ‘ = _ an” 2 H. o “J” .16 3e sin(Q/2) '( )6 (6 9) h 11 I) 1 Quasi/2) 6170 W 616 "( )_§ sin(Q/2) ( ' ) (c) From Eq. (6.169) IH Q I lH Q | 1 sin(3Q/2) ( )_ "( )_ 3 sin(o/2) d 0 Q —0 when H,.(Q) > 0 h Wm] an ( > _ —o + 77 when 1—1,.(o) < 0 mt Inlnml which are sketched in Fig. 6-26(b). We see that the system is a low-pass FIR ﬁlter with linear phase. M a l @ Consider a causal discrete—time FIR ﬁlter described by the impulse response " Jllhrm Mn] = {2,2,—2,—2} (a) Sketch the impulse response h[n] of the ﬁlter. (b) Find the frequency response H (0) of the ﬁlter. mmm‘mlﬂ (c) Sketch the magnitude response |H(Q)I and the phase response 6(9) of the ﬁlter. in. (a) The impulse response h[n] is sketched in Fig. 6-27(a). Note that h[n] satisfies the autumn,“ condition (6.164) with N = 4. (b) By deﬁnition (6.27) mutan w HUI) = Z h[n] fin" = 2 + 2(3an — Ze‘jm —— 26—130 mm»le n: -00 Nrmmmm = 2(1 —‘e"j3n) + 2(6'in — e‘jm) : 26~j30/2(ej3Q/2 _ e—j3Q/2) + Ze—j3Q/2(ejQ/2 , gem/2) ‘.,_ 1 , 9 3o ‘ i = J'wm/2(sinE + sin—2—) = H,.(o) e’[(”/2)_(3n/2)] (6.171) .1 “In” 1 H 0 ' Q . 30 wrere ,.( )—sm(—2—) +s1n(7) (c) From Eq. (6.171) D. 39 IH(Q)I = IH,.(Q)| = sin(3) + sin(—E-) i 'n' 2 — in H,. o > 0 E 6m) 2 / 2 3 ( ) z . —qT/2—§o HAD) <0 which are sketched in Fig. 6-27(b). We see that the system is a bandpass FIR ﬁlter with linear phase. CHAP, 6] FOURIER ANALYSIS OF DISCRETE—TIME SIGNALS AND SYSTEMS 337 |H(Q)| 1.54 A (b) Fig. 6-27 SIMULATION 6.43. Consider the RC low—pass ﬁlter shown in Fig. 6~28(a) with RC = 1. (0) Construct a discrete—time ﬁlter such that hdin] I l/lc(t)|t=nfif\~ = hc(l’lTs) where hc(t) is the impulse response of the RC ﬁlter, hl,[n] is the impulse response of the discrete-time ﬁlter, and T. is a positive number to be chosen as part of the design procedures. l (b) Plot the magnitude response ch(w)l of the RC ﬁlter and the magnitude response |Hd(w7})l of the discrete—time ﬁlter for Ts = l and TS = 0.1. (6.172) ’ (a) The system function Hc(s) of the RC ﬁlter is given by (Prob. 3.23) 1 H, .= 6.173 C(s) 8 +1 ( ) and the impulse response MO) is hc(t)=e"u(t) (6.174) By Eq. (6.172) the corresponding MM] is given by /1(,[rL] = (3_"’r-"Lt[l’l.] = (6—7i")l'Lt[n] (6.175) [In M 'W. ...
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