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A#11_Solutions

# A#11_Solutions - Assignment#11 — Solutions p.1...

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Unformatted text preview: Assignment #11 —- Solutions - p.1 ECSE—2410 Signals & Systems — Spring 2009 Due Fri 02/27/09 Use the properties and transform Tables (no integrations of the definition) to solve the following: 1 1(10). Use duality to find the Fourier transform of x(t) = 1+t2' Peak Teﬂ‘ 5%.4‘1 3i» Mfg \$5917) 5m} 21,5 4H 2__ 2: M We») '-‘- 2“ NH e Hm Hi i @‘F m=§m=ie HEZdzéEt’uﬁ): W32 Hr’tz‘ g 4'9 776 1+ 132“ 2(10). Find the Fourier transform of x(t) = [31"21 ~2itl M a Q 182+ ﬁxer»): e H imp—CH“ Tim ~3Ii‘r24l N .a 2w M):- Mia) 2a M Emamele a? mg 1320 kw): (:26 Assignment #11 — Solutions - p.2 ECSE-2410 Signals & Systems — Spring 2009 Due Fri 02/27/09 3(8). Find the Fourier transform of x(t) = sinc(t)sinc(2t) [2+ Ma: Sim (:9 a? X25): 5,516,129 mg“ X60: Mﬂxzéﬂ wﬁim) == igl‘glw) *3???) Assignment #11 — Solutions — p.3 ECSE—2410 Signals & Systems — Spring 2009 Due Fri 02/27/09 4(10). Find the Fourier transform of x(t) = sinc(t) * E-sinc(2t)] 75 7: T317 1“3) M 9L1£~9=§~§~ sake/0c) 4-» Mﬂi‘h a.) 1 I21”) gall-9:??? stack—é) 94> ,Ebgco 2.1 2_ Ewan: Mi? 2* he) MZM 23115663 EM 210) L U 5(8). Find the inverse transform of X = 1116:: . J ._ t + me 73446 Egg“): 9 X360 = 8 W“) H13 ‘43 Assignment #11 —- Solutions - p.4 ECSE—2410 Signals & Systems — Spring 2009 Due Fri 02/27/09 _jw 6(8). Find the inverse transform of X = e 1+j(a)—1)' at Paw—m Tabbz Ema): {if} 4—» 1295+): e w) .... w. .3 t r’x )== 241 d Kain/7‘} "" Xiéw it ‘+\},(w~ﬂ 4"? 21* m “3a) 3.2)“) Oder): M2661) Lilac): 8 I200)” Wmﬁ) M A Leak wise/\ch gamma ﬁr!) mtéei) 353“) My): grater): 9966*!)6 a 6: Mi“! 8 7(10). Find the inverse transform of X 2 2004.260) _ 1+}a) “we FNQM ‘mbie 350).:4,“ 4.; 99m: 8 at?) ‘ HJLU ‘tﬁw “2m _‘2q; 3 a ) Empxwzwstzwt =EM 8 +8 32nd “you ; Ewe + Ewe?» w {ﬂag 2 Qt; (+52.) ’1‘ “£20044 .. t—z) % wt: gamma) + 6 “W23 Assignment #11 - Solutions — p.5 ECSE-2410 Signals & Systems — Spring 2009 Due Fri 02/27/09 8(8). Find the inverse transform of X = 68inc(3(a)— 27F». ’X;&) Li”? :22;sz -=-— Q \$1AC<%}W m 13: X5 3 TEAM bjzﬁ‘t 31w) =Xiﬁw~ an“) a z; “4 st-zm) 4-» Ma : go, we, X69 aim was M £2??? j , ~ «ea-=- (3, z . 5%“ I 0) else 9(10).Find the inverse transform of . Uow ~36, L 31‘: ! mm): ~2izm+0+ mm)» mat—2. raga + 2mm V...»— N§ “3+ ‘ 3‘ 63?. \$8): ZFLﬁ<M r: \M 4 Eiféin‘cbé) SM (:5); 2:) 3,515+) W" 3%) Magda :9 we) Mam; 3115614. Assignment #11 — Solutions — p.6 ECSE—2410 Signals & Systems — Spring 2009 Due Fri 02/27/09 w is x1 (t)=—72;sinc(2t). 10(8). We know that the inverse transform of -2 -1 0 1 2 However, if we think of this pulse as made up of two adjacent pulses of width 2, i. 2., as gsinc(t)cos(t). then the inverse Fourier transform is x2 (t) -2 1g 2 Show that x1 (t) = x2 Estavt mi 0914i == %Q§wagé%) (is fi‘g’éwji eragzsfkéb). 1? TT 1— Mb 3% “6 we 8° 5 *g—Cééjx : ) W ‘2—4‘ ‘91- “J” “id/6" W” J 7rth {\J D waz‘é a #C 833%.“ 6;.) J W ZT‘t 2.3 2.1-. swat) We ._ .L— 2M " Tl” 2t ‘2 3: glﬁCCEt) :2 yit+2~ IT @‘g/D Assignment #11 - Solutions - 13.7 ECSE—2410 Signals & Systems - Spring 2009 Due Fri 02/27/09 11(10). If the impulse response of a LTI system is MI) 2 2sinc(t) cos(4t), find output y(t) , when the input is x(t) 2 1+ cos(t) + sin(4t) =-.. ~ L 31" . Le+ mi awe—+54»me % “m ‘1: ant ‘ T53“ like): gamma—t): Nﬁ<€ﬂ+ad J“ 29696) waste w my») :2 Xﬁwmai] 4 EwH) New iéw}, 5125} 3% Lord; Fiﬁ-i: ~E~Sékcﬂﬂﬁv is “if II? «J +5111! 32 / ru—lzﬁ anew.» ﬂ; :1 sh“ "‘ Kitﬁz‘wrﬁ): \$3.145 H W :7 -Z_"'i W 2- " a 5 ’ E ‘2? -3 My 2%}: H- 6950) + SM‘H} a» EM 2 mgwpyrﬁgéw-mgzwﬂi-J @év ) M} Egg) ...
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A#11_Solutions - Assignment#11 — Solutions p.1...

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